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The goal of this course is to show some fundamental
elements of numerical approximation of partial differential
equations. Linear evolution equations involving transport and diffusion phenomenons which can be observed in most of
mechanical problems are discretized by two methods: finite differences and finite volumes.
Main notions like accuracy, stability, convergence, numerical diffusion and dispersion,
are analyzed. Implementation is carefully studied. The numerical methods presented in the course are studied with
exercices and programming sessions.
Plan :
PDEs in Mechanics
Finite difference method in 1D.
Finite volume method in 1D.
Finite volume method in 2D: cartesian meshes, unstructured meshes.
Additionnal features on linear hyperbolic systems and high order finite volume methods. | 677.169 | 1 |
Top positive review
5.0 out of 5 starsElementary Mathematics for Teachers - Textbook - Required reading for education program - Cover fundamental well
ByRatherLiveInKeyWeston October 11, 2012Top critical reviewThere was a problem loading comments right now. Please try again later.
The book is simple, logical and intuitive. The sequence of topics through the chapters is a careful description from concrete (using manipulatives i.e. number disks, place value charts, etc.), then moving to pictorial (where pictures represent numbers) working in the abstract. Every topic in the book is functional in the process how to add. Every new concept gives more sense of the previously introduced ones in feedback. For example starting from the friendly numbers to the grouping of the numbers (to start multiplication, the connection of addition and subtraction and the connection of addition to multiplication is very consistently built in the text). With the consistent building the students are ready to start complete algorithms by third grade (or middle of second grade). There is a reason as of which number bonds to start with, which addition to teach first, likewise the order of multiplication and skip counting taught and this follows with the introduction of fractions and elements of algebra. The book is highly recommended. The writer of this review taught in-service teachers from this book, as well as parents joining the group. They all said, they wished that mathematics was introduced to them like in this book- it would have made sense and they probably liked mathematics (and reflected the love of mathematics more). The teachers liked the suggestions for constructing their own notes for their teaching.
Please note: You may purchase the book from The publisher (Sefton - Ash Publishing) for a more reasonable price.
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Horribly written and hard to follow. Several grammatical errors as well. The book often "yells" at you to "read the text". The homework at the end of each section is not appropriate practice, it is time consuming and not beneficial to improve your knowledge of the material. I wish my university used a better textbook. This book forces you to buy more books that go along with it and it is frustrating flipping back and forth between them. This book ishould be replaced with another in all schools. I am in awe that my major university chooses to use this text. | 677.169 | 1 |
Find a Plandome, NYApplications of Second Order Differential
Equations including Electric Circuit Problems and Vibration of a Mass on a Spring.
5. Numerical Methods for solving Differential Equations such as Euler's Method and the Runge-Kutta method.
6. Series solutions of Differential Equations including the method of Frobenius.
7. | 677.169 | 1 |
Advanced Problem Solving Module 15
Current
Advanced Problem Solving Module 15
For many students, STEP questions on areas of Pure Mathematics will be most familiar. However, the applied topics provide a brilliant chance to increase the number of questions you can attempt come the big day. Here, you are invited to review the important concepts you will need to be able to employ to tackle STEP Statistics questions and read through some hints and tips, before finally having a
go at some STEP questions yourself.
Related | 677.169 | 1 |
Details about Precalculus:
This text, intended for a graphing calculator-required precalculus course, shows students when and how to use concepts, and promotes real understanding, m not just rote memorization. The graphing calculator is used as a tool to help explain ideas rather than merely to find answers
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Rent Precalculus 5th edition today, or search our site for other textbooks by Franklin D. Demana | 677.169 | 1 |
Math Study Skillsoutlines good study habits and provides students with study strategies and tips to improve in areas such as time management, organization, and test-taking skills. With a friendly and relatable voice, Alan Bass addresses the misgivings and challenges many students face in a math class, and offers techniques to improve their study skills, as well as opportunities to practice and assess these techniques. This math study skills workbook is short enough to be used as a supplement in a math course, but can also be used as a main text in a study skills class. | 677.169 | 1 |
I am Jim Fowler.
Jim Fowler is a mathematician at The Ohio State University. His research interests broadly include geometry and topology, and more specifically focus on the topology of high-dimensional manifolds and geometric group theory. In other words, he thinks in depth about highly symmetric geometric objects. He's fond of using computational techniques to attack problems in pure mathematics.
Office hours This online course is a first and friendly introduction to calculus, suitable for someone who has never seen the subject before, or for someone who has seen some calculus but wants to review the concepts and practice applying those concepts to solve problems.
Sequences and Series will challenge us to think very carefully about "infinity." What does it mean to add up an unending list of numbers? How can an infinite task result in a finite answer? These questions lead us to some very deep concepts—but also to some powerful computational tools which are used not only in math but in many quantitative disciplines.
This course is a first introduction to sequences, infinite series, convergence tests, and Taylor series. It is suitable for someone who has seen just a bit of calculus before.
M2O2C2 is an invitation to think carefully about how one thing changing affects something else. What is the "derivative" of a function of many variables? How can a curved object be approximated by a flat plane? What does the chain rule look like when many things are affecting many other things? How do we find an input which maximizes a function of many variables?
This is a course in multivariable differential calculus, but we will also introduce a ton of linear algebra. The result is a course targeted at a student who has seen a bit of calculus and is willing to learn about matrices and vectors to provide the best possible vantage point from which to understand derivatives of functions of many variables. | 677.169 | 1 |
Details about College Algebra:
Focusing on helping students to develop the conceptual understanding and analytical skills necessary to experience success in mathematics, this book presents mathematical topics using a learning system to actively engage students in the learning process. It addresses the diverse needs of students through an open design and other helpful features.
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Rent College Algebra 9th edition today, or search our site for other textbooks by Margaret L. Lial. Every textbook comes with a 21-day "Any Reason" guarantee. Published by Pearson. | 677.169 | 1 |
Related software: 04/06/2015 21:35:15 Introduction to Linear Algebra, Fourth Edition [Gilbert Strang] on Amazon.com. *FREE* shipping on qualifying offers. Gilbert Strang's textbooks have changed the ...WELCOME TO the Seventh Edition of Introduction to Programming Using Java, a free, on-line textbook on introductory programming, which uses Java ...The Math Book: From Pythagoras to the 57th Dimension, 250 Milestones in the History of MathematicsPython for Kids: A Playful Introduction to Programming is the book that fills the hole in my Python library. It starts with a nice introduction and installation ...This example clarifies the purpose of order of operations: to have ONE way to interpret a mathematical statement.What single book is the best introduction to your field (or specialization within your field) for laypeople? September 8, 2007 5:40 PM SubscribeBig Ideas MATH: A Common Core Curriculum for Middle School and High School Mathematics Written by Ron Larson and Laurie Boswell.Print worksheets for preschool-sixth grade students by grade or subject area-math, English, handwriting, history, and more plus coloring pages and puzzles.Connected Mathematics Project is a problem-centered curriculum promoting an inquiry-based teaching-learning classroom environment. Mathematical ideas are identified ...500 libros digitales PDF gratis matematica algebra lineal analisis funcional probabilidades topologia teoria de numeros estadistica calculo intro to algebra book | 677.169 | 1 |
At this page you find some selected links ordered by topic,
where we have laid emphasis on interactive learning material
which needs no download or additional software, although some
spreadsheets (requiring Microsoft Excel to run on your computer)
and other media have been included.
For a refined search on this page use your browser's search functionality
(Menu Edit®Find in Page or the
key combination Ctrl F)!
Graphing Linear Equations
(applet) Learning program on linear functions and their graphical representation
as straight lines in a plane. Includes tutorial, test, quiz, crossword puzzle
and - as tools - the possibility to plot graphs of functions and a calculator.
You may pose questions to the teachers and discuss with pupils of the
Bremen High School, USA.
Graph Applet (applet) A straight line may interactively be moved, the
coefficients m and b of its equation are shown.
Trigonometry (some applets)
A number of units to illustrate definitions, graphs and simple properties
of the trigonometric functions, including the
Sum Formulae, the Law of Sines and the Law of Cosines.
In particular see two applets about the graphs of the
Sine and the
Cosine function
and a further applet by which the definitions of the
various trigonometric functions may be compared.
Curves
Famous Curves (graphics and applets)
More than sixty famous plane curves or families of curves
are discussed and represented graphically.
In the applet versions their parameters may be changed interactively.
See also the
Famous Curves Applet Index. (School of Mathematical and Computational Sciences,
University of St Andrews, Scotland).
Realtime
Fourier sound synthesis (spreadsheet) Sound generator:
Determine the intensities of overtones by means of a scroll bar and listen to the according sound.
(Even without sound utility, the graphical side of this spreadsheet is a nice
illustration of the Fourier series). | 677.169 | 1 |
vard_University_-_Linear_Al
1.
USE OF LINEAR ALGEBRA I Math 21b, O. Knill
This is not a list of topics covered in the course. It is rather a lose selection of subjects, in which linear algebra is
useful or relevant. The aim is to convince you that it is worth learning this subject. Most likely, some of this
handout does not make much sense yet to you. Look at this page at the end of the course again, some of the
content will become more interesting then.
GRAPHS, NETWORKS. Linear al-
gebra can be used to understand
networks which is a collection of
nodes connected by edges. Networks
4 are also called graphs. The adja- Application: An is the number of n-
ij
step walks in the graph which start
cency matrix of a graph is defined
by Aij = 1 if there is an edge from at the vertex i and end at the vertex
1 2 3 node i to node j in the graph. Oth- j.
erwise the entry is zero. A prob-
lem using such matrices appeared on
a blackboard at MIT in the movie
"Good will hunting".
CHEMISTRY, MECHANICS
Complicated objects like a The solution x(t) = exp(At) of the
bridge (the picture shows Storrow differential equation x = Ax can be
˙
Drive connection bridge which is understood and computed by find-
part of the "big dig"), or a molecule ing the eigenvalues of the matrix A.
(i.e. a protein) can be modeled by Knowing these frequencies is impor-
finitely many parts (bridge elements tant for the design of a mechani-
or atoms) coupled with attractive cal object because the engineer can
and repelling forces. The vibrations damp dangerous frequencies. In
of the system are described by a chemistry or medicine, the knowl-
differential equation x = Ax, where
˙ edge of the vibration resonances al-
x(t) is a vector which depends on lows to determine the shape of a
time. Differential equations are an molecule.
important part of this course.
QUANTUM COMPUTING A
quantum computer is a quantum
Theoretically, quantum computa-
mechanical system which is used to
tions could speed up conventional
perform computations. The state x
computations significantly. They
of a machine is no more a sequence
could be used for example for cryp-
of bits like in a classical computer
tological purposes. Freely available
but a sequence of qubits, where
quantum computer language (QCL)
each qubit is a vector. The memory
interpreters can simulate quantum
of the computer can be represented
computers with an arbitrary num-
as a vector. Each computation step
ber of qubits.
is a multiplication x → Ax with a
suitable matrix A.
CHAOS THEORY. Dynamical
systems theory deals with the
iteration of maps or the analysis of
Examples of dynamical systems are
solutions of differential equations.
our solar system or the stars in a
At each time t, one has a map
galaxy, electrons in a plasma or par-
T (t) on the vector space. The
ticles in a fluid. The theoretical
linear approximation DT (t) is
study is intrinsically linked to linear
called Jacobean is a matrix. If the
algebra because stability properties
largest eigenvalue of DT (t) grows
often depends on linear approxima-
exponentially in t, then the system
tions.
shows "sensitive dependence on
initial conditions" which is also
called "chaos".
2.
USE OF LINEAR ALGEBRA II Math 21b, O. Knill
CODING, ERROR CORRECTION
Coding theory is used for encryp-
tion or error correction. In the first
case, the data x are maped by a
map T into code y=Tx. For a good Linear algebra enters in different
code, T is a "trapdoor function" in ways, often directly because the ob-
the sense that it is hard to get x jects are vectors but also indirectly
back when y is known. In the sec- like for example in algorithms which
ond case, a code is a linear subspace aim at cracking encryption schemes.
X of a vector space and T is a map
describing the transmission with er-
rors. The projection of T x onto the
subspace X corrects the error.
DATA COMPRESSION Image- Typically, a picture, a sound or
(i.e. JPG), video- (MPG4) and a movie is cut into smaller junks.
sound compression algorithms These parts are represented by vec-
(i.e. MP3) make use of linear tors. If U denotes the Fourier trans-
transformations like the Fourier form and P is a cutoff function, then
transform. In all cases, the com- y = P U x is transferred or stored on
pression makes use of the fact that a CD or DVD. The receiver obtains
in the Fourier space, information back U T y which is close to x in the
can be cut away without disturbing sense that the human eye or ear does
the main information. not notice a big difference.
SOLVING SYSTEMS OR EQUA- Solving systems of nonlinear equa-
TIONS When extremizing a func- tions can be tricky. Already for sys-
tion f on data which satisfy a con- tems of polynomial equations, one
straint g(x) = 0, the method of has to work with linear spaces of
Lagrange multipliers asks to solve polynomials. Even if the Lagrange
a nonlinear system of equations system is a linear system, the task
f (x) = λ g(x), g(x) = 0 for the of solving it can be done more ef-
(n + 1) unknowns (x, l), where f ficiently using a solid foundation of
is the gradient of f . linear algebra.
Rotations are represented by or-
thogonal matrices. For example,
GAMES Moving around in a world if an object located at (0, 0, 0),
described in a computer game re- turning around the y-axes by an
quires rotations and translations to angle φ, every point in the ob-
be implemented efficiently. Hard- ject gets transformed by matrix
the
ware acceleration can help to handle
cos(φ) 0 sin(φ)
this. 0 1 0
− sin(φ) 0 cos(φ)
3.
USE OF LINEAR ALGEBRA (III) Math 21b, Oliver Knill
STATISTICS When analyzing data
statistically, one often is interested
in the correlation matrix Aij = For example, if the random variables
E[Yi Yj ] of a random vector X = have a Gaussian (=Bell shaped) dis-
(X1 , . . . , Xn ) with Yi = Xi − E[Xi ]. tribution, the correlation matrix to-
This matrix is derived from the data gether with the expectation E[Xi ]
and determines often the random determines the random variables.
variables when the type of the dis-
tribution is fixed.
A famous example is the prisoner
dilemma. Each player has the
choice to corporate or to cheat.. The
game is described by a 2x2 matrix
3 0
like for example . If a
5 1
GAME THEORY Abstract Games player cooperates and his partner
are often represented by pay-off ma- also, both get 3 points. If his partner
trices. These matrices tell the out- cheats and he cooperates, he gets 5
come when the decisions of each points. If both cheat, both get 1
player are known. point. More generally, in a game
with two players where each player
can chose from n strategies, the pay-
off matrix is a n times n matrix
A. A Nash equilibrium is a vector
p ∈ S = { i pi = 1, pi ≥ 0 } for
which qAp ≤ pAp for all q ∈ S.
NEURAL NETWORK In part of
neural network theory, for exam-
ple Hopfield networks, the state
space is a 2n-dimensional vector
space. Every state of the network
is given by a vector x, where each
For example, if Wij = xi yj , then x
component takes the values −1 or
is a fixed point of the learning map.
1. If W is a symmetric nxn matrix,
one can define a "learning map"
T : x → signW x, where the sign
is taken component wise. The en-
ergy of the state is the dot prod-
uct −(x, W x)/2. One is interested
in fixed points of the map.
4.
USE OF LINEAR ALGEBRA (IV) Math 21b, Oliver Knill
MARKOV. Suppose we have three
bags with 10 balls each. Every time The problem defines a Markov
we throw a dice and a 5 shows up, chain described
by a matrix
we move a ball from bag 1 to bag 2, 5/6 1/6 0
if the dice shows 1 or 2, we move a 0 2/3 1/3 .
ball from bag 2 to bag 3, if 3 or 4 1/6 1/6 2/3
turns up, we move a ball from bag 3 From this matrix, the equilibrium
to bag 1 and a ball from bag 3 to bag distribution can be read off as an
2. What distribution of balls will we eigenvector of a matrix.
see in average?
SPLINES In computer aided de-
sign (CAD) used for example to
If we write down the conditions, we
construct cars, one wants to inter-
will have to solve a system of 4 equa-
polate points with smooth curves.
tions for four unknowns. Graphic
One example: assume you want to
artists (i.e. at the company "Pixar")
find a curve connecting two points
need to have linear algebra skills also
P and Q and the direction is given
at many other topics in computer
at each point. Find a cubic function
graphics.
f (x, y) = ax3 + bx2 y + cxy 2 + dy 3
which interpolates.
SYMBOLIC DYNAMICS Assume
that a system can be in three dif-
The dynamics of the system is
ferent states a, b, c and that tran-
coded with a symbolic dynamical
sitions a → b, b → a, b → c,
system. The transition matrix is
c → c, c → a are allowed. A
0 1 0
a c possible evolution of the system is
then a, b, a, b, a, c, c, c, a, b, c, a... One
1 0 1 .
1 0 1
calls this a description of the system
Information theoretical quantities
with symbolic dynamics. This
like the "entropy" can be read off
language is used in information the-
b ory or in dynamical systems theory.
from this matrix.
INVERSE PROBLEMS The recon-
struction of a density function from Toy problem: We have 4 containers
projections along lines reduces to with density a, b, c, d arranged in a
the solution of the Radon trans- square. We are able and measure
q r form. Studied first in 1917, it is to- the light absorption by by sending
day a basic tool in applications like light through it. Like this, we get
a b medical diagnosis, tokamak moni- o = a + b,p = c + d,q = a + c and
o
toring, in plasma physics or for as- r = b + d. The problem is to re-
trophysical applications. The re- cover a, b, c, d. The system of equa-
c d construction is also called tomogra-
p tions is equivalent to Ax = b, with
phy. Mathematical tools developed x = (a, b, c, d) and b = (o, p, q, r) and
for the solution of this problem lead 1 1 0 0
to the construction of sophisticated 0 0 1 1
A= 1 0 1 0 .
scanners. It is important that the
inversion h = R(f ) → f is fast, ac- 0 1 0 1
curate, robust and requires as few
datas as possible.
5.
ICE: A BLACKBOARD PROBLEM Math 21b, O. Knill
In the movie
"Good Will
Hunting", the
main character
Will Hunting
(Matt Damon)
solves a black-
board challenge
problem, which
is given as a
challenge to a
linear algebra
class.
THE "WILL HUNTING" PROBLEM.
4
G is the graph
1 2 3
Find.
1) the adjacency matrix A.
2) the matrix giving the number of 3 step walks.
3) the generating function for walks from point i → j.
4) the generating function for walks from points 1 → 3.
This problem belongs to linear algebra and calculus evenso the problem origins from graph the-
ory or combinatorics. For a calculus student who has never seen the connection between graph
theory, calculus and linear algebra, the assignment is actually hard – probably too hard - as the
movie correctly indicates. The problem was posed in the last part of a linear algebra course.
An explanation of some terms:
THE ADJACENCY MATRIX. The structure of the graph can be encoded with a 4 × 4 array
which encodes how many paths of length 1, one can take in the graph from one node to an
other:
no one no one which can more conve- 0 1 0 1
1 0 2 1
one none two one niently be written as L=
no two no no an array of numbers 0 2 0 0
one one no no called a matrix: 1 1 0 0
Problem 2 asks to find the matrix which encodes all possible paths of length 3.
GENERATING FUNCTION. To any graph one can assign for every pair of nodes i, j a series
f (z) = ∞ a(ij) z n , where a(ij) is the number of possible walks from node i to node j with n
n=0 n n
steps. Problem 3) asks for an explicit expression of f (z) and problem 4) asks for an explicit
expression in the special case i = 1, j = 3.
Linear algebra has many relations to other fields in mathematics. It is not true that linear
algebra is just about solving systems of linear equations.
7.
LINEAR EQUATIONS Math 21b, O. Knill
SYSTEM OF LINEAR EQUATIONS. A collection of linear equations is called a system of linear equations.
An example is
3x − y − z = 0
−x + 2y − z = 0 .
−x − y + 3z = 9
It consists of three equations for three unknowns x, y, z. Linear means that no nonlinear terms like
x2 , x3 , xy, yz 3 , sin(x) etc. appear. A formal definition of linearity will be given later.
LINEAR EQUATION. More precicely, ax + by = c is the general linear equation in two variables. ax +
by + cz = d is the general linear equation in three variables. The general linear equation in n variables is
a1 x1 + a2 x2 + ... + an xn = 0 Finitely many such equations form a system of linear equations.
SOLVING BY ELIMINATION.
Eliminate variables. In the example the first equation gives z = 3x − y. Putting this into the second and
third equation gives
−x + 2y − (3x − y) =0
−x − y + 3(3x − y) =9
or
−4x + 3y = 0
.
8x − 4y = 9
The first equation gives y = 4/3x and plugging this into the other equation gives 8x − 16/3x = 9 or 8x = 27
which means x = 27/8. The other values y = 9/2, z = 45/8 can now be obtained.
SOLVE BY SUITABLE SUBTRACTION.
Addition of equations. If we subtract the third equation from the second, we get 3y − 4z = −9 and add
three times the second equation to the first, we get 5y − 4z = 0. Subtracting this equation to the previous one
gives −2y = −9 or y = 2/9.
SOLVE BY COMPUTER.
Use the computer. In Mathematica:
Solve[{3x − y − z == 0, −x + 2y − z == 0, −x − y + 3z == 9}, {x, y, z}] .
But what did Mathematica do to solve this equation? We will look in this course at some efficient algorithms.
GEOMETRIC SOLUTION.
Each of the three equations represents a plane in
three-dimensional space. Points on the first plane
satisfy the first equation. The second plane is the
solution set to the second equation. To satisfy the
first two equations means to be on the intersection
of these two planes which is here a line. To satisfy
all three equations, we have to intersect the line with
the plane representing the third equation which is a
point.
LINES, PLANES, HYPERPLANES.
The set of points in the plane satisfying ax + by = c form a line.
The set of points in space satisfying ax + by + cd = d form a plane.
The set of points in n-dimensional space satisfying a 1 x1 + ... + an xn = a0 define a set called a hyperplane.
8.
RIDDLES:
"15 kids have bicycles or tricycles. Together they Solution. With x bicycles and y tricycles, then x +
count 37 wheels. How many have bicycles?" y = 15, 2x + 3y = 37. The solution is x = 8, y = 7.
Solution If there are x brothers and y systers, then
"Tom, the brother of Carry has twice as many sisters
Tom has y sisters and x − 1 brothers while Carry has
as brothers while Carry has equal number of systers
x brothers and y − 1 sisters. We know y = 2(x −
and brothers. How many kids is there in total in this
1), x = y − 1 so that x + 1 = 2(x − 1) and so x =
family?"
3, y = 4.
INTERPOLATION.
Solution. Assume the parabola is
Find the equation of the ax2 + bx + c = 0. So, c = −1, a + b + c =
parabola which passes through 4, 4a+2b+c = 13. Elimination of c gives
the points P = (0, −1), a + b = 5, 4a + 2b = 14 so that 2b = 6
Q = (1, 4) and R = (2, 13). and b = 3, a = 2. The parabola has the
equation 2x2 + 3x − 1 = 0
TOMOGRAPHY
Here is a toy example of a problem one has to solve for magnetic
resonance imaging (MRI), which makes use of the absorption and
emission of energy in the radio frequency range of the electromag-
netic spectrum.
q r
Assume we have 4 hydrogen atoms whose nuclei are excited with
intensity a, b, c, d. We measure the spin echo intensity in 4 different a b
directions. 3 = a + b,7 = c + d,5 = a + c and 5 = b + d. What o
is a, b, c, d? Solution: a = 2, b = 1, c = 3, d = 4. However,
also a = 0, b = 3, c = 5, d = 2 solves the problem. This system
has not a unique solution even so there are 4 equations and 4 c d
unknowns. A good introduction into MRI can be found online at p
(
INCONSISTENT. x − y = 4, y + z = 5, x + z = 6 is a system with no solutions. It is called inconsistent.
EQUILIBRIUM. As an example of a system with many
variables, consider a drum modeled by a fine net. The
heights at each interior node needs the average the
a11 a12 a13 a14 heights of the 4 neighboring nodes. The height at the
boundary is fixed. With n2 nodes in the interior, we
a21 x11 x12 a24 have to solve a system of n2 equations. For exam-
ple, for n = 2 (see left), the n2 = 4 equations are
x11 = a21 + a12 + x21 + x12 , x12 = x11 + x13 + x22 + x22 ,
a31 x21 x22 a34
x21 = x31 +x11 +x22 +a43 , x22 = x12 +x21 +a43 +a34 . To
the right, we see the solution to a problem with n = 300,
a41 a42 a43 a44 where the computer had to solve a system with 90 000
variables.
LINEAR OR NONLINEAR?
a) The ideal gas law P V = nKT for the P, V, T , the pressure, volume and temperature of the gas.
b) The Hook law F = k(x − x0 ) relates the force F pulling a string at position x which is relaxed at x 0 .
9.
MATRICES AND GAUSS-JORDAN Math 21b, O. Knill
MATRIX FORMULATION. Consider the sys- The system can be written as Ax = b, where A is a matrix
tem of linear equations (called coefficient matrix) and and x and b are vectors.
3x − y − z = 0 3 −1 −1 x 0
−x + 2y − z = 0 A = −1 2 −1 , x = y , b = 0 .
−x − y + 3z = 9 −1 −1 3 z 9
((Ax)i is the dot product
of the i'th row with x).
3 −1 −1 | 0
We also look at the augmented matrix B = −1 2 −1 | 0 .
where one puts separators for clarity reasons. −1 −1 3 | 9
MATRIX "JARGON". A rectangular array of numbers is called a
matrix. If the matrix has m rows and n columns, it is called a
m × n matrix. A matrix with one column only is called a column m
vector, a matrix with one row a row vector. The entries of a
matrix are denoted by aij , where i is the row and j is the column.
In the case of the linear equation above, the matrix A is a square
matrix and the augmented matrix B above is a 3 × 4 matrix.
n
GAUSS-JORDAN ELIMINATION. Gauss-Jordan Elimination is a process, where successive subtraction of
multiples of other rows or scaling brings the matrix into reduced row echelon form. The elimination process
consists of three possible steps which are called elementary row operations:
• Swap two rows.
• Divide a row by a scalar
• Subtract a multiple of a row from an other row.
The process transfers a given matrix A into a new matrix rref(A)
REDUCED ECHELON FORM. A matrix is called in reduced row echelon form
1) if a row has nonzero entries, then the first nonzero entry is 1. (leading one)
2) if a column contains a leading 1, then the other column entries are 0.
3) if a row has a leading 1, then every row above has leading 1's to the left.
Pro memoriam: Leaders like to be number one, are lonely and want other leaders above to their left.
RANK. The number of leading 1 in rref(A) is called the rank of A.
SOLUTIONS OF LINEAR EQUATIONS. If Ax = b is a linear system of equations with m equations and n
unknowns, then A is a m × n matrix. We have the following three possibilities:
• Exactly one solution. There is a leading 1 in each row but not in the last row.
• Zero solutions. There is a leading 1 in the last row.
• Infinitely many solutions. There are rows without leading 1 and no leading 1 is in the last row.
JIUZHANG SUANSHU. The technique of successively
eliminating variables from systems of linear equations
is called Gauss elimination or Gauss Jordan
elimination and appeared already in the Chinese
manuscript "Jiuzhang Suanshu" ('Nine Chapters on the
Mathematical art'). The manuscript appeared around
200 BC in the Han dynasty and was probably used as
a textbook. For more history of Chinese Mathematics,
see
˜
djoyce/mathhist/china.html.
12.
(exactly one solution) (no solution) (infinitely many solutions)
MURPHYS LAW.
"If anything can go wrong, it will go wrong".
"If you are feeling good, don't worry, you will get over it!"
"Whenever you do Gauss-Jordan elimination, you screw up
during the first couple of steps."
MURPHYS LAW IS TRUE. Two equations could contradict each other. Geometrically this means that
the two planes do not intersect. This is possible if they are parallel. Even without two planes be-
ing parallel, it is possible that there is no intersection between all three of them. Also possible is that
we don't have enough equations (for example because two equations are the same) and that there are
many solutions. Furthermore, we can have too many equations and the four planes would not intersect.
RELEVANCE OF EXCEPTIONAL CASES. There are important applications, where "unusual" situations
happen: For example in medical tomography, systems of equations appear which are "ill posed". In this case
one has to be careful with the method.
The linear equations are then obtained from a method called the Radon
transform. The task for finding a good method had led to a Nobel
prize in Medicis 1979 for Allan Cormack. Cormack had sabbaticals at
Harvard and probably has done part of his work on tomography here.
Tomography helps today for example for cancer treatment.
MATRIX ALGEBRA. Matrices can be added, subtracted if they have the same size:
a11 a12 · · · a1n b11 b12 · · · b1n a11 + b11 a12 + b12 · · · a1n + b1n
a21 a22 · · · a2n b21 b22 · · · b2n a21 + b21 a22 + b22 · · · a2n + b2n
A+B =
···
+ =
··· ··· ··· ··· ··· ··· ··· ··· ··· ··· ···
am1 am2 · · · amn bm1 bm2 · · · bmn am1 + bm2 am2 + bm2 · · · amn + bmn
They can also be scaled by a scalar λ:
a11 a12 · · · a1n λa11 λa12 · · · λa1n
a21 a22 · · · a2n λa21 λa22 · · · λa2n
λA = λ
···
=
··· ··· ··· ··· ··· ··· ···
am1 am2 · · · amn λam1 λam2 · · · λamn
13.
LINEAR TRANSFORMATIONS Math 21b, O. Knill
TRANSFORMATIONS. A transformation T from a set X to a set Y is a rule, which assigns to every element
in X an element y = T (x) in Y . One calls X the domain and Y the codomain. A transformation is also called
a map.
LINEAR TRANSFORMATION. A map T from Rn to Rm is called a linear transformation if there is a
m × n matrix A such that
T (x) = Ax .
EXAMPLES.
3 4
• To the linear transformation T (x, y) = (3x+4y, x+5y) belongs the matrix . This transformation
1 5
maps the plane onto itself.
• T (x) = −3x is a linear transformation from the real line onto itself. The matrix is A = [−3].
• To T (x) = y · x from R3 to R belongs the matrix A = y = y1 y2 y3 . This 1 × 3 matrix is also called
a row vector. If the codomain is the real axes, one calls the map also a function. function defined on
space.
y1
• T (x) = xy from R to R3 . A = y = y2 is a 3 × 1 matrix which is also called a column vector. The
y3
map defines a line in space.
1 0
• T (x, y, z) = (x, y) from R3 to R2 , A is the 2 × 3 matrix A = 0 1 . The map projects space onto a
0 0
plane.
1 1 2
• To the map T (x, y) = (x + y, x − y, 2x − 3y) belongs the 3 × 2 matrix A = . The image
1 −1 −3
of the map is a plane in three dimensional space.
• If T (x) = x, then T is called the identity transformation.
PROPERTIES OF LINEAR TRANSFORMATIONS. T (0) = 0 T (x + y) = T (x) + T (y) T (λx) = λT (x)
In words: Linear transformations are compatible with addition and scalar multiplication. It does not matter,
whether we add two vectors before the transformation or add the transformed vectors.
ON LINEAR TRANSFORMATIONS. Linear transformations generalize the scaling transformation x → ax in
one dimensions.
They are important in
• geometry (i.e. rotations, dilations, projections or reflections)
• art (i.e. perspective, coordinate transformations),
• CAD applications (i.e. projections),
• physics (i.e. Lorentz transformations),
• dynamics (linearizations of general maps are linear maps),
• compression (i.e. using Fourier transform or Wavelet trans-
form),
• coding (many codes are linear codes),
• probability (i.e. Markov processes).
• linear equations (inversion is solving the equation)
14.
LINEAR TRANSFORMATION OR NOT? (The square to the right is the image of the square to the left):
| | ··· |
COLUMN VECTORS. A linear transformation T (x) = Ax with A = v1 v2 · · · vn has the property
| | ·· |
·
1 0 0
· · ·
that the column vector v1 , vi , vn are the images of the standard vectors e1 = ·
. e i = 1 . e n = · .
· · ·
0 0 1
In order to find the matrix of a linear transformation, look at the
image of the standard vectors and use those to build the columns
of the matrix.
QUIZ.
a) Find the matrix belonging to the linear transformation, which rotates a cube around the diagonal (1, 1, 1)
by 120 degrees (2π/3).
b) Find the linear transformation, which reflects a vector at the line containing the vector (1, 1, 1).
INVERSE OF A TRANSFORMATION. If S is a second transformation such that S(T x) = x, for every x, then
S is called the inverse of T . We will discuss this more later.
SOLVING A LINEAR SYSTEM OF EQUATIONS. Ax = b means to invert the linear transformation x → Ax.
If the linear system has exactly one solution, then an inverse exists. We will write x = A −1 b and see that the
inverse of a linear transformation is again a linear transformation.
THE BRETSCHER CODE. Otto Bretschers book contains as a motivation a
"code", where the encryption happens with the linear map T (x, y) = (x + 3y, 2x +
5y). The map has the inverse T −1 (x, y) = (−5x + 3y, 2x − y).
Cryptologists use often the following approach to crack a encryption. If one knows the input and output of
some data, one often can decode the key. Assume we know, the enemy uses a Bretscher code and we know that
a b
T (1, 1) = (3, 5) and T (2, 1) = (7, 5). How do we get the code? The problem is to find the matrix A = .
c d
2x2 MATRIX. It is useful to decode the Bretscher code in general If ax + by = X and cx + dy = Y , then
a b
x = (dX − bY )/(ad − bc), y = (cX − aY )/(ad − bc). This is a linear transformation with matrix A =
c d
−1 d −b
and the corresponding matrix is A = /(ad − bc).
−c a
"Switch diagonally, negate the wings and scale with a cross".
16.
ROTATION:
−1 0 cos(α) − sin(α)
A= A=
0 −1 sin(α) cos(α)
Any rotation has the form of the matrix to the right.
ROTATION-DILATION:
2 −3 a −b
A= A=
3 2 b a
A rotation dilation is a composition of a rotation by angle arctan(y/x) and a dilation by a factor x2 + y 2 . If
z = x + iy and w = a + ib and T (x, y) = (X, Y ), then X + iY = zw. So a rotation dilation is tied to the process
of the multiplication with a complex number.
BOOST:
The boost is a basic Lorentz transformation
cosh(α) sinh(α) in special relativity. It acts on vectors (x, ct),
A=
sinh(α) cosh(α) where t is time, c is the speed of light and x is
space.
Unlike in Galileo transformation (x, t) → (x + vt, t) (which is a shear), time t also changes during the
transformation. The transformation has the effect that it changes length (Lorentz contraction). The angle α is
related to v by tanh(α) = v/c. One can write also A(x, ct) = ((x + vt)/γ, t + (v/c 2 )/γx), with γ = 1 − v 2 /c2 .
ROTATION IN SPACE. Rotations in space are defined by an axes of rotation
and an angle. A rotation by 120◦ around a line containing (0, 0, 0) and (1, 1, 1)
0 0 1
blongs to A = 1 0 0 which permutes e1 → e2 → e3 .
0 1 0
REFLECTION AT PLANE. To a reflection at the xy-plane belongs the matrix
1 0 0
A = 0 1 0 as can be seen by looking at the images of ei . The picture
0 0 −1
to the right shows the textbook and reflections of it at two different mirrors.
PROJECTION ONTO SPACE. To project a 4d-object into xyz-space, use
1 0 0 0
0 1 0 0
for example the matrix A =
0 0 1 0 . The picture shows the pro-
0 0 0 0
jection of the four dimensional cube (tesseract, hypercube) with 16 edges
(±1, ±1, ±1, ±1). The tesseract is the theme of the horror movie "hypercube".
20.
FRACTALS. Closely related to linear maps are affine maps x → Ax + b. They are compositions of a linear
map with a translation. It is not a linear map if B(0) = 0. Affine maps can be disguised as linear maps
x A b Ax + b
in the following way: let y = and defne the (n+1)∗(n+1) matrix B = . Then By = .
1 0 1 1
Fractals can be constructed by taking for example 3 affine maps R, S, T which contract area. For a given object
Y0 define Y1 = R(Y0 ) ∪ S(Y0 ) ∪ T (Y0 ) and recursively Yk = R(Yk−1 ) ∪ S(Yk−1 ) ∪ T (Yk−1 ). The above picture
shows Yk after some iterations. In the limit, for example if R(Y 0 ), S(Y0 ) and T (Y0 ) are disjoint, the sets Yk
converge to a fractal, an object with dimension strictly between 1 and 2.
x 2x + 2 sin(x) − y
CHAOS. Consider a map in the plane like T : → We apply this map again and
y x
again and follow the points (x1 , y1 ) = T (x, y), (x2 , y2 ) = T (T (x, y)), etc. One writes T n for the n-th iteration
of the map and (xn , yn ) for the image of (x, y) under the map T n . The linear approximation of the map at a
2 + 2 cos(x) − 1 x f (x, y)
point (x, y) is the matrix DT (x, y) = . (If T = , then the row vectors of
1 y g(x, y)
DT (x, y) are just the gradients of f and g). T is called chaotic at (x, y), if the entries of D(T n )(x, y) grow
exponentially fast with n. By the chain rule, D(T n ) is the product of matrices DT (xi , yi ). For example, T is
chaotic at (0, 0). If there is a positive probability to hit a chaotic point, then T is called chaotic.
FALSE COLORS. Any color can be represented as a vector (r, g, b), where r ∈ [0, 1] is the red g ∈ [0, 1] is the
green and b ∈ [0, 1] is the blue component. Changing colors in a picture means applying a transformation on the
cube. Let T : (r, g, b) → (g, b, r) and S : (r, g, b) → (r, g, 0). What is the composition of these two linear maps?
OPTICS. Matrices help to calculate the motion of light rays through lenses. A
light ray y(s) = x + ms in the plane is described by a vector (x, m). Following
the light ray over a distance of length L corresponds to the map (x, m) →
(x + mL, m). In the lens, the ray is bent depending on the height x. The
transformation in the lens is (x, m) → (x, m − kx), where k is the strength of
the lense.
x x 1 L x x x 1 0 x
→ AL = , → Bk = .
m m 0 1 m m m −k 1 m
Examples:
1) Eye looking far: AR Bk . 2) Eye looking at distance L: AR Bk AL .
3) Telescope: Bk2 AL Bk1 . (More about it in problem 80 in section 2.4).
21.
IMAGE AND KERNEL Math 21b, O. Knill
IMAGE. If T : Rn → Rm is a linear transformation, then {T (x) | x ∈ Rn } is called the image of T . If
T (x) = Ax, then the image of T is also called the image of A. We write im(A) or im(T ).
EXAMPLES.
x 1 0 0 x
1) If T (x, y, z) = (x, y, 0), then T (x) = A y = 0 1 0 y . The image of T is the x − y plane.
z 0 0 0 z
2) If T (x, y)(cos(φ)x − sin(φ)y, sin(φ)x + cos(φ)y) is a rotation in the plane, then the image of T is the whole
plane.
3) If T (x, y, z) = x + y + z, then the image of T is R.
SPAN. The span of vectors v1 , . . . , vk in Rn is the set of all combinations c1 v1 + . . . ck vk , where ci are real
numbers.
PROPERTIES.
The image of a linear transformation x → Ax is the span of the column vectors of A.
The image of a linear transformation contains 0 and is closed under addition and scalar multiplication.
KERNEL. If T : Rn → Rm is a linear transformation, then the set {x | T (x) = 0 } is called the kernel of T .
If T (x) = Ax, then the kernel of T is also called the kernel of A. We write ker(A) or ker(T ).
EXAMPLES. (The same examples as above)
1) The kernel is the z-axes. Every vector (0, 0, z) is mapped to 0.
2) The kernel consists only of the point (0, 0, 0).
3) The kernel consists of all vector (x, y, z) for which x + y + z = 0. The kernel is a plane.
PROPERTIES.
The kernel of a linear transformation contains 0 and is closed under addition and scalar multiplication.
IMAGE AND KERNEL OF INVERTIBLE MAPS. A linear map x → Ax, Rn → Rn is invertible if and only
if ker(A) = {0} if and only if im(A) = Rn .
HOW DO WE COMPUTE THE IMAGE? The rank of rref(A) is the dimension of the image. The column
vectors of A span the image. (Dimension will be discussed later in detail).
EXAMPLES. (The
same examples as above)
1 0
cos(φ) sin(φ)
1) 0 and 1 2) and 3) The 1D vector 1 spans the
− sin(φ) cos(φ)
0 0 image.
span the image.
span the image.
HOW DO WE COMPUTE THE KERNEL? Just solve Ax = 0. Form rref(A). For every column without
leading 1 we can introduce a free variable si . If x is the solution to Axi = 0, where all sj are zero except si = 1,
then x = j sj xj is a general vector in the kernel.
1 3 0
2 6 5
EXAMPLE. Find the kernel of the linear map R3 → R4 , x → Ax with A = 3
. Gauss-Jordan
9 1
−2 −6 0
1 3 0
0 0 1
elimination gives: B = rref(A) =
0 0 0 . We see one column without leading 1 (the second one). The
0 0 0
equation Bx = 0 is equivalent to the system x + 3y = 0, z = 0. After fixing z = 0, chose y = t freely and
can
−3
obtain from the first equation x = −3t. Therefore, the kernel consists of vectors t 1 . In the book, you
0
have a detailed calculation, in a case, where the kernel is 2 dimensional.
22.
kernel
image
domain
codomain
WHY DO WE LOOK AT THE KERNEL?
WHY DO WE LOOK AT THE IMAGE?
• It is useful to understand linear maps. To which
• A solution Ax = b can be solved if and only if b
degree are they non-invertible?
is in the image of A.
• Helpful to understand quantitatively how many
• Knowing about the kernel and the image is use-
solutions a linear equation Ax = b has. If x is
ful in the similar way that it is useful to know
a solution and y is in the kernel of A, then also
about the domain and range of a general map
A(x + y) = b, so that x + y solves the system
and to understand the graph of the map.
also.
In general, the abstraction helps to understand topics like error correcing codes (Problem 53/54 in Bretschers
book), where two matrices H, M with the property that ker(H) = im(M ) appear. The encoding x → M x is
robust in the sense that adding an error e to the result M x → M x + e can be corrected: H(M x + e) = He
allows to find e and so M x. This allows to recover x = P M x with a projection P .
PROBLEM. Find ker(A) and im(A) for the 1 × 3 matrix A = [5, 1, 4], a row vector.
ANSWER. A · x = Ax = 5x + y + 4z = 0 shows that the kernel is a plane with normal vector [5, 1, 4] through
the origin. The image is the codomain, which is R.
PROBLEM. Find ker(A) and im(A) of the linear map x → v × x, (the cross product with v.
ANSWER. The kernel consists of the line spanned by v, the image is the plane orthogonal to v.
PROBLEM. Fix a vector w in space. Find ker(A) and image im(A) of the linear map from R 6 to R3 given by
x, y → [x, v, y] = (x × y) · w.
ANSWER. The kernel consist of all (x, y) such that their cross product orthogonal to w. This means that the
plane spanned by x, y contains w.
PROBLEM Find ker(T ) and im(T ) if T is a composition of a rotation R by 90 degrees around the z-axes with
with a projection onto the x-z plane.
ANSWER. The kernel of the projection is the y axes. The x axes is rotated into the y axes and therefore the
kernel of T . The image is the x-z plane.
PROBLEM. Can the kernel of a square matrix A be trivial if A 2 = 0, where 0 is the matrix containing only 0?
ANSWER. No: if the kernel were trivial, then A were invertible and A 2 were invertible and be different from 0.
PROBLEM. Is it possible that a 3 × 3 matrix A satisfies ker(A) = R 3 without A = 0?
ANSWER. No, if A = 0, then A contains a nonzero entry and therefore, a column vector which is nonzero.
PROBLEM. What is the kernel and image of a projection onto the plane Σ : x − y + 2z = 0?
ANSWER. The kernel consists of all vectors orthogonal to Σ, the image is the plane Σ.
PROBLEM. Given two square matrices A, B and assume AB = BA. You know ker(A) and ker(B). What can
you say about ker(AB)?
ANSWER. ker(A) is contained in ker(BA). Similar ker(B) is contained in ker(AB). Because AB = BA, the
0 1
kernel of AB contains both ker(A) and ker(B). (It can be bigger: A = B = .)
0 0
A 0
PROBLEM. What is the kernel of the partitioned matrix if ker(A) and ker(B) are known?
0 B
ANSWER. The kernel consists of all vectors (x, y), where x in ker(A) and y ∈ ker(B).
23.
BASIS Math 21b, O. Knill
LINEAR SUBSPACE. A subset X of Rn which is closed under addition and scalar multiplication is called a
linear subspace of Rn .
WHICH OF THE FOLLOWING SETS ARE LINEAR SPACES?
a) The kernel of a linear map. d) the line x + y = 0.
b) The image of a linear map. e) The plane x + y + z = 1.
c) The upper half plane. f) The unit circle.
BASIS. A set of vectors v1 , . . . , vm is a basis of a linear subspace X of Rn if they are
linear independent and if they span the space X. Linear independent means that
there are no nontrivial linear relations ai v1 + . . . + am vm = 0. Spanning the space
means that very vector v can be written as a linear combination v = a 1 v1 +. . .+am vm
of basis vectors. A linear subspace is a set containing {0} which is closed under
addition and scaling.
1 0 1
EXAMPLE 1) The vectors v1 = 1 , v2 = 1 , v3 = 0 form a basis in the three dimensional space.
0 1 1
4
If v = 3 , then v = v1 + 2v2 + 3v3 and this representation is unique. We can find the coefficients by solving
5
1 0 1 x 4
Ax = v, where A has the vi as column vectors. In our case, A = 1 1 0 y = 3 had the unique
0 1 1 z 5
solution x = 1, y = 2, z = 3 leading to v = v1 + 2v2 + 3v3 .
EXAMPLE 2) Two nonzero vectors in the plane which are not parallel form a basis.
EXAMPLE 3) Three vectors in R3 which are in a plane form not a basis.
EXAMPLE 4) Two vectors in R3 do not form a basis.
EXAMPLE 5) Three nonzero vectors in R3 which are not contained in a single plane form a basis in R 3 .
EXAMPLE 6) The columns of an invertible n × n matrix form a basis in R n as we will see.
REASON. There is at least one representation because the vectors
FACT. If v1 , ..., vn is a basis, then every
vi span the space. If there were two different representations v =
vector v can be represented uniquely
a1 v1 +. . .+an vn and v = b1 v1 +. . .+bn vn , then subtraction would
as a linear combination of the vi .
lead to 0 = (a1 − b1 )v1 + ... + (an − bn )vn . This nontrivial linear
v = a 1 v1 + . . . + a n vn .
relation of the vi is forbidden by assumption.
FACT. If n vectors v1 , ..., vn span a space and w1 , ..., wm are linear independent, then m ≤ n.
REASON. This is intuitively clear in dimensions up to 3. You can not have more then 4 vectors in space which
are linearly independent. We will give a precise reason later.
| | |
A BASIS DEFINES AN INVERTIBLE MATRIX. The n × n matrix A = v1 v2 . . . vn is invertible if
| | |
and only if v1 , . . . , vn define a basis in Rn .
EXAMPLE. In the example 1), the 3 × 3 matrix A is invertible.
FINDING A BASIS FOR THE KERNEL. To solve Ax = 0, we bring the matrix A into the reduced row echelon
form rref(A). For every non-leading entry in rref(A), we will get a free variable t i . Writing the system Ax = 0
with these free variables gives us an equation x = i ti vi . The vectors vi form a basis of the kernel of A.
24.
REMARK. The problem to find a basis for all vectors wi which are orthogonal to a given set of vectors, is
equivalent to the problem to find a basis for the kernel of the matrix which has the vectors w i in its rows.
FINDING A BASIS FOR THE IMAGE. Bring the m × n matrix A into the form rref(A). Call a column a
pivot column, if it contains a leading 1. The corresponding set of column vectors of the original matrix A
form a basis for the image because they are linearly independent and are in the image. Assume there are k of
them. They span the image because there are (k − n) non-leading entries in the matrix.
REMARK. The problem to find a basis of the subspace generated by v 1 , . . . , vn , is the problem to find a basis
for the image of the matrix A with column vectors v 1 , ..., vn .
EXAMPLES.
1) Two vectors on a line are linear dependent. One is a multiple of the other.
2) Three vectors in the plane are linear dependent. One can find a relation av 1 + bv2 = v3 by changing the size
of the lengths of the vectors v1 , v2 until v3 becomes the diagonal of the parallelogram spanned by v 1 , v2 .
3) Four vectors in three dimensional space are linearly dependent. As in the plane one can change the length
of the vectors to make v4 a diagonal of the parallelepiped spanned by v 1 , v2 , v3 .
0 0 1 1 1 0
EXAMPLE. Let A be the matrix A = 1 1 0 . In reduced row echelon form is B = rref(A) = 0 0 1 .
1 1 1 0 0 0
To determine a basis of the kernel we write Bx = 0 as a system of linear equations: x + y = 0, z = 0. The
variable. With y t, x = −t is fixed. The linear system rref(A)x = 0 is solved by
variable yis the free =
x −1 −1
x = y = t 1 . So, v = 1 is a basis of the kernel.
z 0 0
EXAMPLE. Because the first and third vectors in rref(A) are columns with leading 1's, the first and third
0 1
columns v1 = 1 , v2 = 0 of A form a basis of the image of A.
1 1
WHY DO WE INTRODUCE BASIS VECTORS? Wouldn't it be just
easier to look at the standard basis vectors e1 , . . . , en only? The rea-
son for more general basis vectors is that they allow a more flexible
adaptation to the situation. A person in Paris prefers a different set
of basis vectors than a person in Boston. We will also see that in many
applications, problems can be solved easier with the right basis.
For example, to describe the reflection of a ray at a
plane or at a curve, it is preferable to use basis vec-
tors which are tangent or orthogonal. When looking
at a rotation, it is good to have one basis vector in
the axis of rotation, the other two orthogonal to the
axis. Choosing the right basis will be especially im-
portant when studying differential equations.
1 2 3
A PROBLEM. Let A = 1 1 1 . Find a basis for ker(A) and im(A).
0 1 2
1 0 −1 1
SOLUTION. From rref(A) = 0 1 2 we see that = v = −2 is in the kernel. The two column vectors
0 0 0 1
1
1 , 2, 1, 1 of A form a basis of the image because the first and third column are pivot columns.
0
25.
DIMENSION Math 21b, O. Knill
REVIEW LINEAR SPACE. X linear space: 0 ∈ X REVIEW BASIS. B = {v1 , . . . , vn } ⊂ X
closed under addition and scalar multiplication. B linear independent: c1 v1 +...+cn vn = 0 implies
c1 = ... = cn = 0.
Examples: Rn , X = ker(A), X = im(A) are linear B span X: v ∈ X then v = a1 v1 + ... + an vn .
spaces. B basis: both linear independent and span.
BASIS: ENOUGH BUT NOT TOO MUCH. The spanning
condition for a basis assures that there are enough vectors
to represent any other vector, the linear independence condi-
tion assures that there are not too many vectors. A basis is,
where J.Lo meets A.Hi: Left: J.Lopez in "Enough", right "The
man who new too much" by A.Hitchcock
AN UNUSUAL EXAMPLE. Let X be the space of polynomials up to degree 4. For example
p(x) = 3x4 + 2x3 + x + 5 is an element in this space. It is straightforward to check that X is a
linear space. The "zero vector" is the function f (x) = 0 which is zero everywhere. We claim that
e1 (x) = 1, e2 (x) = x, e3 (x) = x2 , e4 (x) = x3 and e5 (x) = x4 form a basis in X.
PROOF. The vectors span the space: every polynomial f (x) = c 0 + c1 x + c2 x2 + c3 x3 + c4 x4 is a sum f =
c0 e1 + c1 e2 + c2 e3 + c3 e4 + c4 e5 of basis elements.
The vectors are linearly independent: a nontrivial relation 0 = c 0 e1 + c1 e2 + c2 e3 + c3 e4 + c4 e5 would mean that
c0 + c1 x + c2 x2 + c3 x3 + c4 x4 = 0 for all x which is not possible unless all cj are zero.
DIMENSION. The number of elements in a basis of UNIQUE REPRESENTATION. v1 , . . . , vn ∈ X ba-
X is independent of the choice of the basis. It is sis ⇒ every v ∈ X can be written uniquely as a sum
called the dimension of X. v = a 1 v1 + . . . + a n vn .
EXAMPLES. The dimension of {0} is zero. The dimension of any line 1. The dimension of a plane is 2, the
dimension of three dimensional space is 3. The dimension is independent on where the space is embedded in.
For example: a line in the plane and a line in space have dimension 1.
IN THE UNUSUAL EXAMPLE. The set of polynomials of degree ≤ 4 form a linear space of dimension 5.
REVIEW: KERNEL AND IMAGE. We can construct a basis of the kernel and image of a linear transformation
T (x) = Ax by forming B = rrefA. The set of Pivot columns in A form a basis of the image of T , a basis for
the kernel is obtained by solving Bx = 0 and introducing free variables for each non-pivot column.
EXAMPLE. Let X the linear space from above. Define the linear transformation T (f )(x) = f (x). For
example: T (x3 + 2x4 ) = 3x2 + 8x3 . Find a basis for the kernel and image of this transformation.
SOLUTION.
Because T (e1 ) = 0, T (e2 ) = e1 , T (e3 ) = 2e2 , T (e4 ) = 3e3 , T (e5 ) = 4e4 , the matrix of T is
0 1 0 0 0
0 0 2 0 0 which is almost in row reduced echelon form. You see that the last four columns
are pivot columns. The kernel is spanned by e1 which corresponds to the con-
A= 0 0
0 3 0
stant function f (x) = 1. The image is the 4 dimensional space of polynomials
0 0 0 0 4
of degree ≤ 3.
0 0 0 0 0
Mathematicians call a fact a "lemma" if it is used to prove a theorem and if does not deserve the be honored by
the name "theorem": | 677.169 | 1 |
"Connect the process of problem solving with the content of the Common Core. Mathematics educators have long recognized the importance of helping students to develop problem-solving skills. More recently, they have searched for the best ways to provide their students with the knowledge encompassed in the Common Core State Standards (CCSS). This volume is one in a series from NCTM that equips classroom teachers with targeted, highly effective problems for achieving both goals at once. The 44 problems and tasks for students in this book are organized into the major areas of the high school Common Core: algebra, functions, geometry, statistics and probability, and number and quantity. Examples of modeling, the other main CCSS area, are incorporated throughout. Every domain that is required of all mathematics students is represented. For each task, teachers will find a rich, engaging problem or set of problems to use as a lesson starting point. An accompanying discussion ties these tasks to the specific Common Core domains and clusters they help to explore. Follow-up sections highlight the relevant CCSS Standards for Mathematical Practice that students will engage in as they work on these problems. This book provides high school mathematics teachers with dozens of problems they can use as is, adapt for their classrooms, or be inspired by while creating related problems on other topics. - Publisher."
Matemáticas Estudio y enseñanza (Secundaria | 677.169 | 1 |
Precalculus, 5th Edition
Description
Bob Blitzer has inspired thousands of students with his engaging approach to mathematics, making this beloved series the #1 in the market. Blitzer draws on his unique background in mathematics and behavioral science to present the full scope of mathematics with vivid applications in real-life situations. Students stay engaged because Blitzer often uses pop-culture and up-to-date references to connect math to students' lives, showing that their world is profoundly mathematical.
Table of Contents
P. Prerequisite
P Prerequisites: Fundamental Concepts of Algebra
P.1 Algebraic Expressions, Mathematical Models, and Real Numbers
P.2 Exponents and Scientific Notation
P.3 Radicals and Rational Exponents
P.4 Polynomials
P.5 Factoring Polynomials
P.6 Rational Expressions
1. Functions and Graphs
1.1 Graphs and Graphing Utilities
1.2 Basics of Functions and Their Graphs
1.3 More on Functions and Their Graphs
1.4 Linear Functions and Slope
1.5 More on Slope
1.6 Transformations of Functions
1.7 Combinations of Functions; Composite Functions
1.8 Inverse Functions
1.9 Distance and Midpoint Formulas; Circles
2. Polynomial and Rational Functions
2.1 Complex Numbers
2.2 Quadratic Functions
2.3 Polynomial Functions and Their Graphs
2.4 Dividing Polynomials; Remainder and Factor Theorems
2.5 Zeros of Polynomial Functions
2.6 Rational Functions and Their Graphs
2.7 Polynomial and Rational Inequalities
2.8 Modeling Using Variation
3. Exponential and Logarithmic Functions
3.1 Exponential Functions
3.2 Logarithmic Functions
3.3 Properties of Logarithms
3.4 Exponential and Logarithmic Equations
3.5 Exponential Growth and Decay; Modeling Data
4. Trigonometric Functions
4.1 Angles and Radian Measure
4.2 Trigonometric Functions: The Unit Circle
4.3 Right Triangle Trigonometry
4.4 Trigonometric Functions of Any Angle
4.5 Graphs of Sine and Cosine Functions
4.6 Graphs of Other Trigonometric Functions
4.7 Inverse Trigonometric Functions
4.8 Applications of Trigonometric Functions
5. Analytic Trigonometry
5.1 Verifying Trigonometric Identities
5.2 Sum and Difference Formulas
5.3 Double-Angle, Power-Reducing, and Half-Angle Formulas
5.4 Product-to-Sum and Sum-to-Product Formulas
5.5 Trigonometric Equations
6. Additional Topics in Trigonometry
6.1 The Law of Sines
6.2 The Law of Cosines
6.3 Polar Coordinates
6.4 Graphs of Polar Equations
6.5 Complex Numbers in Polar Form; DeMoivre's Theorem
6.6 Vectors
6.7 The Dot Product
7. Systems of Equations and Inequalities
7.1 Systems of Linear Equations in Two Variables
7.2 Systems of Linear Equations in Three Variables
7.3 Partial Fractions
7.4 Systems of Nonlinear Equations in Two Variables
7.5 Systems of Inequalities
7.6 Linear Programming
8. Matrices and Determinants
8.1 Matrix Solutions to Linear Systems
8.2 Inconsistent and Dependent Systems and Their Applications
8.3 Matrix Operations and Their Applications
8.4 Multiplicative Inverses of Matrices and Matrix Equations
8.5 Determinants and Cramer's Rule
9. Conic Sections and Analytic Geometry
9.1 The Ellipse
9.2 The Hyperbola
9.3 The Parabola
9.4 Rotation of Axes
9.5 Parametric Equations
9.6 Conic Sections in Polar Coordinates
10. Sequences, Induction, and Probability
10.1 Sequences and Summation Notation
10.2 Arithmetic Sequences
10.3 Geometric Sequences and Series
10.4 Mathematical Induction
10.5 The Binomial Theorem
10.6 Counting Principles, Permutations, and Combinations
10.7 Probability
11. Introduction to Calculus
11.1 Finding Limits Using Tables and Graphs
11.2 Finding Limits Using Properties of Limits
11.3 Limits and Continuity
11.4 Introduction to Derivatives
Enhance your learning experience with text-specific study materials.
This title is also sold in the various packages listed below. Before purchasing one of these packages, speak with your professor about which one will help you be successful in your course. | 677.169 | 1 |
Research
Numerical linear algebra, computational mathematics, applied/computational
multivariate statistical analysis and random matrix theory, computational
algebra.
In particular, the
development of efficient algorithms for accurate computations as they
pertain to the these fields. | 677.169 | 1 |
This ebook is available for the following devices:
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Mathematica is today's most advanced technical computing system. It features a rich programming environment, two-and three-dimensional graphics capabilities and hundreds of sophisticated, powerful programming and mathematical functions using state-of-the-art algorithms. Combined with a user-friendly interface, and a complete mathematical typesetting system, Mathematica offers an intuitive easy-to-handle environment of great power and utility.
The Mathematica GuideBook for Symbolics (code and text fully tailored for Mathematica 5.1) deals with Mathematica's symbolic mathematical capabilities. Structural and mathematical operations on single and systems of polynomials are fundamental to many symbolic calculations and they are covered in considerable detail. The solution of equations and differential equations, as well as the classical calculus operations (differentiation, integration, summation, series expansion, limits) are exhaustively treated. Generalized functions and their uses are discussed. In addition, this volume discusses and employs the classical orthogonal polynomials and special functions of mathematical physics. To demonstrate the symbolic mathematics power, a large variety of problems from mathematics and phyics are discussed.
Unique Features: Familiarizes the reader with symbolic mathematics functions in Mathematica for algebra, analysis, as well as orthogonal polynomials and the special functions of mathematical physics and shows how to use them effectively
Detailed discussions of the most frequent symbolic operations: equation solving, differentiation, series expansion, integration and organizing and performing symbolic calculations in mathematica, as compared to paper-and-pencil calculations
Numerous examples from mathematics, physics, and computer science
Clear organization, complete topic coverage, and accessible exposition for both novices and experts
Website for book with additional materials and updates:
Accompanying DVD contains all material in the form of hyperlinked Mathematica notebooks that can be edited and manipulated; striking color graphics and animations are included on the DVD
Michael Trott is a symbolic computation and computer graphics expert. He holds a Ph.D. in theoretical physics and joined the R&D team at Wolfram Research in 1994, the creators of Mathematica. Since 1998, he has been leading the development of the Wolfram Functions Site which features more that 10,000 visualizations and 85,000 formulas and identities, and also allows for semantical searches. | 677.169 | 1 |
Details about The Mathematics of Projectiles in Sport:
The mathematical theory underlying many sporting activities is of considerable interest to both applied mathematicians and sporting enthusiasts alike. Here Professor de Mestre presents a rigorous account of the techniques applied to the motion of projectiles. This equips the reader for the final section of the book in which an enlightening collection of sporting applications is considered, ranging from the high jump to frisbees and soccer to table tennis. The presentation should be accessible to most undergraduate science students and provides an ideal setting for the development of mathematical modeling techniques.
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Rent The Mathematics of Projectiles in Sport 1st edition today, or search our site for other textbooks by Neville de Mestre. Every textbook comes with a 21-day "Any Reason" guarantee. Published by Cambridge University Press.
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Description:
Located at the University of Illinois at Urbana-Champaign, The Office for Mathematics, Science, and Technology Education is primarily interested in creating resources for educators working on these topics. First-time visitors should definitely start by looking through the "Resources" area, as they will find classroom tested exercises that cover basic topics in algebra, trigonometry, and a number of related fields in math. Moving along, the site also features teaching modules that will help educators explain different concepts in technology, which can be most useful, particularly for beginning students. Finally, if visitors to the site have questions, there is a contact form and a place to make suggestions about future material that might be covered in future projects and modules. | 677.169 | 1 |
Browse related Subjects ...
Read More to learn the most up-to-date and efficient algorithms for particular classes of problems. The first sections of each chapter are expository and therefore accessible to master's level graduate students. However, the chapters also contain advanced material on current topics of interest to researchers. For instance there are chapters which describe the polynomial-time linear programming algorithms of Khachian and Karmarkar and the techniques used to solve combinatorial and integer programming problems, an order of magnitude larger than was possible just a few years ago. Overall a comprehensive yet lively and up-to-date discussion of the state-of-the-art in optimization is presented in this book | 677.169 | 1 |
Details about Algebra for College Students:
This text presents a comprehensive introduction to the topics in intermediate- and college-level algebra. It provides students with the basic skills they'll need to succeed in their next mathematics course. As the book progresses, the mathematical language matures, examples grow less structured, and exercises grow more varied, thus improving students problem solving skills.
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Rent Algebra for College Students 6th edition today, or search our site for other textbooks by R. David Gustafson | 677.169 | 1 |
For many manufacturers, the bulk of product-development dollars go to build, test, and refine prototypes, or possibly make repairs to them when things go wrong. In fact, most companies probably hope things go wrong at this stage so they can be corrected because flawed products that make it to market can bring class-action lawsuits.
This is why manufacturers more and more turn to "virtual prototyping" tools that let engineers analyze and visualize certain aspects of designs before they're built. In these tools, the math that describes and runs the simulations resides deep inside the software as rich sets of models and algorithms that represent real systems.
But what about all the work and thought that go into a design at the conceptual stage? Why do engineers still reach for a notepad, pen, and calculator when making critical design decisions? For many, the answer is there are no viable alternatives. Specifically, designers don't believe computers "do math" — equation formulation, algebraic manipulations, simplifications, and so forth — the same way they do.
This is true only if one views the use of spreadsheets or programming languages as "doing math" on computers. And for many " computersavvy" engineers, this is exactly what has happened. I have personally encountered several engineering projects where supporting calculations become locked inside cryptic spreadsheet formulae such as ((B12 + 2*$A$1)/A12)*2.3128.
These formula-based calculation tools tend to contort engineering information into inefficient, if not dangerous, forms. Admiral Harold Gehman, chairman of the Columbia [Space Shuttle] Accident Investigation Board, stated, "Neither NASA nor the board are satisfied with the model. It's essentially an Excel spreadsheet," noting that critical engineering calculations were performed with tools never designed for engineering work.
So-called interactive math systems such as Maple get around the problem. They explicitly manage math information in a way that more closely resembles how engineers work. Such packages have been referred to as "spreadsheets on steroids" for the depth and speed of their math solvers. This is a gross oversimplification because interactive systems differ fundamentally from spreadsheets and most other packages used by engineers to do math calculations. And it is these differences that help engineers be more productive.
Engineers are under ever-greater pressure to bring new products to market better, faster, and cheaper. The ability to enter mathematical problems in an intuitive way and get readable answers can cut the time needed for design calculations from months to days. A platform that has the computational breadth and depth to manage any level of engineering calculation, including dimensions and units, eliminates mathematical errors and lowers the risk of production delays and associated costs.
Increasingly, innovation starts with a mathematical premise that is then developed into a conceptual design. It is impossible in these cases to predetermine the analytical needs of the user because he still doesn't know all the parameters. Interactive math tools give computational headroom for such an approach.
In some ways, productivity gains have less to do with solving a math problem and more with the formulation, management, and deployment of the problem and answer. Engineers do more than calculate and simulate: they explore, collaborate, communicate, test, present and manage a diverse range of information.
It's been our experience that interactive math systems applied to the design process give up to 20X more bang for the buck than in any other stage of product development. And subsequent projects may see even greater benefit when design knowledge can be readily reused.
Maplesoft (maplesoft.com) is a developer of mathematical and analytical software. | 677.169 | 1 |
Related software: 04/13/2015 09:59:23 Computer Science Discrete Mathematics Ebook & Notes Preview of the attached file Discrete mathematics is the study of mathematical structures that are fundamentally ...Looking for books on Discrete Mathematics? Check our section of free e-books and guides on Discrete Mathematics now! This page contains list of freely available E ...Discrete Mathematics [Norman L. Biggs] on Amazon.com. *FREE* shipping on qualifying offers. The long-awaited second edition of Norman Bigg's best-selling Discrete ...Susanna Epp's DISCRETE MATHEMATICS WITH APPLICATIONS, FOURTH EDITION provides a clear introduction to discrete mathematics. Renowned for her lucid, …Buy Number Theory and Discrete Mathematics journals, books & electronic media online at Springer. Choose from a large range of academic titles in the Mathematics ...Discrete Mathematics with Algorithms by M. O. Albertson and J. P. Hutchinson Downloadable Version Chapters 1 through 8, as well as the Solutions to Questions …Discrete Mathematics by W W L Chen - free book at E-Books Directory - download hereviii Discrete Mathematics Demystified 3.4 Further Ideas in Elementary Set Theory 47 Exercises 49 CHAPTER 4 Functions and Relations 51 4.1 A Word About Number …Discrete Mathematics could be a scoring subject for GATE computer science aspirants .Its not a kind of subject which you can start studying 48 hours in advance .From the reviews: "Pace's Mathematics of Discrete Structures for Computer Science is a delightful book, written to be read and enjoyed as it moves from justifying ... discrete mathematics ebook | 677.169 | 1 |
Find a Kingwood, TX GeometryThomas and received an A in the course. Linear Algebra is the study of matrices and their properties. The applications for linear algebra are far reaching whether you want to continue studying advanced algebra or computer science | 677.169 | 1 |
MAT0056 Foundations of College Mathematics I, Module Course
View in Catalog 2credit hour(s)Fall | Spring | Summer Prerequisite(s): Appropriate placement and initial diagnostic assessment scores, or voluntary enrollment. Students are highly encouraged to consult with an academic advisor before choosing their developmental mathematics course. This is a self-paced, computer-mediated course designed to allow students who have shown mastery of most of the first level of developmental mathematics the opportunity to accelerate the course. Students are given the ALEKS Initial Assessment to identify skills in the developmental math sequence not yet mastered. An individual learning plan is established and students are assigned to the module(s) containing the competencies not yet mastered. The individualized program is administered through the ALEKS computer-based learning and assessment platform. Through individualized learning plans, students show mastery of fundamental arithmetic and algebraic concepts necessary for
. Successful completion of this course requires students complete all prescribed modules and achieve a score of 70% or greater on the course assessment. Successful completion of the course leads to one of the second-level developmental mathematics course options. Does not count toward A.A. degree or A.S | 677.169 | 1 |
Details about Elements of Mathematics for Economics and Finance:
This book equips first-year undergraduates with the mathematical skills, facts and terminology required for degrees in economics, finance, management and business studies. It is especially suitable for those who did not progress past GCSE and who have had a break of at least two years from mathematics; such students often lack confidence in handling mathematical concepts so the aim of this book is to provide a basic text that focuses strongly on examples, while giving sufficient attention to the exposition of the principal constructions and theoretical results.The text starts with basic principles and leads as far as constrained optimisation, with several entry points to accommodate students with differing mathematical backgrounds. The fundamental ideas are described in the simplest mathematical terms and developed at an easy pace; the text touches on ideas, introduces them gently and then uses basic illustrative examples and exercises (with solutions) to show how these ideas may be brought to bear on problems in economics and finance.This text will serve as a handbook of mathematical techniques for first-year undergraduate in economics, finance, management science and business studies, but it will also be a useful reference for students on MBA courses.
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Rent Elements of Mathematics for Economics and Finance 1st edition today, or search our site for other textbooks by Tim Phillips. Every textbook comes with a 21-day "Any Reason" guarantee. Published by Springer.
Need help ASAP? We have you covered with 24/7 instant online tutoring. Connect with one of our tutors now. | 677.169 | 1 |
Quick Algebra
Quick Algebra is a math application that you can use to solve simple linear algebra problems, such as analyzing matrices or systems of equations, without saying no to a user-friendly interface and nice aesthetics.
What it can do:
* Find the determinant of a square matrix
* Find the rank of a matrix
* Find the inverse of a square matrix
* Solve a system of linear equations
* Analyze an isometry in a vector or Euclidean space
* Simplify (through Gauss' algorithm) a matrix or a system of linear equations | 677.169 | 1 |
Mathematics: Classification of Algebra
Universal algebra (sometimes called general algebra) is the field of mathematics that studies algebraic structures themselves, not examples ("models") of algebraic structures. For instance, rather than take particular groups as the object of study, in universal algebra one takes "the theory of groups" as an object of study. Subject area topics include:
Basic idea
Varieties
Basic construction
Some basic theorems
Motivations and applications
Algebraic geometry is a branch of mathematics which combines techniques of abstract algebra, especially commutative algebra, with the language and the problems of geometry. It occupies a central place in modern mathematics and has multiple conceptual connections with such diverse fields as complex analysis, topology and number theory. Subject area topics include:
Zeros of simultaneous polynomials
Affine varieties
Regular functions
The category of affine varieties
Projective space
The modern viewpoint
History
Applications
Functional analysis is the branch of mathematical analysis concerned with the study of the vector spaces in which limit processes can be defined, and the linear operators acting upon these spaces that are (in some way) compatible with these limits. Subject area topics include:
Normed linear spaces
Banach spaces
Hilbert spaces
Banach algebras
Normed algebras
Topological algebras
Topological groups
Elementary algebra is a fundamental and relatively basic form of algebra taught to students who are presumed to have little or no formal knowledge of mathematics beyond arithmetic. The major difference between algebra and arithmetic is the inclusion of variables. While in arithmetic only numbers and their arithmetical operations (such as +, -, ×, ÷) occur, in algebra, one also uses symbols such as x and y, or a and b to denote variables. Subject area topics include:
Real number system
Constants
Variables
Mathematical expressions
Equations
Intermediary algebra
College-level algebra
Abstract algebra is the subject area of mathematics that studies algebraic structures, such as groups, rings, fields, modules, vector spaces, and algebras. The phrase abstract algebra was coined at the turn of the 20th century to distinguish this area from what was normally referred to as algebra, the study of the rules for manipulating formulae and algebraic expressions involving unknowns and real or complex numbers, often now called elementary algebra. Subject area topics include:
Algebraic structures
Groups
Rings
Fields
Axiomatically
Linear algebra is a branch of mathematics concerned with the study of vectors, with families of vectors called vector spaces or linear spaces, and with functions that input one vector and output another, according to certain rules. These functions are called linear maps (or linear transformations or linear operators) and are often represented by matrices. Linear algebra is central to modern mathematics and its applications. Subject area topics include:
Vector spaces
Matrices
Algebraic combinatorics is an area of mathematics that employs methods of abstract algebra, notably group theory and representation theory, in various combinatorial contexts and, conversely, applies combinatorial techniques to problems in algebra. Subject are topics include:
Enumerative
Matroids
Polytopes
Partially ordered sets
Finite geometries
Algebraic number theory is a major branch of number theory which studies algebraic structures related to algebraic integers. This is generally accomplished by considering a ring of algebraic integers O in an algebraic number field K/Q, and studying their algebraic properties such as factorization, the behavior of ideals, and field extensions. In this setting, the familiar features of the integers – such as unique factorization – need not hold. The virtue of the primary machinery employed – Galois theory, group cohomology, group representations, and L-functions – is that it allows one to deal with new phenomena and yet partially recover the behavior of the usual integers. Subject area topics include: | 677.169 | 1 |
Lesson Overview: As you've already read, this is a self-contained module for calculus students. One-dimensional kinematics is typically covered in beginning calculus courses, whether the level is high-school honors, AP, or collegiate. This module could take 1-2 days depending on the readiness of the students and their allotted time. I've included a guided worksheet for the students to complete as they go through the module. See the next page for copies of this, as well as the key.
The module's primary emphasis is on the notions of rate of change, critical points, and a graphical understanding of the differences between position, velocity, and acceleration. Moreover, because this module was completed as a result of a research experience, I've dedicated a portion of the site explaining how applied math connects with real-world phenomena, as well as other sciences. While this content isn't necessary for an understanding of the essential ideas, I believe that it's important because students should be exposed to the nature of research in math and science! For more information regarding the nature of mathematics, I encourage you to check out the official benchmarks.
2.03, Interpret the derivative as a function: translate between verbal and algebraic descriptions of equations involving derivatives.
2.05, Interpret the second derivative: identify the corresponding characteristics of the graphs of ƒ, ƒ', and ƒ".
2.06, Apply the derivative in graphing and modeling contexts: interpret the derivative as a rate of change in varied applied contexts, including velocity, speed, and acceleration.
Offline Lessons: If you don't have access to the necessary technology, I've also included complete lesson-plans that form a unit on 1-D kinematics. Both lessons are designed for a 90-minute block schedule, and each includes a student note-sheet. | 677.169 | 1 |
Products in Intermediate
Algebra presents the essentials of algebra with some applications. The emphasis is on practical skills, problem solving, and computational techniques. Topics covered range from equations and inequalities to functions and graphs, polynomial and rational functions, and exponentials and logarithms. Trigonometric functions and complex numbers are also considered, together with exponentials and logarithms.
Comprised of eight chapters, this book begins with a discussion on the fundamentals of algebra, each topic explained, illustrated, and accompanied by an ample set of exercises. The proper use of...
Intermediate Algebra focuses on the principles, operations, and approaches involved in intermediate algebra.
The book first elaborates on basic properties and definitions, first-degree equations and inequalities, and exponents and polynomials. Discussions focus on the greatest common factor and factoring by grouping, factoring trinomials, special factoring, equations with absolute value, inequalities involving absolute value, formulas, first-degree equations, graphing simple and compound inequalities, and properties of real numbers. The text then takes a look at rational expressions, rational...
College Algebra and Trigonometry, Second Edition provides a comprehensive approach to the fundamental concepts and techniques of college algebra and trigonometry.
The book incorporates improvements from the previous edition to provide a better learning experience. It contains chapters that are devoted to various mathematical concepts, such as the real number system, the theory of polynomial equations, trigonometric functions, and the geometric definition of each conic section. Progress checks, warnings, and features are inserted. Every chapter contains a summary, including terms and symbols with...
Beginning Algebra: A Text/Workbook, Second Edition focuses on the principles, operations, and approaches involved in algebra.
The publication first elaborates on the basics, linear equations and inequalities, and graphing and linear systems. Discussions focus on solving linear systems by graphing, elimination method, graphing ordered pairs and straight lines, linear and compound inequalities, addition and subtraction of real numbers, and properties of real numbers. The text then examines exponents and polynomials, factoring, and rational expressions. Topics include multiplication and division...
Algebra for College Students, Revised and Expanded Edition is a complete and self-contained presentation of the fundamentals of algebra which has been designed for use by the student.
The book provides sufficient materials for use in many courses in college algebra. It contains chapters that are devoted to various mathematical concepts, such as the real number system, sets and set notation, matrices and their application in solving linear systems, and notation of functions. The theory of polynomial equations, formulas for factoring a sum and a difference of cubes, roots of polynomials, and the...
Elementary Algebra, Third Edition focuses on the basic principles, operations, and approaches involved in elementary algebra.
The book first ponders on the basics, linear equations and inequalities, and graphing and linear systems. Discussions focus on the elimination method, solving linear systems by graphing, word problems, addition property of equality, solving linear equations, linear inequalities, addition and subtraction of real numbers, and properties of real numbers. The text then takes a look at exponents and polynomials, factoring, and rational expressions. Topics include reducing rational...
Essentials of Psychology introduces contemporary psychological research and caters to the varied needs of students and instructors. The book is composed of 14 basic chapters, which provide comprehensive coverage of theories and research within each of the traditional areas of psychology. Chapters are dedicated to topics that discuss the major divisions of psychology; the physiological basis of behavior; the ways people change and the ways they stay the same over time; personality and behavior assessment; and treatment of psychological problems. Psychologists, students, and teachers of psychology...
With the help of this quick study guide, your teen should be able to breeze through 12th grade algebra. There will be principles explained in an easy-to-understand manner as well as plenty of examples to instill concepts in the memory. Using this guide, expect to see A's the next time your teen brings home his/her test paper. Grab a copy today!
College algebra is just one of those things that have kept students in college longer than expected because it's mind-boggling. But it can be made solvable with the use of this quick study guide. With its brilliant explanations and numerical representations, this book will likely take your understanding of the subject to a whole new level. Secure a copy now.
Foundations of Galois Theory is an introduction to group theory, field theory, and the basic concepts of abstract algebra. The text is divided into two parts. Part I presents the elements of Galois Theory, in which chapters are devoted to the presentation of the elements of field theory, facts from the theory of groups, and the applications of Galois Theory. Part II focuses on the development of general Galois Theory and its use in the solution of equations by radicals. Equations that are solvable by radicals; the construction of equations solvable by radicals; and the unsolvability by radicals...
Lie Algebras is based on lectures given by the author at the Institute of Mathematics, Academia Sinica. This book discusses the fundamentals of the Lie algebras theory formulated by S. Lie. The author explains that Lie algebras are algebraic structures employed when one studies Lie groups. The book also explains Engel's theorem, nilpotent linear Lie algebras, as well as the existence of Cartan subalgebras and their conjugacy. The text also addresses the Cartan decompositions and root systems of semi-simple Lie algebras and the dependence of structure of semi-simple Lie algebras on root systems....
Lectures in General Algebra is a translation from the Russian and is based on lectures on specialized courses in general algebra at Moscow University. The book starts with the basics of algebra. The text briefly describes the theory of sets, binary relations, equivalence relations, partial ordering, minimum condition, and theorems equivalent to the axiom of choice. The text gives the definition of binary algebraic operation and the concepts of groups, groupoids, and semigroups. The book examines the parallelism between the theory of groups and the theory of rings; such examinations show the...
Introduction to Abstract Algebra provides insight into the methods of abstract algebra. This book provides information pertinent to the fundamental concepts of abstract algebra.
Organized into five chapters, this book begins with an overview of the study of natural numbers that are used historically for the purpose of counting the objects in different assemblages. This text then examines the concepts of set and elements of a set. Other chapters contain an intuitive survey of the different kinds of real numbers, with the inclusion of many very important results on integers. This book presents...
Intended for a serious first course or a second course, this textbook will carry students beyond eigenvalues and eigenvectors to the classification of bilinear forms, to normal matrices, to spectral decompositions, and to the Jordan form. The authors approach their subject in a comprehensive and accessible manner, presenting notation and terminology clearly and concisely, and providing smooth transitions between topics. The examples and exercises are well designed and will aid diligent students in understanding both computational and theoretical aspects. In all, the straightest, smoothest path...
A Selection of Problems in the Theory of Numbers focuses on mathematical problems within the boundaries of geometry and arithmetic, including an introduction to prime numbers. This book discusses the conjecture of Goldbach; hypothesis of Gilbreath; decomposition of a natural number into prime factors; simple theorem of Fermat; and Lagrange's theorem. The decomposition of a prime number into the sum of two squares; quadratic residues; Mersenne numbers; solution of equations in prime numbers; and magic squares formed from prime numbers are also elaborated in this text. This publication is a good...
Mathematical Analysis: Differentiation and Integration is devoted to two basic operations of mathematical analysis, differentiation and integration. The problems directly connected with the operations of differentiation and integration of functions of one or several variables are discussed, together with elementary generalizations of these operations. This volume is comprised of seven chapters and begins by considering the differentiation of functions of one variable and of n variables, paying particular attention to derivatives and differentials as well as their properties. The next chapter... | 677.169 | 1 |
Find a Great Neck Algebra 1...Included in the course were:
1. Methodsefficients and Variation of Parameters.
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Content
This unit will enable students to understand the basic and more advanced mathematics and statistics involved in the finance world. It introduces financial mathematical concepts which underpin the financial decision making process. In addition the unit covers financial markets, portfolio management and econometric methods | 677.169 | 1 |
Students should know that it doesn't take much to put the "fun" in "function." Really, all it takes is writing "ction" afterwards.
The Functions standards introduce students to the idea of an input-output function, different examples of functions (from linear to trigonometric), and various ways of expressing functions (verbally, algebraically, and graphically). | 677.169 | 1 |
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As has been noted, this book dates from an era (mid 70s to early 80s) when A Level Maths was a more uncompromising preposition than it is today. It is an excellent book, but most of today's students will struggle with it. It doesn't hold your hand or teach the basics, they are assumed. I recently saw it (along with sister book "Further Pure Mathematics") as one of a few books recommended by Cambridge University at this level. Needless to say that some of the topics are no longer examined, but if you are looking for a rigorous book, that with ability and persistence will give you an excellent grounding, this is it. A bonus is the 1970s A Level questions..compare with today's questions!
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A sound resource for maths A Level. Everything explained very well... Gets you used to understanding and writing properly with identities rather than equals all over the place... Gets you used to understanding proper mathematical terms... An old book with a preface from 1981, but old doesn't mean bad! It's a book that, if you enjoy maths, you'll pick-up to have a read just because you want to find out something new in a concise manner. Great book! I'd recommend it to anyone studying A Level and anybody looking to jump into higher level mathematics!
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Since purchasing this book I have begun, where I left off, prior to losing my original book in transit to my new house, in completing the exercises after each section, in full, and showing the workings of each mathematical problem. I am not trying to outdo the authors, but after one or two examples, each problem sets the students mind working, where am I going wrong, or how did he arrive at this answer, (preview on answers at the end of the book),and I feel, as a mature student, at the end of the day, it would be more practical to show all workings and answers to every exercise problem, with comments, than wander around in the dark thinking, "How did he do that, and where am I going wrong?"
The book itself is a treasure, and an absolute must for the teacher or engineer to be!!
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I had this book when I sat A level maths in the mid 80's. I had no teacher for almost a year, and a various stream of supply teachers who knew less than we did for the remainder of the course. This book was priceless. Comprehensive and clearly written, it is very easy to follow and the examples for explanation were excellent. While it does assume a sound level of basic knowledge, I would have to assume that most students embarking on A-level maths, would have that anyway! So for those of you who have, you will not be disappointed.
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This book does covers the A level syllabus. It is very thorough and if competed and understood will cover the first year of most degree courses. Even though not explained in the book the subject matter has practical uses and is used in Science, Engineering and Finance, so it is well worth the effort in doing the work in the book. This book also has some weaknesses that you need to be aware of when using it. Much of the content has to be tackled in the order of chapters in the book. This is because the techniques used in earlier chapters are not identified or cross-referenced so you have to know them from earlier in the book. It is not possible to dip in and out of the book. The exceptions being the chapters on permutations and combinations and vectors, which can be taken in isolation. The index is poor and many of the terms and not explained i.e. real number, integer, and order of equation. The book needs a glossary. The examples in the book do not explain what is going on i.e. diving by a common factor or by 2 or rearranging an equation. If you find mathematics difficult then you will have to spend hours figuring out what is going on for yourself. The best way to read this book is to start at the beginning. Read and make notes very slowing i.e. one page per hour or less. Do every example, missing nothing out and go to the end of the book. This will take about a year and a half to do. It is necessary to also have a good teacher with this book to explain what is going on. I would recommend that you buy anther 2 books i.e. Pure Mathematics: A First course and A Second Course 3rd Edition series: Longman, authors: John K Backhouse / Peter P Houldsworth / Bay E Cooper / Peter Horril. Pure mathematics was the forerunner to this book and retains a personal approach. This series is aimed at self taught students and has answers to all the questions. The books might not be available on Amazon.
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This is an excellent text covering 'basic' pure maths and is eminently suitable for those who want to sharpen their maths skills without having to pay for a formal course of study. If being used as preparation for AS or A level Maths exams it could usefully be supplemented by the Collins Student Support Materials series.
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This bok is recommended reading for Maths courses at (in particular) Warwick University. It is (accoriding to information recieved) prefered to the new version, Core Maths for A-Level. However, it is not required reading
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this is pretty much 'pure mathematics' vol 1 by the same authors. i just wanted to mention, if you are not actually studying for the exam, this book is perfectly usable. otherwise it's best to stick with books matched to the current A level syllabus. | 677.169 | 1 |
Practice and hone important addition skills with this first book. Select one of twenty math problems with complete solutions that educate the student in the addition process. The book also includes four bonus word problems with complete explanations and answers. Easily navigate the links from the problem list to view the solution. Most appropriate for 4th and 5th grade students Miscellaneous.
This book is for everyone. Starting from 5th standard up to Post Graduation, everyone will find something useful. The Book clears "Fundamentals" and "Basics" of Mathematics with logic and hence strengthens the base. You can do calculations at lightning speed with 100% accuracy. It is extremely useful for beginners and experts alike. A book for students preparing for Quantitative Aptitude test.
Practice adding and subtracting decimal numbers with this secondThis eBook introduces the subject of addition and subtraction practice, the number bonds of 15 and 20, arithmetic codebooks and crosswords, written addition and subtraction (including decimals), adding three numbers and dealing with money introduces the subject of algebra to the student encompassing, inverse operators, equations, the order of precedence, algebraic conventions, BODMAS, expressions, formulae, factorising, rearranging and solving linear, quadratic and simultaneous equations as well as inequalities.
This eBook introduces the subject of multiplication and division practice, reviews some multiplication facts, looks at multiplication grids, and covers the written methods of multiplication and division.
This eBook introduces the subject of differentiation, across this wide-ranging subject, starting with definitions and first principles to developing an understanding and appreciation of the first and second order differentials of the equation y = xn through a development of the equations of the gradient and normal to a curve at a particular point as well as a thorough review of maximum, minimum ..
EasyA Maths Edition: Higher Level is a complete summary of the maths higher level Junior Certificate exam. It contains all the information that students need to do well in their Junior Certificate exam. The information is condensed and easier to understand and it saves students time as they dont have to search a maths books for the information they want to know subjects of addition and subtraction and its practice, numbers from 0 to 999, place value, number lines, counting on and back in 1s, 2s, 3s, 5s and 10s, counting sequences and patterns, addition, subtraction, number machines, number facts, number bonds, what's the difference, adding three numbers and working with money.
This eBook introduces transformation as translations, reflections, rotations and enlargements as individual or composite operations as well as congruence and similarity. We illustrate each translation using right-angled triangles, but the principles developed extend to all 2D shapes as well as to 3D shapes using extensions. Congruence and similarity is similarly explored.
This eBook reviews some advanced topics in algebra, including exploring the nature of polynomials, functions, equations and identity's, examining the mathematical nomenclature used in multiplication and division. We consider multiplying out brackets, taking out common factors, manipulating algebraic fractions and simplifying expressions. We include an extensive selection of questions.
what do you want to known about how to learn multiplication and division in china , you may look this .Just a multiplication table. It will make your ability improve in math .You will find calculation is geting easier. again introduces the student to number patterns and sequences, including odd and even numbers, square numbers, square roots, cube numbers, cube roots, prime numbers, linear sequences, square number and cube number sequences, Fibonacci number sequences, triangular number sequences and sequences of the powers of 2, 3, 4, 5 and 10. | 677.169 | 1 |
Mathematics Preparation Program
Sponsored by the College of Mathematics, Natural Sciences, and Technology
FOR INCOMING FRESHMEN AND HIGH SCHOOL STUDENTS:
Get a head start! Register for MP2 and get the support you need to be successful in college.
As an incoming freshman, you are eligible to register for the Mathematics Preparation Program (MP2) before you officially begin your college career.
MP2 offers an on-line mathematics course that will prepare students for college-level mathematics using a mastery approach to learning mathematics. The mastery approach allows students to "test out" of content areas they already know. When students experience roadblocks, they will receive various forms of support, such as video tutorials, guided assistance for homework problems and on-line tutoring, before being provided additional opportunities to "test in" to new content areas. Additionally, students will participate in on-line discussion with a mathematics instructor using video conference capabilities, have access to peer tutors on-line, and will take 2 in-person exams. The exams can be taken on the Dover Campus on a specified date, or can be taken at their former high school (pending approval). These activities will help students build a solid foundation of fundamental mathematical concepts and procedures necessary for success in college mathematics courses and college courses with a quantitative reasoning emphasis. Students may earn college credits toward graduation depending on the content areas they successfully completed on-line.
Students attending the MP2 summer course will…
Build effective study skills and learn the mathematics necessary to be successful in their college courses
Build competence and confidence as they learn to be independent and responsible while using the self-paced on-line approach
Improve their understanding of mathematical rules, formulas, and equations.
Program Options
MP2 offers the convenience of flexible registration options.
See application for program cost and details.
MP2 Summer Course Information
8-week on-line Summer Program for any incoming DSU freshmen, or
5-week on-campus Summer Program for any local Delaware incoming DSU freshmen
Application deadlines…
See application for deadlines
Space is limited, so apply early.
About the Lecture Series
The Dean's Distinguished Lecturer Series occurs several times throughout the academic year. Featuring foremost speakers and researchers, these talks are sponsored by the Office of the Dean of the College of Mathematics, Natural Sciences and Technology and are free and open to the public. A more specialized, technical lecture, geared toward science majors, faculty, professionals (and other interested persons) is generally scheduled to take place during the morning hours on the day of the public, evening talk.
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Dr. Gary L. Collins
College of Mathematics, Natural Sciences & Technology
Teaching Assistant Program
A program to increase retention and graduation rates of students in the STEM (Science, Technology, Engineering and Mathematics) disciplines
The Science and Mathematics Skills-Development Program, a program initially funded by Title III, supports the incorporation of teaching assistants in several of the gatekeeper courses. The aim of the program supports the College's initiative to add 1 - to - 2 hours of student exposure to mathematics and science coursework. The goal of these additional recitations and study sessions is to enhance students' understanding of the course materials through an enhancement of their problem-solving skills and scientific reading and writing skills. Offered during the fall and spring semesters, these sessions are offered in small-class settings (averaging about 10 students, with an upper limit of 15 students) and are supervised by qualified instructors who have demonstrated their mastery of the material to be taught.
Teaching Assistants (TA) for the program consist of a mix of adjunct professors, graduate students, and/or talented junior and senior undergraduate students. TAs receive compensation, enhanced teaching skills and are afforded opportunities for professional development. For more information about this program, please contact your department chairperson or department designee, or Ms. Vanessa D. Nesbit.
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Check with your instructor to find out if a TA is available for your course
Student support programs for science majors at DSU
2016-2017 Scholarships & Programs Guide/Application
(download here)
Funding your college education, and emerging from the college experience fully prepared for the 21st century STEM workplace, is, by far, one of the most difficult challenges you will face. There are many decisions to make that will potentially impact your future, in a very short amount of time.
You already know that getting good grades is important, but what else does it take to be "competitive" in the current economic climate? What more should you be doing to better position yourself to further your education? What should you be doing to develop professionally, in your field? What scholarships, awards and internships should you actively pursue?
At Delaware State University, your advisor, along with the faculty and staff of the CMNST, are part of a TEAM that is ready and willing to advise and help guide you in what the best possible path is for YOU.
Our College offers information on several support programs for students of STEM (Science, Technology, Engineering and Mathematics) disciplines seeking college funds and research training opportunities:
AMRC — Applied Mathematics Research Center Research opportunities in applied mathematics, scholarships, enrichment programs (Contact the Mathematical Sciences Dept for details.)
Graduate Partnership Award Graduate student tuition and stipend support for individuals pursuing doctoral degrees in select courses of study in biomedical sciences
OSCAR (Optical Science Center for Applied Research)/CREOSA (Center for Research and Education in Optical Sciences and Applications) Student research program with an emphasis on lasers and optics; scholarships, internships available
Delaware EPSCoR RII — Experimental Program to Stimulate Competitive Research Student research opportunities in biological and environmental sciences for conducting year-round or summer research working with faculty research mentors in both the College of Agriculture & Related Sciences and the College of Mathematics, Natural Sciences and Technology
Bridge to the Doctorate: Greater Philadelphia AMP — Alliance for Minority Participation in the Sciences Scholarships for graduate/doctoral studies, career preparation, networking opportunities. (Contact Dr. Mazen Shahin for details.)
HSRC — Hydrogen Storage Research Center Studies in hydrogen storage and ecologically-friendly fuel storage options, scholarships and internships available. (Contact the Chemistry Dept for details.)
INBRE — Delaware INBRE - IDeA Network of Biomedical Research Excellence Undergraduate and graduate student stipends available in biomedical science-related majors, research opportunities, enrichment programs
MARC U*STAR - Maximizing Access to Research Careers Undergraduate research training program for rising juniors and seniors interested in obtaining PhD or PhD/MD degrees in biomedical research disciplines. Annual stipend; paid, intensive on-campus and external research experiences; participation in seminars on professional development; opportunities to present; travel funding to ABRCMS.
Mathematics Preparation Program (MP2) Preparatory course open to all incoming freshmen prior to taking college-level mathematics.
Title III Revitalizing Education and Excellence in Mathematics and Science (REEMS) Program Teaching assistantships for adjuncts, graduate and talented undergraduate students to support students taking STEM gatekeeper courses. By recommendation of chair and graduate director only.
Robert Noyce Teacher Scholarship Program Scholarships to provide financial support to STEM juniors for three (3) years. Concentration must be in teaching K-12 STEM subjects.
Additional Resources:
ROTC — Army ROTC/ Air Force ROTC Program Leadership training in preparation for Army, scholarships and stipends available
Work-study/Financial Aid Needs-based criteria, based on FAFSA eligibility, as determined by Financial Aid Office*
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* Scholarships: Administered through Financial Aid Office
Science Programs Just for Youth K-12!
"Every great advance in science has issued from a new audacity of imagination."
~John Dewey, The Quest for Certainty, 1929
The College of Mathematics, Natural Sciences and Technology is committed to providing a high-quality science-based education -- not just for university-enrolled students!
We are faculty, staff and students actively involved in inspiring and enriching the lives of the youth in our immediate (and extended) community aspiring towards careers in the natural and physical sciences through various outreach efforts. The programs listed below are available "just for youth K-12." They range from open-enrollment notices for summer camps, to information about new programs and scholarships available to graduating high school students specifically considering the sciences at DSU. Please click on the links to download information or feel free to contact the program administrators directly with inquiries.
Science Events (not necessarily during the summer)
Summer Programs
$$$ for College
Science Events (not necessarily during the summer)
Are you a student who has an interest in science, technology, engineering and/or mathematics? Then you won't want to miss this! Join us for the 4th annual Kent County Delaware Science Fair! Senator Chris Coons and a special guest speaker will be on hand to welcome and celebrate our students grades 5-12 who will show off their aptitude for science research and compete for cool prizes.
Whether you would like to compete in the research expo or you just want to come see what your peers are up to, we hope to see you there! Visit for more details.
Summer Programs
What will YOU do this summer??
Now accepting applications, the Research and Engineering Program (REAP) is a 5-wk, non-residential research program open to h.s. students (grades 10-12) interested in science, technology, engineering and mathematics. Download flier.
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STARS at DSU!
The Science and Technology Academy for Residence Scholars (STARS) program is a four-week, residential enrichment program on the campus of Delaware State University (DSU) for rising 10th – 12th grade high school students. The STARS program is designed to stimulate and extend the interest of high school students in the fields of mathematics, science, and information technology. Students will work in small group settings in science labs, computer labs, and classrooms at DSU, participating in hands-on activities in Mathematics, Science, Information Technology, English, and SAT Preparation. A limited number of scholarships are available. Dr. Mazen Shahin, Program Director.
Click here to visit website and apply. Download flier.
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College $$$ for Science Students
Scholarship in Mathematics and Science (SIMS) Available to Incoming Freshmen
Scholarships are available to qualified incoming freshmen at DSU and offer two years of support to students who continue to meet eligibility requirements. Support is offered to meet the financial need of students defined by the Department of Education. (Cost of Attendance – Estimated Family Contribution) up to $8000. Click here to download application.
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Science and Technology Academy for Residence Scholars
STARS 2016
The Science and Technology Academy for Residence Scholars (STARS) program is a two-week residential enrichment program on the campus of Delaware State University (DSU) for rising 10th – 12th grade high school students. The STARS program is designed to stimulate and extend the interest of high school students in the fields of mathematics, science and information technology. The science labs, computer labs and classrooms at DSU will be the setting for small group, hands-on activities in Mathematics, Science, Information Technology, and SAT Preparation. The majority of STARS instructors are DSU professors.
The main objectives of the STARS program are:
To increase science and mathematics-related knowledge, skills and confidence in a group of high school students.
To increase the participants' interest in and engagement with mathematics, science and technology and increase the likelihood that they will major in a science field in college.
To increase the student's familiarity with the large variety of careers in science.
Hear what people are saying about STARS!
See more from our students at: STARS 2014 Blog
2015 STARS was partially funded by:
*Tensor-SUMMA of Mathematical Association of America
*I Could Do Great Things Foundation
Eligible applicants
We are recruiting students who...
(1) Are rising 10th, 11th, and 12th grade students;
(2) Have a GPA average of B or above;
(3) Have an interest in math and science;
(4) Are motivated and enthusiastic;
(5) Would like to experience college life; and
(6) Would enjoy working with a diverse group of faculty and peers.
Program Highlights:
Cost: $1,000 per student
On-campus living arrangements
Cafeteria meal plan
Small group, hands-on activities in Mathematics, Science, Information Technology, and SAT Preparation
Field trips
Collegiate experience
Opportunity to qualify for future scholarships to attend Delaware State University
Program dates:
The program runs Sunday, July 10 - Saturday, July 23, 2016. Apply by April 1st.
Printable application is available here.
Program dates: July 10, 2015 – July 23, 2016
Applications due April 1, 2016
Download flier here
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STARS Program Contact:
Program Overview
Thank you for your interest in Delaware State University's Graduate Partnership Award Program. This program, sponsored by the National Institutes of Health Bridge-to-the-Doctorate program, establishes a "bridge," helping students to obtain a Ph.D. in Biomedical Science by providing the financial support to pursue graduate level studies, providing faculty mentors to guide students academically and professionally, and surrounding students with a support network of staff assisting them in accomplishing their goals. The program seeks to encourage students' interest in graduate study by removing the financial barriers associated with continued study and hesitancy surrounding admission pre-requisites.
The successful applicant will plan to pursue a Ph.D. degree in Biomedical Science, as indicated on the list of sanctioned programs of study (see below). Applicants to the program are evaluated on the basis of their record of achievement, as well as their research experiences, motivation, and connection to their field of study.
We look forward to receiving your application for graduate admission to the Graduate Partnership Award Program. Please forward your inquires to the Project Director, Dr. Melissa Harrington, 302.857.7117 or email: mharrington@desu.edu.
Sanctioned Programs of Study
Master of Science
Biological Sciences
Chemistry
Molecular and Cellular Neuroscience
Natural Resources
Plant Science
Food Science
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Project Staff
Program Overview
DSU's MBRS RISE-Scholar program provides research training and financial support to students majoring in science or mathematics who are interested in pursuing a career in biomedical science.
Academically-promising students selected as RISE-scholars will be awarded $2400 per semester and will be partnered with a faculty member to carry out a research project during the academic year and participate in a paid summer internship. RISE Scholars will become part of a family of research-active faculty, students and graduate students.
RISE-Scholars meet weekly with their research mentors and dedicate 10 to 15 hours per week to their research project. RISE-Scholars must also attend a weekly "Introduction to Laboratory Research" class and complete weekly assignments. During the summer,
RISE-Scholars participate in a 9-week, paid research internship either at DSU or another partnering University.
Selection Criteria & Process
RISE-Scholars must:
Be freshmen or sophomores at DSU in fall 2010
Major in science, mathematics or agricultural science
Have a high school or college GPA > 3.0
Maintain a GPA > 2.8 while in the program
Submit completed applications to:
Dr. Cynthia van Golen, Program Director
Delaware State University
Science Center South 120
1200 N. Dupont Hwy
Dover, DE 19901
Phone: 302-857-7463
Email: cvangolen@desu.edu
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Prentice Hall Math Course 2 Study Guide and Practice Workbook 2004c9New:
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About the Book
A math text creates a path for students - one that should be easy to navigate, with clearly marked signposts, built-in footholds, and places to stop and assess progress along the way. Research-based and updated for today's classroom, Prentice Hall Mathematics is that well-constructed path. An outstanding author team and unmatched continuity of content combine with timesaving support to help teachers guide students along the road to success. | 677.169 | 1 |
A-level Core Mathematics Curriculum
Stage: 5 Challenge Level:
A-level mathematics is currently modular giving students a choice in the maths they study for their AS (1 year course) or A level (extending the AS). Modules are Core pure maths (C1-4), Statistics (S1, S2), Mechanics (M1, M2) and Decision (D1, D2 being algorithms and so on). There are higher modules for further mathematics candidates. These Further Pure modules would be very handy for engineers.
GCSE content
Relevant content from GCSE/KS3 is
Basic number work
Ratios
Units
Arithmetic with fractions, simplest form etc.
Linear equations and graphs
Scientific notion
Areas, volumes, perimeter
Basic data handling / representation
Students who have only done GCSE are likely to struggle with indices, notation and formal manipulation.
AS Mathematics
All AS maths students will do C1 and C2 and one of S1, M1 and D1
For potential biology students, the strong recommendation is to choose the S1 option.
For potential physics or engineering students, the strong recommendation is to choose M1 | 677.169 | 1 |
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Find Us on Social
There are several starting points in the department curriculum. Most students start in the calculus sequence, discrete math, linear algebra, statistics, or in a computer science course.
CALCULUS
We have a three course calculus sequence that is different from AP Calculus and different from the traditional Calc I, II, III sequence found at most colleges and universities. In particular, all three of our courses use functions of multiple variables.
Where should I start in the calculus sequence?
Math 135, Applied Multivariable Calculus I: This course is appropriate for students with no calculus background; however, it is different from a traditional calculus 1 and may be an appropriate course for students with a calculus background who are interested in developing mathematical modeling skills. Course Description.
Math 137, Applied Multivariable Calculus II: This course is recommend for students who have had a successful year of high school calculus (as merely one example, AB calculus with a score of 4 or better on the AP exam). Course Description.
Math 237, Applied Multivariable Calculus III: Students who enter Macalester having taken BC Calculus with a score of 4 or higher on the AP exam are encouraged to start here. Course Description.
How far do I need to go in the calculus sequence?
Mathematics, Physics, and Chemistry majors need to complete the calculus sequence through Math 237. Economics and most Biology students need to complete any one of Math 135, Math 137, or Math 237. Biology majors with a Biochemistry emphasis need to compete Math 137. We encourage you to see the departmental web pages (Biology, Chemistry, Economics, Physics and Astronomy) for these other majors and speak with a member of the department to verify their mathematics requirements.
OTHER STARTING POINTS IN MATHEMATICS
Math 155: Introduction to Statistical Modeling is our introductory statistics course. It is required for the Math major, the AMS major, the Statistics minor, and other majors on campus (including Biology and Economics). Math 155 is a course unique to Macalester, with an emphasis on multivariate modeling, and it cannot be replaced by AP Statistics credits. Course Description.
Math 136: Discrete Mathematics is a good starting point for students who are interested in mathematics and want to try out areas of mathematics that are different than calculus. This course is required for the mathematics major and the computer science major but not the applied mathematics and statistics major. Course Description.
Math 236: Linear Algebra is a good starting point for students who have already completed calculus at the level of Math 137. This course is required for the mathematics major and the applied mathematics and statistics major. Course Description.
STARTING POINTS IN COMPUTER SCIENCE
Comp 123: Core Concepts in Computer Science is the most common introductory course in computer science. It is suitable for students with little or no background in computing, programming, or computer science. This course serves both as a first course in the major and minor as well as an introduction to computer science for those not planning to take further coursework. This course is offered every semester. Course Description.
Comp 120: Computing and Society is a first year seminar course that is offered most fall semesters. It replaces Comp 123 in the major and minor. Course Description.
Comp 124: Object-Oriented Programming and Data Structures: Students who have significant prior experience in computer science may choose to start in Comp 124 with permission of the instructor. Course Description. | 677.169 | 1 |
Details about Complex Analysis:
An introduction to complex analysis for students with some knowledge of complex numbers from high school. It contains sixteen chapters, the first eleven of which are aimed at an upper division undergraduate audience. The remaining five chapters are designed to complete the coverage of all background necessary for passing PhD qualifying exams in complex analysis. Topics studied include Julia sets and the Mandelbrot set, Dirichlet series and the prime number theorem, and the uniformization theorem for Riemann surfaces, with emphasis placed on the three geometries: spherical, euclidean, and hyperbolic. Throughout, exercises range from the very simple to the challenging. The book is based on lectures given by the author at several universities, including UCLA, Brown University, La Plata, Buenos Aires, and the Universidad Autonomo de Valencia, Spain.
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Solving Math Problems: Some Practical Remarks
Overview
This paper describes a method for solving math problems. The basic idea is to combine two things: • First, a simple method for making handwritten notes while thinking about a problem. This method is aimed at supporting – a step-by-step approach to problem solving and – reflective thinking: Better understand and control what you do while solving a problem. • Second, a densely packed cheat sheet with broad advice on math problem solving. At present, this sheet focuses on general methods for problem solving. Later versions may contain material on specific domains like calculus or algebra. You can download this paper • as a .pdf file: • as a .tex file: This is useful if you want to adapt the cheat sheet.
Making Notes
Layout for the notes
Here are some ideas on how to use the notemaking process. • Use a two column layout. (Read on! This is not the two-column proof you may know from geometry lessons.) The paper (in portrait format) is divided by a vertical line. The left part is about two thirds of the the page width and is used for the main body of notes. The right part is used for reflection about what you are doing. You can use the problem name as a headline. • Organize your notes by process stages. As we will see a few paragraphs later, the cheat sheet suggests several process stages, like "Getting Started", "Make a plan" and so on. For each process stage a collection of useful problem solving tools is given - for the "Getting Started" stage you can use tools like "make a diagram", "introduce useful notation" or "make a table of special cases for small numbers". • Use abbreviations for important stages and tools. For example, use "signposts" like "gs" for "Getting Started", "rep" for "Representations" and so on. • Use the reflection column. Here you can add what you think about the material in the main left column. E.g., are you stuck? What can you do about it? What are alternatives to your current approach? The cheat sheet contains a number of useful tools for reflection. You can use tool abbreviations again - a question mark for obstacles or an exclamation mark for insights. Don't worry whether your notes really belong in the left or right column - remember these are your work notes and not a final presentation of your results. The separate reflection column is merely a vehicle to give you better control over what you are doing in your problem solving. • Use a hierarchic layout, linking ideas by short lines. This may be a matter of liking, but again it works well for me. With these lines, I can better see connections between ideas, especially if I add an idea later.
Some details
Here are some remarks about special aspects of notemaking. • Use reflections at least at the end of each stage, and whenever you feel confused. Writing up is often a great help. • Dealing with parallel approaches. Often enough a problem can be tackled via different roads - e.g. using induction or contradiction. You could note both approaches and examine them one after the other, starting with the most promising one, or you can try a parallel strategy and start a separate sheet for each approach. For complex problems, this may be the better process.
• Referencing. Sometimes you have to make a reference to another part of your notes. A simple way to do this is via numbers. Number your pages to manage references between sheets, e.g. 5-3: page 5, remark number 3. Finally, here are two trivial things that work well for me. • Write small. Getting large amounts of information on one sheet is very useful. Perhaps you'll find that larger sheets of paper and finer pencils work well for you. • Write lightly. This makes erasing much easier.
An example
The picture on the next page shows a number of layout essentials.
Remarks on the cheat sheet
• A focus on heuristics. The cheat sheet in its present state contains only heuristic advice. You'll find almost no domain specific information, e.g. about primes or about calculus. • Use of endnotes. The cheat sheet is meant to contain as much information as reasonable and legible (hrmpf?), so it consists mainly of short descriptions of concepts that are by no means self-explanatory. More (but arguably still insufficient) information about some of the cheat sheet items can be found in the endnotes. • Redundant information. You'll find that some items appear more than once. This is to increase chances that you find it quickly in different contexts. • References. A simple ">" points to other paragraphs of the cheat sheet. • Adapt. This is by far the most important point: Adapt this cheat sheet for your own purposes. (Sharing and discussing the result with others might be a good idea.) • Document structure. A This document was designed for easy adaptation, so I tried to keep the L TEX-aspects of the document easy. There are a few parameters you can change to manipulate the cheat sheet, especially the font size and the number of columns. • Download. You can download the .tex file for this document at • More Cheat Sheets. You find more cheat sheets on or or or On the following page you find the cheat sheet.
Advice from tricki.org
Don't start from scratch55 Hunt for analogies56 Mathematicians need to be metamathematicians57 Think about the converse58 Try to prove the opposite59 Look for related problems60 Work on clusters of problems61 Look at small cases62 Try to prove a stronger result63 Prove a consequence first64 Think axiomatically even about concrete objects65 Temporarily suspend rigor66 Turn off all but one of the difficulties67 Simplify your problem by generalizing it68 If you don't know how to make a decision, then don't make it69 If an argument looks promising but needs some technical hypothesis, try assuming that hypothesis for now, but aim to remove it later.70
Use Your Knowledge
use relevant theorems use solved problems use their results use their methods — how to find such material: from memory via the unknown: which material has the same unknown? via explicit search process gather information > — utilize the material: make material applicable to problem modify your problem modify your material
Make a Plan
construct a proof hierarchy:10 proof strategy11 proof tactics proof details — construction tools: ask repeatedly: "How can I do this?"12 work top-down work bottom-up construct intermediate elements of the proof hierarchy construct a penultimate step13 use wishful thinking / make it easier14 — use forward search use backward search — generate seminal ideas for a proof look at related problems > use your knowledge > use important principles > find new approaches >
Stay Functional
talk to people — eat / drink... exercise, physical activity breathe deeply and calmly take a break; sleep work in a new setting — music - make or listen nonmath activity math activity outside your domain flood yourself with new ideas
Persist
work on for just 15 minutes (and repeat this) use coping self talk imagine the work done remember previous successes
Carry Out the Plan
work with care correctness proved? correctness evident? be critical
Reflect
use the reflection column — reflect on the way:15 ask "So what?" collect questions what's the problem / obstacle? what's the conflict? what's your aim? what's your plan? what can you do? can you do something better? — reflect at the end: what worked? what didn't work? and why? use results elsewhere use methods elsewhere — check list of Common Errors>
Common Errors
thinking that is... hasty narrow fuzzy sprawling53 — working without aim working without plan errors in carrying out a plan lack of reflection — check lists of errors54
Notes
1 What is unknown? What are the data? What is the condition? - These fundamental questions are of course due to George Polya's "How to Solve It", Princeton 1988. 2 Use "pr" not only for the initial problem, but for later sub-problems that arise in the course of your work. 3 Use this for collecting the options you have in a given situation. 4 For me, this is an especially useful tool. 5 Cf. the section on representations. 6 Use this to formulate a hypothesis. Then try to proof or disproof it. 7 You can highlight these signposts by drawing a little circle around them. 8 The General Process and much of the rest of this sheet is of course masssively influenced by Polya's "How to Solve It". 9 E.g., replace nasty things with nice ones. Cf. the "wishful thinking tool" below. 10 Build a hierarchy of steps to prove something. Example - Induction: On the top level, we have the induction principle. On the level below, we have the base case (often for n = 0 or n = 1) and the induction step. Then we have arguments for proving the base case and arguments for the induction step. In most cases however, a proof cannot be constructed in such a neat top-down manner - we have to assemble it from several building blocks, using a combination of top-down and bottom-up strategies. To get a better impression of the proof hierarchy idea - Leslie Lamport's article on writing structured proofs is a worthwhile read: 11 The three hierarchy levels strategy, tactics and details are not quite sharp, but nevertheless useful. 12 This and the next tool are of course closely releated to the "working backwards" tool. 13 Ask yourself what might be the last step in your argument, the one that will yield the conclusion. (From: Paul Zeitz, Art and Craft of Problem Solving, New York 1999, chapter 2.2). 14 Try to make difficulties in your problem disappear - "[...] if the problem involves big, ugly numbers, make them small and pretty" - again taken from Paul Zeitz' book. 15 Use the reflection column for applying these items. 16 Looking at more general problems is sometimes useful. In induction for example, you have more to prove - but also more to build on. 17 There are countless websites that illustrate math concepts. For starters, have a look at or 18 This list is largely taken from 19 Major source for this list of "Important Principles": Christian Hesse: Das kleine Einmaleineins des klaren Denkens, Munich 2009 20 Try to reduce the problem to another problem that is already solved. 21 Count something by counting something else. Little Gauss' summation of 1 + ... + 100 is a famous example. 22 Consider even and odd numbers, or more abstract: Try to establish two nonoverlapping classes and extract information from this. 23 If n + 1 items are put into n boxes, there must be at least one box with two or more items. 24 |A ∪ B| = |A| + |B| − |A ∩ B| and the general case for this. 25 Assume the opposite of what is to be shown and develop a contradiction. 26 Besides classical induction, think of more elaborate versions like downward induction. 27 Try to solve a more general problem. 28 Consider special cases. 29 Get insight into the problem by varying several aspects of it. 30 Construct something that remains unchanged under certain transformations. 31 Construct something that can only increase under certain transformations. 32 Tool for contradiction proofs: Show that if a certain solution exists, there should be a smaller one, and then another even smaller, but that such an infinite descent isn't possible for the given problem. 33 Look for symmetries in a given system. 34 Look at extremal elements. 35 Can you reduce a problem to a simpler version of itself? 36 Colour your problem and derive information from this. Example: Missing corners in a checkerboard.
elements of chance into your problem to make it simpler. backwards. 39 Divide the problem into smaller parts, solve them and combine this for a solution of the initial problem. 40 Check all possible solutions. 41 If you have to construct something via an algorithm, take the most you can get in every step, or more abstract: Use locally optimal solutions to construct a global optimum. Cf. Arthur Engel's "Problem Solving Strategies", New York 1998. 42 For example, build a physical model of a spatial construct. 43 Try to guess a solution and check it. 44 45 46 47 Purpose and scope of the sites differ. Read the introductions and FAQs carefully. 48 49 50 - for research level math questions. 51 52 53 This diagnosis is taken from David Perkins, Outsmarting IQ 54 For a list of common erros in undergraduate mathematics, have a look at 55 A common mistake people make when trying to answer a mathematical question is to work from first principles: it is almost always easier to modify something you already know. This article illustrates the point with examples that range from simple arithmetic to problems from the forefront of research. 56 It is surprising how often the following general approach to problem-solving is successful: you have a problem you don't yet know how to solve; you think of a somewhat similar context where you can formulate an analogous problem that you do know how to solve; you then work out what the corresponding solution ought to be in the context you started with. Even quite loose analogies can do a wonderful job of guiding you in the right direction. 57 If you want to prove a theorem, then one way of looking at your task is to regard it as a search, amongst the huge space of potential arguments, for one that will actually work. One can often considerably narrow down this formidable search by thinking hard about properties that a successful argument would have to have. In other words, it is a good idea to focus not just on the mathematical ideas associated with your hoped-for theorem, but also on the properties of different kinds of proofs. This very important principle is best illustrated with some examples. 58 If you are trying to prove a mathematical statement, it is often a good idea to think about its converse, especially if the converse is not obvious. This is particularly useful if the statement you are trying to prove is a lemma that you would like to use to prove something else. Some examples will help to explain why. More obviously, it is useful if you are trying to prove a result that would, if true, be best possible. 59 If you want to prove a mathematical statement, try proving the negation of that statement. Very often it gives you an insight into why the original statement is true, and sometimes you discover that it is not true. This tip can be iterated. 60 When you are trying to solve a problem, it can be very helpful to formulate similar-looking problems and think about those too. Sometimes they turn out to be interesting in themselves, and sometimes they lead you to ideas that are useful for the original problem. 61 It is not usually a good research strategy to think about one isolated problem, unless you are already some way to solving it. Solving a problem involves a certain degree of luck, so your chances of success are much greater if you look at a cluster of related problems. 62 Can't see how to solve a problem? Then see if you can solve it in special cases. Some special cases will be too easy to give you a good idea of how to approach the main problem, and some will be more or less as difficult as the main problem, but if you search for the boundary between these two extremes, you will often discover where the true difficulty lies and what it is. And that is progress. 63 As a student, one is asked to prove many statements that have been carefully designed so that their hypotheses are exactly the appropriate ones for deducing the conclusion. This makes it possible to design "trick" questions with unnecessarily strong hypotheses: these questions can be hard if one tries to use the hypotheses as they stand, since their full strength is irrelevant. The problems that arise when one is doing research are often of the "trick" variety: one has not been set them by a benign professor in the sky. So it is a good idea to investigate whether weaker hypotheses will suffice. As with looking at small cases, this can help one to locate the true point of difficulty of a problem. Similarly, if you are trying to find an example of a mathematical structure X that has a certain property P ,
38 Work
37 Introduce
it may be easier to look for an X that has a stronger property Q. And even if you fail, you are likely to understand much better what is required to find an X that satisfies P . 64 This is the flip side of "Try to prove a stronger result"; if one wants to prove some statement, it can be useful to first prove a weaker consequence of that statement, and then use that weaker result as a stepping stone to the full result. 65 If you are trying to prove a fact about the exponential function, it may be easier not to use any of the common definitions of this function, but to use instead a few properties that it has, of which the most important is that exp(x + y) = exp(x) exp(y). In general, it is often possible to turn concrete problems into abstract ones in this way, and doing so can considerably clarify the problems and their solutions. 66 The final proof of a result should of course be fully rigorous. But this certainly does not prevent one from suspending rigor in order to locate the right proof strategy to pursue. For instance, if one needs to compute some complicated integral expression, one can temporarily suspend concerns about whether operations such as interchange of integrals is actually justified, and go ahead and perform these operations anyway in order to find a plausible answer. One can always go back later and try to make the argument more rigorous. 67 A problem may have several independent difficulties plaguing it; for instance one may need to establish an estimate which is uniform both with respect to a large parameter N , and an independent small parameter ε. In that case, one can often proceed by passing to a special case in which only one of the difficulties is "active", solving each of these basic special cases, and then try to merge the arguments together. For instance, one could set N = 1 and get an argument which is uniform as ε → 0, then set ε = 1 and get an argument which is uniform as N → ∞, then try to put them together. 68 Sometimes if you generalize a statement, the result is easier to prove. There are several reasons for this, discussed in separate articles linked to from this page. 69 Very often a proof requires one to choose some object that will make the rest of the proof work. And very often it is far from obvious how to make the choice. Often a good way of getting round the problem is to take an arbitrary object of the given type, give it a name X, and continue with the proof as if you had chosen X. Along the way, you will find that you need to assume certain properties P1 , . . . , Pk of X. Then your original problem is reduced to the question "Does there exist an object of the given type with properties P1 , . . . , Pk ?" Often, this is a much more straightforward question than the main problem you were trying to solve. 70 This entire section is an unchanged quotation from: If there are any copyright problems, please let me know.
Some Practical Remarks on Solving Math Problems
Description
The paper shows a method of solving math problems, combining
a simple layout for handwritten notes that supports reflective intelligence and
a densely packed cheat sheet with a large number of to...
The paper shows a method of solving math problems, combining a simple layout for handwritten notes that supports reflective intelligence and a densely packed cheat sheet with a large number of tools for problem solving. | 677.169 | 1 |
Module 1 embodies critical changes in Geometry as outlined by the Common Core. The heart of the module is the study of transformations and the role transformations play in defining congruence. The topic of transformations is introduced in a primarily experiential manner in Grade 8 and is formalized in Grade 10 with the use of precise language. The need for clear use of language is emphasized through vocabulary, the process of writing steps to perform constructions, and ultimately as part of the proof-writing process.
The student materials consist of the student pages for each lesson in Module 1.
The copy ready materials are a collection of the module assessments, lesson exit tickets and fluency exercises from the teacher materials. | 677.169 | 1 |
How to Help Your Child Excel in Math 8vo-over 7¾"-9¾" tall. Wraps have only light wear, spine unbent. Pages are clean, text is unmarked.
Top Notch Books
TX, USA
$8.00
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About the Book
The book is an alphabetical dictionary and handbook that gives parents of elementary, middle school, and high school students what they need to know to help their children understand the math they're learning. The book can also be used by students themselves and is suitable for anybody who is reviewing math to take standardized tests or other exams. Foreign students, whose English-language mathematics vocabulary needs to be strengthened, will also benefit from this book. | 677.169 | 1 |
books.google.com - Explains how to use a slide rule to perform various mathematical operations.... Easy Introduction to the Slide Rule | 677.169 | 1 |
Based on the best-selling series by the Aufmann team, this hardcover text for the intermediate algebra course adheres to the formula that has made the Aufmann developmental texts so reliable for both students and instructors. The text's clear writing style, emphasis on problem-solving strategies, and proven Aufmann Interactive Method--in an objective-based framework--offer guided learning for both lecture and self-paced courses. The completely integrated learning system is organized by objectives. Each chapter begins with a list of learning objectives, which are woven throughout the text, in Exercises, Chapter Tests, and Cumulative Reviews, as well as through the print and multimedia ancillaries. The result is a seamless, easy-to-follow learning system. This special MEDIA ENHANCED EDITION now comes with Enhanced WebAssign and flash videos for every end of chapter test question available online through the student website.
Table of Contents
Review of Real Numbers
Introduction to Real Numbers
Operations on Rational Numbers
Variable Expressions
Verbal Expressions and Variable Expressions
First Degree Equations and Inequalities
Equations in One Variable
Coin, Stamp, and Integer Problems
Value Mixture and Motion Problems
Applications: Problems Involving Percent
Inequalities in One Variable
Absolute Value Equations and Inequalities
Linear Functions and Inequalities in Two Variables
The Rectangular Coordinate System
Introduction to Functions
Linear Functions
Slope of a Straight Line
Finding Equations of Lines
Parallel and Perpendicular Lines
Inequalities in Two Variables
Systems of Equations and Inequalities
Solving Systems of Linear Equations by Graphing and by the Substitution Method
Solving Systems of Linear Equations by the Addition Method
Solving Systems of Equations by Using Determinants and by Using Matrices
Application Problems
Solving Systems of Linear Inequalities
Polynomials and Exponents
Exponential Expressions
Introduction to Polynomials
Multiplication of Polynomials
Division of Polynomials
Factoring Polynomials
Special Factoring
Solving Equations by Factoring
Rational Expressions
Introduction to Rational Functions
Operations on Rational Expressions
Complex Fractions
Rational Equations
Proportions and Variation
Literal Equations
Rational Exponents and Radicals
Rational Exponents and Radical Expressions
Operations on Radical Expressions
Radical Functions
Solving Equations Containing Radical Expressions
Complex Numbers
Quadratic Equations and Inequalities
Solving Quadratic Equations by Factoring or by Taking Square Roots
Solving Quadratic Equations by Completing the Square and by Using the Quadratic Formula | 677.169 | 1 |
Transition to Advanced Mathematics / Edition 7
Overview analyze a situation, extract pertinent facts, and draw appropriate conclusions. With their proven approach, Smith, Eggen, and St. Andre introduce students to rigorous thinking about sets, relations, functions and cardinality. The text also includes introductions to modern algebra and analysis with sufficient depth to capture some of their spirit and characteristics. | 677.169 | 1 |
Find a West BrentwoodNumerical Methods for solving Differential Equations such as Euler's Method and the Runge-Kutta method.
6. Series solutions of Differential Equations including the method of Frobenius.
7. Solving systems of Linear Differential Equations. | 677.169 | 1 |
An electrical and computer engineering graduate student researcher shares his insights from academics and industry about how he uses Mathematica and the Wolfram Language, and how it compares to Matlab. | 677.169 | 1 |
This video explains the second half of row reduction, a basic algorithm in linear algebra used to solve systems of linear equations. Parameters are introduced corresponding to non-leading columns of the augmented matrix of the system.
This is the 14th lecture of this course on Linear Algebra given by Assoc Prof N J Wildberger. | 677.169 | 1 |
An electrical and computer engineering graduate student researcher shares his insights from academics and industry about how he uses Mathematica and the Wolfram Language, and how it compares to Matlab. | 677.169 | 1 |
Topics in this book include linear equations, inequalities and absolute values, systems of linear equations, powers, exponents, quadratic equations and factoring, rational expressions, and more This comprehensive resource supports NCTM standards and also includes…
"A work of considerable substance." — The New Yorker . In this marvelous oral history, the words of such legends as Louis Armstrong, Fats Waller, Jelly Roll Morton, Duke Ellington, and Billy Holiday trace the birth, growth, and changes in jazz over the…
Connect students in grades 7 and up with science using Science Tutor: Chemistry. This effective 48-page resource provides additional concept reinforcement for students who struggle in chemistry. Each lesson in this book contains an Absorb section to instruct and…
This strikingly comprehensive manual effectively shows the who, what, when, why and how of comedy improvisation. Unlike other improv books, this text provides the tools you need to start an improvisational team or club at your school. The book presents a complete… | 677.169 | 1 |
Lynwood CalculusVectors and Applications
8. Complex | 677.169 | 1 |
Description Course fulfills the second half in fundamentals of plane geometry, covering the vocabulary and concept of geometry through the use of formal proof and algebra and coordinate geometry. Completion of the geometry sequence prepares students for higher level mathematics courses and for those science courses requiring a working knowledge of geometry.
Intended Learning Outcomes
Evaluate financial responsibility of business ownership based on ratios and percents. | 677.169 | 1 |
Summary
This third edition ofTeaching Mathematics for the 21st Centurycontinues to help teachers let the secret outto open up to their students the wonderful discoveries and challenges of the pattern-making and problem-solving aspects of a fascinating subject: mathematics. The rationale remains the sameto enable prospective and current teachers to access and use tools and strategies to effectively teach mathematics to contemporary students. Changing demographics, knowledge of how people learn, and technology all impact the way we educate our young people. This edition incorporates lessons and strategies from programs that have proven success in many types of classrooms. Many of these examples help students connect mathematics to real life situations and communicate their understanding of the underlying concepts. Although technology is constantly being upgraded, ways to increase student motivation through its application remains a goal. For example--since applets can enhance a lesson whether the teacher uses a computer projector, a "smart" board, or has students work individually on computers--we have identified several sources of mathematics applets that can be correlated to various lessons. Research citations and summaries have been updated to reflect current information on teaching and learning.For future teachers.
Table of Contents
Prior to Entering the Classroom
History and Introduction to Mathematics Education Reform
Learning, Motivation, and Basic Management Skills
Concrete to Abstract With Tools, Manipulatives, Computer Programs, and Calculators | 677.169 | 1 |
Suitland AlgebraCurrently, I am studying for my Computer Science and Electrical Engineering degrees. Typically most algebra two skills are learned after basic geometry skills. I was fortunate to have an exceptional geometry teacherMarlene L.
...So work the first 6 or 7 problems as quickly as possible, and bubble in your choice. Elementary math includes the 4 basic operations with whole numbers and fractions. The fraction work includes common fractions and decimal fractions. | 677.169 | 1 |
Synopses & Reviews
Publisher Comments
REAs Pre-Calculus reviews sets, numbers, operations and properties, coordinate geometry, fundamental algebraic topics, solving equations and inequalities, functions, trigonometry, exponents and logarithms, conic sections, matrices and determinants.
Synopsis
REAs Essentials is an insightful series of practical and informative study guides covering nearly 100 subjects. Our Pre-Calculus study guide is packed with student-friendly topic reviews that explain everything about Pre-Calculus. Its a handy resource when preparing for Pre-Calculus exams or doing homework, and it makes a great textbook companion. The Essentials of Pre-Calculus is also an invaluable resource for math teachers outlining a course curriculum, writing exams, and developing classroom assignments. | 677.169 | 1 |
Description
Business Math helps students achieve success by incorporating Algebra I, Algebra II, and Geometry topics into practical business and personal finance contexts. Students see algebra at work within the most critical areas of finance. Students learn about investments, credit, automobile expenses, insurance, income tax, household budgeting, and more while gaining confidence in working with common algebraic functions.
Intended Learning Outcomes
Distinguish between two-dimensional and three-dimensional figures and design algebraic equations from their properties.
Solve algebraic equations by utilizing the properties and relationships between angles and lines. | 677.169 | 1 |
Description:
Chris Caldwell of the University of Tennessee at Martin provides the Graph Theory Tutorials Website. Sections included at the site are Introduction to Graph Theory, Euler Circuits and Paths, Coloring Problems, and Adjacency Matrices (under construction). Each section consists of an interactive tutorial discussing the basic concepts of graph theory. Registration (press the REGISTER button at the bottom of first page of each tutorial) is required for each tutorial. The user must either pass a quiz in the tutorial section or write a comment before continuing to the next page. Links to related resources are also provided at the site. This site is useful for high school students and is definitely worth a visit. | 677.169 | 1 |
This class presents the fundamental probability and statistical concepts used in elementary data analysis. It will be taught at an introductory level for students with junior or senior college-level mathematical training including a working knowledge of calculus. A small amount of linear algebra and programming are useful for the class, but not required.
This course provides an introduction to complex analysis which is the theory of complex functions of a complex variable. We will start by introducing the complex plane, along with the algebra and geometry of complex numbers, and then we will make our way via differentiation, integration, complex dynamics, power series representation and Laurent series into territories at the edge of what is known todayIn this course you will learn a whole lot of modern physics (classical and quantum) from basic computer programs that you will download, generalize, or write from scratch, discuss, and then hand in. Join in if you are curious (but not necessarily knowledgeable) about algorithms, and about the deep insights into science that you can obtain by the algorithmic approach.
This course will cover the mathematical theory and analysis of simple games without chance moves. This course explores the mathematical theory of two player games without chance moves. We will cover simplifying games, determining when games are equivalent to numbers, and impartial games | 677.169 | 1 |
Intermediate Algebra : Functions and Authentic ApplicationsUnique and enthusiastic in its approach, Lehmann's book is a rich combination of important skills, concepts, and applications. This book captivates readers'interest as they use curve fitting to model current, compelling, and authentic situations. The curve fitting approach emphasizes concepts related to functions in a natural, substantial way and encourages readers to view functions graphically, numerically, and symbolically as well as to verbally describe concepts related to functions. The examples in the test demonstrate both how to perform skills and how to investigate concepts. Students learn why they perform skills to solve problems as well as how to solve the problems. Explorations deepen users'understanding as they investigate mathematics with graphing calculator and pencil and paper activities.Topics discussed include linear functions, modeling with linear functions, systems of linear functions, exponential functions, logarithmic functions, polynomial functions, using quadratic functions to model data, rational functions, radical functions, and modeling with sequences and series.For individuals interested in deepening their understanding of algebra. | 677.169 | 1 |
9780395888650-level text in an innovative workbook series, Introductory Algebra: An Integrated Approach is ideal for the first-year developmental arithmetic and pre-algebra instructor seeking to accommodate individual teaching and learning styles. Aufmann and Lockwood present math as a cohesive subject by weaving the themes of number sense, logic, geometry, statistics, and probability throughout the text at increasingly sophisticated levels. These themes are illustrated by applications from more than 100 disciplines | 677.169 | 1 |
Lathrop, CA PrecalculusThese words are represented by variables in an equation which adds to the frustration because many new ideas are presented quickly. By making charts and using illustrations I will try to present the new definitions and their associated variables in an orderly manner so that memorization of formulas is easier. Then the application of the math is a simple replacement of variables by numbers.
Kwa D.
...They skills comprise computing the volumes and surface areas of prisms, pyramids, cylinders, cones, and spheres; committing to memory the formulas for prisms, pyramids, and cylinders. Pre algebra paves a path for students going on with Algebra 1 and 2 smoothly and confidently. Therefore, by the... | 677.169 | 1 |
10 Answers
10
This guided discovery approach goes by other names, as well. One such name is "Inquiry Based Learning" or IBL. A list of guided discovery problems is often referred to as an "IBL script". Many such scripts are available from the Journal of Inquiry Based Learning in Mathematics (JIBLM):
@Jon Bannon: The JIBLM site is great and seems full of great books (I have yet to look through all of them). I'd up vote if I could (my reputation is <15). Thanks!
– Théophile CantelobreJan 27 '13 at 4:25
You may be interested in learning about the Moore Method. The idea is to "encourages students to solve problems using their own skills of critical analysis and creativity" without relying on textbooks. HERE you can find some references.
My favourite is Alexandre Kirillov's "Elements of the Theory of Representations" Grundlehren der Mathematischen Wissenschaften, Springer, vol 220. A lot of representation theory is worked out through examples and exercises.
I have not read (or, in this case, worked) through the book, but Jeffrey Strom's ``Modern Classical Homotopy Theory" guides the reader through the proofs of all the theorems stated in the book (as opposed to proving them himself). To my very limited knowledge, this is the first "IBL-type" book in algebraic topology.
I'm impressed with two books by Dr R. P. Burn that seem to be in the spirit of your question:-
Groups: a path to geometry, CUP, 1985, 0-521-30037-1
A pathway to number theory, CUP, 2nd ed., 1997, 978-0-521-57540-9
Each consists of an ordered sequence of problems (answers provided):-
... to enable students to participate in the formulation of central mathematical ideas before a formal treatment (which, suitably introduced, they may well be able to provide themselves)
Source: a preface to A pathway to number theory)
They are aimed at advanced high school, or early undergraduate level students. The sequence starts by getting the reader to initially explore special cases and then work towards a generalisation, usually a theorem. The books include references to selected standard texts that are recommended to be read concurrently.
This is not to criticise you as I know it is not very well known, which is why I say it: if you think a question should be in CW mode, you could on the one hand give your answer right away in CW mode. And, this is the main point, if you do not or there are already other answers in non-CW mode, you should better flag for moderator attention with request to turn into CW (as opposed to asking the OP since they can just CW the question, which then leaves your and other answers in non-CW mode) [I now already flagged for mods so no action is required, this is just for info.]
– user9072Jan 23 '13 at 10:51
You may find this one interesting: Number Theory Through Inquiry (MAA textbooks). I have used it three times. First time, which I strictly followed the method, we just coverd the first four chapters. Second time, I have relaxed myself a bit and we covered the first six chapters. Last time (current term), I have used all the teaching methods I know (including modified Moore method), we are nearly covering all chapters!
You may also find this paper interesting: " and Less" (PRIMUS,22(7):509-524, 2012) where I told the story of using a very modified Moore method in a Multivariable Calculus Course.
I've just put up such a text for an Introduction to Proofs course, here. It is Free, including LaTeX source. (I've only taught out of it one time so no doubt there are typos, places that could use refinement, etc.) | 677.169 | 1 |
Details about Prentice Hall Mathematics, Algebra 2:
Our comprehensive table of contents allows teachers to easily include trigonometry, statistics, or precalculus readiness in the Algebra 2 course along with more traditional topics.
Content accessible to all
Abundant exercises graded by difficulty allow teachers to meet the needs of an increasingly wide range of Algebra 2 students.
Algebra 1 reviewed
Key Algebra 1 concepts and skills are reviewed in Chapter 1 so that all students can be successful moving on to more advanced content. Throughout the text, key skills are reviewed and reinforced where needed.
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Video tutorial and screenshots
Editorial review
The following review written by one of the Software Informer contributors applies to version 2.0
Graphmatica is a small program designed for students and teachers of Mathematics. With it, you can graph your own equation just typing it over a text field and pressing enter.
Graphmatica is very easy-to-use and have a lot of interesting options.
You can easily switch the graph paper (ex. to show polar coordinates instead of the standard x-y graph). Moreover you can change any graph color: the background, grids, and function line colors can be altered to easily match your needs.
Looking at the math features, after you draw a line over the graph you can easily (I mean 1-click) find derivatives for such equation. The same occurs with numerical integrations: just select the area and the program makes the rest.
Personally, I found this program very interesting. The only drawback is that you could use it just for 30 days, after that you have to pay for it. Another issue to mention is that you have no help at all when typing the equations. If you misspelled something, then you get just an error message. It will be crucial to read carefully the help file before to start using it seriously.
Pros
Very fast drawing.
Complete set of graphs.
Cons
The errors on equations are not very descriptive.
Comments on Graphmatica | 677.169 | 1 |
Courses
Fundamentals of Math
FOM is a transitional course in our math curriculum designed to automate the upper level skills of arithmetic while introducing abstract arithmetic concepts.
Math Models
This course is designed for cadets who need a review or introduction to basic arithmetic operations before continuing to a higher level. Cadets in this course will build their skills in fractions, decimals, percentages, signed numbers, powers and roots. After mastering the basics, they will continue on to learn algebraic expressions and formulas, solving equations and inequalities, polynomials and factoring in preparation for higher-level classes like algebra II and pre-calculus. Additionally, cadets will build their geometry skills through angles, triangles, planes and solid figures. Before concluding the class, cadets will learn how to use graphs, maps and charts.
Algebra I
This course focuses on mathematical problem solving or the application of mathematical concepts in new situations. Cadets learn by working carefully designed problems that lead to the use of productive thought patterns utilizing concepts.
Algebra II
This course is designed to prepare cadets for pre-calculus. Cadets will continue to study area and volume and be introduced to more abstract geometric concepts, review prior algebra concepts and proceed through trigonometric ratios, rectangular and polar coordinates and on to changing from one coordinate system to the other.
Geometry
One of the main objectives of this course is to enhance cadets' logical thinking skills. The course begins with an introduction to the basic concepts of geometry, such as lines, planes and points, then progresses to the study of various geometric shapes and their properties, and concludes with proofs.
Pre-Calculus
This course provides in-depth coverage of trigonometry, logarithms, analytic geometry and upper level algebraic concepts. Cadets expand their understanding of mathematical concepts through the use of symbolic reasoning and analytical methods to represent mathematical situations, to express generalizations and study the relationships between them.
Advanced Mathematics
With a teacher's recommendation, cadets accelerating in math may take the first year (pre-calculus) and then proceed to calculus. All other cadets will continue with advanced math. Pre-calculus and advanced math provide in-depth coverage of trigonometry, logarithms, analytic geometry and upper level algebraic concepts.
Calculus
An advanced placement (AP) course in calculus consists of a full high school academic year, which is comparable to college calculus coursework. It is expected that cadets who take AP calculus will seek college credit, college placement, or both, from institutions of higher learning.
Statistics
This course introduces cadets to the major concepts and tools for collecting, analyzing and drawing conclusions from data. Cadets will be exposed to four major concepts: exploring data, planning a study, probability as it relates to distribution of data and inferential reasoning and modeling.
Honors and Advanced Placement
The advanced placement program prepares cadets to take the College Board Advanced Placement Exam for each of the courses offered. Currently, the MMA Mathematics Department offers statistics and calculus AP opportunities.
Faculty
Rudy Parast Department Chair
Rudy Parast has over 25 years of experience teaching and has taught at both the high school and college level. Mr. Parast attended the University of Washington in Seattle where he received his Bachelor of Science in Chemical Engineering in 1982 with a minor in chemistry and mathematics. The following year, he and his wife moved to the Rio Grande Valley, where she accepted a position with the Marine Military Academy. In 2001, he returned to school and earned his master's degree in mathematics from the University of Texas in Edinburg. | 677.169 | 1 |
Intermediate Algebra
Overview essays that open each chapter connect algebra to real world problem solving and can be used to stimulate class discussions and promote collaborative learning. Functions and graphing are introduced early (Chapter Three) and are then integrated throughout the rest of the text. This approach allows for visual interpretation of the mathematical concepts,which in turn encourages students to develop an intuitive understanding of equations and their graphs. Also by introducing these topics early,students become familiar and comfortable with concepts that are critical to their success in future math courses. Intermediate Algebra includes marginal notes and examples that indicate how technology can enhance the study of algebra through exploration,visualization,and geometric interpretation. These examples fall at the end of section discussions and may be omitted if a graphing tool is not being used. The text is written in a clear,concise style with numerous examples which are connected by thoughtful transitions that either reinforce the student's understanding of the previous concepts or prepare them for the next example. Each example is followed by a "Check Yourself' exercise that facilitates the student's active involvement in the learning process. | 677.169 | 1 |
suffer from an inferiority complex where mathematics is concerned, regarding figures and equations with a fear based on bewilderment and inexperience. This book dispels some of the subject's alarming aspects, starting at the very beginning and assuming no mathematical education.Written in a witty and engaging style, the text contains an illustrative example for every point, as well as absorbing glimpses into mathematical history and philosophy. Topics include the system of tens and other number systems; symbols and commands; first steps in algebra and algebraic notation; common fractions and equations; irrational numbers; algebraic functions; analytical geometry; differentials and integrals; the binomial theorem; maxima and minima; logarithms; and much more. Upon reaching the conclusion, readers will possess the fundamentals of mathematical operations, and will undoubtedly appreciate the compelling magic behind a subject they once dreaded.
Top Customer Reviews
I love math and frequently read math books the way some people read novels. I found this one to be very clear. An excellent introduction to a number of mathematical topics. His coverage of basic combinatorics is the clearest I have ever seen!
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This book is an absolutely fantastic introduction to the basic concepts of Mathematics. I currently work as a Mathematics tutor and implement many of the techniques he uses in this book. It's surprising to see how much math-phobia really just translates to a lack of understanding of key mathematical concepts.
This book could teach an intelligent 12 year-old the fundamental concepts of algebra through calculus. The book is far more conceptual than any textbook you would ever read. The exposition is brilliantly clear and well motivated. You start with understanding the basic number systems and the fundamental rules of counting. Then you progress through basic algebra and geometry, and finally reach analytic geometry and calculus.
Connections are made between subjects in every section of the text! One thing you learn from the early chapter is again applied later on. For example, the chapter on combinations heavily influences the chapter on the binomial theorem. The author explains flawlessly how the symmetry of combinations is the driving force behind the binomial theorem!
There are many historical notes and many deep ideas and works from various authors are mentioned! I wish that textbooks like this were used in every mathematical subject! Surely, many more people would appreciate mathematics if there were...
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Designed as a companion to The Economist Style Guide, the best-selling guide to writing style, The Economist Numbers Guide is invaluable to anyone who wants to be competent and able to communicate effectively with numbers.
In addition to general advice on basic numeracy, the guide points out common errors and explains the recognized techniques for solving financial problems, analysing information of any kind, and effective decision making. Over one hundred charts, graphs, tables, and feature boxes highlight key points. Also included is an A–Z dictionary of terms covering everything from amortization to zero-sum game.
Editorial Reviews
From the Publisher
A handy reference to stay on top of global economic trends. In today's global arena it is imperative that business people keep abreast of the economics of nations around the world. Every day features the release of a new barrage of updated economic indicators and figures that carry often hidden messages about the direction of segments of the economy. This pocket reference enables readers to quickly revisit the meaning and impact of late breaking economic news and to make better decisions based on the looming economic terrain.
--This text refers to an out of print or unavailable edition of this title.
From the Inside Flap
Crucial to business success, numerical methods are often viewed as too complex to understand, much less use. They are, in fact, far less complicated, able to be broken down into stepby-step instructions and processed by basic computing devices. This invaluable resource from the publishers of The Economist, the leading international business journal, simplifies and demystifies the numbers game, illustrating just how straightforward—and relatively easy—it really is. Taking you clearly and concisely through numerous fundamental functions, both elementary and advanced, The Economist Numbers Guide arms you with the tools necessary to not only approach numbers with more confidence, but solve financial problems more easily, analyze information more accurately, and make decisions more effectively. Covering finance and investment, forecasting techniques, hypothesis testing, linear programming, and a host of other important topics, it shows you how to handle everything from figuring interest and quantifying risk to projecting inflation and evaluating investment opportunities. In addition to the basic mechanics of numerical techniques, the Guide takes a look at their practical applications, including their role in stock control, simulation, and project management. To help you sidestep potentially costly mistakes, it also highlights common errors to avoid, such as rounding incorrectly and bypassing time series selection. Along with sample calculations, concise definitions, and clear explanations, as well as more than 100 charts, graphs, and tables, The Economist Numbers Guide features an A-to-Z dictionary that encompasses key terms—from autocorrelation to zero sum game—and provides useful reference material on such essentials as conversion factors and formulae for calculating areas and volumes. In-depth and easy-to-use, this is an indispensable reference for business and numbers success.
--This text refers to an out of print or unavailable edition of this title.
Top Customer Reviews
I have read cover-to-cover a previous edition of this book (when it was published by Wiley in 1998) and recently had an opportunity to carefully peruse this current edition (5th ed. by Bloomberg Press???). What I found is that this is a strange case of how a great book (the 1998 edition) turned into merely a good book (this 5th edition). Because of this regression toward the average, I deducted one star from my review (but still feel that it is good enough for 4 stars). As you may have noticed, I really loved the older edition of The Economist Numbers Guide that I thankfully own. It is a great overview and introduction of mathematics as it relates to business. There are a lot of great things about that edition of this book. One of the things I admired about it was the range of topics covered, from interest rates and basic probability/statistics all the way up to Markov Chains, linear programming, and marginal analysis. It is hard to find the breadth of topics covered in that book elsewhere - whether all in one book or in any combination of books. So I found it perplexing that this 5th edition dedacted some materials and topics covered in older editions. Gone are the interesting discussion of descriptive statistics for sets of data that do not easily conform to any of the standard probability distributions (e.g., where median is the best measure of the 'average' and substitutes must be used for the more common parameters such as standard deviation). I have a hard time finding anything coherent much less accessible on those topics elsewhere so it is a shame that they were left off of the 5th edition. The only new material (not previously present) is a short blurb on public-key cryptography.Read more ›
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This book provides concise and clear definitions of business analytics with practical applications. Excellent for the neophyte in business math. Helpful index and glossary to get started. Good guide to use if learning stats or marketing research.
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Study skills can reteach a person's behavioral approach to studying. Without a good study habit, opportunity will pass by without even knowing it. It's the difference between a good student and an outstanding self-learner.
...The second major area for standard calculus classes is to know the formulas that provide the shortcuts for the calculations and how to see when to use each. To understand the formulas, I find it very beneficial to often go through the proof of the formula, as long as it is not overwhelming. In ... | 677.169 | 1 |
Overview—theoretical and practical—and this book seeks to move the debate forward along both dimensions while seeking to relate them where appropriate. A discussion of meaning can start from a theoretical examination of mathematics and how mathematicians over time have made sense of their work. However, from a more practical perspective, anybody involved in teaching mathematics is faced with the need to orchestrate the myriad of meanings derived from multiple sources that students develop of mathematical knowledge.
This book presents a wide variety of theoretical reflections and research results about meaning in mathematics and mathematics education based on long-term and collective reflection by the group of authors as a whole. It is the outcome of the work of the BACOMET (BAsic COmponents of Mathematics Education for Teachers) group who spent several years deliberating on this topic. The ten chapters in this book, both separately and together, provide a substantial contribution to clarifying the complex issue of meaning in mathematics education.
This book is of interest to researchers in mathematics education, graduate students of mathematics education, under graduate students in mathematics, secondary mathematics teachers and primary teachers with an interest in mathematics.
Related Subjects
Table of Contents
Meanings of Meaning of Mathematics.- "Meaning" and School Mathematics.- The Meaning of Conics: Historical and Didactical Dimensions.- Reconstruction of Meaning as a Didactical Task: The Concept of Function as an Example.- Meaning in Mathematics Education.- Collective Meaning and Common Sense.- Mathematics Education and Common Sense.- Communication and Construction of Meaning.- Making Mathematics and Sharing Mathematics: Two Paths to Co-Constructing Meaning?.- The Hidden Role of Diagrams in Students' Construction of Meaning in Geometry.- What's a Best Fit? Construction of Meaning in a Linear Algebra Session.- Discoursing Mathematics Away.- Meaning and Mathematics. | 677.169 | 1 |
65Test with success using the Spectrum Math workbook! This book helps students in grade 7 apply essential math skills to everyday life. The lessons focus on ratio and proportion, fractions, percents, calculating interest, perimeter, volume, and statistics, and the activities help extend problem-solving and analytical abilities. The book features easy-to-understand directions, is aligned to national and state standards, and also includes a complete answer key.
Top Customer Reviews
I purchased grade 7th and 8th. No wonder they stop there because it's the same materials all over again. Just buy either one, 7th or 8th, and use it for both grades because unless you need more practice problems, it's the same topics.
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I bought this Spectrum 7th Grade Math workbook for my eleven year old daughter. I've found it ideal for her, since each subject is introduced with good clear concise explanations, and she hates reading anything long or complicated. As other reviewers have commented, the exercises are plentiful and well structured. My daughters who is good at maths generally, has been whizzing through it. I would say the level is 6th/7th grade.
Since buying this Math workbook, I've bought the 8th grade math workbook and 8th grade enrichment workbook for her to progress onto, as well as the Geometry and Agebra workbooks for more concentrated practice. I've also bought her the Spectrum English books including writing, language arts, phonics and vocabulary. She loves using them all and has made massive progress in her English which was weak, but now probably at an A standard.
I've become such a Spectrum fan that I've also bought the science workbooks (literally superb) and just ordered the test books and Geography.
I'm a homeschooling mom and use a lot of curricula products but Spectrum is hands down best value for money and absolutely top notch. I also use and have used Singapore Math, MEP Math, School Specialty Publishing, A Brighter Child, and Pylon Math, but if I had to choose just one curriculum I'd recommend Spectrum since it covers all the bases and does it well.
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Just what I wanted! It's a wonderful practice book that covers all the basics. Lots of repetition, which is what I felt my child needed to really get comfortable with operations. I'm using it to keep his skills from backsliding over the summer.
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My home schooled 7th grader is using this book for math this year. It seems to cover all the basics, but teaching some of the concepts will take extra effort. The book does not teach; it only tests for understanding. The beginning lessons are very easy reviews, but they do increase in difficulty.
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I bought this book to keep my child fresh on math during the summer. The work sheets were great. As this was a review of what she learned during the school year she didn't need direction. If you are looking for a book to teach math, this isn't it- but it was PERFECT for our needs.
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Only basic problems were given. Sample problems were too easy with no challenging problems. Grade 8 Spectrum Math covers same topics but has better sample problems. If your child is 7th grader and you are getting this to prepare him/her for the school, get Grade 8 Spectrum Math.
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I used this book to help my son review math over the summer. It does not have much in the way of illustrations or higher theoretical concepts, but it does have clear explanations, and plenty of problems for practice. We generally did about half of each lesson, which left plenty for further work if necessary. There was at least one mistaken answer on the answer key, so check the work when in doubt.
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need for improved mathematics education at the high school and college levels has never been more apparent than in the 1990s. As early as the 1960s, I. M. Gel'fand and his colleagues in the USSR thought hard about this same question and developed a style for presenting basic mathematics in a clear and simple form that engaged the curiosity and intellectual interest of thousands of high school and college students. These same ideas, this same content, unchanged by over thirty years of experience and mathematical development, are available in the present books to any student who is willing to read, to be stimulated and to learn. "The Method of Coordinates" is a way of transferring geometric images into formulas, a method for describing pictures by numbers and letters denoting constants and variables. It is fundamental to the study of calculus and other mathematical topics. Teachers of mathematics will find here a fresh understanding of the subject and a valuable path to the training of students in mathematical concepts and skills.
Editorial Reviews
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"All through both volumes ['Functions & Graphs' and 'The Methods of Coordinates'], one finds a careful description of the step-by-step thinking process that leads up to the correct definition of a concept or to an argument that clinches in the proof of a theorem. We are ... very fortunate that an account of this caliber has finally made it to printed pages... Anyone who has taken this guided tour will never be intimidated ever again... High school students (or teachers) reading through these two books would learn an enormous amount of good mathematics. More importantly, they would also get a glimpse of how mathematics is done."
--- H. Wu, The Mathematical Intelligencer
"This book is a concise and compact treatment of the essential ideas of coordinate geometry. The authors demonstrate powerfully how geomtric ides may be communicated and studied effectively without the aid of pictures. Graphics are of course of vital importance int he methods of Euclidean geometry. However, the methods of coordinate geometry are able to transform pure geometric ideas into algebraic manipulations where the meaning is very clear once the formalism is learnt. In particular the book demonstrates the value of conveying information in the form of images embedded in formulas. This is very useful in the transmission of information by electronic means. . . This book is a valuable tool for teaching the redimentary concepts of analytical geometry. It contains a number of excellent examples and exercises which go further than a mere introductory programme. the exercises, while not numerous, are very thought-provoking and are bound to pose a serious challenge to the interested student."
Top Customer Reviews
The authors of this slim volume demonstrate the power of coordinate geometry, which they describe as a means of translating geometric figures into algebraic formulas, through their lucid exposition, interesting examples, and well-chosen exercises.
The authors begin with the coordinate geometry of the real line. They discuss absolute value and define what distance means. Next the authors examine the coordinate geometry of the plane. They define distance in the plane, show how relations among the coordinates define geometric figures, and discuss different coordinate systems that can be used in the plane. Their examples illustrate how algebraic methods developed by Rene Descartes make it possible to solve geometric problems efficiently that would be quite difficult to solve using synthetic geometry. The authors then treat the coordinate geometry of three-dimensional space in a similar manner.
The second part of the book begins with a problem concerning lattice points in the plane. The authors use this example and its generalizations to justify exploring the coordinate geometry of four-dimensional space. They carefully treat the example of a four-dimensional unit hypercube, examining its properties by considering its analogues in lower dimensions: the segment [0, 1] of the real number line, the unit square in the coordinate plane, and the unit cube in space.
Since the book was initially written for a correspondence course for high school students in the Soviet Union, it is designed for self-study and accessible to students who have had high school courses in algebra and geometry.Read more ›
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The contents of the book is fine, just what I expected when I purchased it.
However, the binding job was done in a very amateur fashion. These books from Springer are printed and bound upon order, it seems that the people who do the job have no pride in their work. Buy the version from Dover instead.
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Trigonometry the Easy Way
Paperback
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Overview laws of sine and cosine, trigonometric functions and inverse functions, waves, conic sections, polynomial approximation, and much more. The book is filled with instructive exercises and their solutions, plus illustrative drawings, graphs, and diagrams. This new edition contains updated coverage on using graphing calculators and computer spreadsheets for solving trigonometric problems.
Most Helpful Customer Reviews
The book is set up to teach Trig by telling a story. While no one can claim the story is interesting, it does help propel the reader forward in learning the subject matter. It was nice to see some of the mathematical theory set up to be applied in a real world situation. The examples using friction, velocity vectors and tree height stick out the most in my mind. There are plenty of example problems to help the reader apply what was learned with answers in the back to verify the work. In my opinion, this is a great tool to use in conjunction with a trig class and a pretty good tool to learn trig on your own.
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If you are new to the science of trigonometry and are in no hurry to learn it, this book is for you. Guaranateed success and highly entertaining. Get your math tools and rock and roll.
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More than 1 year ago
I thought this book might be interesting since it came with a story about Trig, however after reading about 10 pages I could not stand it any longer. It did not really help in my understanding, but this could be due to the silly story. | 677.169 | 1 |
Algebra is one of the broad parts of mathematics, together with number theory,
geometry and ... The more basic parts of algebra are called elementary algebra,
the more abstract ... As a single word ... | 677.169 | 1 |
Plutarch Publications Inc.
Supplemental materials are not guaranteed for used textbooks or rentals (access codes, DVDs, workbooks)....
Show More in the learning process. Math Grade 1 complies with State Curriculum Standards and includes a chart to assess the level at which a child is able to complete each standard presented. This book was written by teachers to provide grade appropriate standards, practice, and application in a straightforward, easy to understand manner. Appropriate materials and presentation produce comprehension. Practice produces proficiency. Application produces students able to interact with | 677.169 | 1 |
prepare you for the new mathematics subtest of the MTEL General Curriculum (03) Exam. It has over 50 lessons, 600 practice questions, detailed explanations, and quizzes to cover each of the concepts on the new math subtest, focusing on critical thought and a deep understanding of the topics. This book also includes two comprehensive sample tests. The author has expertise in teaching, math, and the MTEL. Mary DeSouza Stephens graduated from MIT with a bachelors and masters in Computer Science and Electrical Engineering and has over 15 years of teaching experience, including teaching MTEL prep courses at UMASS Boston, founding an educational services company in California, designing educational software, and teaching K-adult students in math, science, computers, test prep, and humanities at schools around the world. Please visit to find out about our other preparation materials for MTEL exams.
Editorial Reviews
About the Author
Mary (DeSouza) Stephens graduated from MIT with a bachelor's degree and a master's degree in Computer Science and Electrical Engineering. She has over 15 years of teaching experience, including designing courses and teaching as an adjunct faculty member at UMASS Boston; TAing discrete math at MIT; and developing curriculum for and teaching classes in high school computer science, LSAT, algebra, geometry, science, humanities, and K-8 math and computers. Outside of teaching, Mary worked in engineering, product management, and strategy for companies including Edusoft and Oracle. In the past few years, she has spoken at international education conferences and has served as Professional Math Consultant at Merrimack College, founder of Omega Teaching, and Research and Development Manager at Houghton Mifflin Harcourt. She currently is the CEO of PrepForward ( and can be contacted at mary@prepforward.com.
Top Customer Reviews
I purchased this book to help me study for the so dreaded MTEL math subtest. I haven't done any math for almost 20 years, I also have never been good at it in high school. This book has been so helpful!!! It's very well organized into chapters. Each chapter has about 5 to 6 lessons. The lessons are short and contain only what you need to do the exercises. Then Each lesson has a 10 question quiz with detailed answers. The chapter ends with a 20 question quiz that covers all the lessons studied in the chapter along with the detailed answers. At the end of the book you have 2 full length exams with answers of course and a glossary. The book has also a detailed lesson on how to do the open response. The bonus that you will rarely find with any other author of a study book is that whenever you have a question about the book, you can email Mary DeSouza. Not only does she answer questions, but she also gives encouragements and support. She is fantastic!!! The book is excellent and the price is more than reasonable! Very glad I bought it! A little update: I passed the test with an over 240 score!!! Next adventure: The multi subject test! I also recommend that you check math websites. They are great! A lot of them are free and very helpful. Good luck!
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Mary DeSouza has a basic, start at the beginning, approach for this course. I liked it since it had been years and years since I took math. Every topic has 4-5 sections. Each section has a test (which you can re-take if you are struggling). Each topic has a test. So, you build on your knowledge.
The math test I took had an essay question on right triangles, areas and the Pythagorean theory. Eeegads! But I had it nailed down because it was in this program.
Like every course - you get what you give. If you are motivated and want to pass the test, you can use this as a tool. It helped me. I failed the first time, took the course and then passed the second time up.
The most helpful tool for me was the "assessment test" which I saw that I did very poorly on areas and measurement. So I went back and studied that section. And, sure enough, it was on the MTEL!
Good luck to you all. I liked the material and found it very helpful.
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Ok so the book covers every math topic we need to know for the MTEL, great. There are many exercises for us to practice, great too. There are open response questions, that by the way are super hard on this book, which is awesome for practice! But... There are many many mistakes in it. Many exercises give you the wrong letter as the answer. For example, you have to choose one answer: A. 1 B. 2 C. 3 D. 4 You choose letter D, 4, but the short answer says A, 1. When you look at the complete explanation for the exercise you might see "therefore the answer is 4, letter A". (!!!) It's VERY confusing and it happens often.
Also, there are wrong graphs in it, questions where the book mentions the color of the line and the book is black and white.
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I sat down for about 3-4 weeks and went through this book about 2-4 hours a day. I cannot be more thankful for this book! There is no way I would have passed without it. I even passed on my first try! I'm not great at math (and never have been) but thankfully this book is geared for exactly what is on the MTEL. I would go through the book and mark things that I didn't quite understand so that I could go back to them with my boyfriend (who is really a smart cookie!) and then he would help explain things to me in order to help me fully understand. But don't worry, if you don't have someone to further explain the explanations that are written in the book then you may contact Mary Desouza herself. Which is part of the reason why I chose to order this book in the first place! Yes there are some typos, but you'll catch them and you'll be fine. No biggie. What I loved the most is that I can honestly say that as I was sitting through the actual Math MTEL test there were a bunch of questions that when they came up on my screen, immediately a light bulb went off in my head and I could think of the section/page where I did such problems in this book! Also, there were a handful of questions on the test that I would've had NO IDEA how to complete if it weren't for this book. This book is a must-have for anyone taking this test! Thank you so much Mary Desouza!
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I love the format of this book. The author presents 1-2 pages of information in each lesson followed by a 10 question quiz. There are detailed explanations for each of the 10 questions that are valuable to study even if you got the question correct. Each of the lessons is short and focused, and I am working through a lesson or two a day. I looked at several other Mtel Math books at the library, but liked the style of this author the best so I purchased this book.
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This book is extremely helpful. I am NOT good at math and I've failed this MTEL countless times. I rescheduled my test and only had a couple of weeks to study. Her directions are clear, quick, and easy. I found great improvements in my skill after just a few hours of studying.
There are a few errors in the book. I've attached one example below. Sometimes the answer key will say it is one answer, but the explanation will say it is another. So make sure to really check your answers against the explanations, and if you feel you have done it right, make sure to check it again. | 677.169 | 1 |
Mathematics
The Math Department offers a curriculum that ranges from Algebra 1-2 through advanced mathematics concepts. Courses prepare students for future careers. Skills such as reasoning and problem solving are emphasized in all levels of mathematics.
Courses Offered: (For more details about each course check the current course guide.)
Algebra 1-2
Geometry 1
Advanced Algebra 3-4
Advanced Algebra 3-4: Accelerated
Preliminary Math Studies
IB Math Studies SL
IB Math SL 1-2
IB Math SL 3-4
IB Math HL
Probability/Statistics
Possible Sequence | 677.169 | 1 |
Produktbeschreibungen
Pressestimmen
From the reviews:
"This textbook is an introduction to the classical theory of functions of one complex variable. Its distinctive feature are the graphical representations of functions, being the most useful tool in teaching and generally in mathematics. … The self-sufficiency of the textbook and the broad range of graphical examples makes the book useful for students as well as teachers of mathematics. … the book can be warmly recommended both to experts and to a new generation of mathematicians." (Stanislawa Kanas, Zentralblatt MATH, Vol. 1264, 2013)
"Anyone who works with complex variables should read this book. … Visual Complex Functions is a beautiful and careful presentation of an entire advanced introduction to complex analysis based on phase portraits and, where appropriate, other kinds of computer-generated pictures. … My understanding of many ideas and phenomena deepened through reading this book." (Lloyd N. Trefethen, SIAM Review, Vol. 55 (4), 2013)
Buchrückseite
This is an amazing book. In 345 extraordinary pages it presents the foundations of the subject of complex variables -- also known as complex analysis or function theory -- through the powerful medium of the color pictures known as "phase portraits".
Prof. Wegert would like his book to be suitable as an introduction to this field, and perhaps it may be. But for my money the importance of this book is for people who have already been introduced to complex variables and now want to _really_ understand them, to work with them. Every topic is enhanced by this treatment. There is nothing else like this book (apart from Wegert's own three annual calendars to date of "Complex Beauties"). In particular it is very different from Tristan Needham's 1997 text Visual Complex Analysis, which also takes a geometric tack but is not full of actual displays of actual functions.
Wegert's book is so striking that I assumed Birkhauser would have priced it at an absurd level, so that individuals would never buy a copy. Instead it is just around 60 dollars or 40 pounds. Amazing.
If you work with complex variables, buy this book. (- Nick Trefethen, Professor of Numerical Analysis, Oxford University)
I'd recommend this book as a supplement to your favorite complex analysis textbook. It's main feature is the use of "phase portraits" to visualize analytic functions. I'm a huge fan of these phase portraits, and this book inspired me to start creating my own plots on Mathematica (which, for the record, is not that difficult to do). You'll learn (among other things) how to identify singularities, determine the order of zeros (of the function and its derivative) just by seeing the phase portrait. You'll gain a more intimate feel for the various analytic functions: rational functions, trigonometric functions, the Riemann zeta function, Weierstrass p-functions, etc. And it's hard not to be impressed by the sheer beauty of the many pictures.
Wegert's book also tries to double as a complex analysis textbook. In this respect it is less effective, and I'd recommend that you stick to a standard textbook. Wegert also has a slight tendency to "oversell" the book, for example with statements like "the notion of a Riemann surface becomes inevitable once we take the vantage point of phase portraits".
Notwithstanding, the usefulness of these phase portraits is enough to merit a five-star rating. This is an innovation I believe ought to be incorporated in other complex analysis textbooks (even if it requires them to be printed in color!).
Finally, this book might get compared to Needham's "Visual Complex Analysis". I'm less impressed with Needham's book, which introduces a lot of nifty original concepts but gets comparatively little mileage out of it. A beginning undergraduate student might enjoy Needham's book, but I believe that's the extent of the books usefulness. Wegert's book has much to offer for a graduate-level student. | 677.169 | 1 |
gebra guide
2.
. 1
Chapter 1
Getting Started With GeoGebra
In this chapter
• Install and start up GeoGebra
• Examine basic features
• Open and save a GeoGebra file
Starting up GeoGebra
GeoGebra can be downloaded for free at
It can also be accessed through any browser through the above web site (you still click on the "Down-
load" button, and look for the web applet option). This is useful if you are using a public computer or a
computer in a lecture hall and do not have download and install permissions.
This chapter will familiarize you with various basic features of GeoGebra (pronunced Gee-O-gebra).
We will introduce features of GeoGebra as the need arises. When working with the examples given,
you should be at a computer with the Geogebra program open.
Start up the GeoGebra software. The main buttons are shown below and each button has a drop down
menu associated with it..
Algebra View
Geometry View
3.
2 Guide to GeoGebra
Creative Commons Attribution-Noncommercial-No Derivative Works 3.0 United States License.
Do not be concerned if your screen does not exactly look like the above. When we work our examples,
you will be shown which tools must be used and how you can access them via the toolbar.
GeoGebra is a software with a wide variety of capabilities.This guide will focus primarily on how to
work with functions and their properties using GeoGebra.
Graphing a function
The easiest way to see what GeoGebra is about is by graphing a function. Unlike most graphing soft-
ware, GeoGebra allows you to dynamically change the graph with the mouse. That's what sets it
apart.
Example 1 Graph the function
Solution
1 In input bar, type f(x)=3x^2+1
2 Press <Enter> and you'll see the graph.
Moving the graph
GeoGebra is a dynamic geometry software. That is, the graph that's drawn is not static. It has various
properties which can be examined individually. A simple property is to just move the graph. To do this,
follow these steps.
1 Click on the selection tool. . This enables the drag option.
2 Point to the graph, then hold down the left mouse button, and move the graph with the mouse.
3 As you move the graph, note how the expression for f(x) changes.
f x 3x2 1+=
4.
Guide to GeoGebra 3
R. Narasimhan
Changing line styles
To change the line color and/or styles in a plot, right click into the curve you want to change. Click on
Properties, and set the options you wish.
Points on the graph
Example 2 Use the graph of to find various points on the curve and trace along the
curve.
Solution
Steps to trace a point along a graph
1 Click on the point tool and choose new point.
2 Now click on the selection tool.
3 Move mouse until you see the graph highlighted. Left click to place a point on the graph. Move
mouse to the point A on the graph, hold down left mouse button and start moving your mouse. You
should see the point A tracing along the curve. Your window should be similar to the following.
f x 3x2 1+=
5.
4 Guide to GeoGebra
Creative Commons Attribution-Noncommercial-No Derivative Works 3.0 United States License.
Scaling the axes
You can change the scale on the axes by simply clicking on the drawing pad tool . Point the
mouse to the axis you want to scale. By holding down your mouse button you can stretch or shrink the
x or y axes.
You can also use the drop down menu from the drwaing pad tool to zoom in or zoom out.
Saving GeoGebra files
1 Open the File menu and select Save.
2 Select a folder where you want to store the file, or create a new folder.
3 Type in a name for your GeoGebra file.
4 Click Save in order to finish this process. A file with the extension '.ggb' is created. This exten-
sion identifies GeoGebra files and indicates that they can only be opened with GeoGebra.
Avoid using spaces or special symbols in a file name since they can cause unnecessary problems when
transferred to other computers. Instead you can use underscores or upper case letters within the file
name (e.g. First_Drawing.ggb).
Note that the coordinates of the point A are displayed under Dependent Objects. That's because
the point A depends on the function f.
6.
Guide to GeoGebra 5
R. Narasimhan
Printing and exporting graphs
Once you've drawn a graph with GeoGebra, you may want to export the graph so that you can use it
later for another purpose.
To do this, click on File > Export >Graphics View.... Select the option you prefer. You can save as a
graphics file (you will be prompted on the file tye - png, pdf etc.) or save in the Clipboard to paste in a
document..
Some Common Errors
• To move the graph around, you need to have the selection tool highlighted.
7.
. 6
Chapter 2
Lines and Linear Equations
In this chapter
• Graph a line and find its equation given two points
• Explore the slope triangle
• Find distance between two points
• Parallel and perpendicular lines
• Creating dynamoc graphs with sliders
Graphing a line given two points
A simple way to explore the powerful features of GeoGebra is to plot two points, and draw a line.
GeoGebra will automatically determine its equation
Example 1 Plot the points (2,3) and (-4,1) and draw a line through them.
Solution
1 Click on Options > Drawing Pad.
8.
Guide to GeoGebra 7
R. Narasimhan
2 Use the options here to set up your drawing grid.
3 In the Input box, type A=(2,3). Press <ENTER>. Then type B=(-4,1). The two points will be
plotted as follows.
9.
8 Guide to GeoGebra
Creative Commons Attribution-Noncommercial-No Derivative Works 3.0 United States License.
4 Now choose the line tool . Click on Point A. A line will appear. Guide the line to Point B
and click. You should see a graph similar to the one below
5 Save the file as line1.ggb. We will use it later in this chapter.
Check it out
• Click on the selection tool and move the graph. You can translate the graph. See what happens
to the points A and B and the equation.
• You can also rotate the line around a point. The drop down menu from the selection tool has an
option for this.
Using the slope triangle
Using GeoGebra, you can visually indicate the slope of the line that you graphed in Example 1. Fur-
thermore, you can see how the slope changes as you move one or both of the points.
10.
Guide to GeoGebra 9
R. Narasimhan
Example 2 Use GeoGebra to indicate the slope of the line in Example 1.
Solution
1 Open the file line1.ggb, created in Example 1. If you were moving the line around, the points
A and B may have changed in the file. If so, in the algebra window, double click on B to (-4,1).
You can now edit the coordinates of Point B. You can also do this for Point A(2,3).
2 Click at the bottom right of the angle button , and choose the slope button.
3 Move your mouse to highlight the line and click. The slope triangle and the value of the slope will
appear on the graph. and in the algebra window.
4 You can also find the slope by typing m=Slope[a] in the Input box. Here, a is the name of the
line that was created and m is the letter assigned to the slope.
5 You can display the equation of Line a in slope-intercept form by right clicking on its equation and
selecting Equation y=mx+b
6 Save this file under the original name or to a different name.
Check it out
• Click on the selection tool, and move Point B. Keep Point A fixed. What do you notice about
the slope?
• Calculates values of the slope for some different values of Point B and compare to the value
given in GeoGebra.
• In the algebra window, double click on B. You can now edit the coordinates of Point B. You
can also do this for Point A.
11.
10
Finding distance between two points
You can use the Distance command in the Input box to calculate the distance between two points.
Example 2 Use GeoGebra to find the distance between A(-4,1) and B(3,-2).
Solution
1 Start up GeoGebra or open a new GeoGebra workspace by choosing File> New.
2 In the Input box, type A=(-4,1). Press <ENTER> and then type B=(3,-2).
3 In the Input box, type d=Distance[A,B]. Press <ENTER>.The value of the distance is dis-
played in the algebra window.
Check it out
• Use the distance formula to verify that the distance given by
GeoGebra is correct.
• Find the midpoint of points A and B by typing C=Midpoint[A,B] in the Input box
• A good use of the distance function is to help you find the perimeter of a polygon whose verti-
ces are given by coordinates. Plot the points (1,2), (3,-1) and (-1,-2) in GeoGebra. Connect
each pair of points with a line segment by clicking on the Line tool and choosing the segment
option.. Find the perimeter of the resulting triangle.
Parallel and perpendicular lines
Using GeoGebra, you can examine both the geometric and algebraic relationships between two paral-
lel or two perpendicular lines.
d x2 x1–
2
y2 y1–
2
+=
12.
Guide to GeoGebra 11
R. Narasimhan
Example 4 Use Geogebra to sketch the line parallel to the line and passing through
(-1,2).
Solution
1 Start up GeoGebra or open a new GeoGebra workspace by choosing File> New. To see the grid,
make sure the menu item View > Grid is checked.
23 Now type P=(-1,2) in the Input box. Press <ENTER>. Click on the button with perpendicular
lines and then click on the small red arrow. From the submenu, choose Parallel Line.
y 2x– 3+=
13.
12 Guide to GeoGebra
Creative Commons Attribution-Noncommercial-No Derivative Works 3.0 United States License.
4 Now select the Point P and the line a. The parallel line passing through P is now drawn and its
equation is given.
5 Use GeoGebra to find the slope of each line. Type Slope[a] in the Input box. Press
<ENTER>. Then type Slope[b] in the Input box. Press <ENTER>. The slopes and the slope
triangles are now given. Check these values by hand.
Check it out
• What do you observe about the slopes of parallel lines?
• Find the equation of the parallel line by hand, and check with the one from GeoGebra..
14.
Guide to GeoGebra 13
R. Narasimhan
Example 5 Use Geogebra to sketch the line perpendicular to the line and passing
through (-1,2).
Solution
1 Start up GeoGebra or open a new GeoGebra workspace by choosing File> New. To see the grid,
make sure the menu item View > Grid is checked.
2 To get a square grid, right-click in the graphing area. Select Properties and set xaxis:yaxis ratio to
1:1. Otherwise, perpendicular lines will not appear perpendicular!
34 Now type P=(-1,2) in the Input box. Press <ENTER>. Click on the button with perpendicular
lines .
y 2x– 3+=
15.
14 Guide to GeoGebra
Creative Commons Attribution-Noncommercial-No Derivative Works 3.0 United States License.
5 Now select the Point P and the line a. The perpendicular line b, passing through P, is now drawn
and its equation is given.
6 Use GeoGebra to find the slope of each line. Type m1=Slope[a] in the Input box. Press
<ENTER>. Then type m2=Slope[b] in the Input box. Press <ENTER>. The slopes and the
slope triangles are now given. Check these values by hand by examining the slope-intercept forms
of the equations.
Check it out
• Do you observe about the slopes of perpendicular lines are negative reciprocals of each other?
• Find the equation of the perpendicular line by hand, and check with the one from GeoGebra.
16.
Guide to GeoGebra 15
R. Narasimhan
Creating dynamic graphs with sliders
In GeoGebra, you can introduce sliders to make your graph change when its parameters change. For
example, the equation of a line is given by . You can set the slope, m, and the y-intercept, b,
as parameters and then define the line. This allows GeoGebra to change m and b dynamically.
Example 6 Use Geogebra to sketch the line , with m = 3 and b= 4. USe a slider to
vary m and b.
Solution
1 Start up GeoGebra or open a new GeoGebra workspace by choosing File> New. To see the grid,
make sure the menu item View > Grid is checked.
2 In the Input box, type m=3. Press <ENTER>. Then type b=4. Press <ENTER>.
3 Now type y=m*x+b in the Input box.
4 If you look at the left panel, you will see that the line is dependent on the "objects" m and b. Click
on the b = 4 button in the left panel. You will see a slider appear in the graphing area. Do the same
y mx b+=
y mx b+=
17.
16 Guide to GeoGebra
Creative Commons Attribution-Noncommercial-No Derivative Works 3.0 United States License.
for the m = 3 button. Your graph should resemble the following. Now move each slider with your
mouse and see what happens to the graph.
Check it out
• As m varies from negative to zero to positive, what do you observe about the slant of the line?
• What do you observe about the line as b varies from negative to zero to positive?
• You can add a slope triangle to visualize the slope of the line. Refer to Example 2 for instruc-
tions.
• The default range for the sliders is from -5 to 5. You can change this by right-clicking on the
slider and then modifying the range under the slider tab in the properties dialog box. You can
also change its color and other properties. Try it!
NOTE: This guide is a work in progress. Chapters on different functions, including trig func-
tions, and conic section will be added. Your comments are welcome. Please email comments to
reva@mymathspace.net
(Notes on new features to include:
Idea: mx+b : parametrize m and b; list view of points and "guess" at line of best fit) | 677.169 | 1 |
Find a Dania Algebra 2 help students gain mastery of sign operations (interactions of + and -)
As soon as they can grasp it, I make them understand that there are also the realm of negative numbers, the dynamics of the number line, and even the idea of mathematical infinity. Mastery of order of operations is critica... | 677.169 | 1 |
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by James Stewart
excellent conditions!!!! the book arrive on time, look tike new, excellent price.
nucleicacid5
Oct 18, 2012
George
Book is actually pretty good, I like the fact that its first chapter is a review of stuff from intermediate algebra...if you are using this book...and you have just finished intermediate/college algebra...get this book and do each section of ch. 1...this will set you up for a good foundation for the rest of the class. I like the visuals of the book...I have the most recent edition.
Jimmy J L
Jan 12, 2012
Great prep for calculus
A beautiful presentation and treatment of all math required before studying calculus. Comprehensive, and a strong focus on theory. Lots of problems to test yourself. Get through this and then star in your calculus study, as you are now VERY well prepared.
Jonathan R
Oct 20, 2011
the book is great! just what I expected! one of the best math books
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Oct 6, 2011
One of the best Precal Books
This book is one of the best out there in the current markets. Dr. Stewart explains this subject with geometrical shapes to better understand the subject. For example he explains and proves the the phytogorean theorem, laws of sines and cosines, and alot more. Highly recommend this book as well as his other ones like college algebra and calculus. He is an expert in this subjects. The best way to master a theorem in math is to prove and applied its existence and this book aims at teaching this principles | 677.169 | 1 |
Numerical Analysis with CD-ROM / Edition 1/b>
Overview chapter contains a Reality Check, an extended foray into a relevant application area that can be used as a springboard for individual or team projects. MATLAB is used throughout to demonstrate and implement numerical methods.
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Sauer teaches numerical analysis with an emphasis on you using Matlab to learn the concepts. His text is suitable as a first course in this subject, at an undergraduate level. The topics are, by and large, unchanged from those in texts written in the 1970s or even earlier. Like the solving of systems of linear equations. A crucial topic which you should master, because these arise throughout engineering and the physical sciences. So the text goes into the ideas of LU factorisation and Gaussian row reduction with pivoting. Other topics covered included numerical integration and differentiation, and least squares curve fitting. The rigour of the proofs is also well chosen. Enough to satisfy most students. But without being too abstruse to put most off. A strong point of the text is the numerous problems interspersed throughout each chapter. These are usefully divided into two types. The first might be considered traditional problems. Where you solve these by hand, with perhaps only a calculator to plug in a few numbers. The second group consists of using Matlab in a more extended foray into numerical analysis. Here, a mere calculator will not suffice. Doing these problems will improve your facility with Matlab and also hopefully garner a general experience in knowing when to use a maths package. It should also be said that if you already use another package, like Mathematica, then this book can still be germane. Roughly speaking, Mathematica, Matlab and Maple have equivalent functionality. Certainly, this is true at the introductory level of the text. | 677.169 | 1 |
An introduction to Einstein's general theory of relativity. According to Einstein,
gravity is the curvature of space-time; so general relativity is geometry! This course
is geared toward mathematicians with little or no familiarity with physics. Interested
students from other backgrounds are, of course, more than welcome.
Prerequisite: Math 32 and Math 129A or instructor consent. If you are not sure if
you satisfy the prerequisites, please contact the instructor.
Mathematical models used in Biology and other life sciences, discrete and continuous
processes, mathematical methods for analyzing solutions to the above models, interpretation
of mathematical results.
Prerequisite: Math 123, 129A, 133A, or 134 with a grade of C- or better or instructor
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Advanced topics in number theory. Emphasis may be in algebraic number theory (e.g.
Diophantine equations), analytic number theory (e.g. the prime number theorem), and/or
computational number theory (e.g. cryptography). | 677.169 | 1 |
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Deep Dive into Mathematica's Numerics: Applications and Tips
Andrew Moylan
In this course from the Wolfram Mathematica Virtual Conference 2011, you'll learn how to best use Mathematica's numerics functions in advanced settings. Topics include techniques and best practices for using multiple numerics functions together, advanced numeric features, and understanding precision and accuracy.
Jon McLoone, director of technical communications and strategy at Wolfram Research Europe, discusses new developments based on the Wolfram Language, including Wolfram Programming Cloud, Mathematica Online, and the release of ... | 677.169 | 1 |
Find a Plainfield, NJ StatisticsThank you for your consideration. Algebra 1 is a textbook title or the name of a course, but it is not a subject. It is often the course where students become acquainted with symbolic manipulations of quantities | 677.169 | 1 |
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Details about Algebraic Geometry:
An introduction to abstract algebraic geometry, with the only prerequisites being results from commutative algebra, which are stated as needed, and some elementary topology. More than 400 exercises distributed throughout the book offer specific examples as well as more specialised topics not treated in the main text, while three appendices present brief accounts of some areas of current research. This book can thus be used as textbook for an introductory course in algebraic geometry following a basic graduate course in algebra. Robin Hartshorne studied algebraic geometry with Oscar Zariski and David Mumford at Harvard, and with J.-P. Serre and A. Grothendieck in Paris. He is the author of "Residues and Duality", "Foundations of Projective Geometry", "Ample Subvarieties of Algebraic Varieties", and numerous research titles.
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Top positive reviewTop critical review filtering reviews right now. Please try again later.There was a problem loading the comments at the moment. Please try again later.
I hesitated before buying this book, thinking that there would be little that I would understand. I have a degree in mathematics from Cambridge, but that was achieved 50 years ago and although I have kept my interest and read a number of popular books on the subject, I have never continued my studies and have forgotten most of what I once knew.
The book is a delight. Look inside and you will find a wide range of topics, ranging from a series of introductory articles to biographical and historical sections and including descriptions of the branches of the subject currently under investigation and the impact of mathematics on the practical world.
If you have an interest in mathematics and are tempted, do seriously consider buying this book. There will be (almost certainly) some parts that you find very difficult but there will be very many that give you a fascinating insight and great pleasure.
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For the first-time ever, in my 65 years on this planet Timothy Gowers has enabled me to understand and appreciate a mathematical concept i.e. Algebraic geometry. I am so overjoyed, by what, many wiser heads, would take in their stride.
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My formal maths education ended 30 years ago with university level mathematics for engineering. A renewed interest had me looking for a book that explains how the many topics within mathematics are linked together, in a way that limits the use of the formal language of maths. The search is over....this is the book.
It is not a textbook, and you won't earn a college degree by reading it, but anyone reading part 1 will understand fundamentals which will make textbooks much easier to read. Part 2 is also a good path to understanding the history of mathematics, and both why and how it was developed to solve real problems that man has faced down the years. Further sections explain key concepts in language as close to laymans's terms as possible, and give more detailed guides to major subtopics, significant contemporary questions, prominent figures, applications to other area of science and life as a mathematician.
The book is not 'light' reading in any sense. Maths can be inherently difficult, and the contents of the Companion are essays without illustrations. The hard copy is also physically daunting, which is why I would recommend the digital kindle version for each of access and use.
Nonetheless, if you have an interest in mathematics and have some level of maths education then, whether you use it as a guide to the subject or as a companion to deeper study, this is a book you should have.
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This is a great book. Its entries are well written by recognisable experts and at a high enough level so that people attempting research may use them as a starting point or even as a guide to the main people in the field. It can be read by the amateur too. Other good points are: the quality and depth of coverage, the connections made between fields, and the broad perpective it provides on the subject.
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This is a true companion - someone you'd welcome on a trip or sharing an exploration of mathematics. The articles are written by people who are both clear writers and excellent teachers. It helps both the mathematical 'day-tripper' as well as the student of mathematics, as a comprehensive reference.
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This is a brilliant book, everyone who loves maths should own it, and everyone who is curious about whether they could love maths (i.e. recovering from a disasterous experience at school) should read it.
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This is a truly excellent books that I find myself dipping into whenever I have a spare moment. A fair proportion of the material requires a reasonable mathematical background to fully appreciate. Wonderfully insightful articles from world class mathematics.
Many times I read an article and was enlightened about the history, links with other areas or sometimes even the real point of the theory. It reminds me of some sessions with my PhD supervisor when you would come away thinking - Wow! I wish I could think like that! | 677.169 | 1 |
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Part A
Metric and Topological Spaces
1
Lecture A1
Introduction to the Course
This course is an introduction to the basic ideas of topology and metric space
theory for first-year graduate students. Topology studies the idea of "continuity"
in the most general possible context. As a separate subject, it takes its origin
from the pioneering papers of Poincar´ e at the end of the 19th century, and its
development is one of the central stories of 20th century pure mathematics. (The
history by Dieudonn´ e, A History of Algebraic and Differential Topology, 1900 -
1960, is available free to Penn State members online. — but that history begins
roughly where this course leaves off.) But the idea of continuity is so central
that topological methods crop up all over mathematics — in analysis of course
(including its "computational" twin, numerical analysis); in geometry; in pure
algebra; even in number theory. So, if you're a first-year grad student, you most
likely need to take this course (and maybe even Math 528 as well).
In this lecture we'll just describe the course protocol and prerequisites — we'll
get started properly next time.
There is a recommended course textbook —Introduction to Topology by Gamelin
and Greene, which is available in a cheap Dover reprint for 15 bucks. However,
more important than that are the online course materials, which are available in
two ways:
• Through the course website, which is
• Through the course page on ANGEL, Penn State's course management sys-
tem, at
Registered students (i.e., those taking this course for a grade) will want to use
the ANGEL site which will contain homework assignments, quizzes and support-
ing material, which all need to be taken on time in order to receive credit. The AN-
GEL page also contains the detailed course syllabus, which contains University-
required information about exam schedules, academic integrity requirements, of-
fice hours, and suchlike. Please be sure to read this information carefully.
Detailed lecture notes for each lecture will be posted 24 hours in advance on
these sites. To succeed in the course you need to read and study the notes before
the lecture to which they refer. This is very important!
Most lectures will contain one or more in-class exercises. You are expected
to attempt these exercises before the next class session. Students may be called
2
on to present their solutions in class. You grade your own exercises using the
ANGEL system, and this self-assessment will form a (small) component of your
final in-class grade.
Exercise A1.1. Here is your first in-class exercise. Find the corresponding as-
sessment item on ANGEL (it's called "inclassAug23"), and enter your response
("I fully understand this"). You must complete this assignment before tomorrow
evening, when it will expire.
Nowfor some information about the prerequisites for the course, that is, things
that I will assume that you know. There are four groups of these.
1. Number systems and their properties: specifically the natural numbers N,
integers Z, rational numbers Q, real numbers R, and complex numbers C. Courses
in analysis or foundations of mathematics often contain an account of procedures
for "constructing" all or some of these, but we'll just take them as given. The most
important of these number systems for us is the system R of real numbers. The
basic properties of the reals are summarized by
Proposition A1.2. R is an archimedean complete ordered field.
What does this mean?
(a) A field — basic properties of additional, multiplication, subtraction, divi-
sion — commutative, associative, distributive laws.
(b) ordered. There is an order relations < with the expected properties. In
particular x
2
0 for all x.
(c) Archimedean — no 'infinite' or 'infinitesimal' real numbers. There is a
unique (injective) homomorphism of rings Z → R taking 1 to 1. (Proof?)
The archimedean property is that every element of R is smaller than (the
image of) some element of Z.
(d) complete — no 'holes', contrast Q.
There are various ways of expressing completeness. The standard way is the
Least Upper Bound Axiom. Let S be any nonempty set of real numbers. A number
a ∈ R is an upper bound for S if, for all x ∈ S, x a. Let U
S
be the set of all
upper bounds for S. The least upper bound axiom says that, if S and U
S
are both
nonempty, then U
S
has a least member. This is called the least upper bound for S
and written sup S.
3
Example A1.3. Let S = ¦x ∈ R : x
2
< 2¦. Then U
S
= ¦a ∈ R : a > 0, a
2
2¦.
U
S
has a least member, namely sup S =
√
2. The corresponding statement with
R replaced everywhere by Q would be false.
Exercise A1.4. (Cantor's nested interval theorem) Let [a
n
, b
n
], n = 1, 2, . . . be a
sequence of closed intervals in R that are nested in the sense that [a
n+1
, b
n+1
] ⊆
[a
n
, b
n
] and whose lengths b
n
− a
n
tend to zero. Show that the intersection
n
[a
n
, b
n
] contains one and only one real number c.
Exercise A1.5. Use the previous exercise to show that the real numbers are un-
countable, i.e. cannot be listed as c
1
, c
2
, . . .. Hint: Suppose such a listing is
possible. Pick an interval [a
1
, b
1
] of length at most 1 that doesn't contain c
1
, then
a subinterval [a
2
, b
2
] of length at most
1
2
that doesn't contain c
2
, and so on. Apply
the nested interval theorem. What can you say about the limit point c?
4
Lecture A2
Metric Spaces
We continue with our list of prerequisites for the course (we discussed number
1 last time.)
2. Set-theoretic language such as unions (A ∪ B), intersections (A ∩ B),
complements (A¸B), subsets (A ⊆ B) and so on. In topology it is often necessary
to deal with infinite unions and intersections. Let F be a family of sets (that is, a
set whose members are subsets of some other set). The intersection of the family,
written
F∈F
F or just
F, is the set of objects that belong to every member of
the family F: in symbols
x ∈
F ⇔ ∀F ∈ F, x ∈ F.
Similarly the union of the family, written
F∈F
F or just
F, is the set of objects
that belong to some member of the family F: in symbols
x ∈
_
F ⇔ ∃F ∈ F, x ∈ F.
De Morgan's Laws extend to this context: if F is a family of subsets of some set
X, then
X ¸
F∈F
F =
_
F∈F
X ¸ F
and the same with intersection and union reversed.
3. Quantified statements and their proofs. We already met some quantifiers
in the previous section: things like ∀ and ∃. Throughout pure mathematics, but
especially analysis and topology, one meets concepts that are built up of layers of
quantifiers nested together in complicated ways. For instance, a function f : R →
R is continuous at a given point x
0
if
∀ > 0 ∃δ > 0 ∀x ∈ R[x −x
0
[ < δ ⇒ [f(x) −f(x
0
)[ < .
It is necessary that you can grasp what such a complicated statement says, and
that you can understand how the form of the statement dictates the shape of its
proof. For instance the statement above begins "∀ > 0" so its proof has to begin
"Let > 0 be arbitrary", followed by an argument that establishes the rest of the
statement for a given arbitrary value of . You could call a simple pattern like this
a "proof skeleton" corresponding to the "for all" quantifier.
5
Exercise A2.1. Give some examples of other proof skeletons corresponding to
other parts of a statement, e.g. ∃, ⇒, ⇔and so on.
Exercise A2.2. Write down in symbols (as above) what it is for the function f to
be not continuous at x
0
.
4. Axiomatic method This is the "standard operating procedure" of modern
mathematics. We understand specific examples by fitting them into a general
theory. The general theory consists of a collection of axioms, such as those for a
group or a vector space. There are many examples which realize the axioms, and
we develop a theory that applies to all of them.
We will study topology from this point of view. We'll begin with the axioms
for metric spaces.
Definition A2.3. A metric space is a set X equipped with a function d: XX →
R (called a metric or distance function) such that:
(i) d(x, x
) 0 for all x, x
; moreover, d(x, x
) = 0 if and only if x = x
.
(ii) d(x, x
) = d(x
, x) for all x, x
(symmetry).
(iii) d(x, x
) d(x, x
) + d(x
, x
) for all x, x
, x
(triangle inequality).
The motivating example is the metric d(z, w) = [z −w[ on C or R.
Definition A2.4. Let (X, d
X
) and (Y, d
Y
) be metric spaces. An isometry from X
to Y is a bijection f : X → Y such that
d
Y
(f(x
1
), f(x
2
)) = d
X
(x
1
, x
2
)
for all x
1
, x
2
∈ X.
For the usual metric on the plane, the isometries are just the congruences of
Euclidean geometry. Two metric spaces that are related by an isometry are equiv-
alent from the point of view of metric space theory.
Here are some more examples of metric spaces.
Example A2.5. The vector space R
n
can be made into a metric space by defining
the distance between points x = (x
1
, . . . , x
n
) and y = (y
1
, . . . , y
n
) to be
d(x, y) =
_
[x
1
−y
1
[
2
+ +[x
n
−y
n
[
2
_
1/2
.
(The proof that this does indeed satisfy the triangle inequality is a standard ex-
ercise.) The same formula also makes C
n
into a metric space. These are called
Euclidean spaces.
6
Example A2.6. One can also define metrics on R
n
by
d
(x, y) =
_
[x
1
−y
1
[ + +[x
n
−y
n
[
_
and
d
(x, y) = sup
_
[x
1
−y
1
[, . . . , [x
n
−y
n
[
_
.
These are different metrics from the standard one (though, as we'll see later, they
are in a certain sense "equivalent".)
Example A2.7. Let X be any set. The discrete metric on X is defined by
d(x, y) =
_
0 (x = y)
1 (x ,= y)
Example A2.8. Let A be a finite set (the alphabet) and consider the set A
n
of
n-tuples of elements of A, which we think of as n-letter words in the alphabet A.
Define a distance on A
n
by
d(x, y) = #¦i : 1 i n, x
i
,= y
i
¦;
in other words, the number of positions in which the two words differ. This Ham-
ming distance was introduced in 1950 to give a technical foundation to the theory
of error-correcting codes (Hamming, Error-detecting and error-correcting codes,
Bell System Technical Journal 29(1950), 147–160.)
Example A2.9. Let X be any metric space and let Y be a subset of X. Then the
distance function on X, restricted to Y , makes Y into a metric space; this metric
structure is called the subspace metric on Y .
Example A2.10. Here is an infinite-dimensional example. Consider the collection
C[0, 1] of all continuous functions [0, 1] →C. We can define a metric by
d(f, g) = sup¦[f(t) −g(t)[ : t ∈ [0, 1]¦.
Convergence of a sequence of functions in this metric is called uniform conver-
gence.
Definition A2.11. A subset U of a metric space X is open if for every x ∈ U
there is > 0 such that the entire ball
B(x; ) := ¦x
∈ X : d(x, x
) < ¦
is contained in U.
7
The triangle inequality shows that every ball is open, so there are plenty of
open sets. A set F whose complement X ¸ F is open is called closed. Note
carefully that 'closed' does not mean the same as 'not open'. Many sets are neither
open nor closed, and some may be both.
8
Lecture A3
Open and Closed Sets
We finished last time with an important definition: A subset U of a metric
space X is open if for every x ∈ U there is > 0 such that the entire ball
B(x; ) := ¦x
∈ X : d(x, x
) < ¦
is contained in U.
Example A3.1. In a discrete metric space (Example A2.7), the open ball B(x,
1
2
)
is just the set ¦x¦. Consequently, in a discrete space, every subset is open.
Example A3.2. Consider the three metrics on R
n
defined in examples A2.5 and A2.6.
Each open ball in any one of these metrics contains an open ball (with the same
center but possibly different radius) in any one of the other metrics. Consequently,
these three metrics all have exactly the same open sets. That is what is meant by
saying that they are equivalent.
Lemma A3.3. The union of any collection of open sets is open. The intersection
of a finite collection of open sets is open. The empty set ∅, and the entire metric
space X, are open.
Proof. Let F be a collection of open subsets of a metric space X and let U =
F be the union of the family. If x ∈ U, then there is some V ∈ F such that
x ∈ V . Since V is open, there is > 0 such that B(x; ) ⊆ V . Since V ⊆ U,
we also have B(x; ) ⊆ U. Thus for any x ∈ U there exists > 0 such that
B(x; ) ⊆ U; which is to say that U is open.
Now let F = ¦U
1
, . . . , U
n
¦ be a finite collection of open sets and let U =
F = U
1
∩ ∩ U
n
. If x ∈ U, then for each i = 1, . . . , n there is
i
> 0 such
that B(x;
i
) ⊆ U
i
. Let = min¦
1
, . . . ,
n
¦ > 0. Then B(x; ) ⊆ U
i
for all i, and
thus B(x; ) ⊆ U. Thus for any x ∈ U there exists > 0 such that B(x; ) ⊆ U;
which is to say that U is open.
Equivalently, using de Morgan's laws, we have
Lemma A3.4. The intersection of any collection of closed sets is closed. The
union of a finite collection of closed sets is closed. The empty set ∅ and the entire
metric space X are closed.
9
Let X be any metric space, and let S be a subset of X.
Definition A3.5. The interior of S (in X), denoted S
◦
, is the union of all the open
subsets of X that are included in S.
In symbols, we have
S
◦
=
_
_
U : U ⊆ S and U open in X
_
.
The interior of S is an open set (because it is the union of a family of open
sets) and (from the definition) any open subset of X that is included in S is also
included in S
◦
. Thus, the interior of S is just the biggest open subset of X that is
included in X. In particular, S is open if and only if it is equal to its own interior.
Dually, we have
Definition A3.6. The closure of S (in X), denoted S, is the intersection of all the
closed subsets of X that include S.
The closure of S is a closed set, and it is the smallest closed set that includes
S; in particular, S itself is closed iff it is equal to its own closure. Finally,
Definition A3.7. The boundary of S (in X), denoted ∂X, is the set-theoretic
difference ∂S = S ¸ S
◦
.
The boundary of S is the intersection of two closed sets (S and X ¸ S
◦
), and
it is therefore closed.
Example A3.8. Let X = R and let S =
_
(0, 1) ∩ Q
_
∪ (2, 3]. Then S
◦
= (2, 3),
S = [0, 1] ∪ [2, 3], and ∂S = [0, 1] ∪ ¦2, 3¦.
Exercise A3.9. Show that if A, B are subsets of X, and A ⊆ B, then A
◦
⊆ B
◦
and A ⊆ B. Now show that for any subset S of X,
_
_
S
_
◦
_
◦
=
_
S
_
◦
.
One more remark about open sets. Let X be a metric space and let Y be a
subset of X. As we said earlier, Y can be considered as a metric space in its own
right (with the metric that it inherits from X). What is then the relation between
the open subsets of the new space Y and the open subsets of the original space X?
10
Proposition A3.10. Let X be a metric space, Y a metric subspace (as above).
Then a subset V ⊆ Y is open in Y iff it can be written V = U ∩ Y for some
U ⊆ X open in X.
Proof. Notice the following relationship between balls in Y and in X:if y ∈ Y
then
B
Y
(y; r) = B
X
(y; r) ∩ Y.
Let V be open in Y . Then for each y ∈ V there is r
y
> 0 such that B
Y
(y; r
y
) ⊆ V .
Let U =
y∈Y
B
X
(y; r
y
). This is a union of open subsets of X, so open in X,
and
U ∩ Y =
_
_
y∈Y
B
X
(y; r
y
)
_
∩ Y =
_
y∈Y
B
Y
(y; r
y
) = V.
This proves one direction of the "if and only if" statement in the proposition; the
other direction (which is easier) is an exercise.
The definition of continuity is translated in the natural way to the metric space
context.
Definition A3.11. A function f : X → Y between metric spaces is continuous at
x ∈ X if for every > 0 there is δ > 0 such that d(f(x), f(x
)) < whenever
d(x, x
) < δ. It is continuous if it is continuous at every x ∈ X.
But there is a very important alternative characterization in terms of open sets.
Theorem A3.12. Let X and Y be metric spaces. Then f : X → Y is continuous
iff, for every open U ⊆ Y , the inverse image
f
−1
(U) := ¦x ∈ X : f(x) ∈ U¦
is open in X.
Proof. Suppose that f is continuous and let U ⊆ Y be open. Let x ∈ f
−1
(U);
then f(x) ∈ U so by definition of 'open' there is > 0 such that B(f(x); ) ⊆ U.
By definition of 'continuous' there is δ > 0 such that if x
∈ B(x; δ) then f(x
) ∈
B(f(x); ) ⊆ U. But this means that B(x; δ) ⊆ f
−1
(U). Thus the set f
−1
(U) is
open.
Conversely, let x ∈ X, > 0 and suppose that f satisfies the condition in
the theorem. In particular, we may consider U = B(f(x); ), an open set such
that x ∈ f
−1
(U). Our hypothesis tells us that f
−1
(U) is open, which means that
there is a δ > 0 such that B(x; δ) ⊆ f
−1
(U). We have shown that whenever
x
∈ B(x; δ), f(x
) ∈ B(f(x); ). This gives us continuity.
11
Lecture A4
Continuity and sequences
In the previous lecture we saw that a map f : X → Y between metric spaces
is continuous iff f
−1
(U) is open (in X) whenever U is open (in Y ).
Remark A4.1. It is equivalent to say that f
−1
(F) is closed in X whenever F is
closed in Y . But in general nothing can be said about the behavior of the direct
image f(S) of an open or a closed set S under a continuous map.
Definition A4.2. A map f : X → Y between metric spaces is called a homeo-
morphism if it is a bijection and both f and f
−1
are continuous. (Equivalently,
f is a continuous map with a continuous inverse.) If there is a homeomorphism
between X and Y , then we say that these spaces are homeomorphic.
A homeomorphism is a "topological equivalence" (it is easy to check that the
relation of homeomorphism is an equivalence relation in the sense of algebra). In
topology, we consider homeomorphic spaces to be "essentially the same". A fun-
damental question in topology is to devise processes for answering the question,
Are two concretely given spaces homeomorphic or not?
Example A4.3. The map x → x(1 + x
2
)
−
1
2
is a homeomorphism from R to the
open interval 9 −1, 1). Its inverse is y → y(1 −y
2
)
−
1
2
.
Example A4.4. The map t → (cos 2πt, sin 2πt) is a continuous bijection from the
half-open interval (0, 1] onto the unit circle in R
2
, but it is not a homeomorphism.
Example A4.5. Let (X, d) be a metric space and let (X, d
) be the same set
equipped with the discrete metric. Then the identity map (X, d
) → (X, d) is
a continuous bijection, but it is usually not a homeomorphism.
Example A4.6. Are R
n
and R
m
homeomorphic if m ,= n? This and related
questions were problematical in the early days of topology. (Note that there are
continuous maps of R
m
onto R
n
even if m < n — "space-filling curves".) But
the answer, as expected, is indeed "no", as was shown by Brouwer and others at
the beginning of the 20th century.
We'll now study the properties of sequences. A sequence in a metric space X
is a mapping from the natural numbers Nto X: we'll follow the usual abbreviation
of referring to "the sequence (x
n
)" rather than "the sequence which maps the
natural number n to x
n
∈ X."
12
Definition A4.7. Let (x
n
) be a sequence in the metric space X. We say (x
n
)
converges to x ∈ X if for every > 0 there is an integer N such that d(x, x
n
) <
whenever n > N. We write then x = lim
n→∞
x
n
.
Exercise A4.8. Let T be the metric space ¦1,
1
2
,
1
3
, . . . , 0¦ (with the metric it in-
herits as a subspace of R). Show that a sequence (x
n
) in the metric space X is
convergent if and only if there exists a continuous function f : T → X such that
f(1/n) = x
n
; and in that case f(0) = lim
n→∞
x
n
.
Proposition A4.9. A function f : X → Y between metric spaces is continuous
at x if and only if, whenever (x
n
) is a sequence converging to x, the sequence
(f(x
n
)) converges to f(x).
Proof. Suppose that f is continuous at x and let > 0 be given. By definition of
continuity there is δ > 0 such that d(x, x
) < δ implies d(f(x), f(x
)) < . By
definition of convergence there is N such that for all n > N, d(x, x
n
) < δ. Thus
d(f(x), f(x
n
)) < and f(x
n
) converges to f(x).
Suppose that f is not continuous at x. Then there is > 0 such that for all
δ > 0 there is x
with d(x, x
) < δ and d(f(x), f(x
) . Let x
n
be a value of x
corresponding to δ = 1/n. Then x
n
→ x, but d(f(x), f(x
n
)) for all n, hence
f(x
n
) does not converge to f(x).
Remark A4.10. This result illustrates how sequences can be used to probe the
topological properties of metric spaces. At the same time, it shows how the use-
fulness of these "probes" depends on the countable local structure of a metric
space — specifically, on the fact that any ball around x ∈ X contains one of
the countably many balls B(x; 1/n). When we come to consider more general
topological spaces, this "countable local structure" may not be available, and then
sequences will be of less use.
Definition A4.11. Let X be a metric space, A ⊆ X. A point x ∈ X is a limit
point of A if it is the limit of a sequence of distinct points of A.
Lemma A4.12. x is a limit point of A if and only if every open set containing x
also contains infinitely many points of A.
Proof. Suppose that x is a limit point of A. Then there is a sequence (x
n
) of
distinct points of A, converging to x. Let U be an open set containing x. Then
there is some > 0 such that B(x; ) ⊆ U. By definition of convergence, there
13
is N > 0 such that x
n
∈ B(x; ) ⊆ U for all n > N. In particular, U contains
infinitely many of the ¦x
n
¦, all of which are members of A.
Conversely suppose that every open set containing x also contains infinitely
many points of A. Define a sequence (x
n
) inductively as follows: x
1
is any
point in B(x; 1) ∩ A, and, assuming that x
1
, . . . , x
n−1
have been defined, let x
n
be any point in B(x; 1/n) ∩ A that is distinct from the (finitely many!) points
x
1
, . . . , x
n−1
. Then (x
n
) is a sequence of distinct points of A and it converges to
x.
Proposition A4.13. The closure of a subset A is the union of A and the set of
all its limit points. In particular, A is closed if and only if it contains all its limit
points.
Proof. A point x belongs to the closure of Aiff it is not in the interior of X¸A. By
definition of "interior", this is the same as to say that each ball B(x; ) must meet
A. By the previous lemma, then, any limit point of A must belong to A. Con-
versely, suppose that x ∈ A ¸ A. Define a sequence (x
n
) inductively as follows:
x
1
is any point in B(x; 1) ∩A, and, assuming that x
1
, . . . , x
n−1
have been defined,
let x
n
be any point in B(x; δ
n
) ∩ A where δ
n
= min¦d(x
1
, x), . . . , d(x
n−1
, x)¦.
Then (x
n
) is a sequence of distinct points of A and it converges to x, so x is a
limit point of A.
Definition A4.14. An isolated point of A is a point of A that is not a limit point
of A. A closed subset of a metric space is perfect if it has no isolated points.
Exercise A4.15. Show that a ∈ A is isolated if and only if there is > 0 such that
B(a; ) ∩ A = ¦a¦.
Exercise A4.16. A closed ball in a metric space is a set
¯
B(x; r) := ¦x
∈ X : d(x, x
) r¦.
Show that a closed ball is closed. Must every closed ball be the closure of the
open ball of the same center and radius?
14
Lecture A5
Compactness
I'll take for granted the notion of subsequence of a sequence of points in a
matric space. (Formally speaking, if a sequence in X is a map N → X, a subse-
quence of the given sequence is the result of composing it with a strictly mono-
tonic map N →N.)
Lemma A5.1. Every sequence of real numbers has a monotonic subsequence.
Proof. Let (x
n
) be a sequence of real numbers. Suppose that it has no monotonic
increasing subsequence. Then there must be some n such that x
m
< x
n
for all
m > n. (Otherwise, we could build a monotonic increasing subsequence by
induction.)
But now take this n and call it n
1
. Apply the previous argument to the "tail" of
the original sequence beginning at n
1
, i.e. the subsequence x
n
1
+1
, x
n
1
+2
, . . .. This
also has no monotonic increasing subsequence, so by the same argument there is
n
2
> n
1
such that x
m
< x
n
2
for all m > n
2
. Continuing in this way by induction
we get a monotonic decreasing subsequence x
n
1
, x
n
2
, . . ..
It is a standard consequence of completeness that every bounded, monotonic
sequence of real numbers is convergent. Thus we get the Bolzano-Weierstrass
theorem: every bounded sequence of real numbers has a convergent subsequence.
Definition A5.2. A Cauchy sequence in a metric space is a sequence (x
n
) with
the following property: for every > 0 there exists N > 0 such that, for all
n, m > N, d(x
n
, x
m
) < . A metric space is called complete if every Cauchy
sequence in it converges.
A Cauchy sequence is necessarily bounded. Moreover, if a Cauchy sequence
has a convergent subsequence, then the whole sequence is in fact convergent.
These facts combine with the Bolzano-Weierstrass theorem to show that every
Cauchy sequence of real numbers converges: i.e., R is (Cauchy) complete.
Definition A5.3. A metric space X is (sequentially) compact iff every sequence
of points of X has a subsequence that converges in X.
Proposition A5.4. Every closed, bounded subset of R
n
or C
n
is sequentially com-
pact. (A subset is bounded if it is contained in some ball.)
15
Proof. For R this is just the Bolzano-Weierstrass theorem. For R
n
, this follows
from the fact (easily proved) that a sequence of points in R
n
is convergent if and
only if each of its coordinate sequences is convergent.
Proposition A5.5. If A is a subset of any metric space X, and A is compact (in
its own right), then A is bounded and closed in X.
Proof. Let x ∈ X be a limit point of A. Then there is a sequence (a
n
) in A
converging to x. But, by compactness, (a
n
) has a subsequence converging in A.
Thus x ∈ A, and A is closed.
Suppose that A is not bounded. Then one can construct by induction a se-
quence (a
n
) in A such that d(a
n
, a
0
) > 1 + d(a
n−1
, a
0
) for n 1. The triangle
inequality shows that d(a
n
, a
m
) > 1 for n ,= m. Clearly (a
n
) has no convergent
subsequence.
Exercise A5.6. Let f : X → Y be continuous and surjective. Show that if X is
compact then Y is compact also. What has this got to do with the familiar calculus
principle that 'a continuous function on a closed bounded interval is itself bounded
and attains its bounds'?
Despite the above evidence, it's not in general true that 'closed and bounded'
equals 'compact'. (Counterexample?) We shall analyze this in detail.
Definition A5.7. Let X be a metric space. An open cover U for X is a collection
(finite or infinite) of open sets whose union is all of X. A Lebesgue number for
U is a number δ > 0 such that every open ball of radius δ is a subset of some
member of U .
Theorem A5.8. (Lebesgue) Every open cover of a (sequentially) compact metric
space has a Lebesgue number.
Proof. Suppose that U does not have a Lebesgue number. Then for every n
there is x
n
∈ X such that B(x
n
; 1/n) is contained in no member of U . If X is
compact, the sequence (x
n
) has a subsequence that converges, say to x. Now x
belongs to some member U of U , since U is a cover. Thus there is > 0 such
that B(x; ) ⊆ U. There is n > 2/ such that d(x
n
, x) < /2. But then
B(x
n
; 1/n) ⊆ B(x
n
; /2) ⊆ B(x; ) ⊆ U
which is a contradiction.
16
Proposition A5.9. The following conditions on a metric space X are equivalent.
(a) Every sequence in X has a Cauchy subsequence.
(b) For every δ > 0 there is a finite cover of X by balls of radius δ.
In this case we say X is totally bounded (some writers say precompact).
Proof. (a) implies (b): Choose x
1
∈ X, and then inductively choose x
n
∈ X
such that d(x
n
, x
m
) δ when m < n, so long as this is possible. The process
must terminate because if it didn't it would produce a sequence with no Cauchy
subsequence. When it does terminate, with x
N
say, it does so because the balls
B(x
n
; δ), n = 1, . . . , N, cover X.
(b) implies (a): Notice the following implication of (b): given any δ > 0, any
sequence in X has a subsequence all of whose members are separated by at most
δ (call this a 'δ-close subsequence').
Let (x
n
) be any sequence in X. Let (x
1
n
) be a 2
−1
-close subsequence of (x
n
),
let (x
2
n
) be a 2
−2
-close subsequence of (x
1
n
), and so on by induction. Then (x
n
n
) is
a Cauchy subsequence of the original sequence.
A metric space X is said to be (covering) compact if every open cover of X
has a finite subcover.
Proposition A5.10. The following conditions on a metric space X are equivalent:
(a) X is sequentially compact;
(b) X is complete and totally bounded;
(c) X is covering compact.
Proof. It is easy to see that (a) and (b) are equivalent.
Suppose (a). Let U be an open cover of X. Let δ > 0 be a Lebesgue number
for U (which exists because of Theorem A5.8). Since X is totally bounded it has
a finite cover by δ-balls. But each such ball is a subset of a member of U ; so U
has a finite subcover.
In the other direction, suppose (c). Let (x
n
) be a sequence without convergent
subsequence. Then for each x ∈ X there is some
x
such that x
n
/ ∈ B(x;
x
) for
all but finitely many n. The B(x;
x
) form a cover of X. Picking a finite subcover
we obtain the contradiction that x
n
/ ∈ X for all but finitely many n.
17
Remark A5.11. For general topological spaces, the notions 'covering compact'
and 'sequentially compact' are not equivalent, and 'covering compact' is usually
the most appropriate one.
Exercise A5.12. A map f : X → Y between metric spaces is uniformly contin-
uous if for each > 0 there is δ > 0 such that d(f(x), f(x
)) < whenever
d(x, x
) < δ. (The extra information beyond ordinary continuity is that δ does
not depend on x.) Show that if X is compact, every continuous f is uniformly
continuous. Hint: use Theorem A5.8.
18
Lecture A6
Compactness and Completeness
(The lecture will begin with a review of the basic results about compactness,
from the previous section).
Definition A6.1. Let X be a compact metric space. Then C(X) denotes the space
of all continuous functions X →C. It is a metric space equipped with the metric
d(f, g) = sup¦[f(x) −g(x)[ : x ∈ X¦.
(Compare Example A2.10.)
Compactness of X ensures that the supremum exists (why)? We can also
consider the space C
R
(X) of continuous real-valued functions.
Proposition A6.2. For a compact X, the metric spaces C(X) and C
R
(X) are
complete.
Proof. Most completeness proofs proceed in the same three-stage way: Given a
Cauchy sequence, identify a candidate for its limit; show that the candidate is in
the space in question; and show that the sequence approaches the candidate limit
in the metric of the space in question.
Let (f
n
) be a Cauchy sequence in C(X). Then, for each x ∈ X, f
n
(x) is a
Cauchy sequence in C. Since C is complete this sequence converges. Denote its
limit by f(x).
We show that f is a continuous function on X. Fix x
∈ X and let > 0
be given. Because (f
n
) is Cauchy there is N such that for n
, n
N we have
[f
n
(x) − f
n
(x)[ < /3 for all x. Take n
= N and let n
→ ∞ to find that
[f
N
(x) −f(x)[ /3 for all x. Now, f
N
is continuous at x
so there is δ > 0 such
that [f
N
(x) − f
N
(x
)[ < /3 whenever [x − x
[ < δ. It follows that, whenever
[x −x
[ < δ,
[f(x) −f(x
)[
[f(x) −f
N
(x)[ +[f
N
(x) −f
N
(x
)[ +[f
N
(x
) −f(x
)[ < .
This shows that f is continuous.
Finally to show that f
n
→ f in C(X) we must prove that for every > 0
there is N such that [f
n
(x) −f(x)[ < for all x whenever n > N. But in fact we
already proved this in the previous paragraph.
19
It is an interesting and important question what are the compact subsets of
the complete space C(X). The answer is given by the Ascoli-Arzela theorem.
"Closed and bounded" is not enough.
20
Lecture A7
Applications of completeness
Recall that a metric space X is said to be complete if every Cauchy sequence
in X converges. The space (0, 1) is not complete, whereas the homeomorphic
space R is complete. This proves that completeness is not a topological property,
i.e., is not preserved under homeomorphism.
Exercise A7.1. A uniform homeomorphism between metric spaces is a homeo-
morphism f : X → Y such that both f and f
−1
are uniformly continuous. Show
that if X is complete and there is a uniform homeomorphism X → Y , then Y is
complete.
Proposition A7.2. Let X be a complete metric space and let U ⊆ X be an open
set. Then U is homeomorphic to a complete metric space (it is topologically
complete).
Proof. Let f : U →R be the function defined by
f(x) =
1
inf¦d(x, y) : y ∈ X ¸ U¦
.
This is a well-defined, continuous function with the property that f(x
n
) → ∞
whenever x
n
is a sequence in U that converges to some x ∈ X ¸ U.
Now let V be the metric space with the same points as U but with the metric
d
V
(x, x
) = d(x, x
) +[f(x) −f(x
)[.
It is easy to see that this is indeed a metric. Moreover, it is complete: if (x
n
) is
a Cauchy sequence for the metric d
V
, then it is also a Cauchy sequence for the
metric d (so it converges in X, say to x ∈ X) and, in addition, the values f(x
n
)
are bounded (so the limit x in fact belongs to V ).
Finally I claim that the identity map U → V is a homeomorphism. Suppose
x ∈ U and let > 0 be given. Since f is continuous, there is δ
1
> 0 such that
d(x, x
) < δ
1
implies [f(x) − f(x
)[ < /2. Put δ = min¦δ
1
, /2)¦. Then if
d(x, x
) < δ, we have
d
V
(x, x
) <
2
+[f(x) −f(x
)[ < .
So the identity map U → V is continuous, and the continuity of the inverse is
easy.
21
Remark A7.3. A subset of a metric space is called a G
δ
-set (German: Gebeit-
Durschnitt) if it is the intersection of countably many open sets. For example, the
irrational numbers form a G
δ
subset of R. Generalizing the above construction,
it can be shown that any G
δ
subset of a complete metric space is topologically
complete. In particular, the topology on the irrational numbers can be given by a
complete metric. But the same is not true for the rational numbers, as will follow
from the Baire category theorem.
Remark A7.4. A countable union of closed sets is called a F
σ
-set (French: ferm´ e,
somme). This is the only known example of dual mathematical concepts being
described using dual languages.
Definition A7.5. Let X be a metric space. A mapping f : X → X is a (strict)
contraction if there is a constant a < 1 such that d(f(x), f(x
)) ad(x, x
) for
all x, x
∈ X.
Note that a contraction must be continuous.
Theorem A7.6. (Banach) A contraction on a complete metric space has a unique
fixed point (a point x such that f(x) = x).
Proof. A fixed point is unique because if x, x
U
n
is dense.
Remark A7.11. The name 'category theorem' comes from the following (tradi-
tional) terminology. A set A is nowhere dense if the complement of its closure is
dense. A set is of first category if it is the countable union of nowhere dense sets.
Baire's theorem then says that in a complete metric space the complement of a set
of first category (sometimes called a residual set) is dense.
Remark A7.12. Notice that this gives us another proof that Ris uncountable (com-
pare Exercise A1.5). For the complement of a point in Ris a dense open set. More
generally, any complete, perfect (Definition A4.14) metric space is uncountable.
For in a perfect metric space, the complement of any point is dense.
Exercise A7.13. Consider the complete metric space C
R
[0, 1] of continuous real-
valued functions on [0, 1]. For a, b ∈ [0, 1] show that the set of functions f
which are nondecreasing on the interval [a, b] is nowhere dense in C
R
[0, 1]. Using
Baire's theorem, deduce that there exist functions f ∈ C
R
[0, 1] that are nowhere
monotonic, that is, monotonic on no subinterval of [0, 1].
23
Lecture A8
Normed Spaces
Definition A8.1. Let V be a vector space (real or complex). A norm on V is a
function V →R, denoted v → |v|, such that
(a) |v| 0 for all v, and |v| = 0 iff v = 0.
(b) |λv| = [λ[|v|, where λ is a scalar (real or complex).
(c) |v + v
| |v| +|v
|.
Clearly, d(v, v
) = |v − v
| is then a metric. Because of (b), this metric has
an 'affine' structure not present in a general metric.
Example A8.2. Let x = (x
1
, . . . , x
n
) be a vector in R
n
or C
n
. The expressions
|x|
1
= [x
1
[ + +[x
n
[
|x|
2
=
_
[x
1
[
2
+ +[x
n
[
2
_
1/2
|x|
∞
= max¦[x
1
[, . . . , [x
n
[¦
all define norms.
Exercise A8.3. Let a and b be positive real numbers and let p > 1. Show that
inf¦t
1−p
a
p
+ (1 −t)
1−p
b
p
: t ∈ (0, 1)¦ = (a + b)
p
.
Hence (or otherwise) show that the expression
|x|
p
= ([x
1
[
p
+ +[x
n
[
p
)
1/p
is also a norm on R
n
or C
n
.
Example A8.4. If V = C(X), we may define
|f| = sup¦[f(x)[ : x ∈ X¦.
This is a norm, and it gives rise to the usual metric on C(X). Since it is complete
(theorem A6.2) it is a Banach space as defined below.
24
Definition A8.5. If a normed vector space is complete in its metric, it is called a
Banach space.
Proposition A8.6. A linear mapping T : V → W between normed vector spaces
is continuous if and only if there is a constant k such that
|Tv| k|v|
(one then says that T is bounded).
Proof. If T is bounded then |Tu −Tv| k|u −v| so T is continuous.
If T is continuous then there is some δ > 0 such that |u| < δ implies |Tu| <
1. Take k = 1/δ.
Definition A8.7. The best constant k in the above proposition, that is the quantity
sup¦|Tv| : |v| 1¦
is called the norm of T and denoted |T|.
Exercise A8.8. Showthat, with the above norm, the collection /(V, W) of bounded
linear maps from V to W is a normed vector space, and that it is a Banach space
if W is a Banach space.
Exercise A8.9. Showthat the normof linear maps is submultiplicative under com-
position, i.e. |S ◦ T| |S||T|.
25
Lecture A9
Differentiation in Normed Spaces
The basic idea of differentiation is that of 'best linear approximation'. Normed
spaces provide a systematic way to express this.
Notation A9.1. Let f be a function defined on some ball B(0; ) in a normed
space V and having values in another normed space W. We shall write "f(h) =
o(|h|)" to mean that the limit lim
h→0
_
|h|
−1
|f(h)|
_
exists and equals zero. Sim-
ilarly for expressions like "f(h) = g(h) + o(h)".
Exercise A9.2. Suppose that T : V → W is a linear map. Show that T(h) =
o(|h|) if and only if T = 0.
Now let f : V → W be a continuous (need not be linear) map between normed
vector spaces. In fact, we need only suppose f is defined on some open subset U
of V .
Definition A9.3. With above notation, let x ∈ V . We say that f is differentiable
at x if there is a bounded linear map T : V → W such that
f(x + h) = f(x) + T h + o(|h|).
By the exercise, T is unique if it exists. It is called the derivative of f at x and
written Df(x).
Example A9.4. If V = W = R then every linear map V → W is multiplication
by a scalar, i.e. we identify /(V, W) = R. Under this identification our definition
of the derivative corresponds to the usual one from Calculus I.
Example A9.5. If V = R
m
and W = R
n
then /(V ; W) is the space of n m
matrices. The matrix entries of the derivative of f are the partial derivatives of the
components of f as defined in Calculus III. Remark: The existence of the partial
derivatives does not by itself imply differentiability in the sense of our definition
above. This just shows that our definition is right and partials are wrong!
Example A9.6. If V = R and W is arbitrary, the space of (bounded) linear maps
/(V ; W) can be identified with W itself. Under this identification the derivative
of f is given by the usual formula
Df(x) = lim
h→0
f(x + h) −f(x)
h
.
26
Exercise A9.7. Show that the derivative of a linear map is always equal to the
map itself.
Proposition A9.8. (Mean value theorem) Suppose that the function f is differ-
entiable throughout a ball B(x
0
; r) ⊆ V , and that |Df(x)| k for all x ∈
B(x
0
; r). Then
|f(x) −f(x
0
)| k|x −x
0
|
for all x ∈ B(x
0
; r).
Proof. Let > 0. By definition of the derivative each y ∈ B(x
0
; r) has a neigh-
borhood U
y
⊆ B(x
0
; r) such that
|f(y
(t) = F(t, f(t)) for all t ∈ [a, b]. The
value f(a) ∈ V is called the initial condition for the solution.
Theorem A9.13. Suppose that V is a Banach space and suppose also that the
function F is continuous and satisfies a Lipschitz condition, that is, there exists a
constant C > 0 such that
|F(t, v) −F(t
, v
)| C|v −v
|.
Then for any a ∈ R and any v
0
∈ V there exists a nontrivial interval [a, b] on
which there is a solution to the differential equation f
(t) = F(t, f(t)) with initial
condition f(a) = v
0
; moreover, this solution is unique.
28
Proof. We need to use some facts about integration for V -valued functions. It
isn't the job of this course to teach you about integration theory — for that see
Math 501 — so I will just state the various facts that are needed. Moreover, in a
first course on integration these facts are usually proved only for real or complex
valued functions; you will need to take it on faith that the same facts are true for
functions having values in a normed vector space V . Finally, we shall use the
fact that if V is a Banach space (that is, complete) then the space of continuous
functions [a, b] → V , equipped with the sup norm, is also complete. This is a fact
that we have already proved for real or complex valued functions, and the proof
in the general case is exactly the same.
We will use the fundamental theorem of calculus: a function f ∈ C[a, b] is a
solution of the differential equation if and only if it is a solution of the equivalent
integral equation
f(s) = v
0
+
_
s
a
F(t, f(t)) dt, t ∈ [a, b].
We will show that a solution to this equation exists (for b sufficiently close to a)
by applying Banach's contraction mapping theorem (A7.6) to the space C[a, b] of
continuous functions on [a, b] (with values in V , if you like). Let I : C[a, b] →
C[a, b] be defined by the right hand side of the display above, that is,
(If)(s) = v
0
+
_
s
a
F(t, f(t)) dt, t ∈ [a, b].
If f, g ∈ C[a, b] then the integrands in If and Ig differ by C|f − g| at most,
where C is the Lipschitz constant, and therefore
|If −Ig| C(b −a)|f −g|
by standard estimates on integrals. Therefore, if (b − a) < C
−1
, the map I is a
strict contraction and Banach's fixed point theorem provides a unique solution to
If = f, which is a solution to the differential equation with the specified initial
condition.
29
Lecture A10
The inverse function theorem
Let V, V
be normed vector spaces. The space /(V, V
) is then a normed
vector space also. Say T ∈ /(V, V
) is invertible if it has an inverse which is a
bounded linear map from V
= V . Look first at a nbhd of the invertible
operator I. Suppose |S| < 1. For y ∈ V consider the map
φ
y
: V → V, v → y −Sv.
Since |S| < 1 this map is a strict contraction, so it has a unique fixed point x
(Theorem A7.6). This fixed point is an x such that (I + S)x = y. By the triangle
inequality,
|y| (1 −|S|)|x|, so |x| (1 −|S|)
−1
|y|
and the map y → x is bounded. We have shown that if |S| < 1, I+S is invertible.
Moreover
|y −(I + S)
−1
y| = |S(1 + S)
−1
y| |S|(1 −|S|)
−1
|y|
which shows that the map S → (I + S)
−1
is continuous at S = 0.
We have shown that the identity is an interior point of the set of invertibles
and that the inverse is continuous there. However, left multiplication by a fixed
invertible is a homeomorphism from the set of invertibles to itself; so the same
results apply to any point of the set of invertibles.
Exercise A10.2. Let V be a normed space and let W be the normed space /(V, V ).
Show that the mapping i : T → T
−1
is differentiable (where defined) on W, and
that its derivative is
Di(T) H = −T
−1
HT
−1
.
30
Definition A10.3. A map f from an open subset of a normed space V to a normed
space W is continuously differentiable or C
1
if it is differentiable (everywhere)
and the map x → Df(x) is continuous.
The inverse function theorem says that under suitable conditions, if Df(x) is
invertible then f is 'locally invertible' near x.
Theorem A10.4. Let f be as above, defined near x ∈ V , and suppose that f
is continuously differentiable and that Df(x) ∈ /(V, W) is invertible. Suppose
moreover that V, W are Banach spaces. Then there is an open set U containing x
such that f is a bijection of U onto f(U), which is an open set containing f(x),
and such that its inverse g : f(U) → U is also continuously differentiable.
Proof. It is an application of Banach's fixed point theorem A7.6. We will con-
struct the inverse function by looking at the fixed points of a suitable map.
Let A = Df(x)
−1
. For y ∈ W define a map φ
y
: V → V by
φ
y
(z) = z + A (y −f(z)).
A fixed point of φ
y
is a solution to f(z) = y. We have
Dφ
y
(z) = I −A ◦ Df(z) = A ◦
_
Df(x) −Df(z)
_
.
Since Df is continuous there is r > 0 such that if z ∈ U = B(x; r) then
|Dφ
y
(z)| <
1
2
. From the mean value theorem we conclude that if F is a closed
subset of U, and if we know for some reason that φ
y
maps F to F, then φ
y
is
actually a contraction of this complete metric space and so has a unique fixed
point.
Fix z
0
∈ U and let y
0
= f(z
0
). Thus y
0
∈ f(U) and we want to prove
that there is some > 0 such that B(y
0
; ) ⊆ f(U); this will show that f(U) is
open. Choose δ > 0 such that the closed ball B(z
0
; δ) is contained in U. Choose
=
1
2
|A|
−1
δ. If z ∈ B(z
0
; δ) and y ∈ B(y
0
; ) then we may compute
|φ
y
(z) −z
0
| = |φ
y
(z) −φ
y
0
(z
0
)|
|φ
y
(z) −φ
y
(z
0
)| +|φ
y
(z
0
) −φ
y
0
(z
0
)|
1
2
|z −z
0
| +|A(y −y
0
)|
1
2
δ +
1
2
δ = δ
using the mean value theorem A9.8. We conclude that, for these y, the map φ
y
is
a contraction of the complete metric space B(z
0
; δ) and thus has a (unique) fixed
point there. We have shown that each y
0
∈ f(U) has an -neighborhood contained
31
in f(U); thus, f(U) is open. The contraction property of the φ
y
ensures that each
y ∈ f(U) has only one inverse image in U; thus f is a bijection of U onto f(U).
Now to show that the inverse function g = f
−1
: f(U) → U is differentiable
at y = f(x) write g(y + h) − g(y) = u(h), say. The calculations of the previous
paragraph show that |u(h)| 2|A||h| for |h| small. Then
h = f(g(y + h)) −y = f(g(y) + u(h)) −y = Df(x) u(h) + o(|h|).
Apply A = Df(x)
−1
to get
u(h) = A h + o(|h|).
This gives differentiability of g with Dg(f(x)) = Df(x)
−1
. Continuous differen-
tiability now follows from the continuity of the inverse operation on the space of
bounded linear maps (Proposition A10.1).
32
Lecture A11
Topological Spaces
We have been emphasizing that "up to homeomorphism" properties of metric
spaces depend only on their open sets. This motivates a further abstraction which
removes the numerical notion of metric entirely.
Definition A11.1. A topology on a set X is a family T of subsets of X with the
following three properties:
(i) If U
α
is any family of members of T , then the union
α
U
α
is also a member
of T ;
(ii) If U
1
, . . . , U
n
is a finite family of members of T then the intersection
n
i=1
U
i
is also a member of T ;
(iii) The sets ∅ and X are members of T .
A set equipped with a topology is called a topological space. The members of T
are called open subsets of the topological space X.
Notice that the open subsets of a metric space automatically satisfy (i)–(iii);
thus, every metric space carries a topology (called its metric topology). Different
metrics can however give rise to the same topology (as we have already seen);
moreover, there are topologies that do not arise from any metric.
Example A11.2. The discrete topology on X has T = P(X): every subset is
open. This topology arises from the discrete metric.
Example A11.3. The indiscrete topology on X has just two open sets, ∅ and X.
This topology does not arise from a metric as soon as X has more than one point.
(Why not?)
Example A11.4. The cofinite topology on X consists of ∅ together with all the
cofinite subsets of X (a subset is cofinite if its complement is finite).
It is natural to say that a closed set is one whose complement is open, so that
the closed sets in the cofinite topology are just the finite sets together with the
whole space. Notice that this topology has many "fewer" open (or closed) sets
than the topologies with which we are familiar. For example, in the cofinite topol-
ogy on an infinite set, any two nonempty open sets have a nonempty intersection.
A similar but more sophisticated example is the next one.
33
Example A11.5. Let K be an algebraically closed field (for example, the complex
numbers) and let A = K
n
. A subset V of A is called an affine algebraic variety if
there is a set S ⊆ K[X
1
, . . . , X
n
] of polynomials in n variables such that
(x
1
, . . . , x
n
) ∈ S ⇔ p(x
1
, . . . , x
n
) = 0∀p ∈ S;
in other words, a variety is the set of common zeroes of a collection of polynomi-
als. (The Hilbert basis theorem says that S can always be taken to be finite.) It is
a simple exercise (for an algebraist!) to show that any intersection of varieties is a
variety, and a finite union of varieties is a variety. In other words, the varieties can
be considered as the closed sets of a topology on A, called the Zariski topology.
Example A11.6. Let T be a family of topologies on the set X. Then
T is
also a topology on X. In particular, let F be any family of subsets of X at all.
Let T be the family of all topologies T such that T ⊇ F (there always is one
such topology, the discrete topology). The intersection
T is then the smallest
or "weakest" topology containing F. It is called the topology generated by F
(you'll also see F called a subbasis for T ).
Exercise A11.7. Show that the topology on X generated by F can be concretely
expressed as follows: a set U is open in the generated topology if and only if it
can be written as a union
U
α
of some family of sets U
α
, where each U
α
is the
intersection of some finite family (depending on α) of members of F. (The empty
set and the whole space X are included among these sets by convention: we take
the intersection of an empty family to be X and the union of an empty family to
be ∅.)
Remark A11.8. The following language is conventional in this context: the fam-
ily B of subsets of X is called a basis if, for every point x ∈ X, there is some
member of B containing x and every finite intersection of members of B con-
taining x includes some member of B containing x. Thus the topology generated
by a basis is just the collection of all unions of families of members of the basis.
(For example, the metric balls form a basis for the topology of a metric space.)
Moreover, the topology generated by a subbasis F is the topology generated by
the associated basis consisting of finite intersections of members of F.
34
Lecture A12
Compactness and Hausdorffness
Definition A12.1. Let X be a topological space, with topology T , and let Y ⊆ X.
Then the family of subsets of Y given by ¦Y ∩ U : U ∈ T ¦ defines a topology
on Y . It is called the relative topology or subspace topology induced by the given
topology on X.
(Compare Proposition A3.10.)
Any metric space notion which is defined simply in terms of open sets can
be taken over without change to the more general context of topological spaces.
Thus, for example, we can define the interior, closure, and boundary of a subset
just as in the metric space context.
Exercise A12.2. Let X be a topological space and Y a subspace of X, equipped
with the subspace topology. Let S be a subset of X, with closure S. Show that
the closure of S ∩ Y in the subspace topology of Y is a subset of S ∩ Y , and that
these two sets are equal if S itself is a subset of Y ; but that they need not be equal
in general.
Just as in metric spaces we can also define a continuous function f : X → Y
(X and Y being topological spaces) as one which has the property that, for every
open subset U of Y , the inverse image f
−1
(U) is open in X. The proof of the
following lemma is then obvious:
Lemma A12.3. The composite of continuous functions is continuous.
Exercise A12.4. Let f : X → Y be a map of spaces (from now on, "space" will
mean a topological space, unless otherwise specified). Show that f is continuous
as a map from X to Y if and only if it is continuous as a map from X onto f(X),
where f(X) ⊆ Y is equipped with its subspace topology.
Definition A12.5. A homeomorphism is a continuous bijection whose inverse is
also continuous.
Definition A12.6. A space X is compact if every open cover of X has a finite
subcover.
35
This is the same as the definition of "covering compact" that was given in
Lecture 5. We won't make use of the notion "sequentially compact" outside the
realm of metric spaces.
Example A12.7. Any finite topological space is compact. A discrete topological
space is compact if and only if it is finite.
Proposition A12.8. Let f : X → Y be a continuous map. If X is compact, then
f(X) ⊆ Y is also compact.
Proof. By Exercise A12.4 it suffices to consider the case where f(X) = Y . Let
U be an open cover of Y . The sets f
−1
(U), U ∈ U , are then open in X and
they form a cover since for each x ∈ X, the image f(x) belongs to some member
of U . Since X is compact, there is a finite subcover: there is a finite subset
U
1
, . . . , U
n
of U such that the sets f
−1
(U
i
) cover X. But for every y ∈ Y , there
is x ∈ X such that f(x) = y; so x ∈ f
−1
(U
i
) for some i and therefore y ∈ U
i
.
That is, the U
1
, . . . , U
n
cover Y and we have shown that U has a finite subcover
as required.
Proposition A12.9. Let X be a compact space and let Y be a closed subset of X.
Then Y is also compact (in its relative topology).
Proof. We must show that every open cover of Y (in the relative topology) has a
finite subcover. Let U be such a cover. Each open set U ∈ U is (by definition
of the relative topology) of the form V
U
∩ Y , where V
U
is some open subset of
X. The union of all the sets V
U
includes Y ; therefore, the open sets V
U
, together
with the open set X ¸ Y , form an open cover of X. By hypothesis this cover has
a finite subcover, comprising V
U
1
, . . . , V
Un
, together possibly with X ¸ Y . Then
U
1
, . . . , U
n
cover Y .
In metric spaces, compact subsets are necessarily closed (Proposition A5.5).
This need not be true in general topological spaces. Indeed, let X be any topo-
logical space. From Example A12.7, any finite subset of X — in particular, any
one-point subset — must be compact in its subspace topology. But we saw several
examples in the last lecture of topological spaces in which one-point subsets are
not closed.
What is missing is the Hausdorff property.
Definition A12.10. A topological space X is Hausdorff if, for any two distinct
points x, y ∈ X, there exist open sets U, V with x ∈ U, y ∈ V , and U ∩ V = ∅.
36
Any metric space is Hausdorff (take U = B(x, ) and V = B(y, ) for 0 <
< d(x, y)/2). The Hausdorff property ensures a large supply of open sets.
Proposition A12.11. Let X be a Hausdorff space and let Y ⊆ X be compact (in
the subspace topology). Then Y is a closed subset of X.
Proof. We will show that X ¸ Y is open. Let x ∈ X ¸ Y . For each y ∈ Y there
exist disjoint open sets U
y
containing x and V
y
containing y, by the Hausdorff
property. The sets V
y
∩ Y form an open cover of Y (in the subspace topology)
which therefore has a finite subcover, say V
y
1
∩ Y, . . . , V
yn
∩ Y . Let U
x
= U
y
1
∩
∩U
yn
. Then U
x
is an open set, contains x, and does not meet Y . The set X¸Y
itself is the union of all these open sets U
x
, so it is open.
In particular, a one-point subset of a Hausdorff space is closed.
Corollary A12.12. Let f : X → Y be a continuous bijection, where X is a com-
pact space and Y is a Hausdorff space. Then f
−1
is continuous (i.e., f is a
homeomorphism).
Proof. To show that f
−1
is continuous it is enough to show that for any closed
subset K of X, the image f(K) is closed in Y . But K is compact (Proposi-
tion A12.9); therefore f(K) is compact (Proposition A12.8); therefore f(K) is
closed (Proposition A12.11).
37
Lecture A13
Regular and Normal Spaces
Let A and B be disjoint subsets of a topological space X. One says that A and
B are separated by open sets if there exist disjoint open sets U and V such that
A ⊆ U, B ⊆ V . Thus the Hausdorff property says that any two distinct points
can be separated by open sets.
This is in fact just one (the most important one) of a series of separation prop-
erties that a topological space may have.
Definition A13.1. A Hausdorff topological space X is regular if disjoint subsets
A and B of X can be separated by open sets whenever A is a point and B is a
closed set. It is normal if any two disjoint closed subsets can be separated by open
sets.
Remark A13.2. A regular Hausdorff space is also called a T
3
space and a normal
Hausdorff space is a T
4
space; a plain vanilla Hausdorff space is a T
2
space. The
T stands for the German Trennungsaxiom (separation axiom). There are also T
0
,
T
1
, T
2
1
2
, T
3
1
2
, T
5
, and T
6
spaces, and maybe some others that I forgot.
Example A13.3. Every metric space is regular, and indeed normal. To see this we
need to make use of the distance function from a closed set. Let (X, d) be a metric
space and let K be a closed subset. Define a real valued function d
K
: X → R
+
by
d
K
(x) = inf¦d(x, k) : k ∈ K¦.
It is easy to see that d
K
is a continuous function and that d
K
(x) = 0 iff x ∈ K.
Now let Aand B be disjoint closed sets in X and consider the continuous function
f(x) = d
B
(x) −d
A
(x) on X. The sets U = f
−1
((0, ∞)) and V = f
−1
((−∞, 0))
are open (by continuity), disjoint, and A ⊆ U, B ⊆ V . This shows that X is
normal.
Proposition A13.4. Every compact Hausdorff space is regular, and indeed nor-
mal.
Proof. It's best to do this proof in two installments: first regularity, then normality.
Let X be a compact Hausdorff space, p a point of X, and B a closed subset not
containing p. Then B is compact (Proposition A12.9). For each x ∈ B there
38
are disjoint open sets U
x
containing p and V
x
containing x. The V
x
form an open
cover of B, so there is a finite subcover, say V
x
1
, . . . , V
xn
. Then
U
p
= U
x
1
∩ ∩ U
xn
, V
p
= V
x
1
∪ ∪ V
xn
are disjoint open sets, the first containing p, the second including V . Thus X is
regular.
To prove normality, we iterate the argument. Let A and B be disjoint closed
(hence compact) sets. By regularity, for each p ∈ A there are disjoint open U
p
containing p and V
p
including B. The U
p
form an open cover of A and have a
finite subcover say U
p
1
, . . . , U
pn
. Then
U = U
p
1
∪ ∪ U
pn
, V = V
p
1
∩ ∩ V
pn
are disjoint open sets including A and B respectively.
In Example A13.3 we used suitably chosen continuous functions to establish
regularity, and indeed normality, for metric spaces. It's also possible to proceed
in the other direction: starting with normality, construct a supply of continuous
real-valued functions.
Lemma A13.5. Let X be a normal Hausdorff space, F a closed subset of X, and
W an open subset of X including F. Then there is an open subset V of X such
that F ⊆ V ⊆ V ⊆ W.
Proof. Let E = X ¸ W. The closed sets E and F are disjoint, so by normality
there are disjoint open sets U and V with E ⊆ U and F ⊆ V . Since V ⊆ X ¸ U
which is closed, V ⊆ X¸U also. In particular, V doesn't meet E, so it is included
in W.
Theorem A13.6. (Urysohn's lemma) Let E, F be disjoint closed subsets of a nor-
mal Hausdorff space X; then there exists a continuous function f : X → [0, 1]
such that f(x) = 0 for x ∈ E, f(x) = 1 for x ∈ F.
Warning: It is not asserted that the sets E, F are exactly equal to f
−1
¦0¦,
f
−1
¦1¦ respectively; in general this stronger statement is not true. We will use the
following exercise.
Exercise A13.7. In order that a function f : X →R be continuous, it is necessary
and sufficient that for every x ∈ X and every > 0 there exists an open set W
x
containing x such that
f(W
x
) ⊆ (f(x) −, f(x) + ).
39
Proof. It's an induction based on lemma A13.5. Set U
1
= X ¸ F; then U
1
is
an open set containing E. We are going to construct by induction a collection
of open sets U
t
parameterized by dyadic rational numbers t ∈ (0, 1], having the
properties that U
r
⊆ U
s
whenever r < s and E ⊆ U
t
for all t. At the nth stage
of the induction we assume that the sets U
t
have been constructed for t = m2
−n
,
m = 1, . . . , 2
n
. Let us make the convention that U
0
refers to the closed set E
(even though we have not defined a set U
0
). By lemma A13.5, for each m there is
an open set V such that
U
m2
−n ⊆ V ⊆ V ⊆ U
(m+1)2
−n;
call the set V so defined U
(m+
1
2
)2
−n. We have now defined the sets U
t
for all
t = m
2
−(n+1)
and the induction continues.
Now define f : X → [0, 1] by f(x) = inf¦t : x ∈ U
t
¦ ∪ ¦1¦. Clearly f
maps X → [0, 1] with f(E) = ¦0¦ and f(F) = ¦1¦, so we must show that f is
continuous, and we will use Exercise A13.7. Let x ∈ X with f(x) = t; suppose
that 0 < t < 1 (the cases t = 0 and t = 1 are handled by similar arguments). Let
> 0. There exist dyadic rationals r and s such that
t − < r < t < s < t + .
Put W
x
= U
s
¸ U
r
. Then W
x
is an open set and f(W
x
) ⊆ [r, s] ⊆ (t −, t +) as
required.
Of course, [0, 1] can be replaced in the statement of Urysohn's lemma by any
closed interval [a, b].
40
Lecture A14
Spaces of Continuous Functions
Let X be a set, Y ⊆ X, and let f : Y → Z be a function. A function g : X →
Z is an extension of f if g(y) = f(y) for all y ∈ Y ; in other words, if f = g
|Y
.
When X, Y, Z are topological spaces we will naturally be interested in continuous
f and g.
Theorem A14.1. (Tietze extension theorem) Let X be a normal Hausdorff space,
and let Y be a closed subspace. Let f : Y → R be a bounded continuous func-
tion. Then f can be extended to a bounded continuous function g : X → R, with
sup [g[ = sup [f[.
Proof. Let us temporarily call a bounded continuous function h: Y → R ex-
tendible if it extends to a bounded continuous function X → R with the same
sup norm. What we have to prove is that all bounded continuous functions are
extendible.
If h: Y → [−c, c] is a continuous function, let E and F be the disjoint subsets
h
−1
([−c, −c/3]) and h
−1
([(c/3, c]) of Y . They are closed in Y (by continuity),
and Y is closed in X, so they are closed in X (Exercise A12.2). Thus by Urysohn's
lemma there is a continuous function k: X → [−c/3, c/3] with k(E) = ¦−c/3¦,
k(F) = ¦c/3¦. Notice that the supremum norm|h−k| is at most 2c/3. We have
proved the following statement: for each bounded continuous h: Y → R there
exists an extendible bounded continuous k: Y →R with |k|, |h−k| 2|h|/3.
Now we proceed inductively. Let f : Y →Rbe given and suppose inductively
that extendible functions g
1
, . . . , g
n
have been constructed with |g
i
| (2/3)
i
|f|
and
|f −g
1
−. . . −g
n
| (2/3)
n
|f|
. Apply the italicized claim to h = f −g
1
−. . . −g
n
and let g
n+1
be the k that is
constructed. The function g
n+1
then has the desired properties so the induction is
completed.
By the Weierstrass M-test the series
∞
i=1
g
i
converges uniformly on X to a
function g. But the display above shows that, when restricted to Y , f = g. Thus
f is extendible as required.
Recall that a subset of a topological space is dense if its closure is the whole
space. A classical method in analysis is to prove a result first for a dense subset of
some space, and then to extend it to the whole space 'by continuity'.
41
How can we find dense subsets of C(X), the continuous real-valued functions
on a compact Hausdorff space X?
Definition A14.2. Acollection Lof continuous real-valued functions on a set X is
a lattice if, whenever it contains functions f and g, it also contains their pointwise
maximum and minimum — usually written f ∨ g and f ∧ g in this context.
Proposition A14.3. (Stone) Let L be a lattice of continuous functions on a com-
pact space X. If, for all x, x
∈ X and any a, a
∈ R, there is a function f ∈ L
having f(x) = a and f(x
) = a
, then L is dense in C(X).
The property appearing in the statement is called the two point interpolation
property.
Proof. Let h ∈ C(X) be given and let > 0. We are going to approximate h
within by elements of L. By hypothesis, for each x, x
∈ V
x
so they cover X. Take a finite subcover
and let g
x
be the (pointwise) maximum of the corresponding functions f
xx
∈ L.
Because L is a lattice, g
x
∈ L and by construction,
h(y) − < g
x
(y) ∀y, h(x) = g
x
(x).
So we have approximated h from one side by members of L.
Now we play the same trick again from the other direction: let
W
x
= ¦y : g
x
(y) < h(y) + ¦.
Again these form an open cover of X; take a finite subcover and let g be the
(pointwise) minimum of the corresponding g
x
. Then g ∈ L and by construction
h − < g < h + , as required.
Exercise A14.4. Use Stone's theoremto give another proof of the Tietze extension
theorem in the case of compact Hausdorff spaces. (Show that the collection of
extendible functions is a lattice. . . )
There is a more classical formulation which makes use of algebraic rather than
order-theoretic operations.
42
Lemma A14.5. There is a sequence of polynomials p
n
(x) on [−1, 1] converging
uniformly to [x[.
Proof. Writing [x[ =
_
1 + (x
2
−1), this follows from the fact that the binomial
series for (1 + t)
1/2
converges uniformly for t ∈ [−1, 1].
One says that a subset of C(X) is a subalgebra if it is closed under pointwise
addition, subtraction, multiplication of functions, and multiplication by scalars.
Lemma A14.6. A closed subalgebra of C
R
(X) that contains the constant func-
tions is a lattice.
Proof. Let A be the given subalgebra and f, g ∈ A; let h = f − g. There is
no loss of generality in assuming (by rescaling) that [h[ 1 everywhere. Any
polynomial in h belongs to A and hence so does [h[ = limp
n
(h), by closure and
lemma A14.5. But now
f ∧ g =
f + g −[h[
2
, f ∨ g =
f + g +[h[
2
belong to A as well.
A subalgebra of C(X) is said to separate points if for every x, x
∈ X there is
a function in the subalgebra taking different values at x and at x
.
Theorem A14.7. (Stone-Weierstrass) A subalgebra of C
R
(X) which contains the
constants and separates points is dense.
Proof. The closure of the subalgebra is a closed subalgebra, contains the con-
stants, and separates points. Using the algebraic operations, it has the two point
interpolation property. The previous lemma shows that it is a lattice. Hence by
Stone's theorem A14.3 it is all of C(X).
The original result of Weierstrass was
Corollary A14.8. (Weierstrass) Every continuous function on a closed bounded
interval is the uniform limit of polynomials.
Proof. The polynomials form an algebra which contains the constants and sepa-
rates points.
43
Exercise A14.9. In the complex case the Stone-Weierstrass theorem takes the
following form: a ∗-subalgebra of C(X) which contains the constants and sepa-
rates points is dense, where the ∗-condition means that the algebra is closed under
(pointwise) complex conjugation. Prove this. Try to give an example to show that
the complex theorem is not valid without the extra condition (we shall discuss this
in detail later).
44
Lecture A15
Connectedness
What does it mean that a topological space is "all in one piece"?
Proposition A15.1. Let X be a topological space. The following properties of X
are equivalent:
(a) The only subsets of X that are both open and closed are the empty set and X
itself.
(b) The only continuous functions from X to a discrete topological space are
constant.
(c) The only equivalence relation on X with open equivalence classes is the triv-
ial one (every point is equivalent to every other point).
Proof. It is clear that (b) and (c) are equivalent, since if f : X → D is a continuous
function to a discrete space, then the relation "x ∼ y iff f(x) = f(y)" is an
equivalence relation with open equivalence classes, and conversely.
Assuming (c), if Ais an open-and-closed subset of X, then the relation "x ∼ y
iff either none, or both, of x, y belong to A" is an equivalence relation whose
equivalence classes A and X ¸ A are open. Conversely, assuming (a), let ∼ be
an equivalence relation with open equivalence classes, and let A be a nonempty
equivalence class. A is open; but, since the complement of A is a union of equiva-
lence classes and therefore open, it is also true that A is closed. Thus A = X and
the equivalence relation is trivial.
Definition A15.2. A topological space is connected if it satisfies the equivalent
properties of Proposition A15.1. A subset of a topological space is called a con-
nected subset if it is a connected space in the relative topology. A space (or subset)
that is not connected is called disconnected.
Proposition A15.3. Let f : X → Y be continuous and surjective. If X is con-
nected, then Y is connected.
Proof. Let g : Y → D be a continuous map to a discrete space D. Then f ◦
g : X → D is continuous, hence constant because X is connected. But now since
f is surjective, g must be constant itself.
45
Proposition A15.4. Let X be a space and let ¦X
α
¦
α∈A
be a family of subsets
each of which is connected. If
α
X
α
,= ∅ then
α
X
α
is connected.
Proof. Let Y =
α
X
α
and let p ∈
α
X
α
. Let f : Y → D be a continuous map
to a discrete space. Since each X
α
is connected, the restriction of f to X
α
takes
a constant value (say d
α
∈ D); since p ∈ X
α
for every α, d
α
= f(p) for each α.
Thus f takes the constant value f(p) on Y .
Proposition A15.5. Every interval in R is connected, and every connected subset
of R is an interval.
Proof. We begin by proving that every compact interval [a, b] is connected. Sup-
pose that ∼ is an equivalence relation with open equivalence classes; then the
equivalence classes form an open cover U for the compact interval [a, b]. By the
Lebesgue covering theorem (Theorem A5.8)there is δ > 0 such that any ball of
radius less than δ is included in some member of the cover. In particular, any two
points that are less than δ apart belong to the same equivalence class. But any
pair of points in the interval [a, b] can be joined by a chain of points whose suc-
cessive members are less than δ apart; so, any two points in [a, b] are equivalent.
This proves the connectedness of closed intervals; other kinds of intervals can be
written as the increasing union of suitable families of closed intervals, so we may
apply Proposition A15.4.
Conversely, let S be a subset of R that is not an interval. Then there is some
a / ∈ S such that S has members which are less than a and also members that are
greater than a. So, S ∩ (−∞, a) and S ∩ (a, ∞) are disjoint open subsets of S
whose union is S; so S is not connected.
Corollary A15.6. (Intermediate value theorem) Let f : [a, b] →R be continuous.
Then the range of f includes all values between f(a) and f(b).
Proof. The range of f, being the image of a connected space under a continuous
function, is connected. Hence it is an interval. It must contain f(a) and f(b) so,
being an interval, it must also contain everything in between.
If X is a space, the relation "x ∼ y iff there is a connected subset of X
containing both x and y" is an equivalence relation (by Proposition A15.4). The
equivalence classes are called the connected components of X. The connected
component of x is the largest connected subset of X that contains x.
Exercise A15.7. Show that the closure of a connected subset of X is connected,
and deduce that the connected components of X are closed. Give an example
where they are not open.
46
Exercise A15.8. The quasicomponent of x ∈ X is the intersection of all the
open-and-closed subsets that contain x. Show that the connected component of x
is included in the quasicomponent. Can you give an example where they are not
equal?
Definition A15.9. A path in a topological space X is a continuous map [0, 1] →
X. The start of the path is the point γ(0) and the end of the path is the point γ(1).
We say that γ is a path from its start to its end.
Lemma A15.10. Let X be a topological space, γ
1
a path in X from a to b, γ
2
a
path in X from b to c. Then the map γ : [0, 1] → X defined by
γ(t) =
_
γ
1
(2t) (t
1
2
)
γ
2
(2t −1) (t
1
2
)
is a path in X from a to c.
Proof. All that needs to be checked is the continuity of γ at t =
1
2
. Let U be
an open set in X containing b. By continuity of γ
1
, there is δ
1
> 0 such that if
[s−1[ < δ
1
, s ∈ [0, 1] then γ
1
(s) ∈ U; by continuity of γ
2
there is δ
2
> 0 such that
if [s[ < δ
2
, s ∈ [0, 1] then γ
2
(s) ∈ U. Now let δ =
1
2
min¦δ
1
, δ
2
¦; if t −
1
2
[ < δ,
t ∈ [0, 1], then γ(t) ∈ U. This gives the required continuity.
Corollary A15.11. The relation "There is a path in X from a to b" is an equiva-
lence relation.
Proof. The lemma gives transitivity. Reflexivity and symmetry are easy.
The equivalence classes under this equivalence relation are called path com-
ponents of X.
Proposition A15.12. Every path component is connected. Consequently, the path
component of x ∈ X is included in the component of x.
Proof. By Proposition A15.3, the image of a path from a to b is a connected set
containing a and b. Consequently, the path component of X is the union of a
family of connected sets all of which contain x; so it is connected by Proposi-
tion A15.4.
47
A space which has only one path component (i.e., any two points can be joined
by a path) is called path connected. By the preceding proposition, a path con-
nected space is connected. The converse is false, however, as is shown by the
following classical example.
Exercise A15.13. Let Z be the set of points (x, y) ∈ R
2
such that either x = 0
and y ∈ [−1, 1] or 0 < x 1 and y = sin(1/x). Show that Z is a compact
connected metric space with 2 path components.
The same example shows that, in contrast to components, path components
need not be closed in general.
48
Lecture A16
Local Properties of Spaces
Let X be a topological space.
Definition A16.1. An open neighborhood of x ∈ X is an open subset of X that
contains x. A neighborhood of x ∈ X is a subset that includes an open neighbor-
hood of x, i.e. contains x in its interior.
Suppose that P is some kind of topological property (e.g. connectedness).
Definition A16.2. We say that a space X has the property P locally if, for every
x ∈ X, every open neighborhood of x includes another open neighborhood of x
that has property P.
For instance, a space X is locally connected if every open neighborhood of
each x ∈ X includes an open connected neighborhood. Informally we may say
that x has "arbitrarily small connected neighborhoods".
Proposition A16.3. In a locally connected space, the connected components are
open.
(It follows that the connected components are equal to the quasicomponents,
see Exercise A15.8.)
Proof. Let C be a connected component and let x ∈ C. There is a connected
neighborhood N of x. Then N ⊆ C and so N
◦
⊆ C
◦
. Since x ∈ N
◦
, x is an
interior point of C. Since x was arbitrary, C is equal to its own interior and so is
open.
Proposition A16.4. In a locally path-connected space, the connected components
are equal to the path components (and are open).
Proof. The same argument as in the previous proposition shows that the path com-
ponents in a locally path-connected space are open. Since the complement of
each path component is a union of other path components, they are also closed.
Now the connected component of x must be included in any open-and-closed set
that contains x, so in particular it must be included in the path component of x.
The path component is always included in the connected component, so they are
equal.
49
The most important example of a locally connected and locally path-connected
space is furnished by an open subset of R
n
(or any normed vector space).
Exercise A16.5. There is a weaker local path connectedness notion, sometimes
called semilocal path connectedness: X is semilocally path connected if for each
x ∈ X and each open neighborhood V of x, there is an open neighborhood U ⊆ V
of x such that for each y ∈ U there is a path γ : [0, 1] → V starting at x and
ending at y. Show that a space which is locally connected and semilocally path
connected must be locally path connected. Can you give an example of a space
which is semilocally path connected but not locally path connected?
We now turn to consider local compactness. Here it is necessary to modify
the definition somewhat: we say that a space is locally compact if every open
neighborhood of each point contains a compact neighborhood (which need not be
open - there may be very few compact, open sets). But in fact there is a simpler,
equivalent definition.
Proposition A16.6. A Hausdorff space is locally compact iff each point has a
compact neighborhood.
Proof. Local compactness says that each point has "arbitrarily small" compact
neighborhoods, a stronger condition than that of the proposition. Suppose that x ∈
X has a compact neighborhood K, and let N be any neighborhood of x. We need
to find a compact neighborhood that is included in N. Let U = K
◦
∩ N
◦
. Then
U is a neighborhood of x, and U and ∂U are closed subsets of a compact space
(namely K), hence are compact. By regularity (or, more exactly, by the proof of
regularity) there exists an open subset V of U, containing x, whose closure does
not meet the compact set ∂U. Then V ⊆ U is the desirec compact neighborhood.
Let X be a compact Hausdorff space and x
0
∈ X. Then X ¸ ¦x
0
¦ is a locally
compact Hausdorff space (since each point of X distinct from x
0
has a closed,
hence compact, neighborhood that does not meet x
0
). In fact, every locally com-
pact Hausdorff space arises in this way.
Proposition A16.7. Let X be a locally compact Hausdorff space. Let X
+
be the
union of X and a disjoint point, denoted ∞. The following sets form a topology
on X
+
:
(a) The open subsets of X in its original topology.
50
(b) The subsets of the form ¦∞¦ ∪ (X ¸ K) where K ⊆ X is compact.
The topology so defined on X
+
is compact and Hausdorff and the identity map
X → X
+
is a homeomorphism onto its image.
The space X
+
is called the one-point compactification of X. We will need the
following easy exercise.
Exercise A16.8. Let Y be a Hausdorff space. If K, L are compact subspaces of
Y then so is K ∪ L.
Proof. Let T be the family of subsets of X
+
described in the proposition. We
show it is a topology.
(i) Let U
α
be a family of members of T . If they are all of type (a), then their
union is an open subset of X and is of type (a) also. If one of them is of type
(b), say equal to ¦∞¦∪(X¸K), then their union is of the form¦∞¦∪(X¸L)
where L is a closed subset of K, hence compact; so the union is of type (b).
(ii) Let U
1
, . . . , U
n
be a finite family of members of T . If one of them is of
type (a), then their intersection if of type (a). If they are all of type (b), their
intersection is the complement of a finite union of compact sets, which is
compact by Exercise A16.8 above. Thus the intersection id of type (b).
(iii) Clearly ∅ and X
+
belong to T .
Thus T is a topology, and the relative topology that it induces on X is the same
as the original topology of X. It remains to show that this topology is compact
and Hausdorff. To show that T is compact, let W be a family of members of
T covering X
+
. One of them (say W
0
) must contain ∞, and hence must be of
the form ¦∞¦ ∪ (X ¸ K), K ⊆ X compact. Now the remaining W
α
intersect
X in open sets which cover K, so some finite subcollection covers K also; this
subcollection, together with W
0
, forms a finite subcover of W .
Finally to show that T is Hausdorff: Any two points not at infinity can be
separated by open sets of type (a) (by the Hausdorff property of X). Thus it
suffices to show that ∞ can be separated from a point x ∈ X. But by local
compactness, x ∈ X has a compact neighborhood, K say; then K
◦
and ¦∞¦ ∪
(X ¸ K) are disjoint members of T containing x and ∞respectively.
51
Lecture A17
Some set theory
Further topics in our discussion will require some more advanced background
from set theory. We will take a rather "naive" approach to set theory. To learn
about the paradoxes this gives rise to, and how to avoid them, you need a course
in Logic.
Definition A17.1. A partial order on a set S is a relation which is reflexive,
antisymmetric, and transitive: that is, for all x, y, z ∈ S, x x, x y and y x
imply x = y, and x y and y z together imply x z.
A set with a partial order is called a (partially) ordered set. If S is such a set
then
(a) An upper bound for a subset T ⊆ S is an element x such that y x for all
y ∈ T.
(b) S is directed if every two-element subset has an upper bound (equivalently,
every finite subset has an upper bound).
(c) S is totally ordered if for all x, y, either x y or y x.
(d) A subset of S that is totally ordered (in the induced ordering) is called a chain.
(e) S is inductively ordered if every chain in S has an upper bound; it is strictly
inductively ordered if every chain has a least upper bound.
(f) A maximal element for S is an element m ∈ S such that x m implies
x = m (i.e., there isn't anything strictly greater than m.) Notice that this is a
weaker notion than that of upper bound.
(g) S is well-ordered if it is totally ordered and every non-empty subset has a least
member (necessarily unique).
Exercise A17.2. Give simple examples to illustrate all these notions.
Let S be an ordered set. A map f : S → S will be called an inflator (I just
made this word up) if f(x) x for all x ∈ S.
52
Lemma A17.3. (Bourbaki fundamental lemma) Any inflator on a nonempty strictly
inductively ordered set has a fixed point, i.e. a point x such that f(x) = x.
We'll prove this in a moment. For now, let's draw some consequences.
Corollary A17.4. (Weak form of Zorn's lemma) Every nonempty strictly induc-
tively ordered set contains a maximal element.
Proof. Let S be strictly inductively ordered and suppose it contains no maximal
element. Then for each x ∈ S there is a y
x
∈ S with y
x
> x. Consider the map
1
f
sending x to y
x
. Then f is an inflator with no fixed point, contradicting Bourbaki's
lemma.
Corollary A17.5. (Hausdorff maximality principle) Let S be a nonempty partially
ordered set. Each chain in S is included in a maximal chain (i.e., a chain which
is not itself included in any larger chain).
Proof. Let C be the collection of all chains in S, partially ordered by inclusion.
Then C is partially ordered, and indeed is strictly inductively ordered (with least
upper bound for a chain in C given by the union of its members). Now ap-
ply A17.4.
Corollary A17.6. (Zorn's lemma) Every nonempty inductively ordered set (not
necessarily strict) contains a maximal element.
Proof. Let S be such a set and let C be a maximal chain in S (provided by A17.5).
Let m be an upper bound for C. Then m is maximal, since if x > m then C ∪¦x¦
is a chain larger than the supposedly maximal chain C.
Now let us see to proving the Bourbaki lemma A17.3. Let S be nonempty and
strictly inductively ordered and f : S → S an inflator. Pick a point a ∈ S (fixed
throughout the discussion). We'll say that a subset F of S is closed if a ∈ F, if
f(x) ∈ F whenever x ∈ F, and if, whenever C ⊆ F is a chain, the least upper
bound of C belongs to F.
Obviously, the intersection of any family of closed sets is closed. Moreover,
S itself is closed. Thus, there is a minimal closed set M, namely, the intersection
of all the closed sets that there are. The strategy of our proof is to show that M is
totally ordered, i.e., a chain. This will complete the proof because if x is the least
1
For those in the know, this is where we use the axiom of choice.
53
upper bound of M, then x ∈ M and f(x) ∈ M also, hence f(x) x; since f is
an inflator, it follows that f(x) = x.
Note that the collection of all x a is closed. Thus, a is the least element of
the minimal closed set M.
Let us call u ∈ M unavoidable if, whenever x ∈ M and x < u, we have
f(x) u. If u is an unavoidable point, then the set
¦x ∈ M : either x u or f(u) x¦
is readily seen to be closed. Since it is a subset of M, it must be the whole of M,
by minimality.
We can now prove that every point is unavoidable. Let U be the set of unavoid-
able points in M. It contains a. We will show that it is closed; then, by minimality,
it must equal M itself. Suppose then that u is unavoidable and consider f(u). If
x < f(u) then, by the paragraph above, either x < u, or x = u, or f(u) x, and
the last possibility cannot occur. The other two possibilities yield f(x) f(u) by
the unavoidability of u. So, f(U) ⊆ U. A similar argument shows that the least
upper bound of any chain in U belongs to U, so U is closed.
We now know that each point of M is unavoidable. It is easy to deduce that
M is totally ordered. For, let x, y ∈ M. Since x is unavoidable, either y x or
f(x) y, but the latter certainly implies that x y becasue f is an inflator. As
already remarked, this completes the proof of Lemma A17.3.
Proposition A17.7. (Well ordering principle) Every nonempty set can be well-
ordered.
Proof. Let X be a set. Let Y be the set of pairs (S, _) where S is a subset of
X and _ is a well-ordering of S. Y is nonempty and is inductively ordered by
inclusion. By Zorn's lemma, there is a maximal element in Y , say (S, _). But
now we must have S = X; if not, we can choose some x ∈ X ¸ S and well-order
S ∪ ¦x¦ by insisting that x is greater than every element of S. This contradicts
maximality.
54
Lecture A18
Countability and Metrization
We now briefly discuss the theory of cardinal numbers. Let X and Y be sets.
Proposition A18.1. There exists an injection from X to Y if and only if there
exists a surjection from Y to X.
Proof. Without loss of generality assume that X ,= ∅. Let i : X → Y be an
injection. Let I = i(X) ⊆ Y . Fix an element x
0
∈ X. Define s: Y → X as
follows: if y ∈ I, then s(y) is the unique x ∈ X such that i(x) = y; otherwise,
s(y) = x
0
. This is a surjection.
Conversely, suppose that s: Y → X is a surjection. For each x ∈ X choose
some y
x
∈ Y such that s(y
x
) = x. Let i be the map that sends x to y
x
. Then i is
an injection.
Exercise A18.2. The last part of this argument uses the Axiom of Choice directly.
Reformulate it to use Zorn's Lemma.
Theorem A18.3. (Schr¨ oder-Bernstein) If there exists an injection from X to Y
and an injection from Y to X, then there exists a bijection from Y to X.
Proof. (K¨ onig) Assume (wlog) that X and Y are disjoint sets and let Z be their
disjoint union. let f : X → Y and g : Y → X be injections and let h be the union
of these maps, considered as a map Z → Z. A subset S of Z is stable if z ∈ Z
iff h(z) ∈ Z. Let S
z
be the minimal stable set containing z. Explicitly, S
z
is the
set ¦h
n
(z) : n ∈ Z¦, where we note that h
−1
(p) denotes an element q such that
h(q) = p; such an element may not exist, but it is unique if it does exist. By the
injectivity of h, stable sets either coincide or are disjoint; so they partition Z.
It suffices now to establish a bijection S
z
∩X → S
z
∩Y for each S
z
separately.
The stable sets S
z
are of four types:
(a) X-stopping: h
n
(z) is defined only for n = −k, −k + 1, . . . and h
−k
(z) ∈ X;
(b) Y -stopping: h
n
(z) is defined only for n = −k, −k + 1, . . . and h
−k
(z) ∈ Y .
(c) Two way infinite: the h
n
(z) are all distinct for different values of n ∈ Z.
(d) Cyclic: h
m
(z) = z for some m (necessarily even).
55
In the first case, f gives a bijection S
z
∩X → S
z
∩Y . In the second case, g gives
a bijection S
z
∩ Y → S
z
∩ X. In the third and fourth cases, either f or g will
do.
To each set X associate a "number", its cardinality card X. We define card X =
card Y iff there is a bijection between X and Y (one says that X and Y are
equinumerous). We define card X card Y iff there is an injection X → Y . The
Schr¨ oder-Bernstein theorem shows that this defines a partial order on the cardi-
nalities.
Definition A18.4. The cardinality of the natural numbers is denoted ℵ
0
(read:
aleph-zero). A cardinal that is ℵ
0
is called countable. (We also call a set
countable if its cardinality is countable.)
Example A18.5. The integers Z and the rationals Q are countable sets.
Proposition A18.6. For every cardinal there is a strictly larger cardinal.
Proof. Let X be a set and let P(X) (the power set) be the set of all subsets of X.
We will show that there is no surjection X → P(X). Indeed, suppose that f is
such a surjection. Consider
S = ¦x ∈ X : x / ∈ f(x)¦.
This is a subset of X. If S = f(y) for some y ∈ X we obtain (from the definition
of S) the contradiction
y ∈ f(y) ⇔ y / ∈ f(y).
Thus f is not surjective after all. On the other hand, there is an obvious injection
X → P(X) (as one-element subsets). It follows that card(P(X)) > card(X).
We denote the cardinality of P(X) by 2
card X
. More generally, the cardinality
of the set of all maps X → Y is denoted (card Y )
card X
. The cardinal 2
ℵ
0
is
denoted c (the continuum).
Once can do "arithmetic" with cardinals: if X, Y have cardinalities x, y respec-
tively then xy is the cardinality of XY , x +y is the cardinality of X.Y , and so
on. Arithmetic laws like x
y·z
= (x
y
)
z
then follow from corresponding set-theoretic
properties (a map fromY Z to X is the same as a map fromZ to the set of maps
from Y to X).
56
Example A18.7. Let us prove that c
ℵ
0
= c. It's a one-liner:
c
ℵ
0
=
_
2
ℵ
0
_
ℵ
0
= 2
ℵ
0
·ℵ
0
= 2
ℵ
0
= c.
Proposition A18.8. The set of real numbers has cardinality c.
Proof. The set of all sequences of rational numbers has cardinality ℵ
0
ℵ
0
= c. The
subset of convergent sequences must then have cardinality c, and this subset
maps onto R, so card R c. On the other hand, the set of all sequences a = ¦a
n
¦
of zeroes and ones also has cardinality c, and the assignment a →
∞
n=1
3
−n
a
n
is
an injection from this set of sequences to R, so card R c. Schroder-Bernstein
now completes the proof.
Let X be a topological space.
Definition A18.9. X is separable it it has a countable dense subset. It is sec-
ond countable if there is a countable basis for the topology of X: i.e., there is a
countable family B of open sets such that every open set is a union of sets from
B.
Lemma A18.10. A second-countable space is separable. A separable metric
space is second-countable.
However, there are in general examples of separable spaces that are not second
countable. For instance, let X be an uncountable set equipped with the cofinite
topology. Then X is separable (any countable subset of X meets every nonempty
open set) but it is not second countable (a countable family of cofinite sets can
omit only countably many points in total, so there must be some cofinite set that
does not contain any member of the family).
Proof. Let X be a second countable space. Let ¦U
n
¦ be a countable base for the
topology, and select a point x
n
from each U
n
. The set S = ¦x
n
¦ is countable, and
for any x ∈ X and any open set U containing x, there is some U
n
⊆ U containing
x; so U ∩ S contains x
n
and is in particular nonempty. Thus S is dense.
Let X be separable and metrizable, with S a countable dense set. Then the set
B of balls B(s, t) where s ∈ S and t ∈ Q is countable. Moreover, it forms a base
for the topology: if x ∈ X belongs to an open set U, there is a ball B(x, r) ⊆ U;
there is an s ∈ S such that d(x, s) < r/2; there is a rational t with d(x, s) < t <
r/2; the ball B(s, t) belongs to B, contains x, and is included in U.
57
A neighborhood base at a point x ∈ X is a family of open sets containing x
such that each open set containing x must include a member of the family.
Definition A18.11. A topological space X is first countable if each point has a
countable neighborhood base.
Every metric space is first countable. Every second countable space is first
countable but the converse is false: e.g. consider the discrete topology on an
uncountable set.
Exercise A18.12. Show that a compact metric space is separable and second
countable.
Theorem A18.13. (Urysohn metrization theorem) A second countable normal
Hausdorff space is metrizable, i.e. the topology can be given by a metric.
Proof. Let X be such a space and let B be a countable basis for the topology.
Whenever U, V ∈ Bwith U ⊆ V there exists, by Urysohn's lemma, a continuous
function f : X → [0, 1] that is equal to 0 on U and is equal to 1 on X ¸ V . The
collection F of all such functions is countable: denote it ¦f
1
, f
2
, . . .¦.
Define d: X X →R
+
by
d(x, y) =
∞
n=1
2
−n
[f
n
(x) −f
n
(y)[.
We will show that d is a metric defining the topology of X.
It is clear that d(x, x) = 0, that d(x, y) = d(y, x), and that d satisfies the
triangle inequality. If d(x, y) = 0, then f
n
(x) = f
n
(y) for all n. But, if x ,= y,
then there is a V ∈ B containing x but not y (by the Hausdorff property); by
normality there is an open set U containing x with U ⊆ V , and we may take
it that U ∈ B also; so then the function f
n
corresponding to this pair (U, V ) has
f
n
(x) = 0 but f
n
(y) = 1, and this is a contradiction. We conclude that d(x, y) = 0
iff x = y, so d is a metric.
Being the sum of a uniformly convergent series of continuous functions, d is
continuous (relative to the original topology on X). It follows that every d-open
ball is open in the original topology, so every d-open set is open in the original
topology.
It remains to show that every set open in the original topology is d-open. Let
W be open in the original topology, and x ∈ W. There is a set V ∈ B such that
58
x ∈ V and V ⊆ W, and there is a set U ∈ B such that x ∈ U and U ⊆ V . Let f
n
be the function constructed above corresponding to the sets U and V . Let = 2
−n
.
Then, if y ∈ B
d
(x; ), f
n
(y) < 1 and so y ∈ V ; in particular y ∈ W. Thus W
contains a d-ball around each of its points, so it is d-open, as required.
59
Lecture A19
Convergence and Nets
Let ¦x
n
¦ be a sequence in a topological space X. (Okay, I know that I told
you that I was not going to use sequences outside metric spaces, but I lied.) We
say that ¦x
n
¦ converges to a point x ∈ X if, for every open neighborhood U of x,
there is some N such that x
n
∈ U for n N.
If X is first countable then sequences in x contain all the information that
there is about the topology of X. One way to see this is the following
Lemma A19.1. In a first countable space X, a subset A is closed if and only if
the limit of every convergent sequence in A is itself a member of A.
Proof. Suppose that A ⊆ X. A point x belongs to the closure of A iff every open
neighborhood of x contains points of A. But, since X is first countable, this will
be true iff every member of a countable neighborhood base ¦U
n
¦ at x contains
points of A. Picking such a point a
n
∈ A ∩ U
n
for each n = 1, 2, 3 . . . gives a
sequence (a
n
) in A converging to x. The converse statement (if x is the limit of a
convergent sequence in A, then x belongs to the closure of A) is easy.
For another example, try the following exercise.
Exercise A19.2. Let X and Y be first countable. Show that f : X → Y is con-
tinuous if and only if the sequence f(x
n
) → f(x) in Y whenever the sequence
x
n
→ x in X. (Hint: The proof is exactly the same as the one we already dis-
cussed for metric spaces.) Do you need first countability for both spaces, or only
for one of them (and, if so, which)?
But in general sequences are "not enough" to probe the structure of a topolog-
ical space.
Example A19.3. Any totally ordered set X can be equipped with the order topol-
ogy for which a basis is the "half open intervals"
(a, b] := ¦x ∈ X : a < x b¦.
Let X be uncountable, and well-order it (using the axiom of choice). Let Ω ∈ X
be the smallest element that has uncountably many predecessors (a.k.a. the "first
uncountable ordinal"); this exists because of well-ordering. Let C = ¦x ∈ X :
x < Ω¦ be the set of predecessors of Ω. Then Ω belongs to the closure of C, but
nevertheless there is no sequence in C converging to Ω.
60
This motivates the following definition.
Definition A19.4. A net in a set X is a map from a directed set to X. (Recall that
a directed set is a partially ordered set in which any two elements have an upper
bound.)
Let S be a subset of X.
Definition A19.5. A net ¦x
i
¦ is eventually in S if there is i
0
such that x
i
∈ S for
all i i
0
; it is frequently in S if it is not eventually in X ¸ S.
Now suppose that X is a topological space.
Definition A19.6. A net ¦x
i
¦ converges to x ∈ X if it is eventually in every
neighborhood of x.
These definitions match the definitions of "sequence" and "convergent se-
quence" when we take the net simply to be the natural numbers N. On the other
hand, in Example A19.3 above, we could take the collection C itself in its natural
ordering to be a net, and it is then almost a tautology that C converges to Ω even
though we have seen there is no sequence that will do this.
Proposition A19.7. In any topological space X, a subset A is closed if and only
if the limit of every convergent net in A is itself a member of A.
Proof. Emulating the proof of Lemma A19.1, what we have to show is that if x
is a point of the closure of A, then there is some net in A that converges to x. A
net has to parameterized by a directed set. The trick is to pick the collection of all
neighborhoods of x as the directed set! In fact, suppose that U and V are open
neighborhoods of x. We define U V if U ⊇ V (notice that the inclusion is
reversed!). This makes the collection N = ¦U open : x ∈ U¦ a partially ordered
set, and in fact a directed set: an upper bound for U and V is the intersection U∩V .
Since x belongs to the closure of A, for every U ∈ N there exists a
U
∈ A ∩ U.
The mapping
2
from N to A given by U → a
U
is a net. Moreover, it converges
to x by definition: if V is any neighborhood of A then there is an element of N
(namely V itself) such that if U V then a
U
∈ V .
Similarly we have
Proposition A19.8. A map f : X → Y between topological spaces is continuous
if and only if the net f(x
i
) → f(x) in Y whenever the net x
i
→ x in X.
2
Using the axiom of choice here!
61
Proof. The proof that a continuous map has this property is essentially the same
as the corresponding proof for sequences.
Suppose that f has the property in question but is not continuous. Then there
exist a point x ∈ X, y = f(x), and an open neighborhood W of y, such that
f
−1
(W) is not a neighborhood of x. It follows that for every open neighborhood
U of x there exists x
U
∈ U ¸ W. Organize the x
U
into a net parameterized by the
directed set N of neighborhoods of x, as in the previous proposition. Then, by
construction, x
U
converges to x; but f(x
U
) does not converge to y, since it never
lies in the neighborhood W of y.
Exercise A19.9. Suppose that all we know about the topology of X is that it is
generated by some family F of subsets. Show that a net x
i
converges to x iff,
for every U ∈ F which contains x, the net x
i
is eventually in U. (This exercise
should help you understand why we insist that a net should be parameterized by a
directed set.)
One might ask whether we can similarly generalize the definition of com-
pactness, which in the metric space case is expressed as "every sequence has a
convergent subsequence." To do this, we must arrive at the correct definition of
subnet.
Definition A19.10. Let D and D
be directed sets. A function h: D
→ D is
called final if, for any i
0
∈ D, there is j
0
∈ D
such that j j
0
implies h(j) i
0
.
Given a net D → X, the result of composing it with a final function h is called a
subnet or refinement of the original one.
Every subnet of a convergent net is convergent (with the same limit); the defi-
nition of "final function" is cooked up to make this true.
Definition A19.11. A net ¦x
i
¦ in X is called universal if, given any subset S of
X whatsoever, either ¦x
i
¦ is eventually in S or else eventually in X ¸ S.
Lemma A19.12. Every net has a universal subnet.
Proof. This depends unavoidably on the axiom of choice.
Let A : D → X be a net in X, parameterized by a directed set D. A collection
F of subsets of X will be called an A-filter if A is frequently in every member
of F, and if F is closed under finite intersections and the formation of supersets.
Such objects exist: for example, ¦X¦ is an A-filter. The collection of A-filters is
partially ordered by inclusion, and every chain in this partially ordered set has an
62
upper bound (the union). Thus Zorn's Lemma provides a maximal A-filter; call
it F
0
.
Suppose that S ⊆ X has the property that A is frequently in A ∩ S for every
A ∈ F
0
. Then the union of F
0
with the set of all sets A ∩ S, A ∈ F
0
, is again an
A-filter. By maximality we deduce that S itself belongs to F
0
.
We will use this property to construct the desired universal refinement. Let D
be the collection of pairs (A, i) with A ∈ F
0
, i ∈ D, and A(i) ∈ A; it is a directed
set under the partial order
(B, j) (A, i) ⇔ B ⊆ A, j i.
The map (A, i) → i is final so defines a refinement A
of the net A. We claim
that this refinement is universal.
Let S ⊆ X have the property that A
is frequently in S. Let A ∈ F
0
and let
i be arbitrary. By definition, there exist B ∈ F
0
, B ⊆ A, and j i, such that
A(j) = A
(B, j) ∈ B ∩ S ⊆ A∩ S. We conclude that A is frequently in A∩ S
for every A ∈ F
0
and hence, as observed above, that S itself belongs to F
0
.
Now let S be arbitrary. It is not possible that A
be frequently both in S and
in X ¸ S, for then (by the above) both S and X ¸ S would belong to F
0
, and then
their intersection, the empty set, would do so as well, a contradiction. Thus A
fails to be frequently in one of these sets, which is to say that it is eventually in
the other one.
Proposition A19.13. The following properties of a topological space X are equiv-
alent:
(i) X is compact;
(ii) (the finite intersection property) if T is a family of closed subsets of X, and
the intersection of any finite number of members of T is nonempty, then in
fact the intersection of all the members of T is nonempty;
(iii) every universal net in X converges;
(iv) every net in X has a convergent subnet.
Proof. Suppose (i). Let T be a family as in (ii) and let | = ¦X ¸ F : F ∈ T¦.
If
T is empty then | is an open cover of X, hence it has a finite subcover,
so the intersection of some finite subcollection of T is empty. This proves (ii).
Conversely, if we assume (ii) and let | be an open cover that is supposed to have
63
no finite subcover, we reach a similar contradiction by considering the family of
closed sets T = ¦X ¸ U : U ∈ |¦. Thus (i) and (ii) are equivalent.
Suppose (ii), and let ¦x
α
¦ be a universal net in X. If it does not converge, then
each x ∈ X has a neighborhood U
x
for which x
α
is not eventually in U
x
, hence
(by universality) x
α
is eventually in F
x
= X ¸ U
x
. It follows that xα is eventually
in any finite intersection of the F
x
. In particular no finite intersection of the F
x
is
empty. By (ii) the intersection of all the F
x
is nonempty, and this is a contradiction
since x / ∈ F
x
for each x.
(iii) implies (iv) by Lemma A19.12.
Suppose (iv). Let T be a family of closed sets with the finite intersection
property. Let T be the directed set comprised of finite intersections of members
of T (directed by reverse inclusion) and for each D ∈ T let x
D
be a point of D.
Then x
D
is a net in X; choose a subnet convergent to x ∈ X. By definition, any
neighborhood of x meets every F ∈ T; since F is closed, we deduce that x ∈ F.
Thus the intersection of all the F contains x. This proves (ii).
Now let (X
α
) be a family of topological spaces, parameterized by α ∈ A
where A is some index set (possibly infinite or even uncountable). Let X be the
Cartesian product
α
X
α
: that is, an element of X is a mapping x: α → x
α
,
where x
α
∈ X
α
. For each α there is a projection map π
α
: X → X
α
, which sends
the point x to its "coordinate" x
α
.
Definition A19.14. The product topology on X is the topology generated by all
the sets π
−1
α
(U
α
), where α runs over A and U
α
runs over all the open subsets of
X
α
.
Exercise A19.15. Let x
i
be a net in the product X defined above. Show that x
i
converges to x ∈ X if and only if π
α
(x
i
) converges to π
α
(x) for all α ∈ A.
Theorem A19.16. (Tychonoff) The product of any family of compact spaces is
compact.
Proof. We use the characterization (iii) of compactness in Proposition A19.13
above.
Let x
i
be a universal net in X. Then π
α
(x
i
) is a universal net in X
α
, hence
convergent, say to x
α
. Let x ∈ X be the unique point such that π
α
(x) = x
α
.
Then, by Exercise A19.15, x
i
→ x in X. Thus X is compact.
64
Lecture A20
Quotient Spaces
Let X be a set and ∼ an equivalence relation on it. Recall that the notation
X/ ∼ denotes the set of equivalence classes of the equivalence relation. There is
a canonical map π: X → X/ ∼ that sends each x ∈ X to its equivalence class.
Now suppose that X is a topological space.
Definition A20.1. The quotient topology on X/ ∼ is defined by declaring that a
subset U ⊆ X/ ∼ is open iff the inverse image π
−1
(U) is open in X.
Of course, there is something to check here, namely that the sets U identified
by this definition do satisfy the axioms for a topology. But this is easy. Clearly,
the quotient topology makes π continuous, and it is the largest topology (most
open sets) that does so. Moreover, a function f : (X/ ∼) → Y is continuous iff
f ◦ π: X → Y is continuous.
Remark A20.2. Asubset of X is called saturated (with respect to ∼) if it is a union
of equivalence classes. Thus, the open sets of the quotient topology correspond
to the saturated open subsets of X. The saturation of an arbitrary subset S is the
union of all the equivalence classes that meet S, i.e. the smallest saturated set that
includes S.
Example A20.3. Let X = [0, 1] and let ∼ be the equivalence relation that makes
0 and 1 equivalent and whose only other equivalence classes are points. Then
[0, 1]/ ∼ is homeomorphic to the unit circle S
1
. To see this, let g : X → S
1
be
the continuous map g(t) = e
2πit
= (cos 2πt, sin 2πt), and define f : X/ ∼→ S
1
by f(π(t)) = g(t); this is well-defined since g(0) = g(1). By the remark above,
f is a continuous map from X/ ∼ to S
1
, which is in fact bijective. We must show
that f is also open, and for this it suffices to show that f takes each neighborhood
of each point π(x) of X/ ∼ to a neighborhood of f(π(x)) = g(x). That is, we
must show that any saturated open set containing x maps to a neighborhood of
g(x). When x ,∈ ¦0, 1¦ this is easy, since any saturated open set containing x
includes an interval (x − , x + ). What about saturated open sets containing 0
or 1? Any such must include U = [0, ) ∪ (1 − , 1] for some > 0, and then
f(U) = ¦(cos 2πt, sin 2πt) : t ∈ (−, )¦ is a neighborhood of (1, 0) in S
1
. So
we are done.
The ideas underlying this example can be summarized in the next two propo-
sitions. The first is the topological analog of the "first isomorphism theorem" in
65
algebra. Let us call a continuous map f : X → Y ∼-respecting if f(x) = f(x
)
whenever x ∼ x
. Then we have
Proposition A20.4. A continuous g : X → Y is ∼-respecting iff g = f ◦ π for
some continuous f : X/ ∼→ Y .
Proof. If g is ∼-respecting then a function f can be defined by setting f(c), for an
equivalence class c, equal to f(x) for any representative x of c. Then g = f ◦ π,
and f is continuous by definition of the quotient topology.
Proposition A20.5. Let X be a compact space and Y a Hausdorff space and
suppose that g : X → Y is continuous and surjective. Let ∼ be the equivalence
relation on X defined by x ∼ x
iff g(x) = g(x
). Then the map f : X/ ∼→ Y
induced by g is a homeomorphism.
Proof. Tautologically, g is ∼-respecting, so there is a continuous map f as de-
scribed, which is in fact a bijection X/ ∼→ Y . But X/ ∼ is a compact space (as
the image of a compact space under the continuous map π, Proposition A12.8),
and Y is a Hausdorff space, so this bijection is in fact a homeomorphism by Corol-
lary A12.12.
It might be helpful to think of ∼, in this proposition, as the "kernel" of the
map g.
One problem with quotient spaces is that they are frequently not Hausdorff.
An obvious necessary condition for Hausdorffness is that the equivalence classes
should be closed: after all, each equivalence class is the inverse image of a point
in X/ ∼ under the continuous map π, and we know that one-point sets are closed
in a Hausdorff space. But this is by no means a sufficient condition:
Example A20.6. Let X be the unit square [0, 1] [0, 1], which is a compact Haus-
dorff space, and let ∼be the equivalence relation which makes (x
1
, y
1
) equivalent
to (x
2
, y
2
) iff either x
1
= x
2
and y
1
= y
2
or y
1
= y
2
> 0. Then any two saturated
neighborhoods of (0, 0) and (1, 0) must intersect; so the corresponding quotient
space cannot be Hausdorff.
One can even give an example of this kind where all the equivalence classes
except one consist of single points.
Exercise A20.7. Let X be the real line, with the topology generated by all the
open sets of the usual topology together with the complement of the set A =
¦1/n : n = 1, 2, 3, . . .¦. Show that X is a Hausdorff space, but that the quotient
66
space obtained by identifying all the points of A to a single point (this is usually
denoted X/A) is not. (This is also an example of a Hausdorff space that is not
regular.)
We will give some useful conditions for Hausdorffness of quotients. Let ∼ be
an equivalence relation on X. The graph of ∼ is the subset
G = ¦(x
1
, x
2
) : x
1
∼ x
2
¦ ⊆ X X.
Suppose that X/ ∼ is Hausdorff. Then whenever x
1
,∼ x
2
there are disjoint open
subsets U
1
, U
2
of X/ ∼ with π(x
1
) ∈ U
1
and π(x
2
) ∈ U
2
. Thus π
−1
(U
1
)
π
−1
(U
2
) is open in X X, contains (x
1
, x
2
), and does not meet G. We conclude
that if X/ ∼ is Hausdorff, then the graph G is closed. (This statement includes
our previous requirement that the equivalence classes should all be closed, but it
is stronger.)
The previous exercise shows that the converse is not necessarily the case.
However, there are additional conditions under which the closure of the graph
is both necessary and sufficient for the Hausdorffness of the quotient.
Proposition A20.8. Let X be a Hausdorff space and let ∼ be an equivalence
relation on X whose graph is closed. If in addition the projection π: X → X/ ∼
is an open map then X/ ∼ is Hausdorff also.
Proof. Let G denote the graph of ∼. Suppose that π is open. Let x
1
, x
2
∈ X with
π(x
1
) ,= π(x
2
); then the point (x
1
, x
2
) ∈ X X does not meet the closed set
G, so there is an open rectangle of the form U
1
U
2
which contains (x
1
, x
2
) and
does not meet G. But then π(U
1
) and π(U
2
) are disjoint open subsets of X/ ∼
containing x
1
and x
2
respectively.
Proposition A20.9. Let X be a compact Hausdorff space and let ∼be an equiva-
lence relation on X whose graph is closed. Then X/ ∼ is Hausdorff also. More-
over, the projection π: X → X/ ∼ is a closed map (it need not be open).
Proof. We first showthat the saturation of any closed subset of X is closed (equiv-
alently, π: X → X/ ∼ is a closed map). Let C be a closed subset of S. We may
write the saturation S of C as
S = π
2
((C X) ∩ G) ⊆ X,
where π
2
: X X → X is the second projection map of the product space. Now,
C X and G are closed subsets of X X, which is compact (by the Tychonoff
67
theorem). Thus (C X) ∩ G is a closed subset of a compact space, so compact.
It follows that π
2
((C X) ∩G) is a compact subset of a Hausdorff space, so it is
closed. This proves our assertion.
Now suppose that x
1
, x
2
∈ X with π(x
1
) ,= π(x
2
), and let C
1
and C
2
be the
equivalence classes of x
1
and x
2
respectively. Let us show that there are open sets
U
1
, U
2
, including C
1
, C
2
respectively, whose saturations do not meet — that is to
say, (U
1
U
2
) ∩ G = ∅. This is the same argument as the proof that compact
Hausdorff spaces are normal. First, we prove the analog of regularity. Fix x ∈ C
1
.
For each y ∈ C
2
there are open sets P
y
containing x and Q
y
containing y such
that (P
y
Q
y
) ∩ G = ∅. Because C
2
is compact there is a finite subcover of C
2
say by Q
y
1
, . . . , Q
yn
. Then P =
n
j=1
P
y
j
and Q =
n
j=1
Q
y
j
are open sets, one
containing x and one including C
2
, such that P Q∩G = ∅. Repeat the argument
using the compactness of C
1
to obtain the desired open sets U
1
⊇ C
1
, U
2
⊇ C
2
,
with (U
1
U
2
) ∩ G = ∅.
Consider now the closed set X ¸ U
i
, and let S
i
be its saturation; S
i
is closed
(by the first thing we proved above) and does not meet C
i
, so the complement
X ¸ S
i
= V
i
is a saturated open set with C
i
⊆ V
i
⊆ U
i
. Now π(V
i
) and π(V
2
) are
disjoint open subsets containing x
1
and x
2
respectively.
68
Lecture A21
Basics of Topological Groups
Let G be a group. We can think of it as a set provided with a distinguished
element (the identity e) and two distinguished maps, the multiplication map
GG → G, (x, y) → xy
and the inversion map
G → G, x → x
−1
.
Definition A21.1. A topological group is a group provided with a topology for
which the multiplication and inversion maps are continuous.
There are many examples: the (real or complex) numbers under addition,
the nonzero (real or complex) numbers under multiplication, the matrix groups
GL(n, R) and GL(n, C) as well as their various subgroups SL(n, R), SL(n, C),
O(n), U(n), etc. Any group can be thought of as a topological group when en-
dowed with the discrete topology.
In a topological group the left and right multiplication maps by a fixed element
g, defined by L
g
(x) = gx, R
g
(x) = xg, are homeomorphisms G → G. Conse-
quently, every translate gU or Ug of an open subset U is again open (and every
translate of a closed set is closed).
Proposition A21.2. A topological group G is Hausdorff if and only if the one-
point set ¦e¦ is closed.
Proof. Suppose that ¦e¦ is closed. By translation, every one-point set is closed.
If g
1
,= g
2
there exists an open set V = G ¸ ¦g
2
¦ that contains g
1
and does not
contain g
2
. By continuity of the multiplication map at the point (g
1
, e), together
with the definition of the product topology, there are open sets U
1
, U
2
that contain
g
1
and e respectively such that U
1
U
2
⊆ V . But now U
1
and g
2
U
−1
2
are open
neighborhoods of e and g respectively. I claim they don't intersect: if they did,
then u
1
= g
2
u
−1
2
with u
1
∈ U
1
, u
2
∈ U
2
, and then g
2
= u
1
u
2
∈ U
1
U
2
⊆ V , a
contradiction.
This is one example of the "homogeneity" of topological groups.
Let Gbe a topological group and H a subgroup. We denote by G/H the space
of left cosets xH for x ∈ G, or in other words the space of equivalence classes
69
for the equivalence relation x ∼ y ⇔ x
−1
y ∈ H; similarly we denote by H¸G
the space of right cosets Hx, which are equivalence classes for the equivalence
relation x ∼ y ⇔ xy
−1
∈ H. These are called (left and right) homogeneous
spaces of G by H, and we give them the quotient topology.
Exercise A21.3. If H is a normal subgroup of G then the left and right coset
spaces are the same. In this case we know from algebra that G/H has the structure
of a group. We have also just equipped it with a topology. Show that G/H is a
topological group, i.e., multiplication and inversion are continuous.
Proposition A21.4. The quotient maps G → G/H and G → H¸G are open
maps.
Proof. This is equivalent to the statement that the saturation of an open subset of
G is open. But the saturation of an open set U is
_
u∈U
uH =
_
h∈H
Uh;
the second representation exhibits it as a union of open sets, hence open.
Theorem A21.5. Let G be a Hausdorff topological group and let H be a closed
subgroup. Then the homogeneous spaces G/H and H¸G are Hausdorff.
Proof. By A20.8 and A21.4, it suffices to show that the graph Γ of the equivalence
relation x ∼ y ⇔ x
−1
y ∈ H is a closed subset of GG.
Suppose that (x, y) / ∈ Γ; then x
−1
y ∈ G ¸ H which is an open set. By
continuity of multiplication there exist open sets U ¸ x
−1
, V ¸ y with U V ⊆
G¸ H. But then U
−1
V is an open neighborhood of (x, y) that does not meet Γ.
It follows that the complement of Γ is open, so Γ is closed.
Definition A21.6. Let X, Y be topological spaces. A continuous, surjective map
f : X → Y is called a local homeomorphism if every x ∈ X has a neighborhood
U such that the restriction of f to a map U → f(U) is a homeomorphism.
Local homeomorphisms will play a very important role in the study of cov-
ering spaces, a key part of algebraic topology. The classic example of a local
homeomorphism is the exponential map t → e
2πit
from the real line to the circle.
This can be envisaged as a special case of the next result:
70
Proposition A21.7. Let G be a topological group and suppose that H is a closed
subgroup which is discrete (in the induced topology). Then the quotient map
π: G → G/H is a local homeomorphism.
Proof. There is a neighborhood V of e such that H ∩ V = ¦e¦. By the same
kind of argument (continuity of multiplication) that we have already used several
times, there exists a neighborhood U of the identity with U
−1
U ⊆ V . But now
the restriction of π: G → G/H to the set U is injective: if π(u
1
) = π(u
2
) then
u
−1
1
u
2
∈ V ∩ H = ¦e¦, and so u
1
= u
2
. Thus the restriction of π to a map
from U to π(U) is continuous, bijective, and open (A21.4): in other words, it
is a homeomorphism. We have proved that π is a homeomorphism near e; by
translation, the same holds near any point.
A group G acts on a set X if there is given a map G X → X, written
(g, x) → gx, such that ex = x for all x ∈ X and g(hx) = (gh)x for all g, h ∈ G,
x ∈ X. If G is a topological group and X is a topological space we define a
continuous action by requiring that the associated map GX → X be continuous.
An action is transitive if for all x, y ∈ X there exists g ∈ G with gx = y.
Example A21.8. The rotation group SO(n) acts transitively on the unit sphere
S
n−1
.
Proposition A21.9. Suppose that G acts transitively on the Hausdorff space X.
For any x ∈ X let G
x
denote the stabilizer
G
x
= ¦g ∈ G : gx = x¦,
which is a closed subgroup of G. Then the formula
[g] → g x
defines a continuous bijection G/G
x
→ X. It is a homeomorphism if G is com-
pact.
Proof. The continuous map G → X given by g → gx respects the equivalence
relation given by the cosets of G
x
. Thus it passes to a continuous map G/G
x
→
X, which is injective by construction, and is surjective because of the transitivity
of the action. The final statement follows because G/G
x
is compact and X is
Hausdorff.
For example, considering the stabilizer of a single point (the North pole) gives
us the identification S
n−1
∼
= SO(n)/SO(n − 1) (using the fact that SO(n) is
compact).
71
Lecture A22
Homotopy
Hereafter "map" means "continuous map".
Consider the unit square X = I
2
= ¦(x, y) ∈ R
2
: 0 x, y 1¦. Imagine
two continuous paths γ
1
, γ
2
: [0, 1] → X, with γ
1
beginning at (0, 0) and ending
at (1, 1), and γ
2
beginning at (0, 1) and ending at (1, 0). It seems obvious that
these two paths must cross somewhere in the interior of the square. But this is
hard (maybe impossible?) to prove just with the tools that we have developed so
far. We need a little homotopy theory.
Definition A22.1. Let X and Y be topological spaces and let f
0
, f
1
: X → Y be
continuous maps. A homotopy between f
0
and f
1
is a continuous map
h: X [0, 1] → Y
with h(x, 0) = f
0
(x) and h(x, 1) = f
1
(x) for all x ∈ X. If there exists a homo-
topy between f
0
and f
1
we say that these maps are homotopic.
Proposition A22.2. Homotopy is an equivalence relation on the class of maps
X → Y .
Proof. Clearly any map f is homotopic to itself via the constant homotopy h(x, t) =
f(x). If h is a homotopy between f
0
and f
1
, then
k(x, t) = f(x, 1 −t)
is a homotopy between f
1
and f
0
. Finally, homotopies can be concatenated like
paths: if h
is a homotopy from f
0
to f
1
and h
is a homotopy between f
1
and f
2
then
h(x, t) =
_
h
(x, 2t) (t
1
2
)
h
(x, 2t −1) (t
1
2
)
is a homotopy between f
0
and f
2
.
Exercise A22.3. Check in detail that the homotopy h defined in the last part of
this proof is continuous.
72
Remark A22.4. This proof is obviously closely related to the proof that path-
connectedness is an equivalence relation, Lemma A15.10. In fact, path-connectedness
is just homotopy of maps where X is a point. Conversely, one can reduce the no-
tion of homotopy to path-connectedness in the space of maps X → Y if one
defines a suitable topology on this space. But we won't do that here.
Proposition A22.5. If f
0
, f
1
are homotopic maps X → Y , and g
0
, g
1
: Y → Z are
homotopic maps, then g
0
◦ f
0
, g
1
◦ f
1
are homotopic maps X → Z. (In particular,
homotopic maps remain homotopic after composition on the left or the right with
a fixed map.)
There is a related notion, that of relative homotopy. Let X be a space and A
a closed subspace (one says sometimes that (X, A) is a pair). Then a homotopy
relative to A from X to Y is a homotopy h: X [0, 1] → Y such that h(a, t) is
independent of t for all a ∈ A — a homotopy that stays constant on A. This also
defines an equivalence relation by the same reasoning as above.
Example A22.6. Let X be the unit disc in Euclidean space. Then the identity map
X → X is homotopic to the map that sends every point to the origin. A homotopy
between these maps is defined by h(x, t) = (1 − t)x. A similar argument can be
applied to any star-shaped subset of Euclidean space.
Example A22.7. Let X, Y be spaces and let f : X → Y be a continuous map.
The mapping cylinder of f is the space
C
f
= X [0, 1] .
f
Y
obtained from the disjoint union X[0, 1] .Y by identifying (x, 0) with f(x) for
all x ∈ X (and equipped with the quotient topology). As we saw earlier, there is
a continuous map F : C
f
→ Y defined by F(x, s) = f(x), F(y) = y. Moreover,
the identity map C
f
→ C
f
is homotopic, relative to Y , to the map C
f
→ Y → C
f
defined by composing F with the standard inclusion Y → C
f
. A homotopy h
between these maps can be defined by the formula
h((x, s), t) = (x, (1 −t)s), h(y, t) = y.
Notice that the second example actually includes the first: the disc is the map-
ping cylinder of the constant map from a sphere to a point.
The geometric situation defined in these examples occurs often enough to have
a special name.
73
Definition A22.8. Let X be a space and A a closed subspace. A retraction of X
onto A is a continuous map X → A which is the identity on A. A deformation
retraction of X onto A is a homotopy, relative to A, from the identity map on X
to a retraction X → A. If these exist one says that A is a retract or deformation
retract of X.
The examples above establish that a point is a deformation retract of a disk,
and that Y is a deformation retract of the mapping cylinder. Deformation retrac-
tion is a special case of the general notion of homotopy equivalence.
Definition A22.922.10. The relation of homotopy equivalence is, indeed, an equiv-
al74
Lecture A23
Homotopy Equivalence
The examples above establish that a point is a deformation retract of a disk,
and that Y is a deformation retract of the mapping cylinder. Deformation retrac-
tion is a special case of the general notion of homotopy equivalence.
Definition A23.123.2. The relation of homotopy equivalence is, indeed, an equiva-
lDefinition A23.3. A space is called contractible if it is homotopy equivalent to a
point.
The cone CX on a space X is the quotient X [0, 1]/X ¦0¦, or in other
words the mapping cylinder of the constant map from X to a point. Any mapping
cylinder deformation retracts onto the range of the map from which it is defined
(as we have seen), so the cone on any space is contractible.
Homotopy equivalent spaces can be rather different topologically. For exam-
ple, the 1-dimensional spaces given by the "theta curve", the "figure 8", and the
"dumb-bell" are all homotopy equivalent, although no two of themare homeomor-
phic. However, there are topological properties that are preserved by homotopy
equivalence.
Example A23.4. Suppose X and Y are homotopy equivalent. If X is path-
connected, then so is Y .
75
Proof. Let f : X → Y and g : Y → X be mutually inverse homotopy equiva-
lences. Note that for each y ∈ Y , the points y and f(tg(y)) belong to the same
path component: indeed, if h is a homotopy between f ◦ g and the identity, then
t → h(y, , t) defines a path joining y to f(g(y)). Now let y, y
∈ Y and suppose
that X is path connected. Then there is a path γ in X joining g(y) to g(y
). Now
f ◦ γ is a path joining f(g(y)) to f(g(y
)). Thus f(g(y)) and f(g(y
)) are in the
same path component. It follows from the earlier discussion that y and y
are in
the same path component; that is, Y is path connected.
Suppose that X is a space and A a closed subspace. If A is contractible, we
might say that "up to homotopy" Ais the same as a point. If Areally were a point,
the quotient space X/A would be just the same as X. So we might be led to hope
that if Ais contractible, X/Ais "up to homotopy" the same as X, i.e., the quotient
map X → X/A is a homotopy equivalence. Unfortunately this is not true.
Exercise A23.5. Let Z be the "topologist's sine curve" of Exercise A15.13, and
let A ⊆ Z be the interval ¦(0, y) : −1 y 1¦ on the y-axis. Clearly A is
contractible. Show that Z/Ais path-connected. Since Exercise A15.13 shows that
Z itself is not path-connected, this proves that Z and Z/A cannot be homotopy
equivalent.
What is missing in this example is the homotopy extension property (HEP).
Definition A23.6. Let X be a space and A a closed subspace. One says that
(X, A) has the homotopy extension property if any map from the subspace
X ¦0¦ ∪ A [0, 1] ⊆ X [0, 1]
to another space Y can be extended to a map from the whole of X [0, 1] to Y .
The way to think of this is that the initial data consist of a map X → Y
together with a homotopy of the restriction of that map to the subspace A; the
HEP says that this homotopy can be extended to a homotopy defined on the whole
space X. Most "nice" pairs of spaces have the HEP: proofs are often inductive
based on the following example.
Example A23.7. The pair (D
n
, S
n−1
) has the homotopy extension property. To
see this, it is enough to construct geometrically a retraction
r : D
n
[0, 1] → D
n
¦0¦ ∪ S
n−1
[0, 1]
(which can be done for instance by projection from the point (0, 3/2) in R
n+1
);
then the desired extension can be defined by composing with r.
76
Proposition A23.8. Suppose that (X, A) has the HEP and that A is contractible.
Then the quotient map π: X → X/A is a homotopy equivalence.
Proof. The first order of business is to find a potential homotopy inverse. Let
p ∈ A be a point and let h: A [0, 1] → A be a homotopy between the identity
map and the constant map that sends A to p. By the HEP, the homotopy h extends
to a homotopy H between the identity on X and some map G: X → X, such that
G(a) = p for all a ∈ A. Note that G respects the equivalence relation defining the
quotient space, so there is a map g : X/A → X such that g ◦ π = G. This map g
is, I claim, a homotopy inverse for π.
To see this we must show that g ◦ π and π ◦ g are homotopic to the identity on
X and X/A respectively. The first is easy: g ◦ π = G is homotopic to the identity
by construction. What about the other? The homotopy H has H(a, t) ∈ A for
all a ∈ A and all t. Therefore, π ◦ H(a, t) ∈ X/A is constant for all a ∈ A and
t ∈ [0, 1]; that is, π ◦ H respects the equivalence relation. It follows that there is a
map
k: (X/A) [0, 1] → (X/A)
such that k(π(x), t) = π(H(x, t)) for all x ∈ X, t ∈ [0, 1]. In particular, k gives
the desired homotopy between π ◦ g and the identity on X/A.
77
Lecture A24
The fundamental group
For the next few lectures we are going to consider pointed topological spaces.
Such an object is just a topological space X together with a choice of a base point
x
0
∈ X. A map of pointed spaces (X, x
0
) → (Y, y
0
) is just a map from X to Y
that takes x
0
to y
0
. Sometimes, to save space, we'll omit explicit mention of the
base point.
Example A24.1. We regard the unit circle S
1
as the quotient space [0, 1]/¦0, 1¦.
It is a pointed space whose base point ∗ is the equivalence class ¦0, 1¦.
Definition A24.2. A loop in a pointed space (X, x
0
) is a map of pointed spaces
(S
1
, ∗) → (X, x
0
). In other words, it is a path γ : [0, 1] → X such that γ(0) =
γ(1) = x
0
.
The natural notion of homotopy for maps of pointed spaces is based homotopy:
a based homotopy from (X, x
0
) to (Y, y
0
) is a homotopy h: XI → Y such that
h(x
0
, t) = y
0
for all t ∈ I. In particular we can consider the collection of all based
homotopy classes of loops in (X, x
0
). This object is denoted π
1
(X, x
0
) and it is
called the fundamental group of (X, x
0
). As we shall see, there is indeed a natural
way to make it into a group.
Lemma A24.3. Let γ be a based loop in (X, x
0
). Let g : [0, 1] → [0, 1] be a
continuous map with g(0) = 0 and g(1) = 1. Then γ and γ ◦ g represent the same
element of π
1
(X, x
0
). (The composite γ ◦ g is called a reparameterization of γ.)
Proof. The map
h(s, t) = γ((1 −t)s + tg(s))
defines a homotopy between γ and the reparameterized path γ ◦ g.
Let γ
and γ
be paths in X. Recall that the concatenation of these paths is
defined by
γ
γ
(s) =
_
γ
(2s) (s
1
2
)
γ
(2s −1) (s
1
2
)
If γ
and γ
are (based) loops then so is γ
γ
. Moreover, if γ
or γ
is var-
ied by a (based) homotopy, then so is γ
γ
. Thus the operation passes to a
"multiplication" operation on π
1
(X, x
0
).
78
Proposition A24.4. Under this operation, π
1
(X, x
0
) becomes a group.
Proof. There are three things that must be verified: associativity, the existence of
an identity, and the existence of inverses. The proofs of all three make extensive
use of the freedom to deform loops by homotopies and, in particular, to reparam-
eterize them.
For associativity, we just need to note that the paths (γ
1
γ
2
)γ
3
and γ
1
(γ
2
γ
3
)
are related by the reparameterization
s →
_
¸
_
¸
_
2s s
1
4
s +
1
4
1
4
s
1
2
1
2
+
1
2
s
1
2
s 1
The constant loop (e(s) = x
0
for all s ∈ [0, 1]) defines an identity element.
Indeed, it is easy to see that e γ and γ e are simply reparameterizations of γ,
for any loop γ.
Finally suppose that γ is a loop and let γ
−1
be the loop defined by γ
−1
(s) =
γ(1 − s). Then γ γ
−1
is the path s → γ(g(s)) where g(s) = 2s for s
1
2
and
g(s) = 2 −2s for s
1
2
. Then
h(s, t) = γ((1 −t)g(s))
is a homotopy of based loops from γ γ
−1
to the identity path.
The effect of the choice of basepoint is summed up in the following.
Lemma A24.5. Let X be a path-connected space and let x
0
, x
1
∈ X. Then the
groups π
1
(X, x
0
) and π
1
(X, x
1
) are isomorphic.
Proof. Let q : [0, 1] → X be a path from x
0
to x
1
, and let q
−1
(s) = q(1 − s), as
usual. Then, if γ is a loop based at x
0
, the path
φ(γ) := q
−1
γ q
is a loop based at x
1
, and the assignment γ → φ(γ) gives a homomorphism of
groups from π
1
(X, x
0
) to π
1
(X, x
1
). The assignment γ
→ q γ
q
−1
gives the
inverse homomorphism π
1
(X, x
1
) to π
1
(X, x
0
), so these two groups are isomor-
phic.
79
Beware that the isomorphism in the lemma above depends on the choice of the
path q. One says that the groups π
1
(X, x
0
) and π
1
(X, x
1
) are "isomorphic but not
canonically isomorphic".
Let f : (X, x
0
) → (Y, y
0
) be a (basepoint-preserving) map. If γ is a based loop
in X, the composite f ◦ γ is a based loop in Y . Moreover, the assignment [γ] →
[f ◦ γ] passes to homotopy classes and defines a homomorphism π
1
(X, x
0
) →
π
1
(Y, y
0
).
Definition A24.6. The homomorphism so defined is called the induced homomor-
phism of the map f, and it is denoted by f
∗
: π
1
(X, x
0
) → π
1
(Y, y
0
).
Proposition A24.7. The induced homomorphism has the following properties:
(a) The identity map induces the identity homomorphism.
(b) Suppose that f and g are maps (X, x
0
) → (Y, y
0
), which are homotopic by
a homotopy h. Let u ∈ π
1
(Y, y
0
) be defined by the loop t → h(x
0
, t). Then
f
∗
(v) = ug
∗
(v)u
−1
for all v ∈ π
1
(X, x
0
). In particular, based homotopic
maps induce the same homomorphism.
(c) If f : X → Y and g : Y → Z are based maps, then (f ◦ g)
∗
= f
∗
◦ g
∗
.
Item (c) is called functoriality: it is an extremely important property in many
mathematical contexts.
Proof. (a) and (c) are immediate consequences of the definitions. Part (b) would
be easier if we restricted our attention to based homotopies, but we need the gen-
eral case. Consider the map from the unit square to Y defined by
(s, t) → h(γ(s), t).
Going along the bottom and right sides of this square defines the path f u, going
along the left and top sides defines u g. But these are homotopic by a homotopy
that goes diagonally across the square. Explicitly, we want to define
H(r, s) =
_
h(2r(1 −s), 2rs) (r
1
2
)
h((1 −s)(2r −1) + s, s(2r −1) + 1 −s) (r
1
2
)
which gives a homotopy from H(, 0) = f u to H(, 1) = u g.
Corollary A24.8. A homotopy equivalence induces an isomorphism of fundamen-
tal groups.
80
Proof. Let f : X → Y be a homotopy equivalence, with homotopy inverse g.
Then fg and gf are homotopic to the identity, and therefore by (a) and (b) above
they induce isomorphisms on fundamental groups. Thus by (c) above, f
∗
◦ g
∗
and
g
∗
◦ f
∗
are isomorphisms, which implies that f
∗
and g
∗
are isomorphisms too.
In particular
Corollary A24.9. If X is contractible, then its fundamental group is trivial.
81
Lecture A25
Fibrations and the fundamental group
Recall the homotopy extension property for a pair X, A), which we can also
think of as a property of the inclusion map i : A → X. The pair has the HEP (or,
as one also says, the inclusion i is a cofibration) if, given any homotopy of maps
A → Y , together with an extension of the initial data of the homotopy to a map
X → Y , we can always extend the homotopy to a homotopy of maps X → Y ,
respecting the given initial data.
The notion of extension of a map can be expressed in diagrammatic terms
using the inclusion i:
X
A
?
i
OO
//
Y
Here the solid arrows represent the given data and the dotted arrows represent
the extension to be constructed. Dual to the notion of extension is the notion of
lifting: suppose that p: E → B is a surjective map and f : Y → B is any map,
then a lifting of f (over p) is a map g : Y → E that makes the following diagram
commute:
E
p
Y
>>
//
B
in other words p ◦ g = f.
Definition A25.1. We say that p: E → B has the homotopy lifting property or is
a fibration if, given any homotopy of maps Y → B together with a lifting of the
initial data of the homotopy over p, the entire homotopy can be lifted over p.
In other words, p has the HLP if, given the data expressed by the solid arrows
in the diagram below
Y ¦0¦
_
//
E
Y [0, 1]
//
<<
B
82
the dotted arrow can be filled in in such a way as to make the diagram commuta-
tive.
Remark A25.2. The homotopy extension property can be expressed in a similar
diagrammatic way, but to do so we have to consider a homotopy of maps A → Y
as a single map from A to the space Y
I
of maps I → Y . We haven't discussed the
topology of mapping spaces, so we can't write down the full details of this, but it
should not be hard to figure out the general idea.
For now we are not going to worry too much about where fibrations might
come from. Instead we will concentrate on what they might tell us about the
fundamental group.
Definition A25.3. Let X be a space. If X is path-connected and π
1
(X, x
0
) = ¦1¦
for some (and hence any) basepoint x
0
, we will say that X is simply connected.
Contractible spaces, for example, are simply connected (Corollary A24.9).
But there are many other examples.
Exercise A25.4. Show that the n-sphere S
n
is simply connected for n 2. Hints:
First show that it is enough if one knows that any loop in S
n
is homotopic to
one that is not surjective. Then prove this statement by any one of a number
of approximation techniques, e.g. by showing that any loop is homotopic to a
"piecewise linear" loop made up of great circle arcs.
Definition A25.5. Let p: E → B be a fibration, and let B have a fixed basepoint
b
0
. The inverse image F = p
−1
¦b
0
¦ is called the fiber of the fibration.
Suppose now that p: E → B is a fibration and let b
0
∈ B be a base point. Let
γ be a based loop in B and let x ∈ F be a point of the fiber F = p
−1
(b
0
). We may
consider the path γ as a homotopy (of maps of a point into E) and, if we do this,
γ and x together provide exactly the data for a homotopy lifting problem:
¦0¦
_
x
//
E
[0, 1]
γ
//
˜ γ
>>
B
Filling in the dotted arrow provides a lifting ˜ γ of the path γ, but this lifting is
not guaranteed to be a loop! All we know is that p(˜ γ(1)) = γ(1) = b
0
, that is,
83
˜ γ(1) is a point of thee fiber F. It is not uniquely determined by this construction
but we shall see in a moment that the path component of the fiber in which it
lies is uniquely determined. The proof begins to build up the connection between
fibrations and the fundamental group.
Lemma A25.6. Let p: E → B be a fibration, as above, and let γ be a based loop
in B. If γ is nullhomotopic (that is, it represents the identity element in π
1
(B, b
0
)),
then, for any lift ˜ γ : [0, 1] → E of γ, ˜ γ(0) and ˜ γ(1) are in the same path component
of the fiber F.
(It's obvious that ˜ γ(0) and ˜ γ(1) are in the same path component of E — they
are joined by the path ˜ γ, after all! — so the significance of the lemma is that a
path joining them can be found that lies wholly in F.)
Proof. We apply the HLP to a homotopy between γ and the constant path. Let
h(s, t) be such a homotopy with h(s, 0) = γ(s) and h(s, 1) = b
0
. Then h together
with ˜ γ provide the data for a homotopy lifting problem:
[0, 1] ¦0¦
_
˜ γ
//
E
[0, 1] [0, 1]
h
//
H
<<
B
The bottom, right-hand and top sides of the map of the unit square defined by h
are constant maps with value x
0
. Therefore the bottom, right-hand and top sides
of the lifted homotopy H are paths in the fiber F. The concatenation of these three
is a path in F from ˜ γ(0) to ˜ γ(1).
Corollary A25.7. If γ
, γ
are two liftings of the same loop, starting at the same
point γ
(0) = γ
(0), then their end points are in the same path component of F.
Proof. The concatenation (γ
)
−1
γ
is a lifting of a nullhomotopic loop.
To help keep track of this business of path components let's introduce some
notation.
Definition A25.8. For any topological space F, let π
0
(F) denote the set of path
components of F.
84
Suppose that p: E → B is a fibration with fiber F, as discussed above. For
g ∈ π
1
(B, b
0
) and c ∈ π
0
(F), let us define c
g
(read: "c acted on by g") as follows:
choose a based loop γ representing g, and a point x ∈ F representing c, let ˜ γ be
any lift of γ starting at x, and define c
g
to be the path component containing ˜ γ(1).
The lemma and corollary above show that this is well defined.
Proposition A25.9. The construction above defines an action of the fundamental
group π
1
(B, b
0
) on the set π
0
(F). Moreover, if E is path connected, then this
action is transitive; and if in addition E is simply connected, the action is free.
Let us explain the terminology. A (right) action of a group G on a set C is
just a map C G → C, written (c, g) → c
g
, such that c
e
= c and (c
g
)
h
= c
gh
.
An action is free if c
g
= c implies g = e, and it is transitive if for all pairs c, c
of
elements of C, there exists g ∈ G such that c
g
= c
. When a group action is both
free and transitive, the mapping g → c
g
, for fixed c, is a bijection from the group
G to the set C.
Proof. The action law (c
g
)
h
= c
gh
follows from the fact that a lifting of a concate-
nation of loops is a concatenation of liftings of those loops. And c
e
= c follows
from the fact that a constant loop can be lifted to a constant.
Suppose now that E is path connected. Then given any two points x, x
of
the fiber there is a path in E joining them. Composing with p gives a based loop
in B which (tautologically) lifts to a path in E from x to x
. Thus the action is
transitive.
Suppose additionally that E is simply connected. Suppose that c
g
= c for some
c ∈ π
0
(F) and g ∈ π
1
(B, b
0
); this means that g is represented by a based loop
γ in B which lifts to ˜ γ such that ˜ γ(0) and ˜ γ(1) are in the same path component
of F. Construct a loop β in E by concatenating ˜ γ with a path in F from ˜ γ(0) to
˜ γ(1). The composite p ◦ β is the concatenation of γ with a constant path, so it still
represents g ∈ π
1
(B, b
0
). But, on the other hand, β is nullhomotopic in E (say
by a homotopy H) because E is simply connected. Then p ◦ H is a nullhomotopy
of the path p ◦ β, which therefore represents the identity element of π
1
(B, b
0
) as
claimed.
85
Lecture A26
Covering Spaces
Let p: E → B be a surjective map of topological spaces.
Definition A26.1. The map p above is called a covering map (and E is called a
covering space of B) if there is a discrete topological space F (the fiber) such that
each point x ∈ B has an open neighborhood U for which there is a homeomor-
phism p
−1
(U)
∼
= U F making the diagram
p
−1
(U)
∼
=
//
p
##
H
H
H
H
H
H
H
H
H
H
H
H
H
H
H
H
H
U F
pr
1
U
commute.
In other words, a covering map is "locally the projection of a product with
a discrete space". Some authors (e.g. Hatcher) give a definition which appears
more general than this, but it turns out that the definitions are equivalent for path-
connected base spaces B. A neighborhood U with the property described in the
definition will be called a trivializing neighborhood for the fibration.
Example A26.2. A homeomorphism is a covering map with 1-point fiber.
Example A26.3. The map R → S
1
defined by x → x mod 1 (if we think of
the circle as [0, 1] with endpoints identified) or x → e
2πix
(if we think of the unit
circle in C) is a covering map with fiber Z.
Example A26.4. The map S
1
→ S
1
defined by x → nx mod 1 (or z → z
n
in
the complex pictures) is a covering map with fiber an n-point space.
Example A26.5. There are many different possible covering spaces of the "figure
8" space; I'll draw some pictures in class.
Example A26.6. Let B be a compact oriented surface of genus −2 (i.e. a the
surface of a 2-holed donut). Then B can be obtained by a suitable identification
of the edges of a regular octagon. One can tessellate the hyperbolic plane by
such octagons with vertex angle of 45 degrees. From this tessellation we obtain a
covering map from the plane to B.
86
Proposition A26.7. A covering map is a fibration, i.e., it satisfies the homotopy
lifting property.
Proof. Remember what we have to show: for each space Y we can fill in the
"homotopy lifting diagram"
Y ¦0¦
_
f
//
E
Y [0, 1]
h
//
H
<<
B
First of all let's consider the case when Y is a single point (the "path lifting
property"). By definition, B has an open cover consisting of trivializing neigh-
borhoods. The inverse image of this open cover by the map h is then an open
cover of the compact space [0, 1], which has a positive Lebesgue number by
the Lebesgue covering theorem. Partition [0, 1] into finitely many subintervals
0 t
1
t
2
1 of length less than the Lebesgue number. We will
construct the desired lifting by induction over these subintervals.
Suppose then than a lifting H has been constructed on [0, t
k
]. We have h([t
k
, t
k+1
]) ⊆
U for some trivializing neighborhood U, and in particular we may identify p
−1
(U)
with U F where F is the (discrete) fiber. By connectedness, pr
2
(H(t)) must be
constant for any lift H. Therefore the unique continuous lift H of h on [t
k
, t
k+1
]
that agrees with the lift already assumed to have been constructed on [0, t
k
] is
H(t) = (h(t), pr
2
(H(t
k
))) ∈ U F
∼
= p
−1
(U) ⊆ E.
By induction, we can construct H on [0, 1]. Indeed, we have done more than
claimed: we have shown that there is a unique possible choice for H.
Now we return to the case of general Y . Because of the uniqueness of path
lifting, which we just proved, there is no question how to define H for general Y :
namely, for each fixed y ∈ Y , H(y, t) should be the unique lifting of the path t →
h(y, t) that begins at f(y). The only question is whether the function H defined
by this process is continuous as a function of y and t. To see this, fix y
0
∈ Y and
consider the following fact: for each t ∈ [0, 1] there is a product neighborhood
V
t
W
t
of (y
0
, t) such that h(V
t
W
t
) lies in a trivializing neighborhood in B.
By compactness, finitely many of these product neighborhoods V
t
W
t
cover
87
¦y
0
¦ [0, 1]. Let V be the intersection of the finitely many V
t
's that appear and
let be a Lebesgue number for the covering of [0, 1] by the finitely many W
t
's.
Partition [0, 1] into finitely many subintervals 0 t
1
t
2
1 of length
less than . Notice that for each k, h(V [t
k
, t
k+1
]) is contained in a trivializing
neighborhood.
We're going to prove by induction over subintervals that H is continuous on
V [0, 1]. Suppose then that we already know that H is continuous on V [0, t
k
];
in particular, H(y, t
k
) is a continuous function of y for y ∈ V . But by the formula
above, for t ∈ [t
k
, t
k+1
],
H(y, t) = (h(y, t), π
2
(H(y, t
k
))) ∈ U F
∼
= p
−1
(U) ⊆ E,
and this is a continuous function of (y, t). So the induction is complete and we
have shown (in particular) that H is continuous at (y
0
, t) for all t ∈ [0, 1]. Since
y
0
was arbitrary, this suffices to complete the proof.
88
Lecture A27
Applications
So, after all this machinery, we can finally compute π
1
(S
1
).
Theorem A27.1. The group π
1
(S
1
) is isomorphic to Z.
Proof. We make use of the covering space R → S
1
described above (Exam-
ple A26.3). The fiber F = Z is discrete and the total space E = R is contractible,
so according to Proposition A25.9 the action of π
1
(S
1
) on π
0
(F) = F is free and
effective. We can therefore define a bijection π
1
(S
1
) → Z by sending g ∈ G to
0
g
∈ Z. All that is necessary is to show that this is a group isomorphism.
For this, notice that S
1
itself is a topological group. In any topological group
G, the group operation in π
1
(G) can be defined by pointwise multiplying paths
(using the group operation in G), instead of by concatenating them. For the ex-
pression
H(s, t) = γ
to their pointwise product.
However, the lift of a pointwise product of paths (in S
1
) is clearly their pointwise
sum (in R); in particular, the action of π
1
(S
1
) on the fiber Z satisfies 0
gh
= 0
g
+0
h
.
This proves that the bijection π
1
(S
1
) → π
0
(Z) = Z is indeed an isomorphism of
groups.
Exercise A27.2. Using a similar argument to the above, show that if G is any
topological group (not necessarily abelian), then π
1
(G) is an abelian group.
Remark A27.3. The identity map S
1
→ S
1
is a generator of the cyclic group
π
1
(S
1
) = Z, as the above proof shows. The integer n ∈ π
1
(S
1
) is represented by
the map z → z
n
, which "wraps" the unit circle n times around itself.
Here are some (standard) applications of this calculation.
Proposition A27.4. (No-retraction theorem) There is no retraction of the disk
D
2
= ¦z ∈ C : [z[ 1¦ onto its boundary circle S
1
= ¦z ∈ C : [z[ 1¦.
89
Proof. A retraction is a map r : D
2
→ S
1
that is a left inverse to the inclusion
i : S
1
→ D
2
, that is. the diagram
S
1
p
A
A
A
A
A
A
A
A
A
A
A
A
A
A
A
=
//
S
1
D
2
r
OO
should commute. But if this is so then the induced diagram
π
1
(S
1
)
=
//
##
G
G
G
G
G
G
G
G
G
G
G
G
G
G
G
G
π
1
(S
1
)
π
1
(D
2
)
r∗
OO
should commute also, and this is impossible since π
1
(S
1
)
∼
= Z while π
1
(D
2
) is
trivial.
Proposition A27.5. Any continuous map f : D
2
→ D
2
has a fixed point.
Proof. Suppose for a contradiction that f : D
2
→ D
2
has no fixed point. Then
for each x ∈ D
2
, let u = u(x) be the unit vector (x − f(x))/|x − f(x)|. Let
λ = λ(x) be the non-negative root of the quadratic equation
|x + λu|
2
= λ
2
+ 2λu x +|x|
2
= 1.
Then g(x) = x +λu depends continuously on x and it is the point on S
1
obtained
by continuing the ray from f(x) through x until it hits the unit circle. Clearly, if
x ∈ S
1
, then g(x) = x. Thus g is a retraction of D
1
onto S
1
, contradicting the
previous result.
Let a ∈ C and let γ : S
1
→ C ¸ ¦a¦ be a loop in the complex plane that does
not pass through a. The map ν
a
(z) = (z −a)/[z −a[ is a homotopy equivalence
between C ¸ ¦a¦ and the unit circle. The element of π
1
(S
1
) = Z represented by
the composite
ν
a
◦ f : S
1
→ S
1
is called the winding number of f about a, and is denoted n(f; a). It is a homotopy
invariant. (It does not depend on the choice of base point because π
1
(S
1
) = Z is
abelian.)
90
Lemma A27.6. (Rouch´ e's theorem) If f : S
1
→C¸¦0¦, g : S
1
→C, and [g(z)[ <
[f(z)[ for all z ∈ S
1
, then n(f; 0) = n(f + g; 0).
Proof. The map h(z, t) = f(z) + tg(z) is a homotopy between f and f + g in
C ¸ ¦0¦.
Theorem A27.7. (Fundamental theorem of algebra) Every non-constant polyno-
mial over C has a root.
Proof. Suppose not and let p(z) = z
n
+ a
n−1
z
n−1
+ + a
0
be a polynomial
without root. Let p
r
: S
1
→ C ¸ ¦0¦ be the loop defined by p
r
(e
iθ
) = p(re
iθ
).
These are homotopic loops in C ¸ 0. But for r = 0 the winding number n(p
r
, 0)
is zero, whereas for large r we may write
p(re
iθ
) = f(re
iθ
) + g(re
iθ
), f(z) = z
n
and simple estimates show[f(re
iθ
)[ > [g(re
iθ
)[, so by Rouch´ e's theoremn(p
r
, 0) =
n(z
n
, 0) = n. Contradiction.
Lemma A27.8. Let f : S
1
→ S
1
be a loop with is antipodal, that is, f(−z) =
−f(z). Then f represents an odd multiple of the generator of π
1
(S
1
).
Proof. Let g(z) = zf(z); then g(z) = g(−z) so g respects the equivalence rela-
tion z ∼ −z. The quotient of S
1
by this equivalence relation is again a copy of S
1
,
and the quotient map is identified with the squaring map s(z) = z
2
. Consequently,
g factors through s and thus g
∗
factors through s
∗
:
π
1
(S
1
) = Z
s∗
//
π
1
(S
1
) = Z
//
π
1
(S
1
) = Z .
But s
∗
is multiplication by 2, so the winding number of g is even. The winding
number of f is 1 less than that of g, so it is odd.
Theorem A27.9. (Borsuk-Ulam) For every continuous map f : S
2
→ R
2
there
exists x ∈ S
2
such that f(x) = f(−x).
Proof. Suppose not and let f be a counterexample. Then the map
g(x) = (f(x) −f(−x)[/|f(x) −f(−x)|
sends S
2
to S
1
and satisfies g(−x) = −g(x). Consider the restriction of g to the
equator on S
1
. This is an antipodal map S
1
→ S
1
, so it represents an odd multiple
of the generator; in particular it is not nullhomotopic. This is a contradiction since
it factors through the simply connected space S
2
.
91
Corollary A27.10. (Ham sandwich theorem) Any three bounded measurable sub-
sets of R
3
(e.g. ham, bread, cheese) can be simultaneously volume-bisected by a
common plane cut.
Proof. For each unit vector u ∈ S
2
let P
u
be the plane having u as normal vector
that volume-bisects one of the ingredients, say the bread. (The existence of such
a plane follows from the intermediate value theorem; in the general case one has
to maneuver a little to deal with the possible non-uniqueness of such a plane; I'm
sweeping this under the rug.) Define a map S
2
→ R
2
by sending u ∈ S
2
to the
vector (h(u), c(u)), where h(u) and c(u) are the fractions of the ham and cheese,
respectively, that lie on the u side of the plane P
u
. Then h(−u) = 1 − h(u)
and c(−u) = 1 − c(u). The Borsuk-Ulam theorem gives us a point such that
h(u) = h(−u) and c(u) = c(−u). This implies that h(u) = c(u) =
1
2
; thus P
u
perfectly divides the ham and the cheese (as well as the bread).
92
Lecture A1 Introduction to the Course This course is an introduction to the basic ideas of topology and metric space theory for first-year graduate students. Topology studies the idea of "continuity" in the most general possible context. As a separate subject, it takes its origin from the pioneering papers of Poincar´ at the end of the 19th century, and its e development is one of the central stories of 20th century pure mathematics. (The history by Dieudonn´ , A History of Algebraic and Differential Topology, 1900 e 1960, is available free to Penn State members online. — but that history begins roughly where this course leaves off.) But the idea of continuity is so central that topological methods crop up all over mathematics — in analysis of course (including its "computational" twin, numerical analysis); in geometry; in pure algebra; even in number theory. So, if you're a first-year grad student, you most likely need to take this course (and maybe even Math 528 as well). In this lecture we'll just describe the course protocol and prerequisites — we'll get started properly next time. There is a recommended course textbook — Introduction to Topology by Gamelin and Greene, which is available in a cheap Dover reprint for 15 bucks. However, more important than that are the online course materials, which are available in two ways: • Through the course website, which is • Through the course page on ANGEL, Penn State's course management system, at Registered students (i.e., those taking this course for a grade) will want to use the ANGEL site which will contain homework assignments, quizzes and supporting material, which all need to be taken on time in order to receive credit. The ANGEL page also contains the detailed course syllabus, which contains Universityrequired information about exam schedules, academic integrity requirements, office hours, and suchlike. Please be sure to read this information carefully. Detailed lecture notes for each lecture will be posted 24 hours in advance on these sites. To succeed in the course you need to read and study the notes before the lecture to which they refer. This is very important! Most lectures will contain one or more in-class exercises. You are expected to attempt these exercises before the next class session. Students may be called 2
on to present their solutions in class. You grade your own exercises using the ANGEL system, and this self-assessment will form a (small) component of your final in-class grade. Exercise A1.1. Here is your first in-class exercise. Find the corresponding assessment item on ANGEL (it's called "inclassAug23"), and enter your response ("I fully understand this"). You must complete this assignment before tomorrow evening, when it will expire. Now for some information about the prerequisites for the course, that is, things that I will assume that you know. There are four groups of these. 1. Number systems and their properties: specifically the natural numbers N, integers Z, rational numbers Q, real numbers R, and complex numbers C. Courses in analysis or foundations of mathematics often contain an account of procedures for "constructing" all or some of these, but we'll just take them as given. The most important of these number systems for us is the system R of real numbers. The basic properties of the reals are summarized by Proposition A1.2. R is an archimedean complete ordered field. What does this mean? (a) A field — basic properties of additional, multiplication, subtraction, division — commutative, associative, distributive laws. (b) ordered. There is an order relations < with the expected properties. In particular x2 0 for all x. (c) Archimedean — no 'infinite' or 'infinitesimal' real numbers. There is a unique (injective) homomorphism of rings Z → R taking 1 to 1. (Proof?) The archimedean property is that every element of R is smaller than (the image of) some element of Z. (d) complete — no 'holes', contrast Q. There are various ways of expressing completeness. The standard way is the Least Upper Bound Axiom. Let S be any nonempty set of real numbers. A number a ∈ R is an upper bound for S if, for all x ∈ S, x a. Let US be the set of all upper bounds for S. The least upper bound axiom says that, if S and US are both nonempty, then US has a least member. This is called the least upper bound for S and written sup S. 3
Exercise A1. . n = 1.3. be a sequence of closed intervals in R that are nested in the sense that [an+1 . Pick an interval [a1 . √ US has a least member. The corresponding statement with R replaced everywhere by Q would be false. bn+1 ] ⊆ [an . Then US = {a ∈ R : a > 0. bn ] and whose lengths bn − an tend to zero. . then 1 a subinterval [a2 . i.e. . b2 ] of length at most 2 that doesn't contain c2 . Exercise A1. c2 .. and so on. Apply the nested interval theorem. 2. bn ] contains one and only one real number c. namely sup S = 2. Use the previous exercise to show that the real numbers are uncountable. . . What can you say about the limit point c?
4
. .4. Show that the intersection n [an .5. bn ]. Let S = {x ∈ R : x2 < 2}. (Cantor's nested interval theorem) Let [an . cannot be listed as c1 .Example A1. Hint: Suppose such a listing is possible. a2 2}. b1 ] of length at most 1 that doesn't contain c1 .
Set-theoretic language such as unions (A ∪ B). subsets (A ⊆ B) and so on. 3. is the set of objects that belong to every member of the family F : in symbols x∈ F ⇔ ∀F ∈ F . a set whose members are subsets of some other set). Throughout pure mathematics. In topology it is often necessary to deal with infinite unions and intersections. one meets concepts that are built up of layers of quantifiers nested together in complicated ways. 5
.
Similarly the union of the family. written F ∈F F or just F . Quantified statements and their proofs. a function f : R → R is continuous at a given point x0 if ∀ > 0 ∃δ > 0 ∀x ∈ R |x − x0 | < δ ⇒ |f (x) − f (x0 )| < .) 2. and that you can understand how the form of the statement dictates the shape of its proof. but especially analysis and topology. For instance. Let F be a family of sets (that is. intersections (A ∩ B). written F ∈F F or just F . You could call a simple pattern like this a "proof skeleton" corresponding to the "for all" quantifier. We already met some quantifiers in the previous section: things like ∀ and ∃. x ∈ F. is the set of objects that belong to some member of the family F : in symbols x∈ F ⇔ ∃F ∈ F . For instance the statement above begins "∀ > 0" so its proof has to begin "Let > 0 be arbitrary".
De Morgan's Laws extend to this context: if F is a family of subsets of some set X. It is necessary that you can grasp what such a complicated statement says. then X\ F = X \F
F ∈F F ∈F
and the same with intersection and union reversed. x ∈ F. The intersection of the family.Lecture A2 Metric Spaces We continue with our list of prerequisites for the course (we discussed number 1 last time. followed by an argument that establishes the rest of the statement for a given arbitrary value of . complements (A\B).
6
. Definition A2. x) for all x.g. x2 ∈ X. We'll begin with the axioms for metric spaces. Exercise A2. dX ) and (Y. Definition A2. Give some examples of other proof skeletons corresponding to other parts of a statement. The vector space Rn can be made into a metric space by defining the distance between points x = (x1 . x ) = 0 if and only if x = x . . x2 ) for all x1 .
The motivating example is the metric d(z. Two metric spaces that are related by an isometry are equivalent from the point of view of metric space theory. x . x (symmetry).2. w) = |z − w| on C or R. x ) 0 for all x. .
(ii) d(x. There are many examples which realize the axioms. 4.) The same formula also makes Cn into a metric space. xn ) and y = (y1 .Exercise A2. ⇔ and so on.1. . y) = |x1 − y1 |2 + · · · + |xn − yn |2
1/2
.
(The proof that this does indeed satisfy the triangle inequality is a standard exercise. yn ) to be d(x.3. the isometries are just the congruences of Euclidean geometry. Let (X. x (triangle inequality). ∃. For the usual metric on the plane. The general theory consists of a collection of axioms. . . f (x2 )) = dX (x1 . Write down in symbols (as above) what it is for the function f to be not continuous at x0 . . such as those for a group or a vector space.4. x ) + d(x . e. Example A2. moreover. ⇒. d(x. and we develop a theory that applies to all of them. x ) = d(x . x ) d(x. An isometry from X to Y is a bijection f : X → Y such that dY (f (x1 ). x ) for all x. . x . dY ) be metric spaces. . A metric space is a set X equipped with a function d : X ×X → R (called a metric or distance function) such that: (i) d(x. Axiomatic method This is the "standard operating procedure" of modern mathematics. These are called Euclidean spaces. Here are some more examples of metric spaces. We understand specific examples by fitting them into a general theory. (iii) d(x. We will study topology from this point of view.5.
The discrete metric on X is defined by d(x. These are different metrics from the standard one (though. 7
. this metric structure is called the subspace metric on Y .8. Bell System Technical Journal 29(1950). One can also define metrics on Rn by d (x. y) = 0 (x = y) 1 (x = y)
Example A2. they are in a certain sense "equivalent". makes Y into a metric space. y) = #{i : 1 i n. Here is an infinite-dimensional example.7.
in other words. We can define a metric by d(f. x ) < } is contained in U . as we'll see later. 1] → C.) Example A2. which we think of as n-letter words in the alphabet A. the number of positions in which the two words differ. Let X be any set. xi = yi }. ) := {x ∈ X : d(x. y) = sup |x1 − y1 |. Define a distance on An by d(x. 1] of all continuous functions [0. Definition A2.) Example A2.Example A2. 147–160. Let A be a finite set (the alphabet) and consider the set An of n-tuples of elements of A. A subset U of a metric space X is open if for every x ∈ U there is > 0 such that the entire ball B(x. g) = sup{|f (t) − g(t)| : t ∈ [0. Error-detecting and error-correcting codes. . Let X be any metric space and let Y be a subset of X. y) = |x1 − y1 | + · · · + |xn − yn | and d (x. Convergence of a sequence of functions in this metric is called uniform convergence. restricted to Y . This Hamming distance was introduced in 1950 to give a technical foundation to the theory of error-correcting codes (Hamming.9. Consider the collection C[0. 1]}. .10. . .11. Example A2. |xn − yn | . Then the distance function on X.6.
so there are plenty of open sets. Many sets are neither open nor closed. and some may be both. A set F whose complement X \ F is open is called closed.
8
.The triangle inequality shows that every ball is open. Note carefully that 'closed' does not mean the same as 'not open'.
1. .4. which is to say that U is open. and the entire metric space X. . Consider the three metrics on Rn defined in examples A2. i ) ⊆ Ui . 2 ) is just the set {x}. Then B(x. ) ⊆ V .6. . Un } be a finite collection of open sets and let U = F = U1 ∩ · · · ∩ Un . Thus for any x ∈ U there exists > 0 such that B(x. .
Example A3. and thus B(x. . then for each i = 1. Each open ball in any one of these metrics contains an open ball (with the same center but possibly different radius) in any one of the other metrics. Lemma A3. The union of any collection of open sets is open.7). Proof. ) ⊆ Ui for all i.2. ) := {x ∈ X : d(x. x ) < } is contained in U . The empty set ∅ and the entire metric space X are closed.3.5 and A2.
1 Example A3. . Since V ⊆ U . Let = min{ 1 . in a discrete space. . Consequently. The union of a finite collection of closed sets is closed. That is what is meant by saying that they are equivalent. ) ⊆ U . Let F be a collection of open subsets of a metric space X and let U = F be the union of the family. The empty set ∅. Since V is open. ) ⊆ U . . the open ball B(x. Now let F = {U1 . we also have B(x. n } > 0. using de Morgan's laws. then there is some V ∈ F such that x ∈ V . we have Lemma A3. Equivalently. . are open. . 9
. there is > 0 such that B(x. these three metrics all have exactly the same open sets. every subset is open. which is to say that U is open. n there is i > 0 such that B(x.Lecture A3 Open and Closed Sets We finished last time with an important definition: A subset U of a metric space X is open if for every x ∈ U there is > 0 such that the entire ball B(x. The intersection of a finite collection of open sets is open. If x ∈ U . Consequently. If x ∈ U . In a discrete metric space (Example A2. Thus for any x ∈ U there exists > 0 such that B(x. ) ⊆ U . ) ⊆ U . The intersection of any collection of closed sets is closed. . .
3]. 1) ∩ Q ∪ (2. denoted S ◦ . is the set-theoretic difference ∂S = S \ S ◦ . Let X = R and let S = (0. and it is the smallest closed set that includes S. we have S◦ = U : U ⊆ S and U open in X . As we said earlier. 3}. Now show that for any subset S of X. denoted S. is the union of all the open subsets of X that are included in S.8. in particular. S is open if and only if it is equal to its own interior. Definition A3. Finally. Show that if A. In particular. then A◦ ⊆ B ◦ and A ⊆ B. S = [0. Then S ◦ = (2. Y can be considered as a metric space in its own right (with the metric that it inherits from X). B are subsets of X.
One more remark about open sets. 1] ∪ {2.9. 3). What is then the relation between the open subsets of the new space Y and the open subsets of the original space X?
10
.Let X be any metric space. 1] ∪ [2. Example A3. S itself is closed iff it is equal to its own closure.7. and ∂S = [0. The interior of S (in X). The boundary of S is the intersection of two closed sets (S and X \ S ◦ ). and it is therefore closed.
The interior of S is an open set (because it is the union of a family of open sets) and (from the definition) any open subset of X that is included in S is also included in S ◦ . is the intersection of all the closed subsets of X that include S. 3]. and A ⊆ B. The closure of S is a closed set. and let S be a subset of X.6.5. we have Definition A3. Thus. denoted ∂X. Definition A3. The boundary of S (in X). In symbols. the interior of S is just the biggest open subset of X that is included in X. Exercise A3. The closure of S (in X). Dually. Let X be a metric space and let Y be a subset of X. S
◦ ◦
= S
◦
.
Y a metric subspace (as above). But this means that B(x. ry ). Suppose that f is continuous and let U ⊆ Y be open. Theorem A3. Then f : X → Y is continuous iff. we may consider U = B(f (x). > 0 and suppose that f satisfies the condition in the theorem.10. an open set such that x ∈ f −1 (U ). f (x )) < whenever d(x. so open in X. Let V be open in Y . and U ∩Y =
y∈Y
BX (y. x ) < δ. Let x ∈ f −1 (U ). f (x ) ∈ B(f (x).
This proves one direction of the "if and only if" statement in the proposition. Proof. ). It is continuous if it is continuous at every x ∈ X. δ) ⊆ f −1 (U ). δ) ⊆ f −1 (U ). ry ) ⊆ V . Definition A3. Then a subset V ⊆ Y is open in Y iff it can be written V = U ∩ Y for some U ⊆ X open in X. for every open U ⊆ Y . Then for each y ∈ V there is ry > 0 such that BY (y. ). Our hypothesis tells us that f −1 (U ) is open. the inverse image f −1 (U ) := {x ∈ X : f (x) ∈ U } is open in X. This gives us continuity. δ). the other direction (which is easier) is an exercise. δ) then f (x ) ∈ B(f (x). ) ⊆ U . Conversely. A function f : X → Y between metric spaces is continuous at x ∈ X if for every > 0 there is δ > 0 such that d(f (x).12. By definition of 'continuous' there is δ > 0 such that if x ∈ B(x. We have shown that whenever x ∈ B(x. ) ⊆ U .Proposition A3. r) = BX (y. ry )
∩Y =
y∈Y
BY (y. But there is a very important alternative characterization in terms of open sets. r) ∩ Y. This is a union of open subsets of X. Let X be a metric space. The definition of continuity is translated in the natural way to the metric space context. Notice the following relationship between balls in Y and in X:if y ∈ Y then BY (y.11. In particular. let x ∈ X. Proof. Thus the set f −1 (U ) is open. ry ) = V. 11
. which means that there is a δ > 0 such that B(x. Let U = y∈Y BX (y. then f (x) ∈ U so by definition of 'open' there is > 0 such that B(f (x). Let X and Y be metric spaces.
as was shown by Brouwer and others at the beginning of the 20th century.4. We'll now study the properties of sequences. Its inverse is y → y(1 − y 2 )− 2 .) If there is a homeomorphism between X and Y . (Equivalently. A homeomorphism is a "topological equivalence" (it is easy to check that the relation of homeomorphism is an equivalence relation in the sense of algebra). The map t → (cos 2πt.1.5. d) is a continuous bijection. But in general nothing can be said about the behavior of the direct image f (S) of an open or a closed set S under a continuous map. Then the identity map (X. A sequence in a metric space X is a mapping from the natural numbers N to X: we'll follow the usual abbreviation of referring to "the sequence (xn )" rather than "the sequence which maps the natural number n to xn ∈ X.Lecture A4 Continuity and sequences In the previous lecture we saw that a map f : X → Y between metric spaces is continuous iff f −1 (U ) is open (in X) whenever U is open (in Y )." 12
1
. is indeed "no". f is a continuous map with a continuous inverse. Example A4. Example A4. Are two concretely given spaces homeomorphic or not? Example A4. as expected. but it is usually not a homeomorphism. Remark A4.6. then we say that these spaces are homeomorphic. 1). (Note that there are continuous maps of Rm onto Rn even if m < n — "space-filling curves". A fundamental question in topology is to devise processes for answering the question. A map f : X → Y between metric spaces is called a homeomorphism if it is a bijection and both f and f −1 are continuous. It is equivalent to say that f −1 (F ) is closed in X whenever F is closed in Y . we consider homeomorphic spaces to be "essentially the same". Definition A4. 1] onto the unit circle in R2 .2. Are Rn and Rm homeomorphic if m = n? This and related questions were problematical in the early days of topology. sin 2πt) is a continuous bijection from the half-open interval (0. The map x → x(1 + x2 )− 2 is a homeomorphism from R to the 1 open interval 9 − 1. Example A4.3. In topology. Let (X. d ) → (X. but it is not a homeomorphism. d ) be the same set equipped with the discrete metric.) But the answer. d) be a metric space and let (X.
but d(f (x).
Proposition A4. Suppose that f is not continuous at x. By definition of continuity there is δ > 0 such that d(x. f (x )) < . x ) < δ and d(f (x). 0} (with the metric it inherits as a subspace of R). Let X be a metric space. Thus d(f (x). Then there is some > 0 such that B(x. ) ⊆ U . hence f (xn ) does not converge to f (x).Definition A4. When we come to consider more general topological spaces. d(x. Let T be the metric space {1. Suppose that f is continuous at x and let > 0 be given. 3 . f (xn )) for all n. Let xn be a value of x corresponding to δ = 1/n. and in that case f (0) = limn→∞ xn . At the same time.8. the sequence (f (xn )) converges to f (x). whenever (xn ) is a sequence converging to x. Suppose that x is a limit point of A. Then xn → x. x ) < δ implies d(f (x). A ⊆ X. A point x ∈ X is a limit point of A if it is the limit of a sequence of distinct points of A. and then sequences will be of less use. Definition A4. We say (xn ) converges to x ∈ X if for every > 0 there is an integer N such that d(x.11. Then there is a sequence (xn ) of distinct points of A. f (x ) . it shows how the usefulness of these "probes" depends on the countable local structure of a metric space — specifically. This result illustrates how sequences can be used to probe the topological properties of metric spaces. 2 .9. A function f : X → Y between metric spaces is continuous at x if and only if. x is a limit point of A if and only if every open set containing x also contains infinitely many points of A. Remark A4. xn ) < whenever n > N . Then there is > 0 such that for all δ > 0 there is x with d(x. Let U be an open set containing x. there 13
. .12.10. Show that a sequence (xn ) in the metric space X is convergent if and only if there exists a continuous function f : T → X such that f (1/n) = xn . converging to x. Proof. Lemma A4. We write then x = limn→∞ xn . Let (xn ) be a sequence in the metric space X. on the fact that any ball around x ∈ X contains one of the countably many balls B(x. 1/n).
1 1 Exercise A4. Proof. By definition of convergence there is N such that for all n > N . . f (xn )) < and f (xn ) converges to f (x). xn ) < δ. this "countable local structure" may not be available. . .7. By definition of convergence.
and. A point x belongs to the closure of A iff it is not in the interior of X \A. all of which are members of A. 1) ∩ A. ) must meet A. Must every closed ball be the closure of the open ball of the same center and radius?
14
. r) := {x ∈ X : d(x. and. let xn be any point in B(x.16. xn−1 have been defined. U contains infinitely many of the {xn }. ) ∩ A = {a}. A closed ball in a metric space is a set ¯ B(x. Proof.
Show that a closed ball is closed. . . In particular. x ) r}. d(xn−1 . . . ) ⊆ U for all n > N . then. 1) ∩ A. .13. Proposition A4. so x is a limit point of A. A closed subset of a metric space is perfect if it has no isolated points. 1/n) ∩ A that is distinct from the (finitely many!) points x1 . By definition of "interior". A is closed if and only if it contains all its limit points. xn−1 . . let xn be any point in B(x. . Show that a ∈ A is isolated if and only if there is > 0 such that B(a. Define a sequence (xn ) inductively as follows: x1 is any point in B(x. this is the same as to say that each ball B(x. Definition A4. An isolated point of A is a point of A that is not a limit point of A. xn−1 have been defined. x). . Conversely suppose that every open set containing x also contains infinitely many points of A. δn ) ∩ A where δn = min{d(x1 . x)}. Then (xn ) is a sequence of distinct points of A and it converges to x. . The closure of a subset A is the union of A and the set of all its limit points. Define a sequence (xn ) inductively as follows: x1 is any point in B(x. assuming that x1 . In particular. . any limit point of A must belong to A. suppose that x ∈ A \ A. assuming that x1 .is N > 0 such that xn ∈ B(x. Exercise A4. . .15. By the previous lemma. . Exercise A4.14. Conversely. . . Then (xn ) is a sequence of distinct points of A and it converges to x. .
m > N . Then there must be some n such that xm < xn for all m > n. Continuing in this way by induction we get a monotonic decreasing subsequence xn1 . the subsequence xn1 +1 . . if a sequence in X is a map N → X. It is a standard consequence of completeness that every bounded. (Formally speaking. Let (xn ) be a sequence of real numbers. Proof. Every closed. i. monotonic sequence of real numbers is convergent. .. a subsequence of the given sequence is the result of composing it with a strictly monotonic map N → N. we could build a monotonic increasing subsequence by induction.) Lemma A5. xn1 +2 . .e. xm ) < .4.2. then the whole sequence is in fact convergent. Moreover. so by the same argument there is n2 > n1 such that xm < xn2 for all m > n2 . for all n. (Otherwise.3. if a Cauchy sequence has a convergent subsequence. R is (Cauchy) complete. (A subset is bounded if it is contained in some ball.Lecture A5 Compactness I'll take for granted the notion of subsequence of a sequence of points in a matric space. . . These facts combine with the Bolzano-Weierstrass theorem to show that every Cauchy sequence of real numbers converges: i. Proposition A5.) But now take this n and call it n1 . A metric space X is (sequentially) compact iff every sequence of points of X has a subsequence that converges in X. Apply the previous argument to the "tail" of the original sequence beginning at n1 . A Cauchy sequence is necessarily bounded. . A Cauchy sequence in a metric space is a sequence (xn ) with the following property: for every > 0 there exists N > 0 such that. d(xn . bounded subset of Rn or Cn is sequentially compact. This also has no monotonic increasing subsequence. xn2 . Definition A5.) 15
. Suppose that it has no monotonic increasing subsequence. A metric space is called complete if every Cauchy sequence in it converges.e..1.. Thus we get the Bolzano-Weierstrass theorem: every bounded sequence of real numbers has a convergent subsequence. Definition A5. Every sequence of real numbers has a monotonic subsequence.
The triangle inequality shows that d(an . What has this got to do with the familiar calculus principle that 'a continuous function on a closed bounded interval is itself bounded and attains its bounds'? Despite the above evidence.6. this follows from the fact (easily proved) that a sequence of points in Rn is convergent if and only if each of its coordinate sequences is convergent. it's not in general true that 'closed and bounded' equals 'compact'. by compactness. and A is compact (in its own right). say to x. a0 ) > 1 + d(an−1 . since U is a cover. There is n > 2/ such that d(xn . Show that if X is compact then Y is compact also. 1/n) is contained in no member of U . Then one can construct by induction a sequence (an ) in A such that d(an . Proof. ) ⊆ U which is a contradiction. then A is bounded and closed in X. 16
. 1/n) ⊆ B(xn . Suppose that U does not have a Lebesgue number. Proposition A5. Definition A5. am ) > 1 for n = m.7. (an ) has a subsequence converging in A. But then B(xn . Let x ∈ X be a limit point of A. the sequence (xn ) has a subsequence that converges. a0 ) for n 1. ) ⊆ U . x) < /2. Then for every n there is xn ∈ X such that B(xn . An open cover U for X is a collection (finite or infinite) of open sets whose union is all of X.5. Thus there is > 0 such that B(x. Suppose that A is not bounded. If A is a subset of any metric space X.Proof. Proof. Now x belongs to some member U of U . If X is compact. (Counterexample?) We shall analyze this in detail. Thus x ∈ A. For R this is just the Bolzano-Weierstrass theorem.8. Exercise A5. Let f : X → Y be continuous and surjective. Then there is a sequence (an ) in A converging to x. Let X be a metric space. Theorem A5. (Lebesgue) Every open cover of a (sequentially) compact metric space has a Lebesgue number. Clearly (an ) has no convergent subsequence. But. For Rn . and A is closed. A Lebesgue number for U is a number δ > 0 such that every open ball of radius δ is a subset of some member of U . /2) ⊆ B(x.
Since X is totally bounded it has a finite cover by δ-balls. δ). The B(x. (c) X is covering compact. Then (xn ) is n n n a Cauchy subsequence of the original sequence. When it does terminate. (b) For every δ > 0 there is a finite cover of X by balls of radius δ. (b) implies (a): Notice the following implication of (b): given any δ > 0. xm ) δ when m < n. any sequence in X has a subsequence all of whose members are separated by at most δ (call this a 'δ-close subsequence'). Proof. In the other direction. (b) X is complete and totally bounded. / 17
. . In this case we say X is totally bounded (some writers say precompact).10. (a) Every sequence in X has a Cauchy subsequence. and then inductively choose xn ∈ X such that d(xn . n = 1. Let U be an open cover of X. it does so because the balls B(xn . The process must terminate because if it didn't it would produce a sequence with no Cauchy subsequence. x ) for / all but finitely many n.9. with xN say. The following conditions on a metric space X are equivalent: (a) X is sequentially compact. Let (xn ) be any sequence in X. Let δ > 0 be a Lebesgue number for U (which exists because of Theorem A5. Suppose (a). . so long as this is possible. . . Picking a finite subcover we obtain the contradiction that xn ∈ X for all but finitely many n. and so on by induction.8). A metric space X is said to be (covering) compact if every open cover of X has a finite subcover. But each such ball is a subset of a member of U .Proposition A5. Let (xn ) be a sequence without convergent subsequence. so U has a finite subcover. Proof. (a) implies (b): Choose x1 ∈ X. n let (x2 ) be a 2−2 -close subsequence of (x1 ). The following conditions on a metric space X are equivalent. cover X. Proposition A5. It is easy to see that (a) and (b) are equivalent. Then for each x ∈ X there is some x such that xn ∈ B(x. x ) form a cover of X. suppose (c). Let (x1 ) be a 2−1 -close subsequence of (xn ). N .
Hint: use Theorem A5.Remark A5. f (x )) < whenever d(x.) Show that if X is compact. every continuous f is uniformly continuous.8. (The extra information beyond ordinary continuity is that δ does not depend on x. Exercise A5. and 'covering compact' is usually the most appropriate one.12. For general topological spaces. x ) < δ.11. the notions 'covering compact' and 'sequentially compact' are not equivalent. A map f : X → Y between metric spaces is uniformly continuous if for each > 0 there is δ > 0 such that d(f (x).
18
.
We show that f is a continuous function on X. Proposition A6.Lecture A6 Compactness and Completeness (The lecture will begin with a review of the basic results about compactness.1.2. Let X be a compact metric space. identify a candidate for its limit. Fix x ∈ X and let > 0 be given. Take n = N and let n → ∞ to find that |fN (x) − f (x)| /3 for all x. from the previous section). Then. Proof. It follows that.10. Let (fn ) be a Cauchy sequence in C(X). Finally to show that fn → f in C(X) we must prove that for every > 0 there is N such that |fn (x) − f (x)| < for all x whenever n > N . Then C(X) denotes the space of all continuous functions X → C. whenever |x − x | < δ. fn (x) is a Cauchy sequence in C. But in fact we already proved this in the previous paragraph. Since C is complete this sequence converges. Definition A6. 19
. (Compare Example A2. Denote its limit by f (x). fN is continuous at x so there is δ > 0 such that |fN (x) − fN (x )| < /3 whenever |x − x | < δ. and show that the sequence approaches the candidate limit in the metric of the space in question. the metric spaces C(X) and CR (X) are complete. Because (fn ) is Cauchy there is N such that for n . g) = sup{|f (x) − g(x)| : x ∈ X}. Now. This shows that f is continuous. Most completeness proofs proceed in the same three-stage way: Given a Cauchy sequence.) Compactness of X ensures that the supremum exists (why)? We can also consider the space CR (X) of continuous real-valued functions. for each x ∈ X. It is a metric space equipped with the metric d(f. For a compact X. n N we have |fn (x) − fn (x)| < /3 for all x. show that the candidate is in the space in question. |f (x) − f (x )| |f (x) − fN (x)| + |fN (x) − fN (x )| + |fN (x ) − f (x )| < .
"Closed and bounded" is not enough.
20
. The answer is given by the Ascoli-Arzela theorem.It is an interesting and important question what are the compact subsets of the complete space C(X).
Since f is continuous. Now let V be the metric space with the same points as U but with the metric dV (x. Let f : U → R be the function defined by f (x) = 1 . It is easy to see that this is indeed a metric. A uniform homeomorphism between metric spaces is a homeomorphism f : X → Y such that both f and f −1 are uniformly continuous. The space (0. x ) < δ. x ) = d(x.Lecture A7 Applications of completeness Recall that a metric space X is said to be complete if every Cauchy sequence in X converges. Proof. Exercise A7. i. Show that if X is complete and there is a uniform homeomorphism X → Y . x ) < δ1 implies |f (x) − f (x )| < /2. dV (x. This proves that completeness is not a topological property. there is δ1 > 0 such that d(x. Suppose x ∈ U and let > 0 be given. Finally I claim that the identity map U → V is a homeomorphism. Then if d(x. Proposition A7. Let X be a complete metric space and let U ⊆ X be an open set. Then U is homeomorphic to a complete metric space (it is topologically complete). y) : y ∈ X \ U }
This is a well-defined.1. /2)}. 2 So the identity map U → V is continuous.. x ) < 21
.2. in addition. continuous function with the property that f (xn ) → ∞ whenever xn is a sequence in U that converges to some x ∈ X \ U . and the continuity of the inverse is easy. Put δ = min{δ1 . the values f (xn ) are bounded (so the limit x in fact belongs to V ).e. inf{d(x. Moreover. 1) is not complete. x ) + |f (x) − f (x )|. it is complete: if (xn ) is a Cauchy sequence for the metric dV . we have + |f (x) − f (x )| < . whereas the homeomorphic space R is complete. is not preserved under homeomorphism. then Y is complete. say to x ∈ X) and. then it is also a Cauchy sequence for the metric d (so it converges in X.
(Banach) A contraction on a complete metric space has a unique fixed point (a point x such that f (x) = x). the topology on the irrational numbers can be given by a complete metric. which converges to a point x. e somme). Theorem A7. f (x )) ad(x. Proof. But the same is not true for the rational numbers. Another important source of existence theorems. A fixed point is unique because if x. x are two such then d(x. A countable union of closed sets is called a Fσ -set (French: ferm´ . Banach's fixed point theorem is one of the most important sources of existence theorems in analysis. 22
. x ) = 0. then f has a unique fixed point. A subset of a metric space is called a Gδ -set (German: GebeitDurschnitt) if it is the intersection of countably many open sets. Note that a contraction must be continuous. Prove this.4. is the Baire category theorem. xn+1 ) an r and so d(xn . x2 = f (x1 ) and so on. f (x )) ad(x. xn+k ) (an + · · · + an+k−1 )r an r 1−a
which tends to 0 as n → ∞. it can be shown that any Gδ subset of a complete metric space is topologically complete. A mapping f : X → X is a (strict) contraction if there is a constant a < 1 such that d(f (x). start with any x0 ∈ X and define x1 = f (x0 ). x ∈ X. Exercise A7. x ). Let X be a metric space.7. This is the only known example of dual mathematical concepts being described using dual languages. which implies d(x. as will follow from the Baire category theorem. x1 ) = r then d(xn . Banach's fixed point theorem can be extended as follows: if f : X → X is a map and some power f N = f ◦ · · · ◦ f is a strict contraction. In particular.3.Remark A7. the irrational numbers form a Gδ subset of R. To prove existence. Thus (xn ) is a Cauchy sequence. For example. x ) = d(f (x). Definition A7. We have f (x) = lim f (xn ) = lim xn+1 = x so x is a fixed point. If d(x0 . Generalizing the above construction. x ) for all x. which also uses completeness.5.6. Remark A7.
any complete. B(xn . Let x ∈ X and let > 0.12. Let {Un }. 1].) Theorem A7. A set is of first category if it is the countable union of nowhere dense sets. Baire's theorem then says that in a complete metric space the complement of a set of first category (sometimes called a residual set) is dense. rj ) for all j n. 1] of continuous realvalued functions on [0.Definition A7. (It is often important that R has a countable dense subset. rj ) for every j. Remark A7. Consider the complete metric space CR [0. Proof. b ∈ [0.10. Remark A7. Since xn ∈ B(xj .9.
Then (xn ) is a Cauchy sequence. that is. b] is nowhere dense in CR [0. Using Baire's theorem. monotonic on no subinterval of [0.. Since U2 is dense there is x2 ∈ U2 ∩ B(x1 . say converging to a point x . More generally. A space with this property is called separable. 1]. (Baire) In a complete metric space. For a.11. the point x belongs to B(xj . For in a perfect metric space. be dense and open. rn < rn−1 /2.8. and hence to Un . n = 1.14) metric space is uncountable. Notice that this gives us another proof that R is uncountable (compare Exercise A1. without loss of generality take r1 < /4. Since x and were arbitrary. rn ) ⊆ Un .
23
. Moreover. 1] that are nowhere monotonic. 1] show that the set of functions f which are nondecreasing on the interval [a. 2. d(x. /2). deduce that there exist functions f ∈ CR [0. The rational numbers Q are dense in R. the complement of any point is dense. The name 'category theorem' comes from the following (traditional) terminology. rn−1 ). . 1].13. Exercise A7. perfect (Definition A4. x ) < . Un is dense. A set A is nowhere dense if the complement of its closure is dense. Choose x1 ∈ U1 ∩ B(x. r1 ) ⊆ U1 . A subset of a metric space is dense if its closure is the whole space Example A7. Proceed inductively in this way choosing xn and rn such that xn ∈ Un ∩ B(xn−1 . There is r1 such that B(x1 .5). For the complement of a point in R is a dense open set. . r1 ). the intersection of countably many dense open sets is dense. .
1} k v
25
.Definition A8. If T is continuous then there is some δ > 0 such that u < δ implies T u < 1. Proof. W ) of bounded linear maps from V to W is a normed vector space. Exercise A8. Show that.e.6. Proposition A8.5. the collection L(V. with the above norm. Definition A8. i. Show that the norm of linear maps is submultiplicative under composition. If T is bounded then T u − T v k u − v so T is continuous. S ◦ T S T . A linear mapping T : V → W between normed vector spaces is continuous if and only if there is a constant k such that Tv (one then says that T is bounded).7. that is the quantity sup{ T v : v is called the norm of T and denoted T .8. it is called a Banach space. Take k = 1/δ. If a normed vector space is complete in its metric. The best constant k in the above proposition. Exercise A8. and that it is a Banach space if W is a Banach space.9.
let x ∈ V . T is unique if it exists. Example A9. we need only suppose f is defined on some open subset U of V . Remark: The existence of the partial derivatives does not by itself imply differentiability in the sense of our definition above. Exercise A9. If V = W = R then every linear map V → W is multiplication by a scalar. Similarly for expressions like "f (h) = g(h) + o(h)". the space of (bounded) linear maps L(V . Now let f : V → W be a continuous (need not be linear) map between normed vector spaces. Show that T (h) = o( h ) if and only if T = 0. we identify L(V. Definition A9. If V = R and W is arbitrary. With above notation.4. By the exercise. W ) = R. i.3. This just shows that our definition is right and partials are wrong! Example A9. W ) is the space of n × m matrices. Let f be a function defined on some ball B(0.6. If V = Rm and W = Rn then L(V . h→0 h 26
.e. We shall write "f (h) = o( h )" to mean that the limit limh→0 h −1 f (h) exists and equals zero. Example A9. Under this identification the derivative of f is given by the usual formula Df (x) = lim f (x + h) − f (x) . Suppose that T : V → W is a linear map. The matrix entries of the derivative of f are the partial derivatives of the components of f as defined in Calculus III. Under this identification our definition of the derivative corresponds to the usual one from Calculus I. Normed spaces provide a systematic way to express this. We say that f is differentiable at x if there is a bounded linear map T : V → W such that f (x + h) = f (x) + T · h + o( h ). Notation A9. ) in a normed space V and having values in another normed space W . In fact. It is called the derivative of f at x and written Df (x).2. W ) can be identified with W itself.5.1.Lecture A9 Differentiation in Normed Spaces The basic idea of differentiation is that of 'best linear approximation'.
Therefore.Proof. is also complete. which is a solution to the differential equation with the specified initial condition. b] of continuous functions on [a.
t ∈ [a. you will need to take it on faith that the same facts are true for functions having values in a normed vector space V .
s
(I f )(s) = v0 +
a
F (t. complete) then the space of continuous functions [a.
t ∈ [a. g ∈ C[a. b] be defined by the right hand side of the display above. we shall use the fact that if V is a Banach space (that is. Let I : C[a. b] is a solution of the differential equation if and only if it is a solution of the equivalent integral equation
s
f (s) = v0 +
a
F (t. that is. b]. We need to use some facts about integration for V -valued functions. b] → V . in a first course on integration these facts are usually proved only for real or complex valued functions. b] then the integrands in I f and I g differ by C f − g at most. Moreover. We will use the fundamental theorem of calculus: a function f ∈ C[a.
29
. This is a fact that we have already proved for real or complex valued functions. and therefore If −Ig C(b − a) f − g
by standard estimates on integrals.
If f. f (t)) dt. the map I is a strict contraction and Banach's fixed point theorem provides a unique solution to I f = f . f (t)) dt. b] → C[a. b]. equipped with the sup norm. Finally. b] (with values in V . where C is the Lipschitz constant.6) to the space C[a.
We will show that a solution to this equation exists (for b sufficiently close to a) by applying Banach's contraction mapping theorem (A7. and the proof in the general case is exactly the same. if (b − a) < C −1 . It isn't the job of this course to teach you about integration theory — for that see Math 501 — so I will just state the various facts that are needed. if you like).
For y ∈ V consider the map φy : V → V.6). V ) to L(V . V ). V ) is invertible if it has an inverse which is a bounded linear map from V to V . I +S is invertible. We have shown that the identity is an interior point of the set of invertibles and that the inverse is continuous there. This fixed point is an x such that (I + S)x = y. and that its derivative is Di(T ) · H = −T −1 HT −1 . Without loss of generality V = V . V ). By the triangle inequality. V be normed vector spaces. Proposition A10. Exercise A10. Say T ∈ L(V. so it has a unique fixed point x (Theorem A7. V ) is then a normed vector space also. so x (1 − S )−1 y
and the map y → x is bounded. y (1 − S ) x . Then the invertibles form an open subset of L(V.1. Show that the mapping i : T → T −1 is differentiable (where defined) on W .Lecture A10 The inverse function theorem Let V. V ).
Since S < 1 this map is a strict contraction. V be Banach spaces. Moreover y − (I + S)−1 y = S(1 + S)−1 y S (1 − S )−1 y
which shows that the map S → (I + S)−1 is continuous at S = 0. Let V. Proof.2. so the same results apply to any point of the set of invertibles. v → y − Sv. Suppose S < 1. The space L(V. Let V be a normed space and let W be the normed space L(V. left multiplication by a fixed invertible is a homeomorphism from the set of invertibles to itself. Look first at a nbhd of the invertible operator I. However.
30
. We have shown that if S < 1. and the inverse operation T → T −1 is continuous (on this subset) from L(V.
defined near x ∈ V . Theorem A10.Definition A10.3. W ) is invertible. Thus y0 ∈ f (U ) and we want to prove that there is some > 0 such that B(y0 .6. Suppose moreover that V. It is an application of Banach's fixed point theorem A7. Choose 1 = 2 A −1 δ. We have Dφy (z) = I − A ◦ Df (z) = A ◦ Df (x) − Df (z) . δ) and thus has a (unique) fixed point there. Since Df is continuous there is r > 0 such that if z ∈ U = B(x. We will construct the inverse function by looking at the fixed points of a suitable map. and if we know for some reason that φy maps F to F . W are Banach spaces. then φy is actually a contraction of this complete metric space and so has a unique fixed point. From the mean value theorem we conclude that if F is a closed subset of U . The inverse function theorem says that under suitable conditions. if Df (x) is invertible then f is 'locally invertible' near x. If z ∈ B(z0 . δ) is contained in U . and suppose that f is continuously differentiable and that Df (x) ∈ L(V. for these y. Choose δ > 0 such that the closed ball B(z0 . which is an open set containing f (x). Then there is an open set U containing x such that f is a bijection of U onto f (U ). r) then 1 Dφy (z) < 2 . For y ∈ W define a map φy : V → V by φy (z) = z + A · (y − f (z)). Let f be as above. A fixed point of φy is a solution to f (z) = y. ) then we may compute φy (z) − z0 = φy (z) − φy0 (z0 ) φy (z) − φy (z0 ) + φy (z0 ) − φy0 (z0 ) 1 z − z0 + A(y − y0 ) 2
1 δ 2
1 + 2δ = δ
using the mean value theorem A9. the map φy is a contraction of the complete metric space B(z0 . and such that its inverse g : f (U ) → U is also continuously differentiable. this will show that f (U ) is open.8.4. ) ⊆ f (U ). We have shown that each y0 ∈ f (U ) has an -neighborhood contained 31
. Let A = Df (x)−1 . A map f from an open subset of a normed space V to a normed space W is continuously differentiable or C 1 if it is differentiable (everywhere) and the map x → Df (x) is continuous. Proof. Fix z0 ∈ U and let y0 = f (z0 ). We conclude that. δ) and y ∈ B(y0 .
This topology does not arise from a metric as soon as X has more than one point. then the union of T . (iii) The sets ∅ and X are members of T . The cofinite topology on X consists of ∅ together with all the cofinite subsets of X (a subset is cofinite if its complement is finite).2. . A topology on a set X is a family T of subsets of X with the following three properties: (i) If Uα is any family of members of T . Example A11. so that the closed sets in the cofinite topology are just the finite sets together with the whole space. For example. ∅ and X.
Ui
A set equipped with a topology is called a topological space. thus. The indiscrete topology on X has just two open sets. This topology arises from the discrete metric. there are topologies that do not arise from any metric. (Why not?) Example A11. This motivates a further abstraction which removes the numerical notion of metric entirely. Un is a finite family of members of T then the intersection is also a member of T . Notice that this topology has many "fewer" open (or closed) sets than the topologies with which we are familiar.3.Lecture A11 Topological Spaces We have been emphasizing that "up to homeomorphism" properties of metric spaces depend only on their open sets. . It is natural to say that a closed set is one whose complement is open.
α
Uα is also a member
n i=1
(ii) If U1 . . Example A11. any two nonempty open sets have a nonempty intersection. every metric space carries a topology (called its metric topology). . 33
. Notice that the open subsets of a metric space automatically satisfy (i)–(iii). The discrete topology on X has T = P(X): every subset is open. Different metrics can however give rise to the same topology (as we have already seen). in the cofinite topology on an infinite set.4. moreover. A similar but more sophisticated example is the next one.1. Definition A11. The members of T are called open subsets of the topological space X.
34
. Thus the topology generated by a basis is just the collection of all unions of families of members of the basis. Then T is also a topology on X.7. for every point x ∈ X. . . (The empty set and the whole space X are included among these sets by convention: we take the intersection of an empty family to be X and the union of an empty family to be ∅. The intersection T is then the smallest or "weakest" topology containing F .8.6. In other words.) It is a simple exercise (for an algebraist!) to show that any intersection of varieties is a variety. Exercise A11. It is called the topology generated by F (you'll also see F called a subbasis for T ). . the metric balls form a basis for the topology of a metric space.Example A11. . (The Hilbert basis theorem says that S can always be taken to be finite. . xn ) ∈ S ⇔ p(x1 . Let T be the family of all topologies T such that T ⊇ F (there always is one such topology. Let T be a family of topologies on the set X.5. Example A11. let F be any family of subsets of X at all. In particular.
in other words.) Moreover. the complex numbers) and let A = K n . . the discrete topology). . . . . A subset V of A is called an affine algebraic variety if there is a set S ⊆ K[X1 . The following language is conventional in this context: the family B of subsets of X is called a basis if. xn ) = 0∀p ∈ S. Show that the topology on X generated by F can be concretely expressed as follows: a set U is open in the generated topology if and only if it can be written as a union Uα of some family of sets Uα . (For example. a variety is the set of common zeroes of a collection of polynomials. Let K be an algebraically closed field (for example. . the topology generated by a subbasis F is the topology generated by the associated basis consisting of finite intersections of members of F . the varieties can be considered as the closed sets of a topology on A. where each Uα is the intersection of some finite family (depending on α) of members of F . and a finite union of varieties is a variety. there is some member of B containing x and every finite intersection of members of B containing x includes some member of B containing x. . Xn ] of polynomials in n variables such that (x1 . called the Zariski topology.) Remark A11.
A space X is compact if every open cover of X has a finite subcover. Then the family of subsets of Y given by {Y ∩ U : U ∈ T } defines a topology on Y .5. and that these two sets are equal if S itself is a subset of Y . and boundary of a subset just as in the metric space context. Definition A12.Lecture A12 Compactness and Hausdorffness
Definition A12. Let S be a subset of X.3. Let X be a topological space. and let Y ⊆ X. Show that f is continuous as a map from X to Y if and only if it is continuous as a map from X onto f (X). Let X be a topological space and Y a subspace of X.1. we can define the interior. Just as in metric spaces we can also define a continuous function f : X → Y (X and Y being topological spaces) as one which has the property that. Exercise A12. for every open subset U of Y .10. with topology T . closure. Let f : X → Y be a map of spaces (from now on. Thus. where f (X) ⊆ Y is equipped with its subspace topology.4. but that they need not be equal in general. Definition A12. (Compare Proposition A3. the inverse image f −1 (U ) is open in X. Exercise A12. "space" will mean a topological space.6. The composite of continuous functions is continuous. with closure S. equipped with the subspace topology.) Any metric space notion which is defined simply in terms of open sets can be taken over without change to the more general context of topological spaces.2. for example. Show that the closure of S ∩ Y in the subspace topology of Y is a subset of S ∩ Y . unless otherwise specified). 35
. The proof of the following lemma is then obvious: Lemma A12. A homeomorphism is a continuous bijection whose inverse is also continuous. It is called the relative topology or subspace topology induced by the given topology on X.
and U ∩ V = ∅. That is.8. . there exist open sets U.7. together possibly with X \ Y . form an open cover of X. Let f : X → Y be a continuous map.This is the same as the definition of "covering compact" that was given in Lecture 5. . let X be any topological space. 36
. Let X be a compact space and let Y be a closed subset of X. V with x ∈ U . . We must show that every open cover of Y (in the relative topology) has a finite subcover. A discrete topological space is compact if and only if it is finite. then f (X) ⊆ Y is also compact. are then open in X and they form a cover since for each x ∈ X. What is missing is the Hausdorff property.10. Un cover Y . We won't make use of the notion "sequentially compact" outside the realm of metric spaces. Proposition A12. Un cover Y and we have shown that U has a finite subcover as required. Proposition A12. . In metric spaces. . Indeed. A topological space X is Hausdorff if. . If X is compact. Definition A12. Any finite topological space is compact. But we saw several examples in the last lecture of topological spaces in which one-point subsets are not closed. U ∈ U . together with the open set X \ Y . Each open set U ∈ U is (by definition of the relative topology) of the form VU ∩ Y . Then U1 . so x ∈ f −1 (Ui ) for some i and therefore y ∈ Ui . This need not be true in general topological spaces. Proof. there is a finite subcover: there is a finite subset U1 . where VU is some open subset of X. the image f (x) belongs to some member of U .7. . y ∈ V . . . Example A12. Un of U such that the sets f −1 (Ui ) cover X. . From Example A12. By Exercise A12. The union of all the sets VU includes Y . any one-point subset — must be compact in its subspace topology. Since X is compact. Let U be an open cover of Y . The sets f −1 (U ). the U1 . Then Y is also compact (in its relative topology).5). . compact subsets are necessarily closed (Proposition A5.9. the open sets VU .4 it suffices to consider the case where f (X) = Y . . there is x ∈ X such that f (x) = y. . Proof. . VUn . . y ∈ X. But for every y ∈ Y . comprising VU1 . Let U be such a cover. . any finite subset of X — in particular. therefore. By hypothesis this cover has a finite subcover. for any two distinct points x.
37
. . Let U x = Uy1 ∩ · · · ∩ Uyn .11).8). so it is open. Proposition A12. But K is compact (Proposition A12. The sets Vy ∩ Y form an open cover of Y (in the subspace topology) which therefore has a finite subcover. therefore f (K) is compact (Proposition A12. a one-point subset of a Hausdorff space is closed. Then f −1 is continuous (i. To show that f −1 is continuous it is enough to show that for any closed subset K of X.11. For each y ∈ Y there exist disjoint open sets Uy containing x and Vy containing y. ) and V = B(y. The set X \ Y itself is the union of all these open sets U x . contains x.e. y)/2). the image f (K) is closed in Y . ) for 0 < < d(x. Proof. therefore f (K) is closed (Proposition A12.Any metric space is Hausdorff (take U = B(x. The Hausdorff property ensures a large supply of open sets. . where X is a compact space and Y is a Hausdorff space. Let X be a Hausdorff space and let Y ⊆ X be compact (in the subspace topology). Then U x is an open set. Proof.9). . f is a homeomorphism). say Vy1 ∩ Y. Then Y is a closed subset of X.. Vyn ∩ Y . We will show that X \ Y is open. Let f : X → Y be a continuous bijection. . In particular.12. and does not meet Y . Corollary A12. Let x ∈ X \ Y . by the Hausdorff property.
0)) are open (by continuity). It is easy to see that dK is a continuous function and that dK (x) = 0 iff x ∈ K.
2 2
Example A13. k) : k ∈ K}. Proposition A13. T2 1 . The T stands for the German Trennungsaxiom (separation axiom). and maybe some others that I forgot. To see this we need to make use of the distance function from a closed set. A Hausdorff topological space X is regular if disjoint subsets A and B of X can be separated by open sets whenever A is a point and B is a closed set. disjoint. and B a closed subset not containing p.1. ∞)) and V = f −1 ((−∞. d) be a metric space and let K be a closed subset. then normality. and indeed normal.3. B ⊆ V . The sets U = f −1 ((0. One says that A and B are separated by open sets if there exist disjoint open sets U and V such that A ⊆ U . A regular Hausdorff space is also called a T3 space and a normal Hausdorff space is a T4 space. p a point of X.2. Let (X. Now let A and B be disjoint closed sets in X and consider the continuous function f (x) = dB (x) − dA (x) on X.4. and A ⊆ U . This is in fact just one (the most important one) of a series of separation properties that a topological space may have. Thus the Hausdorff property says that any two distinct points can be separated by open sets. B ⊆ V . Definition A13. T3 1 .9). Let X be a compact Hausdorff space. T5 . There are also T0 . and indeed normal.Lecture A13 Regular and Normal Spaces Let A and B be disjoint subsets of a topological space X. It is normal if any two disjoint closed subsets can be separated by open sets. Proof. It's best to do this proof in two installments: first regularity. Every compact Hausdorff space is regular. and T6 spaces. For each x ∈ B there
38
. T1 . Then B is compact (Proposition A12. Every metric space is regular. This shows that X is normal. a plain vanilla Hausdorff space is a T2 space. Define a real valued function dK : X → R+ by dK (x) = inf{d(x. Remark A13.
. 1] such that f (x) = 0 for x ∈ E. say Vx1 . for each p ∈ A there are disjoint open U p containing p and V p including B. . Theorem A13. In order that a function f : X → R be continuous. Then there is an open subset V of X such that F ⊆ V ⊆ V ⊆ W . construct a supply of continuous real-valued functions.6. Then U = U p1 ∪ · · · ∪ U pn . so there is a finite subcover. so it is included in W . in general this stronger statement is not true. Let A and B be disjoint closed (hence compact) sets. The closed sets E and F are disjoint. Warning: It is not asserted that the sets E. and W an open subset of X including F . F are exactly equal to f −1 {0}. f −1 {1} respectively.3 we used suitably chosen continuous functions to establish regularity. In particular. Since V ⊆ X \ U which is closed. . Vxn . . and indeed normality. Then U p = Ux1 ∩ · · · ∩ Uxn . U pn . F be disjoint closed subsets of a normal Hausdorff space X. It's also possible to proceed in the other direction: starting with normality. V = V p1 ∩ · · · ∩ V pn
are disjoint open sets including A and B respectively. To prove normality. 39
. The Vx form an open cover of B. we iterate the argument. V doesn't meet E. F a closed subset of X. . . (Urysohn's lemma) Let E. so by normality there are disjoint open sets U and V with E ⊆ U and F ⊆ V . V p = Vx1 ∪ · · · ∪ Vxn
are disjoint open sets. Exercise A13. In Example A13. . We will use the following exercise. for metric spaces. it is necessary and sufficient that for every x ∈ X and every > 0 there exists an open set Wx containing x such that f (Wx ) ⊆ (f (x) − . By regularity.5.7. Thus X is regular. the first containing p. . f (x) + ). f (x) = 1 for x ∈ F . The U p form an open cover of A and have a finite subcover say U p1 . Let X be a normal Hausdorff space. Lemma A13. then there exists a continuous function f : X → [0. Proof. V ⊆ X \U also. the second including V . Let E = X \ W .are disjoint open sets Ux containing p and Vx containing x.
. . 1] with f (E) = {0} and f (F ) = {1}. Clearly f maps X → [0. 1] can be replaced in the statement of Urysohn's lemma by any closed interval [a. We are going to construct by induction a collection of open sets Ut parameterized by dyadic rational numbers t ∈ (0. By lemma A13. Let x ∈ X with f (x) = t. call the set V so defined U(m+ 1 )2−n . 1]. t + ) as required. m = 1.
40
. 1] by f (x) = inf{t : x ∈ Ut } ∪ {1}. having the properties that Ur ⊆ Us whenever r < s and E ⊆ Ut for all t. Put Wx = Us \ Ur .5. so we must show that f is continuous.7. At the nth stage of the induction we assume that the sets Ut have been constructed for t = m2−n . There exist dyadic rationals r and s such that t− <r <t<s<t+ . suppose that 0 < t < 1 (the cases t = 0 and t = 1 are handled by similar arguments).Proof. [0. Set U1 = X \ F .5. It's an induction based on lemma A13. for each m there is an open set V such that Um2−n ⊆ V ⊆ V ⊆ U(m+1)2−n . . Let us make the convention that U0 refers to the closed set E (even though we have not defined a set U0 ). . Of course. Let > 0. s] ⊆ (t − . then U1 is an open set containing E. Now define f : X → [0. 2n . b]. Then Wx is an open set and f (Wx ) ⊆ [r. We have now defined the sets Ut for all 2 t = m 2−(n+1) and the induction continues. and we will use Exercise A13.
c]) of Y . Then f can be extended to a bounded continuous function g : X → R. . We have proved the following statement: for each bounded continuous h : Y → R there exists an extendible bounded continuous k : Y → R with k . and let f : Y → Z be a function. − gn and let gn+1 be the k that is constructed. so they are closed in X (Exercise A12. . f = g. Let us temporarily call a bounded continuous function h : Y → R extendible if it extends to a bounded continuous function X → R with the same sup norm. and then to extend it to the whole space 'by continuity'. But the display above shows that. . in other words. A classical method in analysis is to prove a result first for a dense subset of some space. . c] is a continuous function. What we have to prove is that all bounded continuous functions are extendible. Z are topological spaces we will naturally be interested in continuous f and g. and let Y be a closed subspace. −c/3]) and h−1 ([(c/3. Notice that the supremum norm h − k is at most 2c/3. Thus f is extendible as required. 41
. Thus by Urysohn's lemma there is a continuous function k : X → [−c/3. When X. .Lecture A14 Spaces of Continuous Functions Let X be a set. with sup |g| = sup |f |. A function g : X → Z is an extension of f if g(y) = f (y) for all y ∈ Y . By the Weierstrass M-test the series ∞ gi converges uniformly on X to a i=1 function g. c/3] with k(E) = {−c/3}. They are closed in Y (by continuity). and Y is closed in X. k(F ) = {c/3}. Theorem A14. let E and F be the disjoint subsets −1 h ([−c. . Y ⊆ X. h − k 2 h /3. Recall that a subset of a topological space is dense if its closure is the whole space. Apply the italicized claim to h = f − g1 − . (Tietze extension theorem) Let X be a normal Hausdorff space. If h : Y → [−c. Proof. Let f : Y → R be given and suppose inductively that extendible functions g1 .2). − gn (2/3)n f . Y. when restricted to Y . . if f = g|Y .1. Now we proceed inductively. The function gn+1 then has the desired properties so the induction is completed. Let f : Y → R be a bounded continuous function. . gn have been constructed with gi (2/3)i f and f − g1 − .
2. . Proof. These sets are open (why?) and x ∈ Vx so they cover X. Then g ∈ L and by construction h − < g < h + . there is a function f ∈ L having f (x) = a and f (x ) = a . for each x. ) There is a more classical formulation which makes use of algebraic rather than order-theoretic operations. By hypothesis.4. Let h ∈ C(X) be given and let > 0. as required. take a finite subcover and let g be the (pointwise) minimum of the corresponding gx .
So we have approximated h from one side by members of L. h(y) − < gx (y) ∀y. x ∈ X there exists fxx ∈ L such that fxx (x) = h(x) and fxx (x ) = h(x ). Proposition A14. then L is dense in C(X). (Stone) Let L be a lattice of continuous functions on a compact space X. it also contains their pointwise maximum and minimum — usually written f ∨ g and f ∧ g in this context. Exercise A14. whenever it contains functions f and g. We are going to approximate h within by elements of L. . (Show that the collection of extendible functions is a lattice. the continuous real-valued functions on a compact Hausdorff space X? Definition A14. Because L is a lattice. Use Stone's theorem to give another proof of the Tietze extension theorem in the case of compact Hausdorff spaces. x ∈ X and any a. 42
.3. Now we play the same trick again from the other direction: let Wx = {y : gx (y) < h(y) + }. Again these form an open cover of X. A collection L of continuous real-valued functions on a set X is a lattice if.How can we find dense subsets of C(X). h(x) = gx (x). The property appearing in the statement is called the two point interpolation property. for all x. let Vx = {y ∈ X : h(y) − < fxx (y)}. gx ∈ L and by construction. Fixing x for a moment. If. Take a finite subcover and let gx be the (pointwise) maximum of the corresponding functions fxx ∈ L. a ∈ R.
let h = f − g. Any polynomial in h belongs to A and hence so does |h| = lim pn (h). The closure of the subalgebra is a closed subalgebra. f + g − |h| . A subalgebra of C(X) is said to separate points if for every x. 1] converging uniformly to |x|. subtraction. The polynomials form an algebra which contains the constants and separates points. Proof. Proof. One says that a subset of C(X) is a subalgebra if it is closed under pointwise addition. Lemma A14. multiplication of functions. The original result of Weierstrass was Corollary A14. and multiplication by scalars. and separates points. it has the two point interpolation property.8. Using the algebraic operations. Hence by Stone's theorem A14. Theorem A14. g ∈ A.3 it is all of C(X).5.5. Proof. The previous lemma shows that it is a lattice. contains the constants.7. by closure and lemma A14. x ∈ X there is a function in the subalgebra taking different values at x and at x .6. 1]. But now f ∧g = belong to A as well. Proof. (Stone-Weierstrass) A subalgebra of CR (X) which contains the constants and separates points is dense. Let A be the given subalgebra and f. Writing |x| = 1 + (x2 − 1). 2 f ∨g = f + g + |h| 2
43
. There is no loss of generality in assuming (by rescaling) that |h| 1 everywhere. this follows from the fact that the binomial series for (1 + t)1/2 converges uniformly for t ∈ [−1. A closed subalgebra of CR (X) that contains the constant functions is a lattice. (Weierstrass) Every continuous function on a closed bounded interval is the uniform limit of polynomials. There is a sequence of polynomials pn (x) on [−1.Lemma A14.
Try to give an example to show that the complex theorem is not valid without the extra condition (we shall discuss this in detail later). In the complex case the Stone-Weierstrass theorem takes the following form: a ∗-subalgebra of C(X) which contains the constants and separates points is dense.
44
.9.Exercise A14. where the ∗-condition means that the algebra is closed under (pointwise) complex conjugation. Prove this.
Let g : Y → D be a continuous map to a discrete space D. The following properties of X are equivalent: (a) The only subsets of X that are both open and closed are the empty set and X itself. hence constant because X is connected. Conversely.Lecture A15 Connectedness What does it mean that a topological space is "all in one piece"? Proposition A15. of x. Let f : X → Y be continuous and surjective.2. But now since f is surjective. It is clear that (b) and (c) are equivalent. and let A be a nonempty equivalence class. assuming (a). Proposition A15. let ∼ be an equivalence relation with open equivalence classes.1. Proof. then the relation "x ∼ y iff f (x) = f (y)" is an equivalence relation with open equivalence classes. (c) The only equivalence relation on X with open equivalence classes is the trivial one (every point is equivalent to every other point). A subset of a topological space is called a connected subset if it is a connected space in the relative topology. Thus A = X and the equivalence relation is trivial. and conversely. Definition A15. it is also true that A is closed. if A is an open-and-closed subset of X. A space (or subset) that is not connected is called disconnected. Then f ◦ g : X → D is continuous. Proof.
45
.1. but. (b) The only continuous functions from X to a discrete topological space are constant. since the complement of A is a union of equivalence classes and therefore open. A topological space is connected if it satisfies the equivalent properties of Proposition A15. y belong to A" is an equivalence relation whose equivalence classes A and X \ A are open. since if f : X → D is a continuous function to a discrete space. or both. Let X be a topological space. Assuming (c). A is open. g must be constant itself.3. then Y is connected. If X is connected. then the relation "x ∼ y iff either none.
Proposition A15.4. Let X be a space and let {Xα }α∈A be a family of subsets each of which is connected. If α Xα = ∅ then α Xα is connected. Proof. Let Y = α Xα and let p ∈ α Xα . Let f : Y → D be a continuous map to a discrete space. Since each Xα is connected, the restriction of f to Xα takes a constant value (say dα ∈ D); since p ∈ Xα for every α, dα = f (p) for each α. Thus f takes the constant value f (p) on Y . Proposition A15.5. Every interval in R is connected, and every connected subset of R is an interval. Proof. We begin by proving that every compact interval [a, b] is connected. Suppose that ∼ is an equivalence relation with open equivalence classes; then the equivalence classes form an open cover U for the compact interval [a, b]. By the Lebesgue covering theorem (Theorem A5.8)there is δ > 0 such that any ball of radius less than δ is included in some member of the cover. In particular, any two points that are less than δ apart belong to the same equivalence class. But any pair of points in the interval [a, b] can be joined by a chain of points whose successive members are less than δ apart; so, any two points in [a, b] are equivalent. This proves the connectedness of closed intervals; other kinds of intervals can be written as the increasing union of suitable families of closed intervals, so we may apply Proposition A15.4. Conversely, let S be a subset of R that is not an interval. Then there is some a ∈ S such that S has members which are less than a and also members that are / greater than a. So, S ∩ (−∞, a) and S ∩ (a, ∞) are disjoint open subsets of S whose union is S; so S is not connected. Corollary A15.6. (Intermediate value theorem) Let f : [a, b] → R be continuous. Then the range of f includes all values between f (a) and f (b). Proof. The range of f , being the image of a connected space under a continuous function, is connected. Hence it is an interval. It must contain f (a) and f (b) so, being an interval, it must also contain everything in between. If X is a space, the relation "x ∼ y iff there is a connected subset of X containing both x and y" is an equivalence relation (by Proposition A15.4). The equivalence classes are called the connected components of X. The connected component of x is the largest connected subset of X that contains x. Exercise A15.7. Show that the closure of a connected subset of X is connected, and deduce that the connected components of X are closed. Give an example where they are not open. 46
Exercise A15.8. The quasicomponent of x ∈ X is the intersection of all the open-and-closed subsets that contain x. Show that the connected component of x is included in the quasicomponent. Can you give an example where they are not equal? Definition A15.9. A path in a topological space X is a continuous map [0, 1] → X. The start of the path is the point γ(0) and the end of the path is the point γ(1). We say that γ is a path from its start to its end. Lemma A15.10. Let X be a topological space, γ1 a path in X from a to b, γ2 a path in X from b to c. Then the map γ : [0, 1] → X defined by γ(t) = is a path in X from a to c. Proof. All that needs to be checked is the continuity of γ at t = 1 . Let U be 2 an open set in X containing b. By continuity of γ1 , there is δ1 > 0 such that if |s−1| < δ1 , s ∈ [0, 1] then γ1 (s) ∈ U ; by continuity of γ2 there is δ2 > 0 such that 1 if |s| < δ2 , s ∈ [0, 1] then γ2 (s) ∈ U . Now let δ = 1 min{δ1 , δ2 }; if t − 2 | < δ, 2 t ∈ [0, 1], then γ(t) ∈ U . This gives the required continuity. Corollary A15.11. The relation "There is a path in X from a to b" is an equivalence relation. Proof. The lemma gives transitivity. Reflexivity and symmetry are easy. The equivalence classes under this equivalence relation are called path components of X. Proposition A15.12. Every path component is connected. Consequently, the path component of x ∈ X is included in the component of x. Proof. By Proposition A15.3, the image of a path from a to b is a connected set containing a and b. Consequently, the path component of X is the union of a family of connected sets all of which contain x; so it is connected by Proposition A15.4. γ1 (2t) γ2 (2t − 1) (t (t
1 ) 2 1 ) 2
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A space which has only one path component (i.e., any two points can be joined by a path) is called path connected. By the preceding proposition, a path connected space is connected. The converse is false, however, as is shown by the following classical example. Exercise A15.13. Let Z be the set of points (x, y) ∈ R2 such that either x = 0 and y ∈ [−1, 1] or 0 < x 1 and y = sin(1/x). Show that Z is a compact connected metric space with 2 path components. The same example shows that, in contrast to components, path components need not be closed in general.
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Definition A16.8.g.2.) Proof. Definition A16. C is equal to its own interior and so is open. Proposition A16.e. In a locally connected space. for every x ∈ X. connectedness). We say that a space X has the property P locally if.1. The same argument as in the previous proposition shows that the path components in a locally path-connected space are open. Since the complement of each path component is a union of other path components. they are also closed. (It follows that the connected components are equal to the quasicomponents. Since x ∈ N ◦ . so they are equal. contains x in its interior. i. Proposition A16. The path component is always included in the connected component. For instance. A neighborhood of x ∈ X is a subset that includes an open neighborhood of x.3.Lecture A16 Local Properties of Spaces Let X be a topological space.4. x is an interior point of C. An open neighborhood of x ∈ X is an open subset of X that contains x. so in particular it must be included in the path component of x. Now the connected component of x must be included in any open-and-closed set that contains x. every open neighborhood of x includes another open neighborhood of x that has property P . Proof. Suppose that P is some kind of topological property (e. Since x was arbitrary. There is a connected neighborhood N of x. In a locally path-connected space. Let C be a connected component and let x ∈ C. 49
. a space X is locally connected if every open neighborhood of each x ∈ X includes an open connected neighborhood. see Exercise A15. Then N ⊆ C and so N ◦ ⊆ C ◦ . Informally we may say that x has "arbitrarily small connected neighborhoods". the connected components are equal to the path components (and are open). the connected components are open.
sometimes called semilocal path connectedness: X is semilocally path connected if for each x ∈ X and each open neighborhood V of x. Local compactness says that each point has "arbitrarily small" compact neighborhoods.5. 50
. containing x. Here it is necessary to modify the definition somewhat: we say that a space is locally compact if every open neighborhood of each point contains a compact neighborhood (which need not be open . But in fact there is a simpler. Then X \ {x0 } is a locally compact Hausdorff space (since each point of X distinct from x0 has a closed. 1] → V starting at x and ending at y. equivalent definition. neighborhood that does not meet x0 ).there may be very few compact. We need to find a compact neighborhood that is included in N . and U and ∂U are closed subsets of a compact space (namely K).7. every locally compact Hausdorff space arises in this way. Suppose that x ∈ X has a compact neighborhood K. A Hausdorff space is locally compact iff each point has a compact neighborhood. Let X be a locally compact Hausdorff space. Let X + be the union of X and a disjoint point. open sets). Then U is a neighborhood of x.The most important example of a locally connected and locally path-connected space is furnished by an open subset of Rn (or any normed vector space). denoted ∞. The following sets form a topology on X + : (a) The open subsets of X in its original topology. whose closure does not meet the compact set ∂U . more exactly. Then V ⊆ U is the desirec compact neighborhood. Exercise A16. Proposition A16. Show that a space which is locally connected and semilocally path connected must be locally path connected. a stronger condition than that of the proposition. Let X be a compact Hausdorff space and x0 ∈ X. there is an open neighborhood U ⊆ V of x such that for each y ∈ U there is a path γ : [0. Proposition A16. In fact. By regularity (or. Let U = K ◦ ∩ N ◦ . and let N be any neighborhood of x. hence compact.6. by the proof of regularity) there exists an open subset V of U . There is a weaker local path connectedness notion. Proof. hence are compact. Can you give an example of a space which is semilocally path connected but not locally path connected? We now turn to consider local compactness.
Let T be the family of subsets of X + described in the proposition. which is compact by Exercise A16. But by local compactness. Thus T is a topology. . Exercise A16. then their union is of the form {∞}∪(X \L) where L is a closed subset of K. Un be a finite family of members of T . One of them (say W0 ) must contain ∞. forms a finite subcover of W .
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. If they are all of type (a). . Proof. Finally to show that T is Hausdorff: Any two points not at infinity can be separated by open sets of type (a) (by the Hausdorff property of X). let W be a family of members of T covering X + .(b) The subsets of the form {∞} ∪ (X \ K) where K ⊆ X is compact. It remains to show that this topology is compact and Hausdorff. L are compact subspaces of Y then so is K ∪ L.8. K ⊆ X compact. this subcollection. together with W0 . The topology so defined on X + is compact and Hausdorff and the identity map X → X + is a homeomorphism onto its image. Thus it suffices to show that ∞ can be separated from a point x ∈ X. hence compact. To show that T is compact. If K. so the union is of type (b). We show it is a topology. (i) Let Uα be a family of members of T . so some finite subcollection covers K also. (ii) Let U1 . then their intersection if of type (a). Now the remaining Wα intersect X in open sets which cover K. say equal to {∞}∪(X \K). their intersection is the complement of a finite union of compact sets.8 above. K say. then their union is an open subset of X and is of type (a) also. (iii) Clearly ∅ and X + belong to T . We will need the following easy exercise. and the relative topology that it induces on X is the same as the original topology of X. x ∈ X has a compact neighborhood. If they are all of type (b). and hence must be of the form {∞} ∪ (X \ K). Let Y be a Hausdorff space. Thus the intersection id of type (b). . then K ◦ and {∞} ∪ (X \ K) are disjoint members of T containing x and ∞ respectively. If one of them is of type (b). The space X + is called the one-point compactification of X. If one of them is of type (a). .
antisymmetric.Lecture A17 Some set theory Further topics in our discussion will require some more advanced background from set theory. To learn about the paradoxes this gives rise to.
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(d) A subset of S that is totally ordered (in the induced ordering) is called a chain. If S is such a set then (a) An upper bound for a subset T ⊆ S is an element x such that y y ∈ T. and transitive: that is. z ∈ S. A partial order on a set S is a relation which is reflexive.. Exercise A17. and x y and y z together imply x z. (e) S is inductively ordered if every chain in S has an upper bound. and how to avoid them. (f) A maximal element for S is an element m ∈ S such that x m implies x = m (i. Give simple examples to illustrate all these notions. for all x.e. either x y or y x. (g) S is well-ordered if it is totally ordered and every non-empty subset has a least member (necessarily unique). x x. (c) S is totally ordered if for all x.) Notice that this is a weaker notion than that of upper bound. A set with a partial order is called a (partially) ordered set. x y and y x imply x = y. every finite subset has an upper bound). there isn't anything strictly greater than m. Let S be an ordered set. it is strictly inductively ordered if every chain has a least upper bound. you need a course in Logic. x for all
(b) S is directed if every two-element subset has an upper bound (equivalently.1. We will take a rather "naive" approach to set theory. y. Definition A17.2. A map f : S → S will be called an inflator (I just made this word up) if f (x) x for all x ∈ S. y.
4.4. i. a chain which is not itself included in any larger chain). (Bourbaki fundamental lemma) Any inflator on a nonempty strictly inductively ordered set has a fixed point. if f (x) ∈ F whenever x ∈ F . contradicting Bourbaki's lemma.e. Let S be such a set and let C be a maximal chain in S (provided by A17. a chain. (Weak form of Zorn's lemma) Every nonempty strictly inductively ordered set contains a maximal element. Pick a point a ∈ S (fixed throughout the discussion). This will complete the proof because if x is the least
1
For those in the know. Corollary A17. Let m be an upper bound for C.e. namely. and if. Then for each x ∈ S there is a yx ∈ S with yx > x. We'll prove this in a moment. The strategy of our proof is to show that M is totally ordered. Consider the map1 f sending x to yx .5).5. Corollary A17. Corollary A17. this is where we use the axiom of choice. the intersection of all the closed sets that there are. a point x such that f (x) = x. Then m is maximal. Each chain in S is included in a maximal chain (i. i. S itself is closed. (Zorn's lemma) Every nonempty inductively ordered set (not necessarily strict) contains a maximal element. let's draw some consequences. the least upper bound of C belongs to F . and indeed is strictly inductively ordered (with least upper bound for a chain in C given by the union of its members).. Then f is an inflator with no fixed point. whenever C ⊆ F is a chain. Now apply A17.6. since if x > m then C ∪ {x} is a chain larger than the supposedly maximal chain C. Proof.
53
. Obviously. Now let us see to proving the Bourbaki lemma A17. (Hausdorff maximality principle) Let S be a nonempty partially ordered set. Then C is partially ordered. Moreover. We'll say that a subset F of S is closed if a ∈ F .e.3. For now. Let S be nonempty and strictly inductively ordered and f : S → S an inflator. Let C be the collection of all chains in S. Proof.3.. there is a minimal closed set M . Proof. partially ordered by inclusion. Thus. Let S be strictly inductively ordered and suppose it contains no maximal element.Lemma A17. the intersection of any family of closed sets is closed.
So. or x = u. y ∈ M . Since it is a subset of M . Suppose then that u is unavoidable and consider f (u). either y x or f (x) y. The other two possibilities yield f (x) f (u) by the unavoidability of u. As already remarked. Since x is unavoidable. Note that the collection of all x a is closed. if not. But now we must have S = X. Y is nonempty and is inductively ordered by inclusion. then the set {x ∈ M : either x u or f (u) x}
is readily seen to be closed. We will show that it is closed. Let U be the set of unavoidable points in M . This contradicts maximality. whenever x ∈ M and x < u. by minimality. Let Y be the set of pairs (S. by minimality. there is a maximal element in Y . Let us call u ∈ M unavoidable if.upper bound of M . We now know that each point of M is unavoidable. by the paragraph above. It contains a. but the latter certainly implies that x y becasue f is an inflator. it must be the whole of M . this completes the proof of Lemma A17.7. f (U ) ⊆ U . A similar argument shows that the least upper bound of any chain in U belongs to U . If u is an unavoidable point. We can now prove that every point is unavoidable. Proof. (Well ordering principle) Every nonempty set can be wellordered. a is the least element of the minimal closed set M . we have f (x) u.
54
. For. let x. hence f (x) x. then. and the last possibility cannot occur. since f is an inflator. or f (u) x. either x < u.3. it follows that f (x) = x. so U is closed. Let X be a set. Thus. ) where S is a subset of X and is a well-ordering of S. Proposition A17. By Zorn's lemma. say (S. it must equal M itself. It is easy to deduce that M is totally ordered. we can choose some x ∈ X \ S and well-order S ∪ {x} by insisting that x is greater than every element of S. then x ∈ M and f (x) ∈ M also. ). If x < f (u) then.
Reformulate it to use Zorn's Lemma. Let I = i(X) ⊆ Y . −k + 1. . There exists an injection from X to Y if and only if there exists a surjection from Y to X. . . Proof. . A subset S of Z is stable if z ∈ Z iff h(z) ∈ Z. then there exists a bijection from Y to X. . The stable sets Sz are of four types: (a) X-stopping: hn (z) is defined only for n = −k. . s(y) = x0 . so they partition Z.2. Explicitly. Let i : X → Y be an injection. (K¨ nig) Assume (wlog) that X and Y are disjoint sets and let Z be their o disjoint union.3. It suffices now to establish a bijection Sz ∩X → Sz ∩Y for each Sz separately. The last part of this argument uses the Axiom of Choice directly. then s(y) is the unique x ∈ X such that i(x) = y. and h−k (z) ∈ Y . considered as a map Z → Z. stable sets either coincide or are disjoint. Proposition A18. let f : X → Y and g : Y → X be injections and let h be the union of these maps. By the injectivity of h. Exercise A18. and h−k (z) ∈ X. Define s : Y → X as follows: if y ∈ I. Without loss of generality assume that X = ∅. but it is unique if it does exist. Theorem A18. For each x ∈ X choose some yx ∈ Y such that s(yx ) = x.Lecture A18 Countability and Metrization We now briefly discuss the theory of cardinal numbers. (Schr¨ der-Bernstein) If there exists an injection from X to Y o and an injection from Y to X. Let X and Y be sets. Fix an element x0 ∈ X. (d) Cyclic: hm (z) = z for some m (necessarily even). 55
.1. suppose that s : Y → X is a surjection. Then i is an injection. Conversely. Proof. Let Sz be the minimal stable set containing z. where we note that h−1 (p) denotes an element q such that h(q) = p. Let i be the map that sends x to yx . (c) Two way infinite: the hn (z) are all distinct for different values of n ∈ Z. such an element may not exist. (b) Y -stopping: hn (z) is defined only for n = −k. This is a surjection. otherwise. Sz is the set {hn (z) : n ∈ Z}. −k + 1.
In the second case.In the first case. Let X be a set and let P(X) (the power set) be the set of all subsets of X. suppose that f is such a surjection. Arithmetic laws like xy·z = (xy )z then follow from corresponding set-theoretic properties (a map from Y × Z to X is the same as a map from Z to the set of maps from Y to X). We define card X = card Y iff there is a bijection between X and Y (one says that X and Y are equinumerous).5. The integers Z and the rationals Q are countable sets. To each set X associate a "number". We will show that there is no surjection X → P(X). the cardinality of the set of all maps X → Y is denoted (card Y )card X .6. either f or g will do. It follows that card(P(X)) > card(X). Proof. The cardinality of the natural numbers is denoted ℵ0 (read: aleph-zero). The cardinal 2ℵ0 is denoted c (the continuum). g gives a bijection Sz ∩ Y → Sz ∩ X. On the other hand. Indeed. Proposition A18. (We also call a set countable if its cardinality is countable.
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. f gives a bijection Sz ∩ X → Sz ∩ Y . A cardinal that is ℵ0 is called countable. We denote the cardinality of P(X) by 2card X . Y have cardinalities x. y respectively then xy is the cardinality of X × Y . In the third and fourth cases. Definition A18. If S = f (y) for some y ∈ X we obtain (from the definition of S) the contradiction y ∈ f (y) ⇔ y ∈ f (y).4. its cardinality card X. More generally. / Thus f is not surjective after all. x + y is the cardinality of X Y .) Example A18. For every cardinal there is a strictly larger cardinal. Consider S = {x ∈ X : x ∈ f (x)}. / This is a subset of X. and so on. We define card X card Y iff there is an injection X → Y . there is an obvious injection X → P(X) (as one-element subsets). Once can do "arithmetic" with cardinals: if X. The Schr¨ der-Bernstein theorem shows that this defines a partial order on the cardio nalities.
The set of all sequences of rational numbers has cardinality ℵ0 ℵ0 = c. and for any x ∈ X and any open set U containing x. Then the set B of balls B(s. t) where s ∈ S and t ∈ Q is countable. so U ∩ S contains xn and is in particular nonempty. For instance. and select a point xn from each Un . Moreover. there are in general examples of separable spaces that are not second countable. and this subset maps onto R. On the other hand. let X be an uncountable set equipped with the cofinite topology. r) ⊆ U . Proof. and is included in U .Example A18. A second-countable space is separable. and the assignment a → ∞ 3−n an is n=1 an injection from this set of sequences to R. contains x.. there is a rational t with d(x. The set S = {xn } is countable. Thus S is dense. It's a one-liner: cℵ0 = 2ℵ0
ℵ0
= 2ℵ0 ·ℵ0 = 2ℵ0 = c. there is a countable family B of open sets such that every open set is a union of sets from B.
Proposition A18. X is separable it it has a countable dense subset. It is second countable if there is a countable basis for the topology of X: i. 57
.9. s) < t < r/2. Definition A18.10. Let X be a topological space. there is a ball B(x. there is some Un ⊆ U containing x. so card R c.e. Let X be a second countable space. it forms a base for the topology: if x ∈ X belongs to an open set U . Then X is separable (any countable subset of X meets every nonempty open set) but it is not second countable (a countable family of cofinite sets can omit only countably many points in total. Let us prove that cℵ0 = c. Let X be separable and metrizable. t) belongs to B. the ball B(s. Let {Un } be a countable base for the topology. Proof.8.7. there is an s ∈ S such that d(x. However. s) < r/2. Schroder-Bernstein now completes the proof. Lemma A18. with S a countable dense set. the set of all sequences a = {an } of zeroes and ones also has cardinality c. The subset of convergent sequences must then have cardinality c. A separable metric space is second-countable. so card R c. The set of real numbers has cardinality c. so there must be some cofinite set that does not contain any member of the family).
13. by Urysohn's lemma. then there is a V ∈ B containing x but not y (by the Hausdorff property). It is clear that d(x. y) = 0 iff x = y. i. There is a set V ∈ B such that 58
. Every metric space is first countable. Let X be such a space and let B be a countable basis for the topology. Whenever U. so d is a metric. Define d : X × X → R+ by
∞
d(x.A neighborhood base at a point x ∈ X is a family of open sets containing x such that each open set containing x must include a member of the family. . A topological space X is first countable if each point has a countable neighborhood base. 1] that is equal to 0 on U and is equal to 1 on X \ V . It remains to show that every set open in the original topology is d-open. so every d-open set is open in the original topology. Being the sum of a uniformly convergent series of continuous functions. Show that a compact metric space is separable and second countable. Definition A18. . . Exercise A18. f2 .}. then fn (x) = fn (y) for all n. so then the function fn corresponding to this pair (U. a continuous function f : X → [0.11. But. y) = 0. If d(x. Every second countable space is first countable but the converse is false: e. y) =
n=1
2−n |fn (x) − fn (y)|.12. by normality there is an open set U containing x with U ⊆ V .
We will show that d is a metric defining the topology of X. The collection F of all such functions is countable: denote it {f1 . x).g.e. V ) has fn (x) = 0 but fn (y) = 1. and we may take it that U ∈ B also. d is continuous (relative to the original topology on X). and x ∈ W . V ∈ B with U ⊆ V there exists. x) = 0. Let W be open in the original topology. the topology can be given by a metric. Proof. y) = d(y. that d(x. and this is a contradiction. It follows that every d-open ball is open in the original topology. if x = y. We conclude that d(x. Theorem A18. (Urysohn metrization theorem) A second countable normal Hausdorff space is metrizable. and that d satisfies the triangle inequality. consider the discrete topology on an uncountable set.
Let = 2−n . Then. fn (y) < 1 and so y ∈ V . ). if y ∈ Bd (x. and there is a set U ∈ B such that x ∈ U and U ⊆ V . as required. so it is d-open. in particular y ∈ W .x ∈ V and V ⊆ W . Let fn be the function constructed above corresponding to the sets U and V . Thus W contains a d-ball around each of its points.
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Lecture A19 Convergence and Nets Let {xn } be a sequence in a topological space X. In a first countable space X. Any totally ordered set X can be equipped with the order topology for which a basis is the "half open intervals" (a. the "first uncountable ordinal"). Suppose that A ⊆ X. If X is first countable then sequences in x contain all the information that there is about the topology of X. but I lied. this will be true iff every member of a countable neighborhood base {Un } at x contains points of A. or only for one of them (and. and well-order it (using the axiom of choice). try the following exercise. then x belongs to the closure of A) is easy. Picking such a point an ∈ A ∩ Un for each n = 1.1.k. One way to see this is the following Lemma A19. 60
. Show that f : X → Y is continuous if and only if the sequence f (xn ) → f (x) in Y whenever the sequence xn → x in X. Exercise A19. Let C = {x ∈ X : x < Ω} be the set of predecessors of Ω. (Hint: The proof is exactly the same as the one we already discussed for metric spaces. a subset A is closed if and only if the limit of every convergent sequence in A is itself a member of A. .3. But. 2. Let Ω ∈ X be the smallest element that has uncountably many predecessors (a.) We say that {xn } converges to a point x ∈ X if. Let X and Y be first countable.
Let X be uncountable. For another example. gives a sequence (an ) in A converging to x. which)? But in general sequences are "not enough" to probe the structure of a topological space. . Proof.2. if so. Then Ω belongs to the closure of C. A point x belongs to the closure of A iff every open neighborhood of x contains points of A. Example A19. The converse statement (if x is the limit of a convergent sequence in A. but nevertheless there is no sequence in C converging to Ω. for every open neighborhood U of x. since X is first countable. (Okay.a. b] := {x ∈ X : a < x b}. this exists because of well-ordering. I know that I told you that I was not going to use sequences outside metric spaces. there is some N such that xn ∈ U for n N . 3 .) Do you need first countability for both spaces.
Since x belongs to the closure of A. and it is then almost a tautology that C converges to Ω even though we have seen there is no sequence that will do this. Similarly we have Proposition A19. in Example A19. A map f : X → Y between topological spaces is continuous if and only if the net f (xi ) → f (x) in Y whenever the net xi → x in X.This motivates the following definition. On the other hand. We define U V if U ⊇ V (notice that the inclusion is reversed!). we could take the collection C itself in its natural ordering to be a net. Proof. and in fact a directed set: an upper bound for U and V is the intersection U ∩V .8. Proposition A19. suppose that U and V are open neighborhoods of x. These definitions match the definitions of "sequence" and "convergent sequence" when we take the net simply to be the natural numbers N. A net has to parameterized by a directed set.3 above. Definition A19. The trick is to pick the collection of all neighborhoods of x as the directed set! In fact.6. a subset A is closed if and only if the limit of every convergent net in A is itself a member of A. A net {xi } converges to x ∈ X if it is eventually in every neighborhood of x. then there is some net in A that converges to x. A net {xi } is eventually in S if there is i0 such that xi ∈ S for all i i0 . Definition A19. it converges to x by definition: if V is any neighborhood of A then there is an element of N (namely V itself) such that if U V then aU ∈ V . it is frequently in S if it is not eventually in X \ S.) Let S be a subset of X. Emulating the proof of Lemma A19. Moreover.5. Definition A19. This makes the collection N = {U open : x ∈ U } a partially ordered set.4. The mapping2 from N to A given by U → aU is a net.
2
Using the axiom of choice here!
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. for every U ∈ N there exists aU ∈ A ∩ U . Now suppose that X is a topological space. what we have to show is that if x is a point of the closure of A.7. In any topological space X. A net in a set X is a map from a directed set to X. (Recall that a directed set is a partially ordered set in which any two elements have an upper bound.1.
xU converges to x.) One might ask whether we can similarly generalize the definition of compactness. by construction. such that f −1 (W ) is not a neighborhood of x. A collection F of subsets of X will be called an N -filter if N is frequently in every member of F .9. for any i0 ∈ D. for every U ∈ F which contains x. Every subnet of a convergent net is convergent (with the same limit). {X} is an N -filter. and an open neighborhood W of y. The collection of N -filters is partially ordered by inclusion. the result of composing it with a final function h is called a subnet or refinement of the original one. given any subset S of X whatsoever.12. Given a net D → X. Definition A19. we must arrive at the correct definition of subnet. there is j0 ∈ D such that j j0 implies h(j) i0 .10. Definition A19. the definition of "final function" is cooked up to make this true. Exercise A19. Then there exist a point x ∈ X.Proof. Then. Let N : D → X be a net in X. The proof that a continuous map has this property is essentially the same as the corresponding proof for sequences. A net {xi } in X is called universal if. Lemma A19. Let D and D be directed sets.11. and if F is closed under finite intersections and the formation of supersets. either {xi } is eventually in S or else eventually in X \ S. Show that a net xi converges to x iff. y = f (x). Such objects exist: for example. Proof. Every net has a universal subnet. which in the metric space case is expressed as "every sequence has a convergent subsequence. and every chain in this partially ordered set has an 62
. It follows that for every open neighborhood U of x there exists xU ∈ U \ W . since it never lies in the neighborhood W of y." To do this. Suppose that all we know about the topology of X is that it is generated by some family F of subsets. (This exercise should help you understand why we insist that a net should be parameterized by a directed set. as in the previous proposition. the net xi is eventually in U . This depends unavoidably on the axiom of choice. Suppose that f has the property in question but is not continuous. parameterized by a directed set D. Organize the xU into a net parameterized by the directed set N of neighborhoods of x. A function h : D → D is called final if. but f (xU ) does not converge to y.
13.upper bound (the union). Proposition A19. i) with A ∈ F0 . Conversely. call it F0 . j) ∈ B ∩ S ⊆ A ∩ S. so the intersection of some finite subcollection of F is empty. Let A ∈ F0 and let i be arbitrary. By maximality we deduce that S itself belongs to F0 . It is not possible that N be frequently both in S and in X \ S. there exist B ∈ F0 . i ∈ D. (ii) (the finite intersection property) if F is a family of closed subsets of X. Proof. hence it has a finite subcover. and N (i) ∈ A. j) (A. The following properties of a topological space X are equivalent: (i) X is compact. then in fact the intersection of all the members of F is nonempty.
The map (A. Let S ⊆ X have the property that N is frequently in S. Thus N fails to be frequently in one of these sets. Now let S be arbitrary. We conclude that N is frequently in A ∩ S for every A ∈ F0 and hence. the empty set. (iv) every net in X has a convergent subnet. a contradiction. is again an N -filter. and then their intersection. i) ⇔ B ⊆ A. would do so as well. j i. Suppose that S ⊆ X has the property that N is frequently in A ∩ S for every A ∈ F0 . A ∈ F0 . If F is empty then U is an open cover of X. and the intersection of any finite number of members of F is nonempty. (iii) every universal net in X converges. that S itself belongs to F0 . B ⊆ A. Let D be the collection of pairs (A. which is to say that it is eventually in the other one. Let F be a family as in (ii) and let U = {X \ F : F ∈ F}. We will use this property to construct the desired universal refinement. Thus Zorn's Lemma provides a maximal N -filter. as observed above. Then the union of F0 with the set of all sets A ∩ S. This proves (ii). i) → i is final so defines a refinement N of the net N . if we assume (ii) and let U be an open cover that is supposed to have 63
. for then (by the above) both S and X \ S would belong to F0 . and j i. such that N (j) = N (B. Suppose (i). We claim that this refinement is universal. By definition. it is a directed set under the partial order (B.
It follows that xα is eventually in any finite intersection of the Fx . Let X be the Cartesian product α Xα : that is. which sends the point x to its "coordinate" xα . By (ii) the intersection of all the Fx is nonempty. The product topology on X is the topology generated by all −1 the sets πα (Uα ). hence convergent. and this is a contradiction since x ∈ Fx for each x. Let x ∈ X be the unique point such that πα (x) = xα . Show that xi converges to x ∈ X if and only if πα (xi ) converges to πα (x) for all α ∈ A . xi → x in X. where xα ∈ Xα . Definition A19. Let xi be a universal net in X. and let {xα } be a universal net in X. Let D be the directed set comprised of finite intersections of members of F (directed by reverse inclusion) and for each D ∈ D let xD be a point of D. For each α there is a projection map πα : X → Xα . Let F be a family of closed sets with the finite intersection property. If it does not converge. Proof. Thus (i) and (ii) are equivalent. we deduce that x ∈ F . then each x ∈ X has a neighborhood Ux for which xα is not eventually in Ux . choose a subnet convergent to x ∈ X.15. We use the characterization (iii) of compactness in Proposition A19. Then xD is a net in X. Thus X is compact. Then πα (xi ) is a universal net in Xα . This proves (ii). Suppose (ii).no finite subcover.15. Theorem A19.16. any neighborhood of x meets every F ∈ F. Exercise A19. an element of X is a mapping x : α → xα . By definition. / (iii) implies (iv) by Lemma A19. say to xα . parameterized by α ∈ A where A is some index set (possibly infinite or even uncountable). Thus the intersection of all the F contains x. Then.12. hence (by universality) xα is eventually in Fx = X \ Ux .13 above. by Exercise A19.14. Let xi be a net in the product X defined above. In particular no finite intersection of the Fx is empty. Suppose (iv). (Tychonoff) The product of any family of compact spaces is compact. Now let (Xα ) be a family of topological spaces.
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. we reach a similar contradiction by considering the family of closed sets F = {X \ U : U ∈ U}. where α runs over A and Uα runs over all the open subsets of Xα . since F is closed.
and it is the largest topology (most open sets) that does so.3. Now suppose that X is a topological space. So we are done. Let X = [0. The quotient topology on X/ ∼ is defined by declaring that a subset U ⊆ X/ ∼ is open iff the inverse image π −1 (U ) is open in X.Lecture A20 Quotient Spaces Let X be a set and ∼ an equivalence relation on it. A subset of X is called saturated (with respect to ∼) if it is a union of equivalence classes. Recall that the notation X/ ∼ denotes the set of equivalence classes of the equivalence relation. There is a canonical map π : X → X/ ∼ that sends each x ∈ X to its equivalence class. 1] for some > 0. namely that the sets U identified by this definition do satisfy the axioms for a topology. the smallest saturated set that includes S. there is something to check here.2. Of course. Moreover. Definition A20. and then f (U ) = {(cos 2πt. ) ∪ (1 − . 1]/ ∼ is homeomorphic to the unit circle S 1 . The saturation of an arbitrary subset S is the union of all the equivalence classes that meet S. Clearly. )} is a neighborhood of (1. we must show that any saturated open set containing x maps to a neighborhood of g(x). f is a continuous map from X/ ∼ to S 1 . To see this.e. and define f : X/ ∼→ S 1 by f (π(t)) = g(t). Thus. Then [0. 1} this is easy. The ideas underlying this example can be summarized in the next two propositions. x + ). By the remark above. let g : X → S 1 be the continuous map g(t) = e2πit = (cos 2πt.1. sin 2πt). The first is the topological analog of the "first isomorphism theorem" in 65
. and for this it suffices to show that f takes each neighborhood of each point π(x) of X/ ∼ to a neighborhood of f (π(x)) = g(x). That is. the open sets of the quotient topology correspond to the saturated open subsets of X. sin 2πt) : t ∈ (− . What about saturated open sets containing 0 or 1? Any such must include U = [0. which is in fact bijective. When x ∈ {0. Remark A20. since any saturated open set containing x includes an interval (x − . We must show that f is also open. 1] and let ∼ be the equivalence relation that makes 0 and 1 equivalent and whose only other equivalence classes are points. 0) in S 1 . But this is easy. this is well-defined since g(0) = g(1). Example A20. the quotient topology makes π continuous. i. a function f : (X/ ∼) → Y is continuous iff f ◦ π : X → Y is continuous.
4. and Y is a Hausdorff space. A continuous g : X → Y is ∼-respecting iff g = f ◦ π for some continuous f : X/ ∼→ Y . Proof. and let ∼ be the equivalence relation which makes (x1 .12. Show that X is a Hausdorff space. Then g = f ◦ π. An obvious necessary condition for Hausdorffness is that the equivalence classes should be closed: after all. y1 ) equivalent to (x2 . Exercise A20. It might be helpful to think of ∼. g is ∼-respecting. 1]×[0. which is in fact a bijection X/ ∼→ Y . in this proposition. 3. as the "kernel" of the map g. y2 ) iff either x1 = x2 and y1 = y2 or y1 = y2 > 0.}. so there is a continuous map f as described. Then we have Proposition A20. Let X be a compact space and Y a Hausdorff space and suppose that g : X → Y is continuous and surjective.algebra. for an equivalence class c. .5. each equivalence class is the inverse image of a point in X/ ∼ under the continuous map π. If g is ∼-respecting then a function f can be defined by setting f (c). which is a compact Hausdorff space. Then the map f : X/ ∼→ Y induced by g is a homeomorphism. 0) must intersect. Let X be the real line. Proposition A12. with the topology generated by all the open sets of the usual topology together with the complement of the set A = {1/n : n = 1. . Proposition A20. Let X be the unit square [0. but that the quotient 66
. 1]. and f is continuous by definition of the quotient topology. But X/ ∼ is a compact space (as the image of a compact space under the continuous map π. Then any two saturated neighborhoods of (0.8). 2. Let ∼ be the equivalence relation on X defined by x ∼ x iff g(x) = g(x ).7. Proof. equal to f (x) for any representative x of c.6. and we know that one-point sets are closed in a Hausdorff space. so this bijection is in fact a homeomorphism by Corollary A12. Tautologically. 0) and (1. But this is by no means a sufficient condition: Example A20. so the corresponding quotient space cannot be Hausdorff. One can even give an example of this kind where all the equivalence classes except one consist of single points. Let us call a continuous map f : X → Y ∼-respecting if f (x) = f (x ) whenever x ∼ x . One problem with quotient spaces is that they are frequently not Hausdorff. .
But then π(U1 ) and π(U2 ) are disjoint open subsets of X/ ∼ containing x1 and x2 respectively. Proposition A20. the projection π : X → X/ ∼ is a closed map (it need not be open). then the point (x1 . Let ∼ be an equivalence relation on X. Now. Let X be a compact Hausdorff space and let ∼ be an equivalence relation on X whose graph is closed. π : X → X/ ∼ is a closed map). Suppose that π is open. Then X/ ∼ is Hausdorff also. Proposition A20. Let G denote the graph of ∼. Thus π −1 (U1 ) × π −1 (U2 ) is open in X × X. and does not meet G. Proof. where π2 : X × X → X is the second projection map of the product space. The graph of ∼ is the subset G = {(x1 . Then whenever x1 ∼ x2 there are disjoint open subsets U1 . Moreover. C × X and G are closed subsets of X × X. there are additional conditions under which the closure of the graph is both necessary and sufficient for the Hausdorffness of the quotient. then the graph G is closed. which is compact (by the Tychonoff 67
.8. x2 ) and does not meet G.) We will give some useful conditions for Hausdorffness of quotients. We conclude that if X/ ∼ is Hausdorff. Suppose that X/ ∼ is Hausdorff.) The previous exercise shows that the converse is not necessarily the case. so there is an open rectangle of the form U1 × U2 which contains (x1 . We first show that the saturation of any closed subset of X is closed (equivalently. We may write the saturation S of C as S = π2 ((C × X) ∩ G) ⊆ X. (This is also an example of a Hausdorff space that is not regular.9. (This statement includes our previous requirement that the equivalence classes should all be closed. Let C be a closed subset of S. Let X be a Hausdorff space and let ∼ be an equivalence relation on X whose graph is closed. x2 ) ∈ X × X does not meet the closed set G. x2 ) : x1 ∼ x2 } ⊆ X × X. U2 of X/ ∼ with π(x1 ) ∈ U1 and π(x2 ) ∈ U2 . Proof. Let x1 .space obtained by identifying all the points of A to a single point (this is usually denoted X/A) is not. contains (x1 . but it is stronger. x2 ∈ X with π(x1 ) = π(x2 ). x2 ). If in addition the projection π : X → X/ ∼ is an open map then X/ ∼ is Hausdorff also. However.
Si is closed (by the first thing we proved above) and does not meet Ci .
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. with (U1 × U2 ) ∩ G = ∅. Qyn . we prove the analog of regularity. C2 respectively. . U2 ⊇ C2 .theorem). and let Si be its saturation. Because C2 is compact there is a finite subcover of C2 say by Qy1 . (U1 × U2 ) ∩ G = ∅. Fix x ∈ C1 . so compact. . For each y ∈ C2 there are open sets Py containing x and Qy containing y such that (Py × Qy ) ∩ G = ∅. whose saturations do not meet — that is to say. Then P = n Pyj and Q = n Qyj are open sets. Consider now the closed set X \ Ui . It follows that π2 ((C × X) ∩ G) is a compact subset of a Hausdorff space. so the complement X \ Si = Vi is a saturated open set with Ci ⊆ Vi ⊆ Ui . Let us show that there are open sets U1 . one j=1 j=1 containing x and one including C2 . Now suppose that x1 . First. Thus (C × X) ∩ G is a closed subset of a compact space. This is the same argument as the proof that compact Hausdorff spaces are normal. . Now π(Vi ) and π(V2 ) are disjoint open subsets containing x1 and x2 respectively. such that P ×Q∩G = ∅. Repeat the argument using the compactness of C1 to obtain the desired open sets U1 ⊇ C1 . x2 ∈ X with π(x1 ) = π(x2 ). . This proves our assertion. and let C1 and C2 be the equivalence classes of x1 and x2 respectively. so it is closed. including C1 . U2 .
Proposition A21. u2 ∈ U2 . the multiplication map G × G → G. or in other words the space of equivalence classes 69 (x. SL(n. etc. Definition A21. O(n). This is one example of the "homogeneity" of topological groups. defined by Lg (x) = gx. are homeomorphisms G → G. every one-point set is closed. then u1 = g2 u−1 with u1 ∈ U1 . Consequently. x → x−1 . Proof. C) as well as their various subgroups SL(n. y) → xy
. and then g2 = u1 u2 ∈ U1 U2 ⊆ V .2. the nonzero (real or complex) numbers under multiplication. We can think of it as a set provided with a distinguished element (the identity e) and two distinguished maps. But now U1 and g2 U2 are open neighborhoods of e and g respectively. By continuity of the multiplication map at the point (g1 . A topological group is a group provided with a topology for which the multiplication and inversion maps are continuous. there are open sets U1 . R) and GL(n. the matrix groups GL(n. e). We denote by G/H the space of left cosets xH for x ∈ G.Lecture A21 Basics of Topological Groups Let G be a group. A topological group G is Hausdorff if and only if the onepoint set {e} is closed. If g1 = g2 there exists an open set V = G \ {g2 } that contains g1 and does not contain g2 . U (n). a 2 contradiction. every translate gU or U g of an open subset U is again open (and every translate of a closed set is closed). Rg (x) = xg. U2 that contain −1 g1 and e respectively such that U1 · U2 ⊆ V . Any group can be thought of as a topological group when endowed with the discrete topology. together with the definition of the product topology. I claim they don't intersect: if they did. Let G be a topological group and H a subgroup. By translation. In a topological group the left and right multiplication maps by a fixed element g. R).1. C). There are many examples: the (real or complex) numbers under addition. Suppose that {e} is closed. and the inversion map G → G.
Proof. Proposition A21.. then x−1 y ∈ G \ H which is an open set. which are equivalence classes for the equivalence relation x ∼ y ⇔ xy −1 ∈ H. and we give them the quotient topology. The quotient maps G → G/H and G → H\G are open maps. y) that does not meet Γ. a key part of algebraic topology. Let G be a Hausdorff topological group and let H be a closed subgroup. Definition A21. If H is a normal subgroup of G then the left and right coset spaces are the same.6. it suffices to show that the graph Γ of the equivalence relation x ∼ y ⇔ x−1 y ∈ H is a closed subset of G × G. A continuous. By A20. In this case we know from algebra that G/H has the structure of a group. It follows that the complement of Γ is open. Suppose that (x. But then U −1 × V is an open neighborhood of (x.for the equivalence relation x ∼ y ⇔ x−1 y ∈ H.8 and A21. so Γ is closed. surjective map f : X → Y is called a local homeomorphism if every x ∈ X has a neighborhood U such that the restriction of f to a map U → f (U ) is a homeomorphism. This can be envisaged as a special case of the next result:
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. Local homeomorphisms will play a very important role in the study of covering spaces. Then the homogeneous spaces G/H and H\G are Hausdorff. y) ∈ Γ. By / continuity of multiplication there exist open sets U x−1 .3. similarly we denote by H\G the space of right cosets Hx. hence open. But the saturation of an open set U is uH =
u∈U h∈H
U h. Exercise A21. The classic example of a local homeomorphism is the exponential map t → e2πit from the real line to the circle. Y be topological spaces. multiplication and inversion are continuous. Show that G/H is a topological group. Theorem A21.
the second representation exhibits it as a union of open sets.4. Proof. This is equivalent to the statement that the saturation of an open subset of G is open.4. V y with U · V ⊆ G \ H. We have also just equipped it with a topology.e. Let X. i.5. These are called (left and right) homogeneous spaces of G by H.
Proposition A21.7. Let G be a topological group and suppose that H is a closed subgroup which is discrete (in the induced topology). Then the quotient map π : G → G/H is a local homeomorphism. Proof. There is a neighborhood V of e such that H ∩ V = {e}. By the same kind of argument (continuity of multiplication) that we have already used several times, there exists a neighborhood U of the identity with U −1 U ⊆ V . But now the restriction of π : G → G/H to the set U is injective: if π(u1 ) = π(u2 ) then u−1 u2 ∈ V ∩ H = {e}, and so u1 = u2 . Thus the restriction of π to a map 1 from U to π(U ) is continuous, bijective, and open (A21.4): in other words, it is a homeomorphism. We have proved that π is a homeomorphism near e; by translation, the same holds near any point. A group G acts on a set X if there is given a map G × X → X, written (g, x) → gx, such that ex = x for all x ∈ X and g(hx) = (gh)x for all g, h ∈ G, x ∈ X. If G is a topological group and X is a topological space we define a continuous action by requiring that the associated map G×X → X be continuous. An action is transitive if for all x, y ∈ X there exists g ∈ G with gx = y. Example A21.8. The rotation group SO(n) acts transitively on the unit sphere S n−1 . Proposition A21.9. Suppose that G acts transitively on the Hausdorff space X. For any x ∈ X let Gx denote the stabilizer Gx = {g ∈ G : gx = x}, which is a closed subgroup of G. Then the formula [g] → g · x defines a continuous bijection G/Gx → X. It is a homeomorphism if G is compact. Proof. The continuous map G → X given by g → gx respects the equivalence relation given by the cosets of Gx . Thus it passes to a continuous map G/Gx → X, which is injective by construction, and is surjective because of the transitivity of the action. The final statement follows because G/Gx is compact and X is Hausdorff. For example, considering the stabilizer of a single point (the North pole) gives us the identification S n−1 ∼ SO(n)/SO(n − 1) (using the fact that SO(n) is = compact). 71
Lecture A22 Homotopy Hereafter "map" means "continuous map". Consider the unit square X = I 2 = {(x, y) ∈ R2 : 0 x, y 1}. Imagine two continuous paths γ1 , γ2 : [0, 1] → X, with γ1 beginning at (0, 0) and ending at (1, 1), and γ2 beginning at (0, 1) and ending at (1, 0). It seems obvious that these two paths must cross somewhere in the interior of the square. But this is hard (maybe impossible?) to prove just with the tools that we have developed so far. We need a little homotopy theory. Definition A22.1. Let X and Y be topological spaces and let f0 , f1 : X → Y be continuous maps. A homotopy between f0 and f1 is a continuous map h : X × [0, 1] → Y with h(x, 0) = f0 (x) and h(x, 1) = f1 (x) for all x ∈ X. If there exists a homotopy between f0 and f1 we say that these maps are homotopic. Proposition A22.2. Homotopy is an equivalence relation on the class of maps X →Y. Proof. Clearly any map f is homotopic to itself via the constant homotopy h(x, t) = f (x). If h is a homotopy between f0 and f1 , then k(x, t) = f (x, 1 − t) is a homotopy between f1 and f0 . Finally, homotopies can be concatenated like paths: if h is a homotopy from f0 to f1 and h is a homotopy between f1 and f2 then h (x, 2t) (t 1 ) 2 h(x, t) = h (x, 2t − 1) (t 1 ) 2 is a homotopy between f0 and f2 . Exercise A22.3. Check in detail that the homotopy h defined in the last part of this proof is continuous.
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Remark A22.4. This proof is obviously closely related to the proof that pathconnectedness is an equivalence relation, Lemma A15.10. In fact, path-connectedness is just homotopy of maps where X is a point. Conversely, one can reduce the notion of homotopy to path-connectedness in the space of maps X → Y if one defines a suitable topology on this space. But we won't do that here. Proposition A22.5. If f0 , f1 are homotopic maps X → Y , and g0 , g1 : Y → Z are homotopic maps, then g0 ◦ f0 , g1 ◦ f1 are homotopic maps X → Z. (In particular, homotopic maps remain homotopic after composition on the left or the right with a fixed map.) There is a related notion, that of relative homotopy. Let X be a space and A a closed subspace (one says sometimes that (X, A) is a pair). Then a homotopy relative to A from X to Y is a homotopy h : X × [0, 1] → Y such that h(a, t) is independent of t for all a ∈ A — a homotopy that stays constant on A. This also defines an equivalence relation by the same reasoning as above. Example A22.6. Let X be the unit disc in Euclidean space. Then the identity map X → X is homotopic to the map that sends every point to the origin. A homotopy between these maps is defined by h(x, t) = (1 − t)x. A similar argument can be applied to any star-shaped subset of Euclidean space. Example A22.7. Let X, Y be spaces and let f : X → Y be a continuous map. The mapping cylinder of f is the space Cf = X × [0, 1]
f
Notice that the second example actually includes the first: the disc is the mapping cylinder of the constant map from a sphere to a point. The geometric situation defined in these examples occurs often enough to have a special name. 73
8. Proposition A22. The relation of homotopy equivalence is. and that Y is a deformation retract of the mapping cylinder. A map that has a homotopy inverse is called a homotopy equivalence. The examples above establish that a point is a deformation retract of a disk. Proof.5 that g ◦ g is a homotopy inverse for f ◦ f .
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.10. relative to A. Suppose that f : X → Y and f : Y → Z have homotopy inverses g and g respectively. Let X be a space and A a closed subspace. and one says that the spaces X and Y are homotopy equivalent if there is indeed a homotopy equivalence between them.Definition A22. so X and A are homotopy equivalent. from the identity map on X to a retraction X → A.9. indeed. A deformation retraction of X onto A is a homotopy. then the inclusion A → X and retraction X → A form a pair of homotopy inverse maps. Let X and Y be spaces and let f : X → Y and g : Y → X be maps. Deformation retraction is a special case of the general notion of homotopy equivalence. Then it follows from Proposition A22. Definition A22. For example. If these exist one says that A is a retract or deformation retract of X. A retraction of X onto A is a continuous map X → A which is the identity on A. if A is a deformation retract of X. an equivalence relation. One says that f and g are homotopy inverses if the composite g ◦ f is homotopic to the identity map on X and f ◦ g is homotopic to the identity map on Y .
and one says that the spaces X and Y are homotopy equivalent if there is indeed a homotopy equivalence between them. indeed. Then it follows from Proposition A22.5 that g ◦ g is a homotopy inverse for f ◦ f . so the cone on any space is contractible. the "figure 8". Any mapping cylinder deformation retracts onto the range of the map from which it is defined (as we have seen). For example. there are topological properties that are preserved by homotopy equivalence. so X and A are homotopy equivalent. and the "dumb-bell" are all homotopy equivalent.2. Example A23. Proof. the 1-dimensional spaces given by the "theta curve". if A is a deformation retract of X. Definition A23. A map that has a homotopy inverse is called a homotopy equivalence. although no two of them are homeomorphic.1. and that Y is a deformation retract of the mapping cylinder. Definition A23. Homotopy equivalent spaces can be rather different topologically. then so is Y . an equivalence relation. The relation of homotopy equivalence is.3. 1]/X × {0}. However. If X is pathconnected. Suppose that f : X → Y and f : Y → Z have homotopy inverses g and g respectively. For example. Proposition A23. The cone CX on a space X is the quotient X × [0. then the inclusion A → X and retraction X → A form a pair of homotopy inverse maps. Deformation retraction is a special case of the general notion of homotopy equivalence. Let X and Y be spaces and let f : X → Y and g : Y → X be maps. Suppose X and Y are homotopy equivalent.Lecture A23 Homotopy Equivalence The examples above establish that a point is a deformation retract of a disk. A space is called contractible if it is homotopy equivalent to a point.4. or in other words the mapping cylinder of the constant map from X to a point. One says that f and g are homotopy inverses if the composite g ◦ f is homotopic to the identity map on X and f ◦ g is homotopic to the identity map on Y . 75
.
If A really were a point. Unfortunately this is not true. 1] ⊆ X × [0. y) : −1 y 1} on the y-axis. that is. Let X be a space and A a closed subspace. then t → h(y. Then there is a path γ in X joining g(y) to g(y ). 1] to Y . One says that (X. Let f : X → Y and g : Y → X be mutually inverse homotopy equivalences.5. It follows from the earlier discussion that y and y are in the same path component.13. Thus f (g(y)) and f (g(y )) are in the same path component. the points y and f (tg(y)) belong to the same path component: indeed. if h is a homotopy between f ◦ g and the identity. Now let y. the quotient space X/A would be just the same as X. What is missing in this example is the homotopy extension property (HEP). 1] → Dn × {0} ∪ S n−1 × [0. t) defines a path joining y to f (g(y)). Y is path connected. then the desired extension can be defined by composing with r.e. y ∈ Y and suppose that X is path connected. Now f ◦ γ is a path joining f (g(y)) to f (g(y )). i. The way to think of this is that the initial data consist of a map X → Y together with a homotopy of the restriction of that map to the subspace A.6. A) has the homotopy extension property if any map from the subspace X × {0} ∪ A × [0. Most "nice" pairs of spaces have the HEP: proofs are often inductive based on the following example. 1] (which can be done for instance by projection from the point (0. To see this. Show that Z/A is path-connected.13 shows that Z itself is not path-connected. 3/2) in Rn+1 ). Suppose that X is a space and A a closed subspace. Note that for each y ∈ Y . If A is contractible. Clearly A is contractible. and let A ⊆ Z be the interval {(0. this proves that Z and Z/A cannot be homotopy equivalent. Example A23.7. we might say that "up to homotopy" A is the same as a point. 76
.Proof. S n−1 ) has the homotopy extension property. The pair (Dn . 1] to another space Y can be extended to a map from the whole of X × [0. it is enough to construct geometrically a retraction r : Dn × [0. Definition A23. . Let Z be the "topologist's sine curve" of Exercise A15. Since Exercise A15. So we might be led to hope that if A is contractible. Exercise A23. the HEP says that this homotopy can be extended to a homotopy defined on the whole space X. the quotient map X → X/A is a homotopy equivalence.. X/A is "up to homotopy" the same as X.
By the HEP. such that G(a) = p for all a ∈ A.Proposition A23. t) ∈ A for all a ∈ A and all t. Let p ∈ A be a point and let h : A × [0.
77
. the homotopy h extends to a homotopy H between the identity on X and some map G : X → X. 1]. that is. t) = π(H(x. In particular. t)) for all x ∈ X. To see this we must show that g ◦ π and π ◦ g are homotopic to the identity on X and X/A respectively. I claim. What about the other? The homotopy H has H(a. so there is a map g : X/A → X such that g ◦ π = G. π ◦ H respects the equivalence relation.8. 1] → (X/A) such that k(π(x). a homotopy inverse for π. The first is easy: g ◦ π = G is homotopic to the identity by construction. Suppose that (X. Proof. Note that G respects the equivalence relation defining the quotient space. It follows that there is a map k : (X/A) × [0. t) ∈ X/A is constant for all a ∈ A and t ∈ [0. A) has the HEP and that A is contractible. The first order of business is to find a potential homotopy inverse. 1]. 1] → A be a homotopy between the identity map and the constant map that sends A to p. Therefore. k gives the desired homotopy between π ◦ g and the identity on X/A. This map g is. t ∈ [0. π ◦ H(a. Then the quotient map π : X → X/A is a homotopy equivalence.
2. The natural notion of homotopy for maps of pointed spaces is based homotopy: a based homotopy from (X. t) = y0 for all t ∈ I. y0 ) is a homotopy h : X × I → Y such that h(x0 .3. x0 ) to (Y. x0 ). ∗) → (X. then so is γ γ . A map of pointed spaces (X. 1}. 1] → X such that γ(0) = γ(1) = x0 . 78
. Then γ and γ ◦ g represent the same element of π1 (X. x0 ) and it is called the fundamental group of (X.) Proof. y0 ) is just a map from X to Y that takes x0 to y0 . This object is denoted π1 (X. x0 ). Lemma A24. it is a path γ : [0. we'll omit explicit mention of the base point.1. Moreover. x0 ). The map h(s.Lecture A24 The fundamental group For the next few lectures we are going to consider pointed topological spaces. x0 ). t) = γ((1 − t)s + tg(s)) defines a homotopy between γ and the reparameterized path γ ◦ g. Example A24. It is a pointed space whose base point ∗ is the equivalence class {0. (The composite γ ◦ g is called a reparameterization of γ. Let g : [0. if γ or γ is varied by a (based) homotopy. Let γ be a based loop in (X. 1] be a continuous map with g(0) = 0 and g(1) = 1. x0 ). Definition A24. In particular we can consider the collection of all based homotopy classes of loops in (X. As we shall see. to save space. Thus the operation passes to a "multiplication" operation on π1 (X. Such an object is just a topological space X together with a choice of a base point x0 ∈ X. A loop in a pointed space (X. 1] → [0. x0 ) → (Y. In other words. Sometimes. Let γ and γ be paths in X. 1}. x0 ). there is indeed a natural way to make it into a group. 1]/{0. x0 ) is a map of pointed spaces (S 1 . We regard the unit circle S 1 as the quotient space [0. Recall that the concatenation of these paths is defined by 1 γ (2s) (s 2 ) γ γ (s) = γ (2s − 1) (s 1 ) 2 If γ and γ are (based) loops then so is γ γ .
as usual. Then. to reparameterize them. 1] → X be a path from x0 to x1 .
79
. Let q : [0. Proof. Then h(s. the path φ(γ) := q −1 γ q is a loop based at x1 .5.Proposition A24. There are three things that must be verified: associativity. x1 ∈ X. x0 ). x1 ) are isomorphic. for any loop γ. The effect of the choice of basepoint is summed up in the following. Then the groups π1 (X. x0 ) and π1 (X. Under this operation. x0 ) becomes a group. 1]) defines an identity element. Indeed. t) = γ((1 − t)g(s)) is a homotopy of based loops from γ γ −1 to the identity path. For associativity. Proof. Finally suppose that γ is a loop and let γ −1 be the loop defined by γ −1 (s) = 1 and γ(1 − s). the existence of an identity. Then γ γ −1 is the path s → γ(g(s)) where g(s) = 2s for s 2 1 g(s) = 2 − 2s for s 2 . Lemma A24. in particular. The assignment γ → q γ q −1 gives the inverse homomorphism π1 (X. Let X be a path-connected space and let x0 . and the existence of inverses. x1 ) to π1 (X. x0 ) to π1 (X. so these two groups are isomorphic. and let q −1 (s) = q(1 − s). and the assignment γ → φ(γ) gives a homomorphism of groups from π1 (X. we just need to note that the paths (γ1 γ2 ) γ3 and γ1 (γ2 γ3 ) are related by the reparameterization 1 2s s 4 1 s→ s+ 1 s 1 4 2 1 4 1 1 + 2s 2 s 1 2 The constant loop (e(s) = x0 for all s ∈ [0. The proofs of all three make extensive use of the freedom to deform loops by homotopies and. it is easy to see that e γ and γ e are simply reparameterizations of γ. x1 ).4. π1 (X. if γ is a loop based at x0 .
8. s) = h(2r(1 − s). 80
. the composite f ◦ γ is a based loop in Y . x0 ) → π1 (Y. (b) Suppose that f and g are maps (X.Beware that the isomorphism in the lemma above depends on the choice of the path q. x0 ). y0 ). x1 ) are "isomorphic but not canonically isomorphic". Moreover. the assignment [γ] → [f ◦ γ] passes to homotopy classes and defines a homomorphism π1 (X. Item (c) is called functoriality: it is an extremely important property in many mathematical contexts. Let f : (X. which are homotopic by a homotopy h. But these are homotopic by a homotopy that goes diagonally across the square. Explicitly. Proof. based homotopic maps induce the same homomorphism. Let u ∈ π1 (Y. y0 ). x0 ) and π1 (X. s(2r − 1) + 1 − s) u to H(·. going along the left and top sides defines u g. Definition A24. y0 ) be defined by the loop t → h(x0 . (c) If f : X → Y and g : Y → Z are based maps. we want to define H(r. 0) = f
Corollary A24.7. t) → h(γ(s). then (f ◦ g)∗ = f∗ ◦ g∗ . x0 ) → π1 (Y. y0 ). t). (r (r
1 ) 2 1 ) 2
which gives a homotopy from H(·. One says that the groups π1 (X. (a) and (c) are immediate consequences of the definitions. Part (b) would be easier if we restricted our attention to based homotopies. x0 ) → (Y.6. Consider the map from the unit square to Y defined by (s. and it is denoted by f∗ : π1 (X. If γ is a based loop in X. Then f∗ (v) = ug∗ (v)u−1 for all v ∈ π1 (X. y0 ) be a (basepoint-preserving) map. 1) = u g. but we need the general case. The homomorphism so defined is called the induced homomorphism of the map f . A homotopy equivalence induces an isomorphism of fundamental groups. In particular. The induced homomorphism has the following properties: (a) The identity map induces the identity homomorphism. t). Going along the bottom and right sides of this square defines the path f u. x0 ) → (Y. 2rs) h((1 − s)(2r − 1) + s. Proposition A24.
f∗ ◦ g∗ and g∗ ◦ f∗ are isomorphisms. In particular Corollary A24. Thus by (c) above.
81
. with homotopy inverse g. which implies that f∗ and g∗ are isomorphisms too. If X is contractible.9.Proof. Let f : X → Y be a homotopy equivalence. then its fundamental group is trivial. Then f g and gf are homotopic to the identity. and therefore by (a) and (b) above they induce isomorphisms on fundamental groups.
1. then a lifting of f (over p) is a map g : Y → E that makes the following diagram commute: >E Y in other words p ◦ g = f . together with an extension of the initial data of the homotopy to a map X → Y . respecting the given initial data. given any homotopy of maps A → Y . which we can also think of as a property of the inclusion map i : A → X. The notion of extension of a map can be expressed in diagrammatic terms using the inclusion i: X O
i
?
A
/
Y
Here the solid arrows represent the given data and the dotted arrows represent the extension to be constructed. we can always extend the homotopy to a homotopy of maps X → Y . A). Dual to the notion of extension is the notion of lifting: suppose that p : E → B is a surjective map and f : Y → B is any map. We say that p : E → B has the homotopy lifting property or is a fibration if. given any homotopy of maps Y → B together with a lifting of the initial data of the homotopy over p. Definition A25. 1] 82
/
B
. as one also says. In other words. the inclusion i is a cofibration) if. p has the HLP if. given the data expressed by the solid arrows in the diagram below /E Y × {0} < _
/B
p
Y × [0. The pair has the HEP (or.Lecture A25 Fibrations and the fundamental group Recall the homotopy extension property for a pair X. the entire homotopy can be lifted over p.
3. x0 ) = {1} for some (and hence any) basepoint x0 . for example.4. We may consider the path γ as a homotopy (of maps of a point into E) and. Let γ be a based loop in B and let x ∈ F be a point of the fiber F = p−1 (b0 ). Remark A25. Definition A25. we will say that X is simply connected. The inverse image F = p−1 {b0 } is called the fiber of the fibration. if we do this.g. Suppose now that p : E → B is a fibration and let b0 ∈ B be a base point. If X is path-connected and π1 (X.2. γ 83
. Let p : E → B be a fibration. e. so we can't write down the full details of this. Then prove this statement by any one of a number of approximation techniques. by showing that any loop is homotopic to a "piecewise linear" loop made up of great circle arcs. 1]
γ
/
B
Filling in the dotted arrow provides a lifting γ of the path γ. But there are many other examples. γ and x together provide exactly the data for a homotopy lifting problem: {0}
_
γ ˜ x
/
>E
[0. but this lifting is ˜ not guaranteed to be a loop! All we know is that p(˜ (1)) = γ(1) = b0 . Contractible spaces. but to do so we have to consider a homotopy of maps A → Y as a single map from A to the space Y I of maps I → Y . Let X be a space. Show that the n-sphere S n is simply connected for n 2. For now we are not going to worry too much about where fibrations might come from. that is.5. We haven't discussed the topology of mapping spaces. Hints: First show that it is enough if one knows that any loop in S n is homotopic to one that is not surjective. Instead we will concentrate on what they might tell us about the fundamental group.the dotted arrow can be filled in in such a way as to make the diagram commutative. Definition A25. Exercise A25.9). but it should not be hard to figure out the general idea. The homotopy extension property can be expressed in a similar diagrammatic way. and let B have a fixed basepoint b0 . are simply connected (Corollary A24.
If γ .6. as above. (It's obvious that γ (0) and γ (1) are in the same path component of E — they ˜ ˜ are joined by the path γ . then. and let γ be a based loop in B. Let h(s. right-hand and top sides of the map of the unit square defined by h are constant maps with value x0 . The concatenation (γ )−1 γ is a lifting of a nullhomotopic loop. Definition A25. It is not uniquely determined by this construction ˜ but we shall see in a moment that the path component of the fiber in which it lies is uniquely determined. Therefore the bottom. then their end points are in the same path component of F . The proof begins to build up the connection between fibrations and the fundamental group. let π0 (F ) denote the set of path components of F . For any topological space F . Proof. ˜ ˜ Corollary A25. 0) = γ(s) and h(s. right-hand and top sides of the lifted homotopy H are paths in the fiber F . γ (0) and γ (1) are in the same path component ˜ ˜ ˜ of the fiber F . b0 )). 1) = b0 . We apply the HLP to a homotopy between γ and the constant path.8. 1] × {0}
_
γ ˜
/ <E
H
[0. after all! — so the significance of the lemma is that a ˜ path joining them can be found that lies wholly in F . 1]
h
/B
The bottom. Then h together with γ provide the data for a homotopy lifting problem: ˜ [0. for any lift γ : [0. The concatenation of these three is a path in F from γ (0) to γ (1). 1] × [0. Let p : E → B be a fibration. 1] → E of γ. 84
.γ (1) is a point of thee fiber F .7. t) be such a homotopy with h(s. starting at the same point γ (0) = γ (0).) Proof. If γ is nullhomotopic (that is. it represents the identity element in π1 (B. γ are two liftings of the same loop. To help keep track of this business of path components let's introduce some notation. Lemma A25.
9. as discussed above. When a group action is both free and transitive. b0 ) on the set π0 (F ). b0 ). and a point x ∈ F representing c. and define c to be the path component containing γ (1). Proposition A25. the action is free. For g ∈ π1 (B. ˜ The lemma and corollary above show that this is well defined. Then p ◦ H is a nullhomotopy of the path p ◦ β. The action law (cg )h = cgh follows from the fact that a lifting of a concatenation of loops is a concatenation of liftings of those loops. let γ be ˜ g any lift of γ starting at x.
85
. g) → cg . A (right) action of a group G on a set C is just a map C × G → C. Then given any two points x. on the other hand. Composing with p gives a based loop in B which (tautologically) lifts to a path in E from x to x . then this action is transitive. Thus the action is transitive. The composite p ◦ β is the concatenation of γ with a constant path. let us define cg (read: "c acted on by g") as follows: choose a based loop γ representing g. Suppose now that E is path connected. The construction above defines an action of the fundamental group π1 (B. there exists g ∈ G such that cg = c . such that ce = c and (cg )h = cgh . for fixed c. And ce = c follows from the fact that a constant loop can be lifted to a constant. Suppose that cg = c for some c ∈ π0 (F ) and g ∈ π1 (B. An action is free if cg = c implies g = e. which therefore represents the identity element of π1 (B. so it still ˜ represents g ∈ π1 (B. and it is transitive if for all pairs c. is a bijection from the group G to the set C. if E is path connected. the mapping g → cg . this means that g is represented by a based loop γ in B which lifts to γ such that γ (0) and γ (1) are in the same path component ˜ ˜ ˜ of F . x of the fiber there is a path in E joining them. β is nullhomotopic in E (say by a homotopy H) because E is simply connected. b0 ) as claimed.Suppose that p : E → B is a fibration with fiber F . Let us explain the terminology. Proof. But. b0 ) and c ∈ π0 (F ). written (c. and if in addition E is simply connected. b0 ). Moreover. Construct a loop β in E by concatenating γ with a path in F from γ (0) to ˜ ˜ γ (1). c of elements of C. Suppose additionally that E is simply connected.
There are many different possible covering spaces of the "figure 8" space.1. Example A26. a covering map is "locally the projection of a product with a discrete space". 1] with endpoints identified) or x → e2πix (if we think of the unit circle in C) is a covering map with fiber Z. I'll draw some pictures in class. a the surface of a 2-holed donut). Definition A26.e. Hatcher) give a definition which appears more general than this. One can tessellate the hyperbolic plane by such octagons with vertex angle of 45 degrees. A neighborhood U with the property described in the definition will be called a trivializing neighborhood for the fibration.6. The map R → S 1 defined by x → x mod 1 (if we think of the circle as [0. The map S 1 → S 1 defined by x → nx mod 1 (or z → z n in the complex pictures) is a covering map with fiber an n-point space. Example A26. In other words. but it turns out that the definitions are equivalent for pathconnected base spaces B. The map p above is called a covering map (and E is called a covering space of B) if there is a discrete topological space F (the fiber) such that each point x ∈ B has an open neighborhood U for which there is a homeomorphism p−1 (U ) ∼ U × F making the diagram = ) p−1 (U H
/U ×F HH HH HH HH pr1 p HHH HH HH H#
∼ =
U
commute.4.3.Lecture A26 Covering Spaces Let p : E → B be a surjective map of topological spaces. Then B can be obtained by a suitable identification of the edges of a regular octagon. Let B be a compact oriented surface of genus −2 (i. Example A26. Example A26.5. Example A26. A homeomorphism is a covering map with 1-point fiber. Some authors (e.2. 86
. From this tessellation we obtain a covering map from the plane to B.g.
. finitely many of these product neighborhoods Vt × Wt cover 87
. By connectedness. i. B has an open cover consisting of trivializing neighborhoods. t) should be the unique lifting of the path t → h(y. 1] into finitely many subintervals 0 t1 t2 ··· 1 of length less than the Lebesgue number. t) that begins at f (y). pr2 (H(tk ))) ∈ U × F ∼ p−1 (U ) ⊆ E. and in particular we may identify p−1 (U ) with U × F where F is the (discrete) fiber. By compactness. 1]
h
/
B
First of all let's consider the case when Y is a single point (the "path lifting property"). Now we return to the case of general Y . Remember what we have to show: for each space Y we can fill in the "homotopy lifting diagram" Y × {0} _
f
/ <
E
H
Y × [0. which we just proved. tk+1 ]) ⊆ U for some trivializing neighborhood U . tk+1 ] that agrees with the lift already assumed to have been constructed on [0.Proposition A26. The inverse image of this open cover by the map h is then an open cover of the compact space [0. Partition [0.7. for each fixed y ∈ Y . = By induction. Suppose then than a lifting H has been constructed on [0. We have h([tk . there is no question how to define H for general Y : namely. which has a positive Lebesgue number by the Lebesgue covering theorem. Indeed. it satisfies the homotopy lifting property. Because of the uniqueness of path lifting. fix y0 ∈ Y and consider the following fact: for each t ∈ [0. we have done more than claimed: we have shown that there is a unique possible choice for H.e. H(y. To see this. By definition. A covering map is a fibration. pr2 (H(t)) must be constant for any lift H. The only question is whether the function H defined by this process is continuous as a function of y and t. 1]. Proof. tk ]. 1] there is a product neighborhood Vt × Wt of (y0 . t) such that h(Vt × Wt ) lies in a trivializing neighborhood in B. Therefore the unique continuous lift H of h on [tk . We will construct the desired lifting by induction over these subintervals. tk ] is H(t) = (h(t). we can construct H on [0. 1].
tk ))) ∈ U × F ∼ p−1 (U ) ⊆ E. t). h(V × [tk . in particular. Partition [0. 1]. t) for all t ∈ [0. this suffices to complete the proof. 1].{y0 } × [0. tk+1 ]) is contained in a trivializing neighborhood. So the induction is complete and we have shown (in particular) that H is continuous at (y0 . We're going to prove by induction over subintervals that H is continuous on V × [0.
88
. 1] into finitely many subintervals 0 t1 t2 ··· 1 of length less than . Notice that for each k. t) = (h(y. Let V be the intersection of the finitely many Vt 's that appear and let be a Lebesgue number for the covering of [0. H(y. 1] by the finitely many Wt 's. tk ) is a continuous function of y for y ∈ V . H(y. = and this is a continuous function of (y. π2 (H(y. tk ]. t). tk+1 ]. Suppose then that we already know that H is continuous on V × [0. Since y0 was arbitrary. 1]. But by the formula above. for t ∈ [tk .
3). in particular. as the above proof shows. Remark A27. 1)) · γ (min(0.4. Using a similar argument to the above. Exercise A27. The fiber F = Z is discrete and the total space E = R is contractible.Lecture A27 Applications So. show that if G is any topological group (not necessarily abelian). then π1 (G) is an abelian group. Proposition A27.2.3. We can therefore define a bijection π1 (S 1 ) → Z by sending g ∈ G to 0g ∈ Z. This proves that the bijection π1 (S 1 ) → π0 (Z) = Z is indeed an isomorphism of groups. All that is necessary is to show that this is a group isomorphism. Proof. notice that S 1 itself is a topological group.1. The integer n ∈ π1 (S 1 ) is represented by the map z → z n . In any topological group G. we can finally compute π1 (S 1 ).9 the action of π1 (S 1 ) on π0 (F ) = F is free and effective. the action of π1 (S 1 ) on the fiber Z satisfies 0gh = 0g +0h . The group π1 (S 1 ) is isomorphic to Z. For the expression H(s. which "wraps" the unit circle n times around itself. after all this machinery. For this. 1 − (2 − t)(1 − s))) defines a homotopy from the concatenation of γ and γ to their pointwise product. However. We make use of the covering space R → S 1 described above (Example A26. t) = γ (max((2 − t)s. instead of by concatenating them. The identity map S 1 → S 1 is a generator of the cyclic group π1 (S 1 ) = Z. so according to Proposition A25. (No-retraction theorem) There is no retraction of the disk D2 = {z ∈ C : |z| 1} onto its boundary circle S 1 = {z ∈ C : |z| 1}. the group operation in π1 (G) can be defined by pointwise multiplying paths (using the group operation in G). the lift of a pointwise product of paths (in S 1 ) is clearly their pointwise sum (in R). Theorem A27. Here are some (standard) applications of this calculation.
89
.
bread.Corollary A27. where h(u) and c(u) are the fractions of the ham and cheese.g. (Ham sandwich theorem) Any three bounded measurable subsets of R3 (e. The Borsuk-Ulam theorem gives us a point such that 1 h(u) = h(−u) and c(u) = c(−u). ham. cheese) can be simultaneously volume-bisected by a common plane cut. c(u)).) Define a map S 2 → R2 by sending u ∈ S 2 to the vector (h(u). that lie on the u side of the plane Pu .
92
. I'm sweeping this under the rug. say the bread. thus Pu perfectly divides the ham and the cheese (as well as the bread). respectively. Then h(−u) = 1 − h(u) and c(−u) = 1 − c(u).10. This implies that h(u) = c(u) = 2 . (The existence of such a plane follows from the intermediate value theorem. For each unit vector u ∈ S 2 let Pu be the plane having u as normal vector that volume-bisects one of the ingredients. Proof. in the general case one has to maneuver a little to deal with the possible non-uniqueness of such a plane. | 677.169 | 1 |
Nation's Math Teachers Introduce 27 New Trig Functions Math Teachers Introduce 27 New Trig Functions
All Graduating Students Must Master Gamsin, Negtan, Cosvnx, 24 Others
WASHINGTON—Adding to the six basic functions that have for years made up the foundation of trigonometry, the nation's mathematics teachers reportedly introduced 27 new functions today that high schoolers will be expected to master. "While the core of the trigonometry curriculum has traditionally consisted solely of sine, cosine, tangent, secant, cosecant, and cotangent, henceforth we will be including gasmin, negtan, cosvnx, and two dozen others, such as tosna and cotosna, that our pupils will need to have down pat in order to pass," Coolidge Senior High School trig teacher Robert Beckman said on behalf of the nation's math educators, emphasizing that students will be required to have full understanding of tofsin, pomen, cocosine, phyxyx, fotsin, and fostin as they apply to the various properties of equilateral, isosceles, and scalene triangles. "Students will also need to know the corresponding graphs for the functions. For example, drin forms a sort of stepladder going up the X and Y axes, while codrin forms a stepladder going down. I can assure you that all of these are absolutely crucial to understanding basic trigonometry, not to mention a requisite for anyone seeking to graduate and move on to college." Beckman added that factoring will be cut from the math curriculum entirely because it's "annoying and too fucking hard sometimes." | 677.169 | 1 |
Course content
Learning outcome
1. Knowledge. The student has knowledge of basic numerical methods for approximation of functions, numerical quadrature, numerical solution of ordinary differential equations, and numerical solution of linear and nonlinear equations. The student masters error analysis of numerical methods and understands the concepts stability and convergence.
2. Skills. The student masters basic techniques for analyzing a large selection of numerical algorithms. The student is able to implement the algorithms in a chosen programming language, set up numerical experiments and interpret the results. The student has established sound routines for quality assurance in implemented algorithms.
3. General competence. The student can describe the chosen scientific method and communicate his or her findings in a written scientific report using precise language.
Learning methods and activities
Lectures and exercises. The exercises require the use of a computer. Portfolio assessment is the basis for the grade awarded in the course. This portfolio comprises a written final examination (70%) and a term project (30%). The results for the constituent parts are to be given in %-points, while the grade for the whole portfolio (course grade) is given by the letter grading system. Retake of examination may be given as an oral examination. All teaching materials are available in English. The lectures will be given in English if they are attended by students from the Master's Programme in Mathematics for International students. Students are free to choose Norwegian or English for written assessments.
Compulsory assignments
Exercises
Specific conditions
Exam registration requires that class registration is approved in the same semester, or that compulsory activities are approved in a previous semester. | 677.169 | 1 |
Math Competitions
Undergraduate
International Mathematics Competition for University Students
This competition, for students completing their first, second, third or fourth year of university education consists of two Sessions of five hours each. Problems come from the fields of algebra, analysis (real and complex) and combinatorics. The working language is English. Over the previous 13 competitions, students from 147 universities from 36 countries have participated.
Math Jeopardy Contest
This contest for undergraduates follows the familiar format of the game show, Jeopardy. Teams of up to four students each provide "the questions" to answers in categories such as calculus, differential equations, discrete mathematics, and linear algebra. It is held annually at the MAA-SE conference (and SIAM-SEAS
when the meeting is joint with MAA-SE).
Morgan Prize for Outstanding Research in Mathematics by an Undergraduate
The Morgan Prize is awarded to an undergraduate student (or students having submitted joint work) for outstanding research in mathematics. Any student who is an undergraduate in a college or university in Canada, Mexico, or the United States or its possessions is eligible to be considered for this $1,000 prize, which is awarded annually. The award is made jointly by the American Mathematical Society, the Mathematical Association of America, and the Society for Industrial and Applied Mathematics.
SIAM Student Paper Competition & SIAM Student Paper Prizes The SIAM Student Paper Prizes are awarded every year to the student author(s) of the most outstanding paper(s) submitted to the SIAM Student Paper Competition. This award is based solely on the merit and content of the student's contribution to the submitted paper. The purpose of the Student Paper Prizes is to recognize outstanding scholarship by students in applied mathematics or computing.
SIAM Student Travel Awards
Any full-time student in good standing is eligible to receive an award plus gratis meeting registration. Top priority will be given to students presenting papers at the meeting, with second priority to students who are co-authors of papers to be presented at the meetings. Only students traveling more than 100 miles to the meetings are eligible for the awards.
High School
Interdisciplinary Contest in Modeling (ICM) ICM is an international contest for high school students and college undergraduates. ICM is an extension of the Mathematical Contest in Modeling (MCM). It is designed to develop and advance interdisciplinary problem-solving skills as well as competence in written communication.
International Mathematical Olympiad
The International Mathematical Olympiad (IMO) is the World Championship Mathematics Competition for High School students and is held annually in a different country. The first IMO was held in 1959 in Romania, with 7 countries participating. It has gradually expanded to over 80 countries from 5 continents.
Math League: Building student interest and confidence in mathematics through solving worthwhile problems.
The Math League specializes in math contests designed to stimulate interest and confidence in mathematics for students from the 4th grade through high school. Over 1 million students participate in Math League contests each year. Contest problems are designed to cover a range of mathematical knowledge for each grade level. All of the problems on each contest require no additional knowledge of mathematics beyond the grade level they test.
Moody's Mega Math (M3) Challenge
The M3 Challenge is an applied math modeling competition for high school juniors and seniors held each year in early March. Students along the entire East Coast, from Maine to Florida are eligible to participate. The Challenge is sponsored by The Moody's Foundation and organized by SIAM.
National Society of Professional Surveyors TrigStar Program Contest
The TrigStar Program Contest is an annual high school mathematics competition sponsored by the National Society of Professional Surveyors based on the practical application of Trigonometry. The program recognizes the best students from high schools throughout the nation.
Middle School
Math Counts: Promoting Middle School Math Achievement
MATHCOUNTS® is a national math enrichment, coaching & competition program that promotes middle school mathematics achievement in every U.S. state & territory. With over 23 years experience, MATHCOUNTS is one of the most successful education partnerships involving volunteers, educators, industry sponsors & students. | 677.169 | 1 |
Math made easier: advice from experts
Many students struggle with various kinds of math, including positive and negative number signs, fractions, factoring, graphing and word problems, instructors in the department of mathematics and statistics said.
In fall 2011, the success rate for college algebra, a core math course, was 59 percent, said Mellisa Hardeman, senior instructor in the department. The success rate dropped anther percentage point the following year, she said.
In fall 2012, 50 to 60 percent of pre-core math students had difficulties solving math problems, said Denise LeGrand, director of the Mac I math lab.
Ike McPhearson, math tutor, explained why students may have trouble comprehending math. One reason is that students may come from a home where education is not valued, he said.
A bad experience with an instructor can also change students' attitudes about math.
"You can't take yourself too seriously as a teacher," said Hardeman. Instructors can never give a student too much help passing math, she said.
Students who took a math course in high school before going to college are less likely to struggle with math, Hardeman said. Some students go to college years after graduating high school, however, and may forget everything they learned in their math classes.
Fortunately, there are a number of strategies that can help students overcome these challenges and develop a better understanding of math.
"In order to make math easy for students, show different ways of how to understand it," said McPherson, who has tutored high school and college students. Another way of making math fun for students is to create different games, he said.
According to LeGrand, the most important way to become better at math is to practice math exercises for 20 to 30 minutes.
"They won't see the results right away," said LeGrand, " but if they go to class and focus on work required, they will be successful and they will build confidence."
In addition, students can get help from tutors at the math lab. Each semester, the lab hires 12 tutors, LeGrand said.
For the math-impaired, there is a new math course called Quantitative and Mathematical Reasoning. The course was designed for students who are not science, technology, engineering or mathematics majors. It focuses on practical math, for example, currency exchange rates. The course fulfills the core math requirement, in place of college algebra.
Pre-core math courses, developmental math courses students take if they do not have the prerequisites for college math classes, are becoming more successful, said Tracy Watson, coordinator for pre-core math. The success rate for those courses rose to 77 percent in fall 2012, she said. Previously, the success rate was 37 percent for a 4-year period, she said.
This semester, there are 80 math majors at the university.
"We all like how math works because it all fits together," Watson said.
"Students who major in math develop a sense of thinking and solving problems," said Thomas McMillan, department chair.
Once students better understand math, they will have the confidence to solve not only math problems, but problems in everyday life as well | 677.169 | 1 |
Instead of presenting the standard theoretical treatments that underlie the various numerical methods used by scientists and engineers, Using R for Numerical Analysis in Science and Engineering shows how to use R and its add-on packages to obtain numerical solutions to the complex mathematical problems commonly faced by scientists and engineers.... more...
This book introduces the basic concepts for the numerical modelling of partial differential equations. It details algorithmic and computer implementation aspects and provides a number of easy-to-use programs. more...
The implicit function theorem is one of the most important theorems in analysis and its many variants are basic tools in partial differential equations and numerical analysis. This second edition of Implicit Functions and Solution Mappings presents an updated and more complete picture of the field by including solutions of problems that have been... more...
This witty and engaging stylebook presents the fundamentals of mathematical operations: number systems, first steps in algebra and algebraic notation, common fractions and equations, and much more. 1958 edition. more...
The Whole Truth About Whole Numbers is an introduction to the field of Number Theory for students in non-math and non-science majors who have studied at least two years of high school algebra. Rather than giving brief introductions to a wide variety of topics, this book provides an in-depth introduction to the field of Number Theory. The topics... more...
From zero to infinity, The Book of Numbers is a handy-sized volume which opens up a new realm of knowledge. Where else in one place could you find out how the illegal numbers racket worked, what makes some people see numbers as colours, why the standard US rail gauge exactly matches the axle width of an ancient Roman chariot, and the numerologic...... more... | 677.169 | 1 |
Trigonometry Lesson Plan and Resources - GCSE
A lesson plan and teaching resources used for an observation lesson on GCSE trigonometry. The lesson takes about 2 hours if you have a mid to high ability group. A very high ability group could possibly complete the lesson in 1 hour. The deciding factor is how much time you want to give up to the mathematical discussion. Pupil assessment and exam question worksheets are included for teaching the lesson.These revision worksheets have been designed to support your preparations in the lead up to exams. Whether you are a teacher or parent, these materials will help support those striving to achieve the best they can.<br /><br />
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In this pack you will receive an initial 'Starter for 10' that checks students understanding of key topics. In total there are 10 marks available across some key topics. The questions are mainly tests of fluency on a particular area of maths. For each question there is a supplementary set of questions that you can use depending on how well a pupil has performed on the initial questions. At the end of the session there is another 'Starter for 10' to check understanding at the end. <br /><br />
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These worksheets will highlight the needs of pupils in your group or your child at home. They could be set as homework or could be used during bespoke revision workshops.<br /><br />
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The sheets build up into a full set; by the end they should have covered all of the key topics at grades D & C. They are a must have resource for any classroom. All of the answers have been provided.<br /><br />
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Topics covered in this pack are<br /><br />
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1. Linear Equations<br /><br />
2. Finding nth term<br /><br />
3. Evaluating indices<br /><br />
4. Rounding (decimal places)<br /><br />
5. Angles (alternate and corresponding) | 677.169 | 1 |
Complex Analysis An Introduction to The Theory of Analytic Functions of One Complex Variable standard source of information of functions of one complex variable, this text has retained its wide popularity in this field by being consistently rigorous without becoming needlessly concerned with advanced or overspecialized material. Difficult points have been clarified, the book has been reviewed for accuracy, and notations and terminology have been modernized. Chapter 2, Complex Functions, features a brief section on the change of length and area under conformal mapping, and much of Chapter 8, Global-Analytic Functions, has been rewritten in order to introduce readers to the terminology of germs and sheaves while still emphasizing that classical concepts are the backbone of the theory. Chapter 4, Complex Integration, now includes a new and simpler proof of the general form of Cauchy's theorem. There is a short section on the Riemann zeta function, showing the use of residues in a more exciting situation than in the computation of definite integrals. | 677.169 | 1 |
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Mathematics
Buckle Down EOI Algebra I
Help your high schoolers master the Oklahoma EOI with Buckle
Down! Our in-depth student texts bring you
skill-strengthening review and EOI test practice that will
benefit every student. You'll reinforce key math concepts
tested on the EOI with student-friendly lessons and examples
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Our two free EOI practice tests, modeled after the actual exam,
will help you establish a starting point for instruction and
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Includes: 1 Workbook, 1 Form A Practice Test, 1 Form B
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Aligned to the PASS Standards and Objectives
Includes coverage of number sense and algebraic
operations, relations and functions, and more | 677.169 | 1 |
There is a newer edition of this item:
This text uses the right triangle approach to college algebra. A graphical perspective, with graphs and coordinates developed in Chapter 2, gives students a visual understanding of concepts. The text may be be used with any graphing utility, or with none at all, with equal ease. Modeling provides students with real-world connections to the problems. Some exercises use real data from the fields of biology, demographics, economics, and ecology. The author is known for his clear writing style and numerous quality exercises and applications.
--This text refers to an out of print or unavailable edition of this title.
--This text refers to an out of print or unavailable edition of this title.
$63David Cohen, a senior lecturer at UCLA, was the original author of the successful, well-respected precalculus series--COLLEGE ALGEBRA, ALGEBRA AND TRIGONOMETRY, PRECALCULUS: A PROBLEMS-ORIENTED APPROACH, and PRECALCULUS: WITH UNIT CIRCLE TRIGONOMETRY.
--This text refers to an out of print or unavailable edition of this title.
Top Customer Reviews
I used this book in my first college math class, which happened to be remedial algebra/precalc. I was truly deficient in math entering college. My high school algebra teachers were more interested in telling stories/talking to the kids they coached in sports than teaching math, so the first and last time I ever went over trig functions during my high school years was in the ACT prep course I took.
That being said, my college instructor for the algebra class (a grad student) wasn't bad, but I really connected to this book. It is written as a precalc book, and in fact, the text was slightly modified and re-branded as "Precalculus" in later editions. It would be useless as "baby's first algebra book" but it does a very competent job of preparing you for calc. Going into the college class I mentioned, my adviser was so concerned by my assessment tests that she thought I might not be able to pass the class, which was required for all majors. Well, I did...and became a math major...and I couldn't have done it without this book.
I still go back to it for reference/refreshers every so often (doing so now, in fact, which inspired me to write this review). Its style is excellent for that. I highly recommend this book for anyone wanting a thorough coverage that will take them through calculus, but it is especially suited to older students or people looking to review/brush up on precalc algebra who find recent editions of textbooks distracting and seemingly formatted with texter attention spans in mind.
This book is the same book my son uses in school. It is not a book you can just pick and learn unless you are a mathematical wizard.We bought it for him to work with a tutor.It is a high school math book.Works for him.
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While this book is excellent for review, it is horrible for trying to learn the material. I suggest buying a different book to learn the concepts and buy this book to practice. The questions are challenging. | 677.169 | 1 |
Synopsis
Facts101 is your complete guide to Discrete Mathematics. In this book, you will learn topics such as The Logic of Quantified Statements, Elementary Number Theory and Methods of Proof, Sequences, Mathematical Induction, and Recursion, and Set Theory | 677.169 | 1 |
1 of 4 videos I custom made for an educator in California for an experimental 1-week video homework program. I have only edited the beginning and ending titles. Although not designed for the general public, you may find them useful.
This video provides a quick general math review of concepts needed to do well in Algebra 1. Covers: Order of Operations, Fraction Arithmetic, Basic Algebraic Concepts and Terminology, Converting Verbal Phrases to Algebraic Expressions and other Miscellaneous Math Concepts. (25:17 | 677.169 | 1 |
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