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Corollary 1. Let \( E \) and the functions \( {F}_{n} : {E}^{1} \rightarrow E \) be as in Theorem 1. Suppose the series\n\n\[ \sum {F}_{n}\left( p\right) \]\n\nconverges for some \( p \in I \), and\n\n\[ \sum {F}_{n}^{\prime } \]\n\nconverges uniformly on \( J - Q \), for each finite subinterval \( J \subseteq I \). \n... | Proof. Let\n\n\[ {s}_{n} = \mathop{\sum }\limits_{{k = 1}}^{n}{F}_{k},\;n = 1,2,\ldots, \]\n\nbe the partial sums of \( \sum {F}_{n} \) . From our assumptions, it then follows that the \( {s}_{n} \) satisfy all conditions of Theorem 1. (Verify!) Thus the conclusions of Theorem 1 hold, with \( {F}_{n} \) replaced by \( ... | No |
Corollary 2. If \( E \) and the \( {f}_{n} \) are as in Theorem 2 and if \( \sum {f}_{n} \) converges uniformly to \( f \) on each finite interval \( J \subseteq I \), then \( \int f \) exists on \( I \), and | \[ {\int }_{p}^{x}f = {\int }_{p}^{x}\mathop{\sum }\limits_{{n = 1}}^{\infty }{f}_{n} = \mathop{\sum }\limits_{{n = 1}}^{\infty }{\int }_{p}^{x}{f}_{n}\text{ for any }p, x \in I. \] | Yes |
Corollary 3. Let a function \( f : {E}^{1} \rightarrow {E}^{m}\left( {{}^{ * }{C}^{m}}\right) \) be relatively continuous on \( \left\lbrack {p,{x}_{0}}\right\rbrack \left( {\text{or}\left\lbrack {{x}_{0}, p}\right\rbrack }\right) ,{x}_{0} \neq p{.}^{4} \) If\n\n\[ f\left( x\right) = \mathop{\sum }\limits_{{n = 0}}^{\i... | The proof is sketched in Problems 4 and 5. | No |
Theorem 1. Let the functions \( f, g, h \) be real or complex (or let \( f, g \) be vector valued and \( h \) scalar valued).\n\nIf they are regulated on \( I \), so are \( f \pm g,{fh} \), and \( \left| f\right| \) ; so also is \( f/h \) if \( h \) is bounded away from 0 on \( I \), i.e., \( \left( {\exists \varepsilo... | The proof, based on the usual limit properties, is left to the reader. | No |
Lemma 1 (Heine-Borel). If a closed interval \( A = \left\lbrack {a, b}\right\rbrack \) in \( {E}^{1} \) (or \( {E}^{n} \) ) is covered by open sets \( {G}_{i}\left( {i \in I}\right) \), i.e., | The proof was sketched in Problem 10 of Chapter 4, §6. | No |
Lemma 2. If a function \( f : {E}^{1} \rightarrow T \) is regulated on \( I = \left\lbrack {a, b}\right\rbrack \), then \( f \) can be uniformly approximated by simple step functions on \( I \) . | Proof. By assumption, \( f\left( {p}^{ - }\right) \) exists for each \( p \in (a, b\rbrack \), and \( f\left( {p}^{ + }\right) \) exists for \( p \in \lbrack a, b) \), all finite.\n\nThus, given \( \varepsilon > 0 \) and any \( p \in I \), there is \( {G}_{p}\left( \delta \right) \) ( \( \delta \) depending on \( p \) ... | Yes |
Theorem 2. If \( f : {E}^{1} \rightarrow E \) is regulated on an interval \( I \subseteq {E}^{1} \) and if \( E \) is complete, then \( \int f \) exists on \( I \), exact at every continuity point of \( f \) in \( {I}^{0} \) . | Proof. In view of Problem 14 of \( §5 \), it suffices to consider closed intervals.\n\nThus let \( I = \left\lbrack {a, b}\right\rbrack, a < b \), in \( {E}^{1} \) . Suppose first that \( f \) is the char-\n\n\n\nFIG... | Yes |
Theorem 4 (second law of the mean). Suppose \( f \) and \( g \) are real, \( f \) is monotone with \( f = \int {f}^{\prime } \) on \( I \), and \( g \) is regulated on \( I;I = \left\lbrack {a, b}\right\rbrack \) . Then\n\n\[{\int }_{a}^{b}{fg} = f\left( a\right) {\int }_{a}^{q}g + f\left( b\right) {\int }_{q}^{b}g\tex... | Proof. To fix ideas, let \( f \uparrow \) ; i.e., \( {f}^{\prime } \geq 0 \) on \( I \) .\n\nThe formula \( f = \int {f}^{\prime } \) means that \( f \) is relatively continuous (hence regulated) on \( I \) and differentiable on \( I - Q \) ( \( Q \) countable). As \( g \) is regulated,\n\n\[{\int }_{a}^{x}g = G\left( ... | Yes |
Theorem 5 (integral test of convergence). If \( f : {E}^{1} \rightarrow {E}^{1} \) is nonnegative and nonincreasing on \( I = \lbrack a, + \infty ) \), then\n\n\[ \n{\int }_{a}^{\infty }f\text{ converges iff }\mathop{\sum }\limits_{{n = 1}}^{\infty }f\left( n\right) \text{ does. } \n\] | Proof. As \( f \downarrow, f \) is regulated, so \( \int f \) exists on \( I = \lbrack a, + \infty ) \) . We fix some natural \( k \geq a \) and define\n\n\[ \nF\left( x\right) = {\int }_{k}^{x}f\text{ for }x \in I. \n\]\n\nBy Theorem 3(iii) in \( §5, F \uparrow \) on \( I \) . Thus by monotonicity,\n\n\[ \n\mathop{\li... | Yes |
Theorem 2. F has the limits\n\n\\[ \nF\\left( {1}^{ - }\\right) = \\frac{\\pi }{2}\\text{ and }F\\left( {-{1}^{ + }}\\right) = - \\frac{\\pi }{2}.\n\\]\n\nThus \\( F \\) becomes relatively continuous on \\( \\bar{I} = \\left\\lbrack {-1,1}\\right\\rbrack \\) if one sets\n\n\\[ \nF\\left( 1\\right) = \\frac{\\pi }{2}\\t... | Proof. We have\n\n\\[ \nF\\left( x\\right) = {\\int }_{0}^{x}f = {\\int }_{0}^{c}f + {\\int }_{c}^{x}f,\\;c = \\sqrt{\\frac{1}{2}}.\n\\]\n\nBy substituting \\( s = \\sqrt{1 - {t}^{2}} \\) in the last integral and setting, for brevity, \\( y = \\) \\( \\sqrt{1 - {x}^{2}} \\), we obtain\n\n\\[ \n{\\int }_{c}^{x}f = {\\in... | No |
Theorem 3.\n\n(i) Each \( {F}_{n} \) is differentiable on \( I = \left( {-1,1}\right) \) and relatively continuous on \( \bar{I} = \left\lbrack {-1,1}\right\rbrack \) .\n\n(ii) \( {F}_{n} \) is increasing on \( \bar{I} \) if \( n \) is even, and decreasing if \( n \) is odd.\n\n(iii) \( {F}_{n}^{\prime }\left( x\right)... | The proof is obvious from (12) and the properties of \( {F}_{0} \) . Assertion (iv) ensures\n\n\n\nFIGURE 26\n\nthat the graphs of the \( {F}_{n} \) add up to one curve. By (ii), each \( {F}_{n} \) is one to one (str... | No |
Theorem 4. The sine and cosine functions (s and c) are differentiable, hence continuous, on all of \( {E}^{1} \), with derivatives \( {s}^{\prime } = c \) and \( {c}^{\prime } = - s \) ; that is,\n\n\[{\left( \sin x\right) }^{\prime } = \cos x\\text{ and }{\left( \cos x\right) }^{\prime } = - \sin x.\] | Proof. It suffices to consider the intervals \( \\overline{{J}_{0}} \) and \( \\overline{{J}_{1}} \), for, by (15), all properties of \( s \) and \( c \) repeat themselves, with period \( {2\\pi } \), on the rest of \( {E}^{1} \). By (13),\n\n\[s = {F}_{0}^{-1}\\text{ on }\\overline{{J}_{0}} = \\left\\lbrack {-\\frac{\... | Yes |
Problem 33. There is yet another bijection that lets us prove that a set of size \( n \) has \( {2}^{n} \) subsets. Namely, for each subset \( S \) of \( \left\lbrack n\right\rbrack = \{ 1,2,\ldots, n\} \), define a function (traditionally denoted by \( {\chi }_{S} \) ) as follows. \( {}^{a} \n\n\[ \n{\chi }_{S}\left( ... | (b) We define a function \( f \) from the set of subsets of \( \left\lbrack n\right\rbrack = \{ 1,2,\ldots, n\} \) to the set of functions from \( \left\lbrack n\right\rbrack \) to \( \{ 0,1\} \) by \( f\left( S\right) = {\chi }_{S} \) . Explain why \( f \) is a bijection.\n\n(c) Why does the fact that \( f \) is a bij... | No |
Theorem 1.2.7. The number of subsets of an n-element set is \( {2}^{n} \) . | The proofs in Problem 28 and Problem 33 use essentially the same bijection, but they interpret sequences of zeros and ones differently, and so end up being different proofs. We will give yet another proof, using bijections similar to those we used in proving the Pascal Equation, at the beginning of Chapter 2. | No |
Explain why it is that the number of bijections from an \( n \) - element set to an \( n \) -element set is equal to \( n \) times the number of bijections from an \( \left( {n - 1}\right) \) -element subset to an \( \left( {n - 1}\right) \) -element set. What does this have to do with Problem 27? | and if \( {b}_{n} \) stands for the number of bijections from an \( n \) -element set to an \( n \) -element set, then\n\n\[ {b}_{n} = n{b}_{n - 1}\text{.} \]\n\n(2.2) | Yes |
Problem 120. Show that the following algorithm (known as Dijkstra's algorithm) applied to a weighted graph whose vertices are labelled 1 to \( n \) gives, for each \( i \), the distance from vertex 1 to \( \mathrm{i} \) as \( d\left( i\right) \) . | 1. Let \( d\left( 1\right) = 0 \) . Let \( d\left( i\right) = \infty \) for all other \( i \) . Let \( v\left( 1\right) = 1 \) . Let \( v\left( j\right) = 0 \) for all other \( j \) . For each \( i \) and \( j \), let \( w\left( {i, j}\right) \) be the minimum weight of an\n\nedge between \( i \) and \( j \), or \( \in... | Yes |
Theorem 3.1.4. If \( 0 \leq k \leq n \), the number of \( k \) -element subsets of an \( n \) -element set is given by\n\n\[ \left( \begin{array}{l} n \\ k \end{array}\right) = \frac{{n}^{\underline{k}}}{k!} = \frac{n!}{k!\left( {n - k}\right) !}. \] | We define \( \left( \begin{array}{l} n \\ k \end{array}\right) \) to be 0 if \( k > n \), because then there are no \( k \) -element subsets of an \( n \) -element set. Notice that this is what the middle term of the formula in the theorem gives us. This explains the entries of row 8 of our table. For now we jump over ... | No |
Problem 123. Suppose we wish to place the books in Problem 122 (satisfying the assumptions we made there) so that each shelf gets at least one book. Now in how many ways may we place the books? (Hint: how can you make sure that each shelf gets at least one book before you start the process described in Problem 122?) (h... | The assignment of which books go to which shelves of a bookcase is simply a function from the books to the shelves. But a function doesn't determine which book sits to the left of which others on the shelf, and this information is part of how the books are arranged on the shelves. In other words, the order in which the... | Yes |
Problem 155. Write down the rows of Stirling's triangle of the first kind for \( k = 0 \) to 6 . | By definition, the Stirling numbers of the first kind are also change of basis coefficients. The Stirling numbers of the first and second kind are change of basis coefficients from the falling factorial powers of \( x \) to the ordinary factorial powers, and vice versa. | No |
Problem 233. We get a more elegant proof if we ask for a picture enumerator for \( {A}_{1} \cup {A}_{2} \cup \cdots \cup {A}_{n} \) . so let us assume \( A \) is a set with a picture function \( P \) defined on it and that each set \( {A}_{i} \) is a subset of \( A \) . | (a) By thinking about how we got the formula for the size of a union, write down instead a conjecture for the picture enumerator of a union. You could use notation like \( {E}_{P}\left( {\mathop{\bigcap }\limits_{{i : i \in S}}{A}_{i}}\right) \) for the picture enumerator of the intersection of the sets \( {A}_{i} \) f... | No |
Problem 234. Frequently when we apply the principle of inclusion and exclusion, we will have a situation like that of part (d) of Problem 231.d. That is, we will have a set \( A \) and subsets \( {A}_{1},{A}_{2},\ldots ,{A}_{n} \) and we will want the size or the probability of the set of elements in \( A \) that are n... | We can find a very elegant way of writing the formula in Problem 234 if we let \( \mathop{\bigcap }\limits_{{i : i \in \varnothing }}{A}_{i} = A \) . for this reason, if we have a family of subsets \( {A}_{i} \) of a set \( A \), we define \( {}^{1}\mathop{\bigcap }\limits_{{i : i \in \varnothing }}{A}_{i} = A \) . | No |
Problem 272. If the list \( \left( {{i\sigma }\left( i\right) {\sigma }^{2}\left( i\right) \ldots {\sigma }^{n}\left( i\right) }\right) \) does not have repeated elements but the list \( \left( {{i\sigma }\left( i\right) {\sigma }^{2}\left( i\right) \ldots {\sigma }^{n}\left( i\right) {\sigma }^{n + 1}\left( i\right) }... | We say \( {\sigma }^{j}\left( i\right) \) is an element of the cycle \( \left( {{i\sigma }\left( i\right) {\sigma }^{2}\left( i\right) \ldots {\sigma }^{n}\left( i\right) }\right) \) . Notice that the case \( j = 0 \) means \( i \) is an element of the cycle. Notice also that if \( j > n,{\sigma }^{j}\left( i\right) = ... | No |
Problem 275. If two cycles of \( \sigma \) have an element in common, what can we say about them? | Problem 275 leads almost immediately to the following theorem. | No |
Theorem 6.2.3. Suppose a group acts on a set \( S \) . The orbits of \( G \) form a partition of \( S \) . | It is probably worth pointing out that this theorem tells us that the orbit \( {Gx} \) is also the orbit \( {Gy} \) for any element \( y \) of \( {Gx} \) . | No |
Problem 290. Complete the proof of Theorem 6.2.3. | Notice that thinking in terms of orbits actually hides some information about the action of our group. When we computed the multiset of all results of acting on \( \{ 1,2\} \) with the elements of \( {D}_{4} \), we got an eight-element multiset containing each side twice. When we computed the multiset of all results of... | No |
Problem 300. A second computation of the result of Problem 299 can be done as follows.\n\n(a) Let \( \widehat{\chi }\left( {\sigma, x}\right) = 1 \) if \( \sigma \left( x\right) = x \) and let \( \widehat{\chi }\left( {\sigma, x}\right) = 0 \) otherwise. Notice that \( \widehat{\chi } \) is different from the \( \chi \... | (b) Reverse the order of the previous summation in order to convert it into a single sum involving the function \( \chi \) given by\n\n\[ \chi \left( \sigma \right) = \text{the number of elements of}S\text{left fixed by}\sigma \text{.} \] | No |
Problem 331. If \( f : S \rightarrow T \) is a function, we say that \( f \) maps \( x \) to \( y \) as another way to say that \( f\left( x\right) = y \) . Suppose \( S = T = \{ 1,2,3\} \) . Give a function from \( S \) to \( T \) that is not onto. | Notice that two different members of \( S \) have mapped to the same element of \( T \) . Thus when we say that \( f \) associates one and only one element of \( T \) to each element of \( S \), it is quite possible that the one and only one element \( f\left( 1\right) \) that \( f \) maps 1 to is exactly the same as t... | No |
If we reverse all the arrows in the digraph of a bijection \( f \), we get the digraph of another function \( g \) . Is \( g \) a bijection? What is \( f\left( {g\left( x\right) }\right) \) ? What is \( g\left( {f\left( x\right) }\right) \) ? | If \( f \) is a function from \( S \) to \( T \), if \( g \) is a function from \( T \) to \( S \), and if \( f\left( {g\left( x\right) }\right) = x \) for each \( x \) in \( T \) and \( g\left( {f\left( x\right) }\right) = x \) for each \( x \) in \( S \), then we say that \( g \) is an inverse of \( f \) (and \( f \)... | No |
Explain why a bijection must have an inverse. | Since a function with an inverse has exactly one inverse \( g \), we call \( g \) the inverse of \( f \) . From now on, when \( f \) has an inverse, we shall denote its inverse by \( {f}^{-1} \) . Thus \( f\left( {{f}^{-1}\left( x\right) }\right) = x \) and \( {f}^{-1}\left( {f\left( x\right) }\right) = x \) . Equivale... | No |
Suppose that \( R \) is an equivalence relation on a set \( X \) and for each \( x \in X \), let \( {C}_{x} = \{ y \mid y \in X \) and \( {yRx}\} \) . If \( {C}_{x} \) and \( {C}_{z} \) have an element \( y \) in common, what can you conclude about \( {C}_{x} \) and \( {C}_{z} \) (besides the fact that they have an ele... | In Problem 352 the sets \( {C}_{x} \) are called equivalence classes of the equivalence relation \( R \) . You have just proved that if \( R \) is an equivalence relation of the set \( X \), then each element of \( X \) is in exactly one equivalence class of \( R \) . Recall that a partition of a set \( X \) is a set o... | No |
What postage do you think we can make with five and six cent stamps? Is there a number \( N \) such that if \( n \geq N \), then we can make \( n \) cents worth of postage? | You probably see that we can make \( n \) cents worth of postage as long as \( n \) is at least 20. However you didn't try to make 26 cents in postage by working with 25 cents; rather you saw that you could get 20 cents and then add six cents to that to get 26 cents. Thus if we want to prove by induction that we are ri... | Yes |
If \( \mathcal{F} \) and \( \mathcal{G} \) are species of subsets of \( X \), how is the EGF for \( \mathcal{F} \cdot \mathcal{G} \) related to the EGFs for \( F \) and \( G \) ? | Prove you are right. (h) | No |
Problem 398. Without giving the proof, how can you compute the EGF \( f\left( x\right) \) for the number of structures using a set of size \( n \) in the species \( {\mathcal{F}}_{1} \cdot {\mathcal{F}}_{2}\cdots \cdot {\mathcal{F}}_{k} \) of structures on \( k \) -tuples of subsets of \( X \) from the EGFs \( {f}_{i}\... | Theorem C.4.1. If \( {\mathcal{F}}_{1},{\mathcal{F}}_{2},\ldots ,{\mathcal{F}}_{n} \) are species of subsets of the set \( X \) and \( {\mathcal{F}}_{i} \) has EGF \( {f}_{i}\left( x\right) \) , then the family of \( k \) -tuple structures \( {\mathcal{F}}_{1} \cdot {\mathcal{F}}_{2}\cdots \cdot {\mathcal{F}}_{n} \) ha... | Yes |
Lemma 8.8. If \( \bigtriangleup {ABC} \) is nondegenerate and its angle bisector at \( A \) intersects \( \left\lbrack {BC}\right\rbrack \) at the point \( D \) . Then\n\n\[ \frac{AB}{AC} = \frac{DB}{DC}. \] | Proof. Applying Claim 20.22, we get that\n\n\n\n\[ \frac{\operatorname{area}\left( {\bigtriangleup {ABD}}\right) }{\operatorname{area}\left( {\bigtriangleup {ACD}}\right) } = \frac{BD}{CD} \]\n\nBy Proposition 8.10 t... | Yes |
Theorem 1.1.2 Let \( A, B \), and \( C \) be subsets of a universal set \( X \) . Then the following hold:\n\n(a) \( A \cup {A}^{c} = X \) ;\n\n(b) \( A \cap {A}^{c} = \varnothing \) ;\n\n(c) \( {\left( {A}^{c}\right) }^{c} = A \) ;\n\n(d) (Distributive law) \( A \cap \left( {B \cup C}\right) = \left( {A \cap B}\right)... | Proof: We prove some of the results and leave the rest for the exercises.\n\n(a) Clearly, \( A \cup {A}^{c} \subset X \) since both \( A \) and \( {A}^{c} \) are subsets of \( X \) . Now let \( x \in X \) . Then either \( x \) is an element of \( A \) or it is not an element of \( A \) . In the first case, \( x \in A \... | No |
Theorem 1.1.3 Let \( \left\{ {{A}_{i} : i \in I}\right\} \) be an indexed family of subsets of a universal set \( X \) and let \( B \) be a subset of \( X \) . Then the following hold:\n\n(a) \( B \cup \left( {\mathop{\bigcap }\limits_{{i \in I}}{A}_{i}}\right) = \mathop{\bigcap }\limits_{{i \in I}}B \cup {A}_{i} \); | Proof of (a): Let \( x \in B \cup \left( {\mathop{\bigcap }\limits_{{i \in I}}{A}_{i}}\right) \) . Then \( x \in B \) or \( x \in \mathop{\bigcap }\limits_{{i \in I}}{A}_{i} \) . If \( x \in B \), then \( x \in B \cup {A}_{i} \) for all \( i \in I \) and, thus, \( x \in \mathop{\bigcap }\limits_{{i \in I}}B \cup {A}_{i... | Yes |
Theorem 1.2.1 Let \( f : X \rightarrow Y \) . If there are two functions \( g : Y \rightarrow X \) and \( h : Y \rightarrow X \) such that \( g\left( {f\left( x\right) }\right) = x \) for every \( x \in X \) and \( f\left( {h\left( y\right) }\right) = y \) for every \( y \in Y \), then \( f \) is bijective and \( g = h... | Proof: First we prove that \( f \) is surjective. Let \( y \in Y \) and set \( x = h\left( y\right) \) . Then, from the assumption on \( h \), we have \( f\left( x\right) = f\left( {h\left( y\right) }\right) = y \) . This shows that \( f \) is surjective.\n\nNext we prove that \( f \) is injective. Let \( x,{x}^{\prime... | Yes |
Theorem 1.2.3 Let \( f : X \rightarrow Y \) be a function, let \( A \) be a subset of \( X \), and let \( B \) be a subset of \( Y \). The following hold:\n\n(a) \( A \subset {f}^{-1}\left( {f\left( A\right) }\right) \).\n\n(b) \( f\left( {{f}^{-1}\left( B\right) }\right) \subset B \). | Proof: We prove (a) and leave (b) as an exercise.\n\n(a) Let \( x \in A \). By the definition of image, \( f\left( x\right) \in f\left( A\right) \). Now, by the definition of preimage, \( x \in {f}^{-1}\left( {f\left( A\right) }\right) \). \( ▱ \) | No |
Theorem 1.2.4 Let \( f : X \rightarrow Y \) be a function, let \( A, B \subset X \), and let \( C, D \subset Y \). The following hold:\n\n(a) If \( C \subset D \), then \( {f}^{-1}\left( C\right) \subset {f}^{-1}\left( D\right) \) ;\n\n(b) \( {f}^{-1}\left( {D \smallsetminus C}\right) = {f}^{-1}\left( D\right) \smallse... | Proof: We prove (b) and leave the other parts as an exercise.\n\n(b) We prove first that \( {f}^{-1}\left( {D \smallsetminus C}\right) \subset {f}^{-1}\left( D\right) \smallsetminus {f}^{-1}\left( C\right) \) . Let \( x \in {f}^{-1}\left( {D \smallsetminus C}\right) \) . Then, from the definition of inverse image, we g... | No |
Theorem 1.2.5 Let \( f : X \rightarrow Y \) be a function, let \( {\left\{ {A}_{\alpha }\right\} }_{\alpha \in I} \) be an indexed family of subsets of \( X \), and let \( {\left\{ {B}_{\beta }\right\} }_{\beta \in J} \) be an indexed family of subsets of \( Y \) . The following hold:\n\n(a) \( f\left( {\mathop{\bigcup... | (a) Let \( y \in f\left( {\mathop{\bigcup }\limits_{{\alpha \in I}}{A}_{\alpha }}\right) \) . From the definition of image of a set, there is \( x \in \mathop{\bigcup }\limits_{{\alpha \in I}}{A}_{\alpha } \) such that \( y = f\left( x\right) \) . From the definition of union of a family of sets, there is \( {\alpha }_... | Yes |
Theorem 1.2.6 Let \( f : X \rightarrow Y \) and \( g : Y \rightarrow Z \) be two functions and let \( B \subset Z \) . The following hold:\n\n(a) \( {\left( g \circ f\right) }^{-1}\left( B\right) = {f}^{-1}\left( {{g}^{-1}\left( B\right) }\right) \) ;\n\n(b) If \( f \) and \( g \) are injective, then \( g \circ f \) is... | Proof: We prove (d) and leave the other parts as an exercise.\n\n(d) Suppose \( g \circ f \) is injective and let \( x,{x}^{\prime } \in X \) be such that \( f\left( x\right) = f\left( {x}^{\prime }\right) \) . Then \( \left( {g \circ f}\right) \left( x\right) = \) \( g\left( {f\left( x\right) }\right) = g\left( {f\lef... | No |
Theorem 1.2.7 Suppose \( A \) is an infinite set. Then there exists a one-to-one function \( f : \mathbb{N} \rightarrow A \) . | Proof: Let \( A \) be an infinite set. We define \( f \) as follows. Choose any element \( {a}_{1} \in A \) and set \( f\left( 1\right) = {a}_{1} \) . Now the set \( A \smallsetminus \left\{ {a}_{1}\right\} \) is again infinite (otherwise \( A = \{ a\} \cup \left( {A \smallsetminus \left\{ {a}_{1}\right\} }\right) \) w... | Yes |
Theorem 1.3.1 — Principle of Mathematical Induction. For each natural number \( n \in \mathbb{N} \) , suppose that \( P\left( n\right) \) denotes a proposition which is either true or false. Let \( A = \{ n \in \mathbb{N} : P\left( n\right) \) is true \( \} \) . Suppose the following conditions hold:\n\n(a) \( 1 \in A ... | Proof: Suppose conditions (a) and (b) hold. Assume, by way of contradiction, that \( A \neq \mathbb{N} \) . Set \( B = \mathbb{N} \smallsetminus A \), that is, \( B = \{ n \in \mathbb{N} : P\left( n\right) \) is false \( \} \) . Then \( B \) is a nonempty subset of \( \mathbb{N} \) . By the Well-Ordering Property of th... | Yes |
Theorem 1.3.2 — Generalized Principle of Mathematical Induction. Let \( {n}_{0} \in \mathbb{N} \) and for each natural \( n \geq {n}_{0} \), suppose that \( P\left( n\right) \) denotes a proposition which is either true or false. Let \( A = \{ n \in \) \( \mathbb{N} : P\left( n\right) \) is true \( \} \) . Suppose the ... | Proof: Suppose conditions (a) and (b) hold. Assume, by way of contradiction, that \( A \nsupseteq \{ k \in \mathbb{N} \) : \( \left. {k \geq {n}_{0}}\right\} \) . Set \( B = \left\{ {n \in \mathbb{N} : n \geq {n}_{0}, P\left( n\right) \text{is false}}\right\} \) . Then \( B \) is a nonempty subset of \( \mathbb{N} \) .... | Yes |
Theorem 1.3.3 — Principle of Strong Induction. For each natural \( n \in \mathbb{N} \) , suppose that \( P\left( n\right) \) denotes a proposition which is either true or false. Let \( A = \{ n \in \mathbb{N} : P\left( n\right) \) is true \( \} \) . Suppose the following two conditions hold:\n\n(a) \( 1 \in A \) .\n\n(... | Remark 1.3.4 Note that the inductive step above says that, in order to prove \( P\left( {k + 1}\right) \) is true, we may assume not only that \( P\left( k\right) \) is true, but also that \( P\left( 1\right), P\left( 2\right) ,\ldots, P\left( {k - 1}\right) \) are true.\n\nThere is also a generalized version of this t... | Yes |
Proposition 1.4.1 For \( x, y, z \in \mathbb{R} \), the following hold:\n\n(a) If \( x + y = x + z \), then \( y = z \) ; | Proof: (a) Suppose \( x + y = x + z \) . Adding \( - x \) (which exists by axiom (2d)) to both sides, we have\n\n\[ \left( {-x}\right) + \left( {x + y}\right) = \left( {-x}\right) + \left( {x + z}\right) . \]\n\nThen axiom (1a) gives\n\n\[ \left\lbrack {\left( {-x}\right) + x}\right\rbrack + y = \left\lbrack {\left( {-... | Yes |
Proposition 1.4.2 Let \( x, y, M \in \mathbb{R} \) and suppose \( M > 0 \) . The following properties hold:\n\n(a) \( \left| x\right| \geq 0 \) ;\n\n(b) \( \left| {-x}\right| = \left| x\right| \)\n\n(c) \( \left| {xy}\right| = \left| x\right| \left| y\right| \)\n\n(d) \( \left| x\right| < M \) if and only if \( - M < x... | Proof: We prove (d) and leave the other parts as an exercise.\n\n(d) Suppose \( \left| x\right| < M \) . In particular, this implies \( M > 0 \) . We consider the two cases separately: \( x \geq 0 \) and \( x < 0 \) . Suppose first \( x \geq 0 \) . Then \( \left| x\right| = x \) and, hence, \( - M < 0 \leq x = \left| x... | No |
Theorem 1.4.3 — Triangle Inequality. Given \( {x, y} \in \mathbb{R} \) , \n\n\[ \left| {x + y}\right| \leq \left| x\right| + \left| y\right| \] | Proof: From the observation above, we have \n\n\[ - \left| x\right| \leq x \leq \left| x\right| \] \n\n\[ - \left| y\right| \leq y \leq \left| y\right| \] \n\nAdding up the inequalities gives \n\n\[ - \left| x\right| - \left| y\right| \leq x + y \leq \left| x\right| + \left| y\right| \] \n\nSince \( - \left| x\right| -... | Yes |
Proposition 1.5.1 Let \( A \) be a nonempty subset of \( \mathbb{R} \) that is bounded above. Then \( \alpha = \sup A \) if and only if\n\n(1’) \( x \leq \alpha \) for all \( x \in A \) ;\n\n(2’) For any \( \varepsilon > 0 \), there exists \( a \in A \) such that \( \alpha - \varepsilon < a \) . | Proof: Suppose first that \( \alpha = \sup A \) . Then clearly (1’) holds (since this is identical to condition (1) in the definition of supremum). Now let \( \varepsilon > 0 \) . Since \( \alpha - \varepsilon < \alpha \), condition (2) in the definition of supremum implies that \( \alpha - \varepsilon \) is not an upp... | Yes |
Theorem 1.5.3 Let \( A \) and \( B \) be nonempty sets and \( A \subset B \). Suppose \( B \) is bounded above. Then \( \sup A \leq \sup B \). | Proof: Let \( M \) be an upper bound for \( B \), then for \( x \in B, x \leq M \). In particular, it is also true that \( x \leq M \) for \( x \in A \). Thus, \( A \) is also bounded above. Now, since \( \sup B \) is an upper bound for \( B \), it is also an upper bound for \( A \). Then, by the second condition in th... | Yes |
Theorem 1.6.1 — The Archimedean Property. The set of natural numbers is unbounded above. | Proof: Let us assume by contradiction that \( \mathbb{N} \) is bounded above. Since \( \mathbb{N} \) is nonempty,\n\n\[ \alpha = \sup \mathbb{N} \]\n\nexists and is a real number. By Proposition 1.5.1 (with \( \varepsilon = 1 \) ), there exists \( n \in \mathbb{N} \) such that\n\n\[ \alpha - 1 < n \leq \alpha \text{.} ... | Yes |
Theorem 1.6.2 The following hold:\n\n(a) For any \( x \in \mathbb{R} \), there exists \( n \in \mathbb{N} \) such that \( x < n \) ; | Proof: (a) Fix any \( x \in \mathbb{R} \) . Since \( \mathbb{N} \) is not bounded above, \( x \) cannot be an upper bound of \( \mathbb{N} \) . Thus, there exists \( n \in \mathbb{N} \) such that \( x < n \) . | Yes |
Theorem 1.6.3 - The Density Property of \( \\mathbb{Q} \) . If \( x \) and \( y \) are two real numbers such that \( x < y \), then there exists a rational number \( r \) such that\n\n\[ x < r < y\\text{.} \] | Proof: We are going to prove that there exist an integer \( m \) and a positive integer \( n \) such that\n\n\[ x < m/n < y \]\n\nor, equivalently,\n\n\[ {nx} < m < {ny} = {nx} + n\\left( {y - x}\\right) . \]\n\nSince \( y - x > 0 \), by Theorem 1.6.2 (3), there exists \( n \\in \\mathbb{N} \) such that \( 1 < n\\left(... | Yes |
Proposition 1.6.4 The number \( \sqrt{2} \) is irrational. | Proof: Suppose, by way of contradiction, that \( \sqrt{2} \in \mathbb{Q} \) . Then there are integers \( r \) and \( s \) with \( s \neq 0 \) , such that\n\n\[ \sqrt{2} = \frac{r}{s} \]\n\nBy canceling out the common factors of \( r \) and \( s \), we may assume that \( r \) and \( s \) have no common factors.\n\nNow, ... | Yes |
Theorem 1.6.5 Let \( x \) and \( y \) be two real numbers such that \( x < y \). Then there exists an irrational number \( t \) such that\n\n\[ x < t < y\text{.} \] | Proof: Since \( x < y \), one has\n\n\[ x - \sqrt{2} < y - \sqrt{2} \]\n\nBy Theorem 1.6.3, there exists a rational number \( r \) such that\n\n\[ x - \sqrt{2} < r < y - \sqrt{2} \]\n\nThis implies\n\n\[ x < r + \sqrt{2} < y \]\n\nSince \( r \) is rational, the number \( t = r + \sqrt{2} \) is irrational (see Exercise ... | Yes |
Lemma 2.1.2 Let \( \ell \geq 0 \) . If \( \ell < \varepsilon \) for all \( \varepsilon > 0 \), then \( \ell = 0 \) . | Proof: This is easily proved by contraposition. If \( \ell > 0 \), then there is a positive number, for example \( \varepsilon = \ell /2 \), such that \( \varepsilon < \ell \) . \( ▱ \) | Yes |
Theorem 2.1.3 A convergent sequence \( \left\{ {a}_{n}\right\} \) has at most one limit. | Proof: Suppose \( \left\{ {a}_{n}\right\} \) converges to \( a \) and \( b \) . Then given \( \varepsilon > 0 \), there exist positive integers \( {N}_{1} \) and \( {N}_{2} \) such that\n\n\[ \left| {{a}_{n} - a}\right| < \varepsilon /2\text{ for all }n \geq {N}_{1} \]\n\nand\n\n\[ \left| {{a}_{n} - b}\right| < \vareps... | Yes |
Lemma 2.1.4 Given real numbers \( a, b \), then \( a \leq b \) if and only if \( a < b + \varepsilon \) for all \( \varepsilon > 0 \) . | Proof: Suppose \( a < b + \varepsilon \) for all \( \varepsilon > 0 \) . And suppose, by way of contradiction, that \( a > b \) . then set \( {\varepsilon }_{0} = a - b \) . Then \( {\varepsilon }_{0} > 0 \) . By assumption, we should have \( a < b + {\varepsilon }_{0} = b + a - b = a \), which is a contradiction. It f... | Yes |
Theorem 2.1.5 — Comparison Theorem. Suppose \( \left\{ {a}_{n}\right\} \) and \( \left\{ {b}_{n}\right\} \) converge to \( a \) and \( b \) , respectively, and \( {a}_{n} \leq {b}_{n} \) for all \( n \in \mathbb{N} \) . Then \( a \leq b \) . | Proof: For any \( \varepsilon > 0 \), there exist \( {N}_{1},{N}_{2} \in \mathbb{N} \) such that\n\n\[ a - \frac{\varepsilon }{2} < {a}_{n} < a + \frac{\varepsilon }{2},\;\text{ for }n \geq {N}_{1}, \]\n\n\[ b - \frac{\varepsilon }{2} < {b}_{n} < b + \frac{\varepsilon }{2},\;\text{ for }n \geq {N}_{2}. \]\n\nChoose \( ... | Yes |
Theorem 2.1.6 - The Squeeze Theorem. Suppose the sequences \( \left\{ {a}_{n}\right\} ,\left\{ {b}_{n}\right\} \), and \( \left\{ {c}_{n}\right\} \) satisfy\n\n\[ \n{a}_{n} \leq {b}_{n} \leq {c}_{n}\text{ for all }n \in \mathbb{N},\n\]\n\nand \( \mathop{\lim }\limits_{{n \rightarrow \infty }}{a}_{n} = \mathop{\lim }\li... | Proof: Fix any \( \varepsilon > 0 \) . Since \( \mathop{\lim }\limits_{{n \rightarrow \infty }}{a}_{n} = \ell \), there exists \( {N}_{1} \in \mathbb{N} \) such that\n\n\[ \n\ell - \varepsilon < {a}_{n} < \ell + \varepsilon\n\]\n\nfor all \( n \geq {N}_{1} \) . Similarly, since \( \mathop{\lim }\limits_{{n \rightarrow ... | Yes |
Theorem 2.1.7 A convergent sequence is bounded. | Proof: Suppose the sequence \( \left\{ {a}_{n}\right\} \) converges to \( a \) . Then, for \( \varepsilon = 1 \), there exists \( N \in \mathbb{N} \) such that\n\n\[ \left| {{a}_{n} - a}\right| < 1\text{ for all }n \geq N. \]\n\nSince \( \left| {a}_{n}\right| - \left| a\right| \leq \left| \right| {a}_{n}\left| -\right|... | Yes |
Lemma 2.1.8 Let \( {\left\{ {n}_{k}\right\} }_{k} \) be a sequence of positive integers with\n\n\[ \n{n}_{1} < {n}_{2} < {n}_{3} < \cdots \n\]\n\nThen \( {n}_{k} \geq k \) for all \( k \in \mathbb{N} \) . | Proof: We use mathematical induction. When \( k = 1 \), it is clear that \( {n}_{1} \geq 1 \) since \( {n}_{1} \) is a positive integer. Assume \( {n}_{k} \geq k \) for some \( k \) . Now \( {n}_{k + 1} > {n}_{k} \) and, since \( {n}_{k} \) and \( {n}_{k + 1} \) are integers, this implies, \( {n}_{k + 1} \geq {n}_{k} +... | Yes |
Theorem 2.1.9 If a sequence \( \left\{ {a}_{n}\right\} \) converges to \( a \), then any subsequence \( \left\{ {a}_{{n}_{k}}\right\} \) of \( \left\{ {a}_{n}\right\} \) also converges to \( a \) . | Proof: Suppose \( \left\{ {a}_{n}\right\} \) converges to \( a \) and let \( \varepsilon > 0 \) be given. Then there exists \( N \) such that\n\n\[ \left| {{a}_{n} - a}\right| < \varepsilon \text{for all}n \geq N\text{.}\]\n\nFor any \( k \geq N \), since \( {n}_{k} \geq k \), we also have\n\n\[ \left| {{a}_{{n}_{k}} -... | Yes |
Theorem 2.3.1 — Monotone Convergence Theorem. Let \( \{ {a}_{n}\} \) be a sequence of real numbers. The following hold:\n\n(a) If \( \left\{ {a}_{n}\right\} \) is increasing and bounded above, then it is convergent.\n\n(b) If \( \left\{ {a}_{n}\right\} \) is decreasing and bounded below, then it is convergent. | Proof: (a) Let \( \left\{ {a}_{n}\right\} \) be an increasing sequence that is bounded above. Define\n\n\[ A = \left\{ {{a}_{n} : n \in \mathbb{N}}\right\} \]\n\nThen \( A \) is a subset of \( \mathbb{R} \) that is nonempty and bounded above and, hence, \( \sup A \) exists. Let \( \ell = \sup A \) and let \( \varepsilo... | Yes |
Theorem 2.3.3 - Nested Intervals Theorem. Let \( {\left\{ {I}_{n}\right\} }_{n = 1}^{\infty } \) be a sequence of nonempty closed bounded intervals satisfying \( {I}_{n + 1} \subset {I}_{n} \) for all \( n \in \mathbb{N} \) . Then the following hold:\n\n(a) \( \mathop{\bigcap }\limits_{{n = 1}}^{\infty }{I}_{n} \neq \v... | Proof: Let \( \left\{ {I}_{n}\right\} \) be as in the statement with \( {I}_{n} = \left\lbrack {{a}_{n},{b}_{n}}\right\rbrack \) . In particular, \( {a}_{n} \leq {b}_{n} \) for all \( n \in \mathbb{N} \) . Given that \( {I}_{n + 1} \subset {I}_{n} \), we have \( {a}_{n} \leq {a}_{n + 1} \) and \( {b}_{n + 1} \leq {b}_{... | Yes |
Theorem 2.3.5 If a sequence \( \left\{ {a}_{n}\right\} \) is increasing and not bounded above, then\n\n\[ \mathop{\lim }\limits_{{n \rightarrow \infty }}{a}_{n} = \infty \] | Proof: Fix any real number \( M \) . Since \( \left\{ {a}_{n}\right\} \) is not bounded above, there exists \( N \in \mathbb{N} \) such that \( {a}_{N} \geq M \) . Then\n\n\[ {a}_{n} \geq {a}_{N} \geq M\text{ for all }n \geq N \]\n\nbecause \( \left\{ {a}_{n}\right\} \) is increasing. Therefore, \( \mathop{\lim }\limit... | No |
Theorem 2.3.6 Let \( \\left\\{ {a}_{n}\\right\\} ,\\left\\{ {b}_{n}\\right\\} \), and \( \\left\\{ {c}_{n}\\right\\} \) be sequences of real numbers and let \( k \) be a constant. Suppose\n\n\[ \n\\mathop{\\lim }\\limits_{{n \\rightarrow \\infty }}{a}_{n} = \\infty ,\\mathop{\\lim }\\limits_{{n \\rightarrow \\infty }}{... | Proof: We provide proofs for (a) and (e) and leave the others as exercises.\n\n(a) Fix any \( M \\in \\mathbb{R} \) . Since \( \\mathop{\\lim }\\limits_{{n \\rightarrow \\infty }}{a}_{n} = \\infty \), there exists \( {N}_{1} \\in \\mathbb{N} \) such that\n\n\[ \n{a}_{n} \\geq \\frac{M}{2}\\text{ for all }n \\geq {N}_{1... | No |
Theorem 2.4.2 A convergent sequence is a Cauchy sequence. | Proof: Let \( \left\{ {a}_{n}\right\} \) be a convergent sequence and let\n\n\[ \mathop{\lim }\limits_{{n \rightarrow \infty }}{a}_{n} = a \]\n\nThen for any \( \varepsilon > 0 \), there exists a positive integer \( N \) such that\n\n\[ \left| {{a}_{n} - a}\right| < \varepsilon /2\text{ for all }n \geq N. \]\n\nFor any... | Yes |
Theorem 2.4.3 A Cauchy sequence is bounded. | Proof: Let \( \left\{ {a}_{n}\right\} \) be a Cauchy sequence. Then for \( \varepsilon = 1 \), there exists a positive integer \( N \) such that\n\n\[ \left| {{a}_{m} - {a}_{n}}\right| < 1\text{for all}m, n \geq N\text{.} \]\n\nIn particular,\n\n\[ \left| {{a}_{n} - {a}_{N}}\right| < 1\text{ for all }n \geq N. \]\n\nLe... | Yes |
Lemma 2.4.4 A Cauchy sequence that has a convergent subsequence is convergent. | Proof: Let \( \left\{ {a}_{n}\right\} \) be a Cauchy sequence that has a convergent subsequence. For any \( \varepsilon > 0 \), there exists a positive integer \( N \) such that\n\n\[ \left| {{a}_{m} - {a}_{n}}\right| \leq \varepsilon /2\text{for all}m, n \geq N\text{.}\]\n\nLet \( \left\{ {a}_{{n}_{k}}\right\} \) be a... | Yes |
Theorem 2.4.5 Any Cauchy sequence of real numbers is convergent. | Proof: Let \( \left\{ {a}_{n}\right\} \) be a Cauchy sequence. Then it is bounded by Theorem 2.4.3. By the Bolzano-Weierstrass theorem, \( \left\{ {a}_{n}\right\} \) has a convergent subsequence. Therefore, it is convergent by Lemma 2.4.4. \( ▱ \) | Yes |
Theorem 2.4.7 Every contractive sequence is convergent. | Proof: By induction, one has\n\n\[ \left| {{a}_{n + 1} - {a}_{n}}\right| \leq {k}^{n - 1}\left| {{a}_{2} - {a}_{1}}\right| \text{ for all }n \in \mathbb{N}. \]\n\nThus,\n\n\[ \left| {{a}_{n + p} - {a}_{n}}\right| \leq \left| {{a}_{n + 1} - {a}_{n}}\right| + \left| {{a}_{n + 2} - {a}_{n + 1}}\right| + \cdots + \left| {{... | Yes |
Proposition 2.5.1 Let \( \\left\\{ {a}_{n}\\right\\} \) be a bounded sequence. Define\n\n\[ \n{s}_{n} = \\sup \\left\\{ {{a}_{k} : k \\geq n}\\right\\} \n\]\n\n(2.8)\n\nand\n\n\[ \n{t}_{n} = \\inf \\left\\{ {{a}_{k} : k \\geq n}\\right\\} \n\]\n\n(2.9)\n\nThen \( \\left\\{ {s}_{n}\\right\\} \) and \( \\left\\{ {t}_{n}\... | Proof: If \( n \\leq m \), then \( \\left\\{ {{a}_{k} : k \\geq m}\\right\\} \\subset \\left\\{ {{a}_{k} : k \\geq n}\\right\\} \) . Therefore, it follows from Theorem 1.5.3 that \( {s}_{n} \\geq {s}_{m} \) and, so, the sequence \( \\left\\{ {s}_{n}\\right\\} \) is decreasing. Since \( \\left\\{ {a}_{n}\\right\\} \) is... | Yes |
Theorem 2.5.2 If \( \\left\\{ {a}_{n}\\right\\} \) is not bounded above, then\n\n\[ \n\\mathop{\\lim }\\limits_{{n \\rightarrow \\infty }}{s}_{n} = \\infty \n\]\n\nwhere \( \\left\\{ {s}_{n}\\right\\} \) is defined in (2.8).\n\nSimilarly, if \( \\left\\{ {a}_{n}\\right\\} \) is not bounded below, then\n\n\[ \n\\mathop{... | Proof: Suppose \( \\left\\{ {a}_{n}\\right\\} \) is not bounded above. Then for any \( k \\in \\mathbb{N} \), the set \( \\left\\{ {{a}_{i} : i \\geq k}\\right\\} \) is also not bounded above. Thus, \( {s}_{k} = \\sup \\left\\{ {{a}_{i} : i \\geq k}\\right\\} = \\infty \) for all \( k \) . Therefore, \( \\mathop{\\lim ... | Yes |
Theorem 2.5.4 Let \( \\left\\{ {a}_{n}\\right\\} \) be a sequence and \( \\ell \\in \\mathbb{R} \) . The following are equivalent:\n\n(a) \( \\lim \\mathop{\\sup }\\limits_{{n \\rightarrow \\infty }}{a}_{n} = \\ell \) .\n\n(b) For any \( \\varepsilon > 0 \), there exists \( N \\in \\mathbb{N} \) such that\n\n\[ \n{a}_{... | Proof: Suppose \( \\mathop{\\limsup }\\limits_{{n \\rightarrow \\infty }}{a}_{n} = \\ell \) . Then \( \\mathop{\\lim }\\limits_{{n \\rightarrow \\infty }}{s}_{n} = \\ell \), where \( {s}_{n} \) is defined as in (2.8). For any \( \\varepsilon > 0 \), there exists \( N \\in \\mathbb{N} \) such that\n\n\[ \n\\ell - \\vare... | Yes |
Theorem 2.5.5 Let \( \\left\\{ {a}_{n}\\right\\} \) be a sequence and \( \\ell \\in \\mathbb{R} \) . The following are equivalent:\n\n(a) \( \\mathop{\\liminf }\\limits_{{n \\rightarrow \\infty }}{a}_{n} = \\ell \) .\n\n(b) For any \( \\varepsilon > 0 \), there exists \( N \\in \\mathbb{N} \) such that\n\n\[ \n{a}_{n} ... | The following corollary follows directly from Theorems 2.5.4 and 2.5.5. | No |
Corollary 2.5.7 Let \( \\left\\{ {a}_{n}\\right\\} \) be a sequence.\n\n(a) Suppose \( \\mathop{\\limsup }\\limits_{{n \\rightarrow \\infty }}{a}_{n} = \\ell \) and \( \\left\\{ {a}_{{n}_{k}}\\right\\} \) is a subsequence of \( \\left\\{ {a}_{n}\\right\\} \) with\n\n\[ \n\\mathop{\\lim }\\limits_{{k \\rightarrow \\inft... | Proof: We prove only (a) because the proof of (b) is similar. By Theorem 2.5.4 and the definition of limits, for any \( \\varepsilon > 0 \), there exists \( N \\in \\mathbb{N} \) such that\n\n\[ \n{a}_{n} < \\ell + \\varepsilon \\text{ and }{\\ell }^{\\prime } - \\varepsilon < {a}_{{n}_{k}} < {\\ell }^{\\prime } + \\va... | Yes |
Theorem 2.5.9 Suppose \( \left\{ {a}_{n}\right\} \) is a sequence such that \( {a}_{n} > 0 \) for every \( n \in \mathbb{N} \) and\n\n\[ \mathop{\limsup }\limits_{{n \rightarrow \infty }}\frac{{a}_{n + 1}}{{a}_{n}} = \ell < 1 \]\n\nThen \( \mathop{\lim }\limits_{{n \rightarrow \infty }}{a}_{n} = 0 \) . | Proof: Choose \( \varepsilon > 0 \) such that \( \ell + \varepsilon < 1 \) . Then there exists \( N \in \mathbb{N} \) such that\n\n\[ \frac{{a}_{n + 1}}{{a}_{n}} < \ell + \varepsilon \text{ for all }n \geq N. \]\n\nLet \( q = \ell + \varepsilon \) . Then \( 0 < q < 1 \) . By induction,\n\n\[ 0 < {a}_{n} \leq {q}^{n - N... | Yes |
Theorem 2.6.1 The following hold:\n\n(a) The subsets \( \varnothing \) and \( \mathbb{R} \) are open.\n\n(b) The union of any collection of open subsets of \( \mathbb{R} \) is open.\n\n(c) The intersection of a finite number of open subsets of \( \mathbb{R} \) is open. | Proof: The proof of (a) is straightforward.\n\n(b) Suppose \( \left\{ {{G}_{\alpha } : \alpha \in I}\right\} \) is an arbitrary collection of open subsets of \( \mathbb{R} \) . That means \( {G}_{\alpha } \) is open for every \( \alpha \in I \) . Let us show that the set\n\n\[ G = \mathop{\bigcup }\limits_{{\alpha \in ... | No |
The following hold:\n\n(a) The sets \( \varnothing \) and \( \mathbb{R} \) are closed.\n\n(b) The intersection of any collection of closed subsets of \( \mathbb{R} \) is closed.\n\n(c) The union of a finite number of closed subsets of \( \mathbb{R} \) is closed. | Proof: The proofs for these are simple using the De Morgan's law. Let us prove, for instance, (b). Let \( \left\{ {{S}_{\alpha } : \alpha \in I}\right\} \) be a collection of closed sets. We will prove that the set\n\n\[ S = \mathop{\bigcap }\limits_{{\alpha \in I}}{S}_{\alpha } \]\n\nis also closed. We have\n\n\[ {S}^... | No |
Theorem 2.6.3 A subset \( A \) of \( \mathbb{R} \) is closed if and only if for any sequence \( \left\{ {a}_{n}\right\} \) in \( A \) that converges to a point \( a \in \mathbb{R} \), it follows that \( a \in A \) . | Proof: Suppose \( A \) is a closed subset of \( \mathbb{R} \) and \( \left\{ {a}_{n}\right\} \) is a sequence in \( A \) that converges to \( a \) . Suppose by contradiction that \( a \notin A \) . Since \( A \) is closed, there exists \( \varepsilon > 0 \) such that \( B\left( {a;\varepsilon }\right) = \left( {a - \va... | Yes |
Theorem 2.6.4 If \( A \) is a nonempty subset of \( \mathbb{R} \) that is closed and bounded above, then \( \max A \) exists. Similarly, if \( A \) is a nonempty subset of \( \mathbb{R} \) that is closed and bounded below, then \( \min A \) exists | Proof: Let \( A \) be a nonempty closed set that is bounded above. Then sup \( A \) exists. Let \( m = \sup A \) . To complete the proof, we will show that \( m \in A \) . Assume by contradiction that \( m \notin A \) . Then \( m \in {A}^{c} \) , which is an open set. So there exists \( \delta > 0 \) such that\n\n\[ \l... | Yes |
Theorem 2.6.5 A subset \( A \) of \( \mathbb{R} \) is compact if and only if it is closed and bounded. | Proof: Suppose \( A \) is a compact subset of \( \mathbb{R} \) . Let us first show that \( A \) is bounded. Suppose, by contradiction, that \( A \) is not bounded. Then for every \( n \in \mathbb{N} \), there exists \( {a}_{n} \in A \) such that\n\n\[ \left| {a}_{n}\right| \geq n \]\n\nSince \( A \) is compact, there e... | Yes |
Theorem 2.6.6 Any infinite bounded subset of \( \mathbb{R} \) has at least one limit point. | Proof: Let \( A \) be an infinite subset of \( \mathbb{R} \) and let \( \left\{ {a}_{n}\right\} \) be a sequence of \( A \) such that\n\n\[ \n{a}_{m} \neq {a}_{n}\text{ for }m \neq n \]\n\n(see Theorem 1.2.7). Since \( \left\{ {a}_{n}\right\} \) is bounded, by the Bolzano-Weierstrass theorem (Theorem 2.4.1), it has a c... | Yes |
Theorem 2.6.7 Let \( D \) be a subset of \( \mathbb{R} \) . A subset \( V \) of \( D \) is open in \( D \) if and only if there exists an open subset \( G \) of \( \mathbb{R} \) such that\n\n\[ V = D \cap G. \] | Proof: Suppose \( V \) is open in \( D \) . By definition, for every \( a \in V \), there exists \( {\delta }_{a} > 0 \) such that\n\n\[ B\left( {a;{\delta }_{a}}\right) \cap D \subset V. \]\n\nDefine\n\n\[ G = { \cup }_{a \in V}B\left( {a;{\delta }_{a}}\right) \]\n\nThen \( G \) is a union of open subsets of \( \mathb... | Yes |
Theorem 2.6.9 Let \( D \) be a subset of \( \mathbb{R} \) . A subset \( K \) of \( D \) is closed in \( D \) if and only if there exists a closed subset \( F \) of \( \mathbb{R} \) such that\n\n\[ K = D \cap F. \] | Proof: Suppose \( K \) is a closed set in \( D \) . Then \( D \smallsetminus K \) is open in \( D \) . By Theorem 2.6.7, there exists an open set \( G \) such that\n\n\[ D \smallsetminus K = D \cap G. \]\n\nIt follows that\n\n\[ K = D \smallsetminus \left( {D \smallsetminus K}\right) = D \smallsetminus \left( {D \cap G... | Yes |
Corollary 2.6.10 Let \( D \) be a subset of \( \mathbb{R} \) . A subset \( K \) of \( D \) is closed in \( D \) if and only if for every sequence \( \left\{ {x}_{k}\right\} \) in \( K \) that converges to a point \( \bar{x} \in D \) it follows that \( \bar{x} \in K \) . | Proof: Let \( D \) be a subset of \( \mathbb{R} \) . Suppose \( K \) is closed in \( D \) . By Theorem 2.6.9, there exists a closed subset \( F \) of \( \mathbb{R} \) such that\n\n\[ K = D \cap F\text{.}\]\n\nLet \( \left\{ {x}_{k}\right\} \) be a sequence in \( K \) that converges to a point \( \bar{x} \in D \) . Sinc... | Yes |
Theorem 3.1.2 — Sequential Characterization of Limits. Let \( f:D \rightarrow \mathbb{R} \) and let \( \bar{x} \) be a limit point of \( D \) . Then\n\n\[ \mathop{\lim }\limits_{{x \rightarrow \bar{x}}}f\left( x\right) = \ell \]\n\n(3.1)\n\nif and only if\n\n\[ \mathop{\lim }\limits_{{n \rightarrow \infty }}f\left( {x}... | Proof: Suppose (3.1) holds. Let \( \left\{ {x}_{n}\right\} \) be a sequence in \( D \) with \( {x}_{n} \neq \bar{x} \) for every \( n \) and such that \( \left\{ {x}_{n}\right\} \) converges to \( \bar{x} \) . Given any \( \varepsilon > 0 \), there exists \( \delta > 0 \) such that \( \left| {f\left( x\right) - \ell }\... | Yes |
Corollary 3.1.3 Let \( f : D \rightarrow \mathbb{R} \) and let \( \bar{x} \) be a limit point of \( D \) . If \( f \) has a limit at \( \bar{x} \), then this limit is unique. | Proof: Suppose by contradiction that \( f \) has two different limits \( {\ell }_{1} \) and \( {\ell }_{2} \) . Let \( \left\{ {x}_{n}\right\} \) be a sequence in \( D \smallsetminus \{ \bar{x}\} \) that converges to \( \bar{x} \) . By Theorem 3.1.2, the sequence \( \left\{ {f\left( {x}_{n}\right) }\right\} \) converge... | Yes |
Corollary 3.1.4 Let \( f : D \rightarrow \mathbb{R} \) and let \( \bar{x} \) be a limit point of \( D \) . Then \( f \) does not have a limit at \( \bar{x} \) if and only if there exists a sequence \( \left\{ {x}_{n}\right\} \) in \( D \) such that \( {x}_{n} \neq \bar{x} \) for every \( n,\left\{ {x}_{n}\right\} \) co... | Example 3.1.5 Consider the Dirichlet function \( f : \mathbb{R} \rightarrow \mathbb{R} \) given by \[ f\left( x\right) = \left\{ \begin{array}{ll} 1, & \text{ if }x \in \mathbb{Q}; \\ 0, & \text{ if }x \in {\mathbb{Q}}^{c}. \end{array}\right. \] Then \( \mathop{\lim }\limits_{{x \rightarrow \bar{x}}}f\left( x\right) \)... | Yes |
Theorem 3.1.5 Let \( f, g : D \rightarrow \mathbb{R} \) and let \( \bar{x} \) be a limit point of \( D \) . Suppose that\n\n\[ \mathop{\lim }\limits_{{x \rightarrow \bar{x}}}f\left( x\right) = {\ell }_{1},\mathop{\lim }\limits_{{x \rightarrow \bar{x}}}g\left( x\right) = {\ell }_{2}, \]\n\nand that there exists \( \delt... | Proof: Let \( \left\{ {x}_{n}\right\} \) be a sequence in \( B\left( {\bar{x};\delta }\right) \cap D = \left( {\bar{x} - \delta ,\bar{x} + \delta }\right) \cap D \) that converges to \( \bar{x} \) and \( {x}_{n} \neq \bar{x} \) for all \( n \) . By Theorem 3.1.2,\n\n\[ \mathop{\lim }\limits_{{n \rightarrow \infty }}f\l... | Yes |
Theorem 3.1.6 Let \( f, g : D \rightarrow \mathbb{R} \) and let \( \bar{x} \) be a limit point of \( D \) . Suppose\n\n\[ \mathop{\lim }\limits_{{x \rightarrow \bar{x}}}f\left( x\right) = {\ell }_{1},\mathop{\lim }\limits_{{x \rightarrow \bar{x}}}g\left( x\right) = {\ell }_{2} \]\n\nand \( {\ell }_{1} < {\ell }_{2} \) ... | Proof: Choose \( \varepsilon > 0 \) such that \( {\ell }_{1} + \varepsilon < {\ell }_{2} - \varepsilon \) (equivalently, such that \( \varepsilon < \frac{{\ell }_{2} - {\ell }_{1}}{2} \) ). Then there exists \( \delta > 0 \) such that\n\n\[ {\ell }_{1} - \varepsilon < f\left( x\right) < {\ell }_{1} + \varepsilon \text{... | Yes |
Theorem 3.1.7 Let \( f, g, h : D \rightarrow \mathbb{R} \) and let \( \bar{x} \) be a limit point of \( D \) . Suppose there exists \( \delta > 0 \) such that \( f\left( x\right) \leq g\left( x\right) \leq h\left( x\right) \) for all \( x \in B\left( {\bar{x};\delta }\right) \cap D, x \neq \bar{x} \) . If \( \mathop{\l... | Proof: The proof is straightforward using Theorem 2.1.6 and Theorem 3.1.2. \( ▱ \) | No |
Theorem 3.2.1 Let \( f, g : D \rightarrow \mathbb{R} \) and let \( c \in \mathbb{R} \) . Suppose \( \bar{x} \) is a limit point of \( D \) and\n\n\[ \mathop{\lim }\limits_{{x \rightarrow \bar{x}}}f\left( x\right) = \ell ,\mathop{\lim }\limits_{{x \rightarrow \bar{x}}}g\left( x\right) = m.\]\n\nThen\n\n(a) \( \mathop{\l... | Proof: Let us first prove (a). Let \( \left\{ {x}_{n}\right\} \) be a sequence in \( D \) that converges to \( \bar{x} \) and \( {x}_{n} \neq \bar{x} \) for every \( n \) . By Theorem 3.1.2,\n\n\[ \mathop{\lim }\limits_{{n \rightarrow \infty }}f\left( {x}_{n}\right) = \ell \text{ and }\mathop{\lim }\limits_{{n \rightar... | Yes |
Theorem 3.2.2 (Cauchy’s criterion) Let \( f : D \rightarrow \mathbb{R} \) and let \( \bar{x} \) be a limit point of \( D \) . Then \( f \) has a limit at \( \bar{x} \) if and only if for any \( \varepsilon > 0 \), there exists \( \delta > 0 \) such that\n\n\[ \left| {f\left( r\right) - f\left( s\right) }\right| < \vare... | Proof: Suppose \( \mathop{\lim }\limits_{{x \rightarrow \bar{x}}}f\left( x\right) = \ell \) . Given \( \varepsilon > 0 \), there exists \( \delta > 0 \) such that\n\n\[ \left| {f\left( x\right) - \ell }\right| < \frac{\varepsilon }{2}\text{ whenever }x \in D\text{ and }0 < \left| {x - \bar{x}}\right| < \delta . \]\n\nT... | Yes |
Theorem 3.2.4 Suppose \( f : \left( {a, b}\right) \rightarrow \mathbb{R} \) is increasing on \( \left( {a, b}\right) \) and \( \bar{x} \in \left( {a, b}\right) \) . Then \( \mathop{\lim }\limits_{{x \rightarrow {\bar{x}}^{ - }}}f\left( x\right) \) and \( \mathop{\lim }\limits_{{x \rightarrow {\bar{x}}^{ + }}}f\left( x\... | Proof: Since \( f\left( x\right) \leq f\left( \bar{x}\right) \) for all \( x \in \left( {a,\bar{x}}\right) \), the set\n\n\[ \{ f\left( x\right) : x \in \left( {a,\bar{x}}\right) \} \]\n\nis nonempty and bounded above. Thus,\n\n\[ \ell = \mathop{\sup }\limits_{{a < x < \bar{x}}}f\left( x\right) \]\n\nis a real number. ... | No |
Theorem 3.3.2 Let \( f : D \rightarrow \mathbb{R} \) and let \( {x}_{0} \in D \) be a limit point of \( D \) . Then \( f \) is continuous at \( {x}_{0} \) if and only if\n\n\[ \mathop{\lim }\limits_{{x \rightarrow {x}_{0}}}f\left( x\right) = f\left( {x}_{0}\right) \] | - Example 3.3.2 Let \( f : \mathbb{R} \rightarrow \mathbb{R} \) be given by \( f\left( x\right) = 3{x}^{2} - {2x} + 1 \) . Fix \( {x}_{0} \in \mathbb{R} \) . Since, from the results of the previous theorem, we have\n\n\[ \mathop{\lim }\limits_{{x \rightarrow {x}_{0}}}f\left( x\right) = \mathop{\lim }\limits_{{x \righta... | Yes |
Theorem 3.3.4 Let \( f, g : D \rightarrow \mathbb{R} \) and let \( {x}_{0} \in D \) . Suppose \( f \) and \( g \) are continuous at \( {x}_{0} \) . Then\n\n(a) \( f + g \) and \( {fg} \) are continuous at \( {x}_{0} \) .\n\n(b) \( {cf} \) is continuous at \( {x}_{0} \) for any constant \( c \) .\n\n(c) If \( g\left( {x... | Proof: We prove (a) and leave the other parts as an exercise. We will use Theorem 3.3.3. Let \( \left\{ {x}_{k}\right\} \) be a sequence in \( D \) that converges to \( {x}_{0} \) . Since \( f \) and \( g \) are continuous at \( {x}_{0} \), by Theorem 3.3.3 we obtain that \( \left\{ {f\left( {x}_{k}\right) }\right\} \)... | No |
Theorem 3.4.1 Let \( D \) be a nonempty compact subset of \( \mathbb{R} \) and let \( f : D \rightarrow \mathbb{R} \) be a continuous function. Then \( f\left( D\right) \) is a compact subset of \( \mathbb{R} \) . In particular, \( f\left( D\right) \) is closed and bounded. | Proof: Take any sequence \( \left\{ {y}_{n}\right\} \) in \( f\left( D\right) \) . Then for each \( n \), there exists \( {a}_{n} \in D \) such that \( {y}_{n} = f\left( {a}_{n}\right) \) . Since \( D \) is compact, there exists a subsequence \( \left\{ {a}_{{n}_{k}}\right\} \) of \( \left\{ {a}_{n}\right\} \) and a po... | Yes |
Theorem 3.4.2 — Extreme Value Theorem. Suppose \( f : D \rightarrow \mathbb{R} \) is continuous and \( D \) is a compact set. Then \( f \) has an absolute minimum and an absolute maximum on \( D \) . | Proof: Since \( D \) is compact, \( A = f\left( D\right) \) is closed and bounded (see Theorem 2.6.5). Let\n\n\[ m = \inf A = \mathop{\inf }\limits_{{x \in D}}f\left( x\right) . \]\n\nIn particular, \( m \in \mathbb{R} \) . For every \( n \in \mathbb{N} \), there exists \( {a}_{n} \in A \) such that\n\n\[ m \leq {a}_{n... | Yes |
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