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Proposition 20.9.21. In a principal ideal domain, any principal ideal is the intersection of prime power ideals. | Proof. This is a more difficult proof and will not be studied in this class. | No |
Example 20.9.22. Show that \( {12}\mathbb{Z} = {2}^{2}\mathbb{Z} \cap 3\mathbb{Z} \) . | Proof. Recall that:\n\n\[ \n{12}\mathbb{Z} = \{ {12n} : n \in \mathbb{Z}\} = \{ \cdots , - {24}, - {12},0,{12},{24},\cdots \} ,\n\]\n\n\[ \n{2}^{2}\mathbb{Z} = 4\mathbb{Z} = \{ {4n} : n \in \mathbb{Z}\} = \{ \cdots , - 8, - 4,0,4,8,\cdots \} ,\text{ and }\n\]\n\n\[ \n3\mathbb{Z} = \{ {3n} : n \in \mathbb{Z}\} = \{ \cdo... | Yes |
Let \( S \) be the set of all real \( 2 \times 2 \) matrices of the form \( \left\lbrack \begin{matrix} a & b \\ - b & a \end{matrix}\right\rbrack \) where \( a, b \in \mathbb{R} \) . Show that \( S \) is a field. | Proof. We know from Exercise 20.1.11 that the set of all \( 2 \times 2 \) matrices form a ring. It remains to show that \( S \) is a division ring with multiplicative commutativity. It will be important in our proof to know that \( S \) has the multiplicative inverse property. Let's show that first.\n\n\[ \text{Let}A \... | Yes |
Proposition 20.9.42. Given \( J \) is an ideal in ring \( R \) and \( a \in J \) . If \( a \) has a multiplicative inverse \( {a}^{-1} \in R \), then \( J = R \) . | Proof. To show that \( J = R \), we can show that \( J \subset R \) and \( R \subset J \) . We already know that \( J \subset R \), by definition of ideal. To show that \( R \subset J \), we must show that every element in \( R \) is also in \( J \) . Consider arbitrary element \( r \in R \) . We will show that \( r \i... | Yes |
Proposition 20.9.43. Given field \( F \), the only two ideals in \( F \) are \( \{ 0\} \) and all of \( F \) . | Proof. Suppose \( J \) is an ideal in field \( F \) . Then either \( J = \{ 0\} \) or \( J \) has a nonzero element \( a \) . By the definition of field, \( a \) must have an inverse; and by Proposition 20.9.42, it follows that \( J = F \) . | Yes |
Proposition 20.10.10. Let \( F \) be a field, and let \( c \) be any element \( F \) . Then \( c \) has at most \( n{n}^{\text{th }} \) roots. | Proof. Given the field \( F \) let \( F\left\lbrack x\right\rbrack \) be the associated polynomial ring over the field \( F \) . The polynomial \( {x}^{n} - c \) is an element of \( F\left\lbrack x\right\rbrack \) . By Proposition 20.10.9, the equation \( {x}^{n} - c = 0 \) has at most \( n \) solutions. This is exactl... | Yes |
Find the roots of \( p\left( x\right) = 2{x}^{2} + {2x} + 5 \) . | Since this is a quadratic polynomial we can use the famous quadratic formula:\n\n\[ x = \frac{-b \pm \sqrt{{b}^{2} - {4ac}}}{2a} \]\n\nIn \( p\left( x\right), a = 2, b = 2 \), and \( c = 5 \) . We substitute those values into the formula and obtain the following:\n\n\[ x = \frac{-2 \pm \sqrt{{2}^{2} - 4 \cdot 2 \cdot 5... | Yes |
Find the roots of \( f\left( x\right) = 3{x}^{3} + {10}{x}^{2} + {11x} + 6 \) . | Since this is a cubic polynomial, we can't use the quadratic formula, at least not to begin with. The coefficients are integers, so we may use Proposition 20.10.13, which says that any rational roots of \( p\left( x\right) \) have numerators that are factors of \( {a}_{0} \) and denominators that are factors of \( {a}_... | Yes |
Proposition 20.10.33. \( \\mathbb{C} \) is an algebraic closure of \( \\mathbb{R} \) . | Proof. Let \( a + {bi} \\in \\mathbb{C} \) be arbitrary and let \( p\\left( x\\right) = \\left( {x - \\left( {a + {bi}}\\right) }\\right) \\left( {x - \\left( {a - {bi}}\\right) }\\right) = \) \( {x}^{2} - {2ax} + {a}^{2} + {b}^{2} \\in \\mathbb{R}\\left\\lbrack x\\right\\rbrack \) . We see that \( a + {bi} \) is a roo... | Yes |
Proposition 21.1.1. For all \( n \in \mathbb{N} \), the following equation (which we denote as \( P\left( n\right) \) ) is true:\n\n\[ \mathop{\sum }\limits_{{i = 1}}^{n}i = \frac{n\left( {n + 1}\right) }{2} \] | Proof. (By induction):\n\nBase case: When \( n = 1 \), the left side of \( P\left( n\right) \) is 1, and the right side is \( 1\left( {1 + 1}\right) /2 = 1 \), so both sides are equal and \( P\left( n\right) \) holds for \( n = 1 \).\n\nInduction step: Let \( k \in \mathbb{N} \) be given and suppose formula \( P\left( ... | Yes |
Proposition 21.4.1. Let \( {a}_{n} \) be the sequence defined by \( {a}_{1} = 1,{a}_{2} = 8 \) , and \( {a}_{n} = {a}_{n - 1} + 2{a}_{n - 2} \) for \( n \geq 3 \) . Then \( {a}_{n} = 3 \cdot {2}^{n - 1} + 2{\left( -1\right) }^{n} \) for all \( n \in \mathbb{N} \) . | Proof. We’ll prove by strong induction that, for all \( n \in \mathbb{N} \) ,\n\n\[ \n{a}_{n} = 3 \cdot {2}^{n - 1} + 2{\left( -1\right) }^{n}.\n\]\n\n\( \left( {P\left( n\right) }\right) \)\n\n\n\nBase case: When \( n = 1 \), the left side of \( P\left( 1\right) \) is \( {a}_{1} = 1 \), and the right side is \( 3 \cdo... | Yes |
- Base step: \( n = 3 \) .\n\n- Induction step: Let \( k \in \mathbb{N} \) with \( k \geq 3 \) be given and assume\n\n\( P\left( n\right) \) is true for \( n = k \) and \( n = k - 1 \) . | - Comment: BAD: When \( k = 3 \) (the first case of the induction step), the induction step requires the cases 3 and 2, but only 2 is covered in the base step. FIX: Add the case \( n = 2 \) to the base step. | No |
- Base step: \( n = 1 \) and \( n = 2 \) .\n\n- Induction step: Let \( k \in \mathbb{N} \) with \( k > 2 \) be given and assume \( P\left( n\right) \) is true for \( n = k \) and \( n = k - 1 \) . | - Comment: BAD. Gap between base case and the first case of the induction step: The first case \( k = 3 \) of the induction step requires the cases 3 and 2 , but the base step only gives the cases 1 and 2 .\n\nFIX: Start induction step at \( k = 2 \) rather than \( k = 3 \) : \ | No |
- Base step: \( n = 1 \) and \( n = 2 \) .\n\n- Induction step: Assume \( P\left( n\right) \) is true for \( n = k \) and \( n = k - 1 \) . Then ... | - Comment: BAD. The variable \( k \) in the induction step is not quantified. FIX: Add \ | No |
For all \( n \in \mathbb{N} \), the following holds:\n\n\[ \text{Any } n \text{-element set has } 2^n \text{ subsets.} \]\n\n\( \left( {P\left( n\right) }\right) \) | Proof.(By induction):\n\nBase case: Since any 1-element set has 2 subsets, namely the empty set and the set itself, and \( 2^1 = 2 \), the statement \( P\left( n\right) \) is true for \( n = 1 \).\n\nInduction step:\n\n- Let \( k \in \mathbb{N} \) be given and suppose \( P\left( k\right) \) is true, i.e., that any \( k... | Yes |
Theorem 0.3.5 (DeMorgan). Let \( A, B, C \) be sets. Then\n\n\[{\left( B \cup C\right) }^{c} = {B}^{c} \cap {C}^{c}\]\n\n\[{\left( B \cap C\right) }^{c} = {B}^{c} \cup {C}^{c}\]\n\nor, more generally,\n\n\[A \smallsetminus \left( {B \cup C}\right) = \left( {A \smallsetminus B}\right) \cap \left( {A \smallsetminus C}\ri... | Proof. The first statement is proved by the second statement if we assume the set \( A \) is our \ | No |
Theorem 0.3.6 (Principle of induction). Let \( P\left( n\right) \) be a statement depending on a natural number \( n \) . Suppose that\n\n(i) (basis statement) \( P\left( 1\right) \) is true,\n\n(ii) (induction step) if \( P\left( n\right) \) is true, then \( P\left( {n + 1}\right) \) is true.\n\nThen \( P\left( n\righ... | Proof. Suppose \( S \) is the set of natural numbers \( m \) for which \( P\left( m\right) \) is not true. Suppose \( S \) is nonempty. Then \( S \) has a least element by the well ordering property. Let us call \( m \) the least element of \( S \) . We know \( 1 \notin S \) by assumption. Therefore \( m > 1 \) and \( ... | Yes |
Proposition 0.3.15. Consider \( f : A \rightarrow B \) . Let \( C, D \) be subsets of \( B \) . Then\n\n\[ \n{f}^{-1}\left( {C \cup D}\right) = {f}^{-1}\left( C\right) \cup {f}^{-1}\left( D\right) \n\]\n\n\[ \n{f}^{-1}\left( {C \cap D}\right) = {f}^{-1}\left( C\right) \cap {f}^{-1}\left( D\right) \n\]\n\n\[ \n{f}^{-1}\... | Proof. Let us start with the union. Suppose \( x \in {f}^{-1}\left( {C \cup D}\right) \) . That means \( x \) maps to \( C \) or \( D \) . Thus \( {f}^{-1}\left( {C \cup D}\right) \subset {f}^{-1}\left( C\right) \cup {f}^{-1}\left( D\right) \) . Conversely if \( x \in {f}^{-1}\left( C\right) \), then \( x \in {f}^{-1}\... | No |
Proposition 0.3.16. Consider \( f : A \rightarrow B \) . Let \( C, D \) be subsets of \( A \) . Then\n\n\[ f\left( {C \cup D}\right) = f\left( C\right) \cup f\left( D\right) \]\n\n\[ f\left( {C \cap D}\right) \subset f\left( C\right) \cap f\left( D\right) . \] | The proof is left as an exercise. | No |
Theorem 0.3.34 (Cantor*). \( \left| A\right| < \left| {\mathcal{P}\left( A\right) }\right| \) . In particular, there exists no surjection from \( A \) onto \( \mathcal{P}\left( A\right) \) . | Proof. There exists an injection \( f : A \rightarrow \mathcal{P}\left( A\right) \) . For any \( x \in A \), define \( f\left( x\right) \mathrel{\text{:=}} \{ x\} \) . Therefore \( \left| A\right| \leq \left| {\mathcal{P}\left( A\right) }\right| \) .\n\nTo finish the proof, we must show that no function \( g : A \right... | Yes |
Proposition 1.1.8. Let \( F \) be an ordered field and \( x, y, z, w \in F \) . Then:\n\n(i) If \( x > 0 \), then \( - x < 0 \) (and vice-versa).\n\n(ii) If \( x > 0 \) and \( y < z \), then \( {xy} < {xz} \) .\n\n(iii) If \( x < 0 \) and \( y < z \), then \( {xy} > {xz} \) .\n\n(iv) If \( x \neq 0 \), then \( {x}^{2} ... | Proof. Let us prove (i). The inequality \( x > 0 \) implies by item (i) of definition of ordered field that \( x + \left( {-x}\right) > 0 + \left( {-x}\right) \) . Now apply the algebraic properties of fields to obtain \( 0 > - x \) . The \ | No |
Proposition 1.1.9. Let \( x, y \in F \) where \( F \) is an ordered field. Suppose \( {xy} > 0 \) . Then either both \( x \) and \( y \) are positive, or both are negative. | Proof. Clearly both of the conclusions can happen. If either \( x \) and \( y \) are zero, then \( {xy} \) is zero and hence not positive. Hence we assume that \( x \) and \( y \) are nonzero, and we simply need to show that if they have opposite signs, then \( {xy} < 0 \) . Without loss of generality suppose \( x > 0 ... | Yes |
Proposition 1.1.11. Let \( F \) be an ordered field with the least-upper-bound property. Let \( A \subset F \) be a nonempty set that is bounded below. Then \( \inf A \) exists. | Proof. Let \( B \mathrel{\text{:=}} \{ - x : x \in A\} \) . Let \( b \in F \) be a lower bound for \( A \) : if \( x \in A \), then \( x \geq b \) . In other words, \( - x \leq - b \) . So \( - b \) is an upper bound for \( B \) . Since \( F \) has the least-upper-bound property, \( c \mathrel{\text{:=}} \sup B \) exis... | Yes |
Proposition 1.2.2. If \( x \in \mathbb{R} \) is such that \( x \leq \varepsilon \) for all \( \varepsilon \in \mathbb{R} \) where \( \varepsilon > 0 \), then \( x \leq 0 \) . | Proof. If \( x > 0 \), then \( 0 < x/2 < x \) (why?). Taking \( \varepsilon = x/2 \) obtains a contradiction. Thus \( x \leq 0 \) . | Yes |
Corollary 1.2.5. \( \inf \{ 1/n : n \in \mathbb{N}\} = 0 \) . | Proof. Let \( A \mathrel{\text{:=}} \{ 1/n : n \in \mathbb{N}\} \) . Obviously \( A \) is not empty. Furthermore, \( 1/n > 0 \) and so 0 is a lower bound, and \( b \mathrel{\text{:=}} \inf A \) exists. As 0 is a lower bound, then \( b \geq 0 \) . Now take an arbitrary \( a > 0 \) . By the Archimedean property there exi... | Yes |
Proposition 1.2.6. Let \( A \subset \mathbb{R} \) be nonempty.\n\n(i) If \( x \in \mathbb{R} \) and \( A \) is bounded above, then \( \sup \left( {x + A}\right) = x + \sup A \) . | Proof. Let us only prove the first statement. The rest are left as exercises.\n\nSuppose \( b \) is an upper bound for \( A \) . That is, \( y \leq b \) for all \( y \in A \) . Then \( x + y \leq x + b \) for all \( y \in A \) , and so \( x + b \) is an upper bound for \( x + A \) . In particular, if \( b = \sup A \), ... | No |
Proposition 1.2.7. Let \( A, B \subset \mathbb{R} \) be nonempty sets such that \( x \leq y \) whenever \( x \in A \) and \( y \in B \). Then \( A \) is bounded above, \( B \) is bounded below, and \( \sup A \leq \inf B \). | Proof. Any \( x \in A \) is a lower bound for \( B \). Therefore \( x \leq \inf B \) for all \( x \in A \), so \( \inf B \) is an upper bound for \( A \). Hence, \( \sup A \leq \inf B \). | "No" |
(i) \( \left| x\right| \geq 0 \), and \( \left| x\right| = 0 \) if and only if \( x = 0 \) . | Proof. (i): If \( x \geq 0 \), then \( \left| x\right| = x \geq 0 \) . Also \( \left| x\right| = x = 0 \) if and only if \( x = 0 \) . If \( x < 0 \), then \( \left| x\right| = - x > 0 \) , which is never zero. | Yes |
Proposition 1.3.2 (Triangle Inequality). \( \left| {x + y}\right| \leq \left| x\right| + \left| y\right| \) for all \( x, y \in \mathbb{R} \) . | Proof. Proposition 1.3.1 gives \( - \left| x\right| \leq x \leq \left| x\right| \) and \( - \left| y\right| \leq y \leq \left| y\right| \) . Add these two inequalities to obtain\n\n\[ \n- \left( {\left| x\right| + \left| y\right| }\right) \leq x + y \leq \left| x\right| + \left| y\right| \n\]\n\nApply Proposition 1.3.1... | Yes |
Corollary 1.3.3. Let \( x, y \in \mathbb{R} \)\n\n(i) (reverse triangle inequality) \( \left| \left( {\left| x\right| - \left| y\right| }\right) \right| \leq \left| {x - y}\right| \) . | Proof. Let us plug in \( x = a - b \) and \( y = b \) into the standard triangle inequality to obtain\n\n\[ \left| a\right| = \left| {a - b + b}\right| \leq \left| {a - b}\right| + \left| b\right| \]\n\nor \( \left| a\right| - \left| b\right| \leq \left| {a - b}\right| \) . Switching the roles of \( a \) and \( b \) we... | Yes |
Corollary 1.3.4. Let \( {x}_{1},{x}_{2},\ldots ,{x}_{n} \in \mathbb{R} \) . Then\n\n\[ \left| {{x}_{1} + {x}_{2} + \cdots + {x}_{n}}\right| \leq \left| {x}_{1}\right| + \left| {x}_{2}\right| + \cdots + \left| {x}_{n}\right| . \] | Proof. We proceed by induction. The conclusion holds trivially for \( n = 1 \), and for \( n = 2 \) it is the standard triangle inequality. Suppose the corollary holds for \( n \) . Take \( n + 1 \) numbers \( {x}_{1},{x}_{2},\ldots ,{x}_{n + 1} \) and first use the standard triangle inequality, then the induction hypo... | Yes |
Proposition 1.3.7. If \( f : D \rightarrow \mathbb{R} \) and \( g : D \rightarrow \mathbb{R} \) ( \( D \) nonempty) are bounded* functions and\n\n\[ f\left( x\right) \leq g\left( x\right) \;\text{ for all }x \in D, \]\n\nthen\n\n\[ \mathop{\sup }\limits_{{x \in D}}f\left( x\right) \leq \mathop{\sup }\limits_{{x \in D}}... | Let us prove this inequality. If \( b \) is an upper bound for \( g\left( D\right) \), then \( f\left( x\right) \leq g\left( x\right) \leq b \) for all \( x \in D \) , and hence \( b \) is an upper bound for \( f\left( D\right) \) . Taking the least upper bound we get that for all \( x \in D \)\n\n\[ f\left( x\right) \... | No |
Proposition 1.5.1.\n\n(i) Every infinite sequence of digits \( 0.{d}_{1}{d}_{2}{d}_{3}\ldots \) represents a unique real number \( x \in \left\lbrack {0,1}\right\rbrack \), and\n\n\[ {D}_{n} \leq x \leq {D}_{n} + \frac{1}{{10}^{n}}\;\text{ for all }n \in \mathbb{N}. \] | Proof. Let us start with the first item. Take an arbitrary infinite sequence of digits \( 0.{d}_{1}{d}_{2}{d}_{3}\ldots \) . Use the geometric sum formula to write\n\n\[ {D}_{n} = \frac{{d}_{1}}{10} + \frac{{d}_{2}}{{10}^{2}} + \frac{{d}_{3}}{{10}^{3}} + \cdots + \frac{{d}_{n}}{{10}^{n}} \leq \frac{9}{10} + \frac{9}{{1... | Yes |
Theorem 1.5.2 (Cantor). The set \( (0,1\rbrack \) is uncountable. | Proof. Let \( X \mathrel{\text{:=}} \left\{ {{x}_{1},{x}_{2},{x}_{3},\ldots }\right\} \) be any countable subset of real numbers in \( (0,1\rbrack \) . We will construct a real number not in \( X \) . Let\n\n\[ \n{x}_{n} = 0.{d}_{1}^{n}{d}_{2}^{n}{d}_{3}^{n}\ldots \n\] \n\nbe the unique representation from the proposit... | Yes |
Proposition 1.5.3. If \( x \in (0,1\rbrack \) is a rational number and \( x = 0.{d}_{1}{d}_{2}{d}_{3}\ldots \), then the decimal digits eventually start repeating. That is, there are positive integers \( N \) and \( P \), such that for all \( n \geq N \) , \( {d}_{n} = {d}_{n + P} \) | Proof. Let \( x = p/q \) for positive integers \( p \) and \( q \) . Suppose \( x \) is a number with a unique representation, as otherwise we have seen above that both its representations are repeating, see also Exercise 1.5.3. This also means that \( x \neq 1 \) so \( p < q \) .\n\nTo compute the first digit we take ... | Yes |
Proposition 2.1.7. A convergent sequence \( \left\{ {x}_{n}\right\} \) is bounded. | Proof. Suppose \( \left\{ {x}_{n}\right\} \) converges to \( x \) . Thus there exists an \( M \in \mathbb{N} \) such that for all \( n \geq M \) we have \( \left| {{x}_{n} - x}\right| < 1 \) . Let \( {B}_{1} \mathrel{\text{:=}} \left| x\right| + 1 \) and note that for \( n \geq M \) we have\n\n\[ \left| {x}_{n}\right| ... | Yes |
Proposition 2.1.10. A monotone sequence \( \left\{ {x}_{n}\right\} \) is bounded if and only if it is convergent. | Proof. Let us suppose the sequence is monotone increasing. Suppose the sequence is bounded, so there exists a \( B \) such that \( {x}_{n} \leq B \) for all \( n \), that is the set \( \left\{ {{x}_{n} : n \in \mathbb{N}}\right\} \) is bounded above. Let\n\n\[ x \mathrel{\text{:=}} \sup \left\{ {{x}_{n} : n \in \mathbb... | No |
Proposition 2.1.15. Let \( {\left\{ {x}_{n}\right\} }_{n = 1}^{\infty } \) be a sequence. Then the following statements are equivalent:\n\n(i) The sequence \( {\left\{ {x}_{n}\right\} }_{n = 1}^{\infty } \) converges.\n\n(ii) The \( K \) -tail \( {\left\{ {x}_{n + K}\right\} }_{n = 1}^{\infty } \) converges for all \( ... | Proof. It is clear that (ii) implies (iii). We will therefore show first that (i) implies (ii), and then we will show that (iii) implies (i). That is,\n\n\n\n(iii)\n\nIn the process we will also show that the limits ar... | Yes |
Proposition 2.1.17. If \( \left\{ {x}_{n}\right\} \) is a convergent sequence, then any subsequence \( \left\{ {x}_{{n}_{i}}\right\} \) is also convergent and\n\n\[ \mathop{\lim }\limits_{{n \rightarrow \infty }}{x}_{n} = \mathop{\lim }\limits_{{i \rightarrow \infty }}{x}_{{n}_{i}} \] | Proof. Suppose \( \mathop{\lim }\limits_{{n \rightarrow \infty }}{x}_{n} = x \) . That means that for every \( \varepsilon > 0 \) we have an \( M \in \mathbb{N} \) such that for all \( n \geq M \)\n\n\[ \left| {{x}_{n} - x}\right| < \varepsilon \]\n\nIt is not hard to prove (do it!) by induction that \( {n}_{i} \geq i ... | No |
Lemma 2.2.1 (Squeeze lemma). Let \( \\left\\{ {a}_{n}\\right\\} ,\\left\\{ {b}_{n}\\right\\} \), and \( \\left\\{ {x}_{n}\\right\\} \) be sequences such that\n\n\[ \n{a}_{n} \\leq {x}_{n} \\leq {b}_{n}\\;\\text{ for all }n \\in \\mathbb{N}.\n\]\n\nSuppose \( \\left\\{ {a}_{n}\\right\\} \) and \( \\left\\{ {b}_{n}\\righ... | Proof. Let \( x \\mathrel{\\text{:=}} \\lim {a}_{n} = \\lim {b}_{n} \) . Let \( \\varepsilon > 0 \) be given. Find an \( {M}_{1} \) such that for all \( n \\geq {M}_{1} \) we have that \( \\left| {{a}_{n} - x}\\right| < \\varepsilon \), and an \( {M}_{2} \) such that for all \( n \\geq {M}_{2} \) we have \( \\left| {{b... | Yes |
Lemma 2.2.3. Let \( \left\{ {x}_{n}\right\} \) and \( \left\{ {y}_{n}\right\} \) be convergent sequences and\n\n\[ \n{x}_{n} \leq {y}_{n} \n\]\n\nfor all \( n \in \mathbb{N} \) . Then\n\n\[ \n\mathop{\lim }\limits_{{n \rightarrow \infty }}{x}_{n} \leq \mathop{\lim }\limits_{{n \rightarrow \infty }}{y}_{n} \n\] | Proof. Let \( x \mathrel{\text{:=}} \lim {x}_{n} \) and \( y \mathrel{\text{:=}} \lim {y}_{n} \) . Let \( \varepsilon > 0 \) be given. Find an \( {M}_{1} \) such that for all \( n \geq {M}_{1} \) we have \( \left| {{x}_{n} - x}\right| < \varepsilon /2 \) . Find an \( {M}_{2} \) such that for all \( n \geq {M}_{2} \) we... | Yes |
Proposition 2.2.5. Let \( \\left\\{ {x}_{n}\\right\\} \) and \( \\left\\{ {y}_{n}\\right\\} \) be convergent sequences.\n\n(i) The sequence \( \\left\\{ {z}_{n}\\right\\} \), where \( {z}_{n} \\mathrel{\\text{:=}} {x}_{n} + {y}_{n} \), converges and\n\n\[\\mathop{\\lim }\\limits_{{n \\rightarrow \\infty }}\\left( {{x}_... | Proof. Let us start with (i). Suppose \( \\left\\{ {x}_{n}\\right\\} \) and \( \\left\\{ {y}_{n}\\right\\} \) are convergent sequences and write \( {z}_{n} \\mathrel{\\text{:=}} {x}_{n} + {y}_{n} \) . Let \( x \\mathrel{\\text{:=}} \\lim {x}_{n}, y \\mathrel{\\text{:=}} \\lim {y}_{n} \), and \( z \\mathrel{\\text{:=}} ... | Yes |
Proposition 2.2.6. Let \( \\left\\{ {x}_{n}\\right\\} \) be a convergent sequence such that \( {x}_{n} \\geq 0 \) . Then\n\n\[ \n\\mathop{\\lim }\\limits_{{n \\rightarrow \\infty }}\\sqrt{{x}_{n}} = \\sqrt{\\mathop{\\lim }\\limits_{{n \\rightarrow \\infty }}{x}_{n}}\n\] | Proof. Let \( \\left\\{ {x}_{n}\\right\\} \) be a convergent sequence and let \( x \\mathrel{\\text{:=}} \\lim {x}_{n} \) . As we just mentioned, \( x \\geq 0 \) .\n\nFirst suppose \( x = 0 \) . Let \( \\varepsilon > 0 \) be given. Then there is an \( M \) such that for all \( n \\geq M \) we have \( {x}_{n} = \\left| ... | No |
Proposition 2.2.7. If \( \left\{ {x}_{n}\right\} \) is a convergent sequence, then \( \left\{ \left| {x}_{n}\right| \right\} \) is convergent and\n\n\[ \mathop{\lim }\limits_{{n \rightarrow \infty }}\left| {x}_{n}\right| = \left| {\mathop{\lim }\limits_{{n \rightarrow \infty }}{x}_{n}}\right| \] | Proof. We simply note the reverse triangle inequality\n\n\[ \left| \right| {x}_{n}\left| -\right| x\left| \right| \leq \left| {{x}_{n} - x}\right| \]\n\nHence if \( \left| {{x}_{n} - x}\right| \) can be made arbitrarily small, so can \( \left| \right| {x}_{n}\left| -\right| x\left| \right| \) . Details are left to the ... | No |
Proposition 2.2.10. Let \( \\left\\{ {x}_{n}\\right\\} \) be a sequence. Suppose there is an \( x \\in \\mathbb{R} \) and a convergent sequence \( \\left\\{ {a}_{n}\\right\\} \) such that\n\n\[ \n\\mathop{\\lim }\\limits_{{n \\rightarrow \\infty }}{a}_{n} = 0\n\]\n\nand\n\n\[ \n\\left| {{x}_{n} - x}\\right| \\leq {a}_{... | Proof. Let \( \\varepsilon > 0 \) be given. Note that \( {a}_{n} \\geq 0 \) for all \( n \) . Find an \( M \\in \\mathbb{N} \) such that for all \( n \\geq M \) we have \( {a}_{n} = \\left| {{a}_{n} - 0}\\right| < \\varepsilon \) . Then, for all \( n \\geq M \) we have\n\n\[ \n\\left| {{x}_{n} - x}\\right| \\leq {a}_{n... | Yes |
Proposition 2.2.11. Let \( c > 0 \) .\n\n(i) If \( c < 1 \), then\n\n\[ \mathop{\lim }\limits_{{n \rightarrow \infty }}{c}^{n} = 0 \]\n\n(ii) If \( c > 1 \), then \( \left\{ {c}^{n}\right\} \) is unbounded. | Proof. First suppose \( c < 1 \) . As \( c > 0 \), then \( {c}^{n} > 0 \) for all \( n \in \mathbb{N} \) by induction. As \( c < 1 \), then \( {c}^{n + 1} < {c}^{n} \) for all \( n \) . So we have a decreasing sequence that is bounded below. Hence, it is convergent. Let \( L \mathrel{\text{:=}} \lim {c}^{n} \) . The 1-... | Yes |
Proposition 2.3.2. Let \( \left\{ {x}_{n}\right\} \) be a bounded sequence. Let \( {a}_{n} \) and \( {b}_{n} \) be as in the definition above.\n\n(i) The sequence \( \left\{ {a}_{n}\right\} \) is bounded monotone decreasing and \( \left\{ {b}_{n}\right\} \) is bounded monotone increasing. In particular, \( \liminf {x}_... | Proof. Let us see why \( \left\{ {a}_{n}\right\} \) is a decreasing sequence. As \( {a}_{n} \) is the least upper bound for \( \left\{ {{x}_{k} : k \geq n}\right\} \) , it is also an upper bound for the subset \( \left\{ {{x}_{k} : k \geq \left( {n + 1}\right) }\right\} \) . Therefore \( {a}_{n + 1} \), the least upper... | No |
Theorem 2.3.4. If \( \left\{ {x}_{n}\right\} \) is a bounded sequence, then there exists a subsequence \( \left\{ {x}_{{n}_{k}}\right\} \) such that\n\n\[ \mathop{\lim }\limits_{{k \rightarrow \infty }}{x}_{{n}_{k}} = \mathop{\limsup }\limits_{{n \rightarrow \infty }}{x}_{n} \]\n\nSimilarly, there exists a (perhaps dif... | Proof. Define \( {a}_{n} \mathrel{\text{:=}} \sup \left\{ {{x}_{k} : k \geq n}\right\} \) . Write \( x \mathrel{\text{:=}} \limsup {x}_{n} = \lim {a}_{n} \) . Define the subsequence as follows. Pick \( {n}_{1} \mathrel{\text{:=}} 1 \) and work inductively. Suppose we have defined the subsequence until \( {n}_{k} \) for... | No |
Proposition 2.3.5. Let \( \left\{ {x}_{n}\right\} \) be a bounded sequence. Then \( \left\{ {x}_{n}\right\} \) converges if and only if\n\n\[ \mathop{\liminf }\limits_{{n \rightarrow \infty }}{x}_{n} = \mathop{\limsup }\limits_{{n \rightarrow \infty }}{x}_{n} \]\n\nFurthermore, if \( \left\{ {x}_{n}\right\} \) converge... | Proof. Let \( {a}_{n} \) and \( {b}_{n} \) be as in Definition 2.3.1. In particular, for all \( n \in \mathbb{N} \) ,\n\n\[ {b}_{n} \leq {x}_{n} \leq {a}_{n} \]\n\nIf \( \liminf {x}_{n} = \limsup {x}_{n} \), then we know that \( \left\{ {a}_{n}\right\} \) and \( \left\{ {b}_{n}\right\} \) have limits and that these two... | Yes |
Proposition 2.3.6. Suppose \( \left\{ {x}_{n}\right\} \) is a bounded sequence and \( \left\{ {x}_{{n}_{k}}\right\} \) is a subsequence. Then\n\n\[ \mathop{\liminf }\limits_{{n \rightarrow \infty }}{x}_{n} \leq \mathop{\liminf }\limits_{{k \rightarrow \infty }}{x}_{{n}_{k}} \leq \mathop{\limsup }\limits_{{k \rightarrow... | Proof. The middle inequality has been proved already. We will prove the third inequality, and leave the first inequality as an exercise.\n\nWe want to prove that \( \lim \sup {x}_{{n}_{k}} \leq \lim \sup {x}_{n} \) . Define \( {a}_{j} \mathrel{\text{:=}} \sup \left\{ {{x}_{k} : k \geq j}\right\} \) as usual. Also defin... | No |
Theorem 2.3.8 (Bolzano-Weierstrass). Suppose a sequence \( \left\{ {x}_{n}\right\} \) of real numbers is bounded. Then there exists a convergent subsequence \( \left\{ {x}_{{n}_{i}}\right\} \) . | Proof. We use Theorem 2.3.4. It says that there exists a subsequence whose limit is \( \limsup {x}_{n} \) . | No |
Proposition 2.3.10. Suppose \( \left\{ {x}_{n}\right\} \) is a monotone unbounded sequence. Then\n\n\[ \mathop{\lim }\limits_{{n \rightarrow \infty }}{x}_{n} = \left\{ \begin{array}{ll} \infty & \text{ if }\left\{ {x}_{n}\right\} \text{ is increasing,} \\ - \infty & \text{ if }\left\{ {x}_{n}\right\} \text{ is decreasi... | Proof. The case of monotone increasing follows from Exercise 2.3.14 part c) below. Let us do monotone decreasing. Suppose \( \left\{ {x}_{n}\right\} \) is decreasing and unbounded, that is, for every \( K \in \mathbb{R} \) , there is an \( M \in \mathbb{N} \) such that \( {x}_{M} < K \) . By monotonicity \( {x}_{n} \le... | No |
Proposition 2.3.13. Let \( \left\{ {x}_{n}\right\} \) be an unbounded sequence. Define \( \left\{ {a}_{n}\right\} \) and \( \left\{ {b}_{n}\right\} \) as above. Then \( \left\{ {a}_{n}\right\} \) is decreasing, and \( \left\{ {b}_{n}\right\} \) is increasing. If \( {a}_{n} \) is a real number for every \( n \), then \(... | Proof. As before, \( {a}_{n} = \sup \left\{ {{x}_{k} : k \geq n}\right\} \geq \sup \left\{ {{x}_{k} : k \geq n + 1}\right\} = {a}_{n + 1} \) . So \( \left\{ {a}_{n}\right\} \) is decreasing. Similarly, \( \left\{ {b}_{n}\right\} \) is increasing.\n\nIf the sequence \( \left\{ {a}_{n}\right\} \) is a sequence of real nu... | Yes |
Proposition 2.4.4. A Cauchy sequence is bounded. | Proof. Suppose \( \left\{ {x}_{n}\right\} \) is Cauchy. Pick \( M \) such that for all \( n, k \geq M \) we have \( \left| {{x}_{n} - {x}_{k}}\right| < 1 \) . In particular, for all \( n \geq M \) ,\n\n\[ \left| {{x}_{n} - {x}_{M}}\right| < 1 \]\n\nBy the reverse triangle inequality, \( \left| {x}_{n}\right| - \left| {... | Yes |
Theorem 2.4.5. A sequence of real numbers is Cauchy if and only if it converges. | Proof. Let \( \varepsilon > 0 \) be given and suppose \( \left\{ {x}_{n}\right\} \) converges to \( x \) . Then there exists an \( M \) such that for \( n \geq M \)\n\n\[ \left| {{x}_{n} - x}\right| < \frac{\varepsilon }{2} \]\n\nHence for \( n \geq M \) and \( k \geq M \), \n\n\[ \left| {{x}_{n} - {x}_{k}}\right| = \l... | Yes |
Proposition 2.5.5. Suppose \( - 1 < r < 1 \) . Then the geometric series \( \mathop{\sum }\limits_{{n = 0}}^{\infty }{r}^{n} \) converges, and\n\n\[ \mathop{\sum }\limits_{{n = 0}}^{\infty }{r}^{n} = \frac{1}{1 - r} \] | Details of the proof are left as an exercise. The proof consists of showing\n\n\[ \mathop{\sum }\limits_{{n = 0}}^{{k - 1}}{r}^{n} = \frac{1 - {r}^{k}}{1 - r} \]\n\nand then taking the limit as \( k \) goes to \( \infty \) . | No |
Proposition 2.5.6. Let \( \sum {x}_{n} \) be a series. Let \( M \in \mathbb{N} \). Then\n\n\[ \mathop{\sum }\limits_{{n = 1}}^{\infty }{x}_{n}\;\text{ converges if and only if }\;\mathop{\sum }\limits_{{n = M}}^{\infty }{x}_{n}\;\text{ converges. } \] | Proof. We look at partial sums of the two series (for \( k \geq M \) )\n\n\[ \mathop{\sum }\limits_{{n = 1}}^{k}{x}_{n} = \left( {\mathop{\sum }\limits_{{n = 1}}^{{M - 1}}{x}_{n}}\right) + \mathop{\sum }\limits_{{n = M}}^{k}{x}_{n} \]\n\nNote that \( \mathop{\sum }\limits_{{n = 1}}^{{M - 1}}{x}_{n} \) is a fixed number... | No |
Proposition 2.5.9. Let \( \sum {x}_{n} \) be a convergent series. Then the sequence \( \left\{ {x}_{n}\right\} \) is convergent and\n\n\[ \mathop{\lim }\limits_{{n \rightarrow \infty }}{x}_{n} = 0 \] | Proof. Let \( \varepsilon > 0 \) be given. As \( \sum {x}_{n} \) is convergent, it is Cauchy. Thus we find an \( M \) such that for every \( n \geq M \) we have\n\n\[ \varepsilon > \left| {\mathop{\sum }\limits_{{j = n + 1}}^{{n + 1}}{x}_{j}}\right| = \left| {x}_{n + 1}\right| \]\n\nHence for every \( n \geq M + 1 \) w... | Yes |
Proposition 2.5.12 (Linearity of series). Let \( \alpha \in \mathbb{R} \) and \( \sum {x}_{n} \) and \( \sum {y}_{n} \) be convergent series. Then\n\n(i) \( \sum \alpha {x}_{n} \) is a convergent series and\n\n\[ \mathop{\sum }\limits_{{n = 1}}^{\infty }\alpha {x}_{n} = \alpha \mathop{\sum }\limits_{{n = 1}}^{\infty }{... | Proof. For the first item, we simply write the \( k \) th partial sum\n\n\[ \mathop{\sum }\limits_{{n = 1}}^{k}\alpha {x}_{n} = \alpha \left( {\mathop{\sum }\limits_{{n = 1}}^{k}{x}_{n}}\right) \]\n\nWe look at the right-hand side and note that the constant multiple of a convergent sequence is convergent. Hence, we tak... | Yes |
Proposition 2.5.13. If \( {x}_{n} \geq 0 \) for all \( n \), then \( \sum {x}_{n} \) converges if and only if the sequence of partial sums is bounded above. | As the limit of a monotone increasing sequence is the supremum, then when \( {x}_{n} \geq 0 \) for all \( n \), we have the inequality\n\n\[ \mathop{\sum }\limits_{{n = 1}}^{k}{x}_{n} \leq \mathop{\sum }\limits_{{n = 1}}^{\infty }{x}_{n} \] | No |
Proposition 2.5.15. If the series \( \sum {x}_{n} \) converges absolutely, then it converges. | Proof. A series is convergent if and only if it is Cauchy. Hence suppose \( \sum \left| {x}_{n}\right| \) is Cauchy. That is, for every \( \varepsilon > 0 \), there exists an \( M \) such that for all \( k \geq M \) and all \( n > k \) we have\n\n\[ \mathop{\sum }\limits_{{j = k + 1}}^{n}\left| {x}_{j}\right| = \left| ... | Yes |
Proposition 2.5.16 (Comparison test). Let \( \sum {x}_{n} \) and \( \sum {y}_{n} \) be series such that \( 0 \leq {x}_{n} \leq {y}_{n} \) for all \( n \in \mathbb{N} \). (i) If \( \sum {y}_{n} \) converges, then so does \( \sum {x}_{n} \). (ii) If \( \sum {x}_{n} \) diverges, then so does \( \sum {y}_{n} \). | Proof. As the terms of the series are all nonnegative, the sequences of partial sums are both monotone increasing. Since \( {x}_{n} \leq {y}_{n} \) for all \( n \), the partial sums satisfy for all \( k \) \[ \mathop{\sum }\limits_{{n = 1}}^{k}{x}_{n} \leq \mathop{\sum }\limits_{{n = 1}}^{k}{y}_{n} \] (2.1) If the seri... | Yes |
Proposition 2.5.17 ( \( p \) -series or the \( p \) -test). For \( p \in \mathbb{R} \), the series\n\n\[ \mathop{\sum }\limits_{{n = 1}}^{\infty }\frac{1}{{n}^{p}} \]\n\nconverges if and only if \( p > 1 \) . | Proof. First suppose \( p \leq 1 \) . As \( n \geq 1 \), we have \( \frac{1}{{n}^{p}} \geq \frac{1}{n} \) . Since \( \sum \frac{1}{n} \) diverges, we see that the \( \sum \frac{1}{{n}^{p}} \) must diverge for all \( p \leq 1 \) by the comparison test.\n\nNow suppose \( p > 1 \) . We proceed as we did for the harmonic s... | Yes |
Proposition 2.5.19 (Ratio test). Let \( \sum {x}_{n} \) be a series, \( {x}_{n} \neq 0 \) for all \( n \), and such that\n\n\[ L \mathrel{\text{:=}} \mathop{\lim }\limits_{{n \rightarrow \infty }}\frac{\left| {x}_{n + 1}\right| }{\left| {x}_{n}\right| } \]\n\nexists. Then\n\n(i) If \( L < 1 \), then \( \sum {x}_{n} \) ... | Proof. If \( L > 1 \), then Lemma 2.2.12 says that the sequence \( \left\{ {x}_{n}\right\} \) diverges. Since it is a necessary condition for the convergence of series that the terms go to zero, we know that \( \sum {x}_{n} \) must diverge.\n\nThus suppose \( L < 1 \) . We will argue that \( \sum \left| {x}_{n}\right| ... | Yes |
Proposition 2.6.1 (Root test). Let \( \sum {x}_{n} \) be a series and let\n\n\[ L \mathrel{\text{:=}} \mathop{\limsup }\limits_{{n \rightarrow \infty }}{\left| {x}_{n}\right| }^{1/n}. \]\n\nThen\n\n(i) If \( L < 1 \), then \( \sum {x}_{n} \) converges absolutely.\n\n(ii) If \( L > 1 \), then \( \sum {x}_{n} \) diverges... | Proof. If \( L > 1 \), then there exists a subsequence \( \left\{ {x}_{{n}_{k}}\right\} \) such that \( L = \mathop{\lim }\limits_{{k \rightarrow \infty }}{\left| {x}_{{n}_{k}}\right| }^{1/{n}_{k}} \) . Let \( r \) be such that \( L > r > 1 \) . There exists an \( M \) such that for all \( k \geq M \), we have \( {\lef... | Yes |
Proposition 2.6.2 (Alternating series). Let \( \left\{ {x}_{n}\right\} \) be a monotone decreasing sequence of positive real numbers such that \( \lim {x}_{n} = 0 \) . Then\n\n\[ \mathop{\sum }\limits_{{n = 1}}^{\infty }{\left( -1\right) }^{n}{x}_{n} \]\n\nconverges. | Proof. Let \( {s}_{m} \mathrel{\text{:=}} \mathop{\sum }\limits_{{k = 1}}^{m}{\left( -1\right) }^{k}{x}_{k} \) be the \( m \) th partial sum. Then write\n\n\[ {s}_{2n} = \mathop{\sum }\limits_{{k = 1}}^{{2n}}{\left( -1\right) }^{k}{x}_{k} = \left( {-{x}_{1} + {x}_{2}}\right) + \cdots + \left( {-{x}_{{2n} - 1} + {x}_{2n... | Yes |
Proposition 2.6.3. Let \( \sum {x}_{n} \) be an absolutely convergent series converging to a number \( x \) . Let \( \sigma : \mathbb{N} \rightarrow \mathbb{N} \) be a bijection. Then \( \sum {x}_{\sigma \left( n\right) } \) is absolutely convergent and converges to \( x \) . | Proof. Let \( \varepsilon > 0 \) be given. As \( \sum {x}_{n} \) is absolutely convergent, take \( M \) such that\n\n\[ \left| {\left( {\mathop{\sum }\limits_{{n = 1}}^{M}{x}_{n}}\right) - x}\right| < \frac{\varepsilon }{2}\;\text{ and }\;\mathop{\sum }\limits_{{n = M + 1}}^{\infty }\left| {x}_{n}\right| < \frac{\varep... | Yes |
Proposition 2.6.10. Let \( \mathop{\sum }\limits_{{n = 0}}^{\infty }{a}_{n}{\left( x - {x}_{0}\right) }^{n} \) be a power series. If the series is convergent, then either it converges at all \( x \in \mathbb{R} \), or there exists a number \( \rho \), such that the series converges absolutely on the interval \( \left( ... | Proof. Write\n\n\[ R \mathrel{\text{:=}} \mathop{\limsup }\limits_{{n \rightarrow \infty }}{\left| {a}_{n}\right| }^{1/n}. \]\n\nWe use the root test to prove the proposition:\n\n\[ L = \mathop{\limsup }\limits_{{n \rightarrow \infty }}{\left| {a}_{n}{\left( x - {x}_{0}\right) }^{n}\right| }^{1/n} = \left| {x - {x}_{0}... | Yes |
Proposition 2.6.12. Let \( \mathop{\sum }\limits_{{n = 0}}^{\infty }{a}_{n}{\left( x - {x}_{0}\right) }^{n} \) and \( \mathop{\sum }\limits_{{n = 0}}^{\infty }{b}_{n}{\left( x - {x}_{0}\right) }^{n} \) be two convergent power series with radius of convergence at least \( \rho > 0 \) and \( \alpha \in \mathbb{R} \) . Th... | That is, after performing the algebraic operations, the radius of convergence of the resulting series is at least \( \rho \) . For all \( x \) with \( \left| {x - {x}_{0}}\right| < \rho \), we have two convergent series so their term by term addition and multiplication by constants follows by what we learned in the las... | No |
Proposition 3.1.2. Let \( S \subset \mathbb{R} \) . Then \( x \in \mathbb{R} \) is a cluster point of \( S \) if and only if there exists a convergent sequence of numbers \( \left\{ {x}_{n}\right\} \) such that \( {x}_{n} \neq x \) and \( {x}_{n} \in S \) for all \( n \), and \( \lim {x}_{n} = x \) . | Proof. First suppose \( x \) is a cluster point of \( S \) . For any \( n \in \mathbb{N} \), we pick \( {x}_{n} \) to be an arbitrary point of \( \left( {x - 1/n, x + 1/n}\right) \cap S \smallsetminus \{ x\} \), which we know is nonempty because \( x \) is a cluster point of \( S \) . Then \( {x}_{n} \) is within \( 1/... | Yes |
Proposition 3.1.4. Let \( c \) be a cluster point of \( S \subset \mathbb{R} \) and let \( f : S \rightarrow \mathbb{R} \) be a function such that \( f\left( x\right) \) converges as \( x \) goes to \( c \) . Then the limit of \( f\left( x\right) \) as \( x \) goes to \( c \) is unique. | Proof. Let \( {L}_{1} \) and \( {L}_{2} \) be two numbers that both satisfy the definition. Take an \( \varepsilon > 0 \) and find a \( {\delta }_{1} > 0 \) such that \( \left| {f\left( x\right) - {L}_{1}}\right| < \varepsilon /2 \) for all \( x \in S \smallsetminus \{ c\} \) with \( \left| {x - c}\right| < {\delta }_{... | Yes |
Lemma 3.1.7. Let \( S \subset \mathbb{R} \) and \( c \) be a cluster point of \( S \) . Let \( f : S \rightarrow \mathbb{R} \) be a function.\n\nThen \( f\left( x\right) \rightarrow L \) as \( x \rightarrow c \) if and only if for every sequence \( \left\{ {x}_{n}\right\} \) of numbers such that \( {x}_{n} \in S \small... | Proof. Suppose \( f\left( x\right) \rightarrow L \) as \( x \rightarrow c \), and \( \left\{ {x}_{n}\right\} \) is a sequence such that \( {x}_{n} \in S \smallsetminus \{ c\} \) and \( \lim {x}_{n} = c \) . We wish to show that \( \left\{ {f\left( {x}_{n}\right) }\right\} \) converges to \( L \) . Let \( \varepsilon > ... | Yes |
Corollary 3.1.9. Let \( S \subset \mathbb{R} \) and \( c \) be a cluster point of \( S \) . Let \( f : S \rightarrow \mathbb{R} \) and \( g : S \rightarrow \mathbb{R} \) be functions. Suppose the limits of \( f\left( x\right) \) and \( g\left( x\right) \) as \( x \) goes to \( c \) both exist, and that\n\n\[ f\left( x\... | Proof. Take \( \left\{ {x}_{n}\right\} \) be a sequence of numbers in \( S \smallsetminus \{ c\} \) that converges to \( c \) . Let\n\n\[ {L}_{1} \mathrel{\text{:=}} \mathop{\lim }\limits_{{x \rightarrow c}}f\left( x\right) ,\;\text{ and }\;{L}_{2} \mathrel{\text{:=}} \mathop{\lim }\limits_{{x \rightarrow c}}g\left( x\... | Yes |
Proposition 3.1.15. Let \( S \subset \mathbb{R}, c \in \mathbb{R} \), and let \( f : S \rightarrow \mathbb{R} \) be a function. Suppose \( A \subset S \) is such that there is some \( \alpha > 0 \) such that \( \left( {A\smallsetminus \{ c\} }\right) \cap \left( {c - \alpha, c + \alpha }\right) = \left( {S\smallsetminu... | Proof. First, let \( c \) be a cluster point of \( A \) . Since \( A \subset S \), then if \( \left( {A\smallsetminus \{ c\} }\right) \cap \left( {c - \varepsilon, c + \varepsilon }\right) \) is nonempty for every \( \varepsilon > 0 \), then \( \left( {S\smallsetminus \{ c\} }\right) \cap \left( {c - \varepsilon, c + \... | Yes |
Proposition 3.2.2. Let \( S \subset \mathbb{R} \), let \( f : S \rightarrow \mathbb{R} \) be a function, and let \( c \in S \) be a point. Then\n\n(i) If \( c \) is not a cluster point of \( S \), then \( f \) is continuous at \( c \) . | Proof. Let us start with the first item. Suppose \( c \) is not a cluster point of \( S \) . Then there exists a \( \delta > 0 \) such that \( S \cap \left( {c - \delta, c + \delta }\right) = \{ c\} \) . Therefore, for any \( \varepsilon > 0 \), simply pick this given delta. The only \( x \in S \) such that \( \left| {... | Yes |
Proposition 3.2.4. Let \( f : \mathbb{R} \rightarrow \mathbb{R} \) be a polynomial. That is\n\n\[ f\left( x\right) = {a}_{d}{x}^{d} + {a}_{d - 1}{x}^{d - 1} + \cdots + {a}_{1}x + {a}_{0}, \]\n\nfor some constants \( {a}_{0},{a}_{1},\ldots ,{a}_{d} \) . Then \( f \) is continuous. | Proof. Fix \( c \in \mathbb{R} \) . Let \( \left\{ {x}_{n}\right\} \) be a sequence such that \( \lim {x}_{n} = c \) . Then\n\n\[ f\left( c\right) = {a}_{d}{c}^{d} + {a}_{d - 1}{c}^{d - 1} + \cdots + {a}_{1}c + {a}_{0} \]\n\n\[ = {a}_{d}{\left( \lim {x}_{n}\right) }^{d} + {a}_{d - 1}{\left( \lim {x}_{n}\right) }^{d - 1... | Yes |
Proposition 3.2.7. Let \( A, B \subset \mathbb{R} \) and \( f : B \rightarrow \mathbb{R} \) and \( g : A \rightarrow B \) be functions. If \( g \) is continuous at \( c \in A \) and \( f \) is continuous at \( g\left( c\right) \), then \( f \circ g : A \rightarrow \mathbb{R} \) is continuous at \( c \) . | Proof. Let \( \left\{ {x}_{n}\right\} \) be a sequence in \( A \) such that \( \lim {x}_{n} = c \) . As \( g \) is continuous at \( c \), then \( \left\{ {g\left( {x}_{n}\right) }\right\} \) converges to \( g\left( c\right) \) . As \( f \) is continuous at \( g\left( c\right) \), then \( \left\{ {f\left( {g\left( {x}_{... | Yes |
Proposition 3.2.9. Let \( f : S \rightarrow \mathbb{R} \) be a function and \( c \in S \) . Suppose there exists a sequence \( \left\{ {x}_{n}\right\} \) , \( {x}_{n} \in S \), and \( \lim {x}_{n} = c \) such that \( \left\{ {f\left( {x}_{n}\right) }\right\} \) does not converge to \( f\left( c\right) \) . Then \( f \)... | Proof: Take the sequence \( \{ - 1/n\} \), which converges to 0 . Then \( f\left( {-1/n}\right) = - 1 \) for every \( n \), and so \( \lim f\left( {-1/n}\right) = - 1 \), but \( f\left( 0\right) = 1 \) . Thus the function is not continuous at 0 . See Figure 3.3. | No |
Lemma 3.3.1. Let \( f : \left\lbrack {a, b}\right\rbrack \rightarrow \mathbb{R} \) be a continuous function. Then \( f \) is bounded. | Proof. Let us prove this claim by contrapositive. Suppose \( f \) is not bounded. Then for each \( n \in \mathbb{N} \) , there is an \( {x}_{n} \in \left\lbrack {a, b}\right\rbrack \), such that\n\n\[ \left| {f\left( {x}_{n}\right) }\right| \geq n \]\n\nThe sequence \( \left\{ {x}_{n}\right\} \) is bounded as \( a \leq... | Yes |
Theorem 3.3.2 (Minimum-maximum theorem). Let \( f : \left\lbrack {a, b}\right\rbrack \rightarrow \mathbb{R} \) be a continuous function. Then \( f \) achieves both an absolute minimum and an absolute maximum on \( \left\lbrack {a, b}\right\rbrack \) . | Proof. The lemma says that \( f \) is bounded, and thus the set \( f\left( \left\lbrack {a, b}\right\rbrack \right) = \{ f\left( x\right) : x \in \left\lbrack {a, b}\right\rbrack \} \) has a supremum and an infimum. There exist sequences in the set \( f\left( \left\lbrack {a, b}\right\rbrack \right) \) that approach it... | No |
Lemma 3.3.7. Let \( f : \left\lbrack {a, b}\right\rbrack \rightarrow \mathbb{R} \) be a continuous function. Suppose \( f\left( a\right) < 0 \) and \( f\left( b\right) > 0 \) . Then there exists a number \( c \in \left( {a, b}\right) \) such that \( f\left( c\right) = 0 \) . | Proof. We define two sequences \( \left\{ {a}_{n}\right\} \) and \( \left\{ {b}_{n}\right\} \) inductively:\n\n(i) Let \( {a}_{1} \mathrel{\text{:=}} a \) and \( {b}_{1} \mathrel{\text{:=}} b \) .\n\n(ii) If \( f\left( \frac{{a}_{n} + {b}_{n}}{2}\right) \geq 0 \), let \( {a}_{n + 1} \mathrel{\text{:=}} {a}_{n} \) and \... | Yes |
Theorem 3.3.8 (Bolzano’s intermediate value theorem). Let \( f : \left\lbrack {a, b}\right\rbrack \rightarrow \mathbb{R} \) be a continuous function. Suppose \( y \in \mathbb{R} \) is such that \( f\left( a\right) < y < f\left( b\right) \) or \( f\left( a\right) > y > f\left( b\right) \) . Then there exists a \( c \in ... | Proof. If \( f\left( a\right) < y < f\left( b\right) \), then define \( g\left( x\right) \mathrel{\text{:=}} f\left( x\right) - y \) . Then we see that \( g\left( a\right) < 0 \) and \( g\left( b\right) > 0 \) and we apply Lemma 3.3.7 to \( g \) to find \( c \) . If \( g\left( c\right) = 0 \), then \( f\left( c\right) ... | Yes |
The polynomial \( f\left( x\right) \mathrel{\text{:=}} {x}^{3} - 2{x}^{2} + x - 1 \) has a real root in \( \left( {1,2}\right) \). | We simply notice that \( f\left( 1\right) = - 1 \) and \( f\left( 2\right) = 1 \) . Hence there must exist a point \( c \in \left( {1,2}\right) \) such that \( f\left( c\right) = 0 \) . To find a better approximation of the root we follow the proof of Lemma 3.3.7. We look at 1.5 and find that \( f\left( {1.5}\right) = ... | Yes |
Proposition 3.3.10. Let \( f\left( x\right) \) be a polynomial of odd degree. Then \( f \) has a real root. | Proof. Suppose \( f \) is a polynomial of odd degree \( d \) . We write\n\n\[ f\left( x\right) = {a}_{d}{x}^{d} + {a}_{d - 1}{x}^{d - 1} + \cdots + {a}_{1}x + {a}_{0}, \]\n\nwhere \( {a}_{d} \neq 0 \) . We divide by \( {a}_{d} \) to obtain a monic polynomial*\n\n\[ g\left( x\right) \mathrel{\text{:=}} {x}^{d} + {b}_{d ... | No |
Theorem 3.4.4. Let \( f : \left\lbrack {a, b}\right\rbrack \rightarrow \mathbb{R} \) be a continuous function. Then \( f \) is uniformly continuous. | Proof. We prove the statement by contrapositive. Suppose \( f \) is not uniformly continuous. We will prove that there is some \( c \in \left\lbrack {a, b}\right\rbrack \) where \( f \) is not continuous. Let us negate the definition of uniformly continuous. There exists an \( \varepsilon > 0 \) such that for every \( ... | Yes |
Lemma 3.4.5. Let \( f : S \rightarrow \mathbb{R} \) be a uniformly continuous function. Let \( \left\{ {x}_{n}\right\} \) be a Cauchy sequence in S. Then \( \left\{ {f\left( {x}_{n}\right) }\right\} \) is Cauchy. | Proof. Let \( \varepsilon > 0 \) be given. There is a \( \delta > 0 \) such that \( \left| {f\left( x\right) - f\left( y\right) }\right| < \varepsilon \) whenever \( x, y \in S \) and \( \left| {x - y}\right| < \delta \) . Find an \( M \in \mathbb{N} \) such that for all \( n, k \geq M \) we have \( \left| {{x}_{n} - {... | Yes |
Proposition 3.4.6. A function \( f : \left( {a, b}\right) \rightarrow \mathbb{R} \) is uniformly continuous if and only if the limits\n\n\[ \n{L}_{a} \mathrel{\text{:=}} \mathop{\lim }\limits_{{x \rightarrow a}}f\left( x\right) \;\text{ and }\;{L}_{b} \mathrel{\text{:=}} \mathop{\lim }\limits_{{x \rightarrow b}}f\left(... | Proof. One direction is not difficult. If \( \widetilde{f} \) is continuous, then it is uniformly continuous by Theorem 3.4.4. As \( f \) is the restriction of \( \widetilde{f} \) to \( \left( {a, b}\right) \), then \( f \) is also uniformly continuous (easy exercise).\n\nNow suppose \( f \) is uniformly continuous. We... | No |
Proposition 3.4.8. A Lipschitz continuous function is uniformly continuous. | Proof. Let \( f : S \rightarrow \mathbb{R} \) be a function and let \( K \) be a constant such that \( \left| {f\left( x\right) - f\left( y\right) }\right| \leq K\left| {x - y}\right| \) for all \( x, y \) in \( S \) . Let \( \varepsilon > 0 \) be given. Take \( \delta \mathrel{\text{:=}} \varepsilon /\kappa \) . For a... | Yes |
Lemma 3.5.5. Suppose \( f : S \rightarrow \mathbb{R} \) is a function, \( \infty \) is a cluster point of \( S \subset \mathbb{R} \), and \( L \in \mathbb{R} \) . Then\n\n\[ \mathop{\lim }\limits_{{x \rightarrow \infty }}f\left( x\right) = L \]\n\nif and only if\n\n\[ \mathop{\lim }\limits_{{n \rightarrow \infty }}f\le... | Proof. First suppose \( f\left( x\right) \rightarrow L \) as \( x \rightarrow \infty \) . Given an \( \varepsilon > 0 \), there exists an \( M \) such that for all \( x \geq M \) we have \( \left| {f\left( x\right) - L}\right| < \varepsilon \) . Let \( \left\{ {x}_{n}\right\} \) be a sequence in \( S \) such that \( \l... | Yes |
Proposition 3.6.2. Let \( S \subset \mathbb{R}, c \in \mathbb{R}, f : S \rightarrow \mathbb{R} \) be increasing, and \( g : S \rightarrow \mathbb{R} \) be decreasing. If \( c \) is a cluster point of \( S \cap \left( {-\infty, c}\right) \), then\n\n\[ \mathop{\lim }\limits_{{x \rightarrow {c}^{ - }}}f\left( x\right) = ... | Proof. Let us assume \( f \) is increasing, and we will show the first equality. The rest of the proof is very similar and is left as an exercise.\n\nLet \( a \mathrel{\text{:=}} \sup \{ f\left( x\right) : x < c, x \in S\} \) . If \( a = \infty \), then given an \( M \in \mathbb{R} \), there exists an \( {x}_{M} \in S,... | No |
Corollary 3.6.3. If \( I \subset \mathbb{R} \) is an interval and \( f : I \rightarrow \mathbb{R} \) is monotone and not constant, then \( f\left( I\right) \) is an interval if and only if \( f \) is continuous. | Proof. Without loss of generality, suppose \( f \) is increasing.\n\nFirst suppose \( f \) is continuous. Take two points \( f\left( {x}_{1}\right) < f\left( {x}_{2}\right) \) in \( f\left( I\right) \) . As \( f \) is increasing, then \( {x}_{1} < {x}_{2} \) . By the intermediate value theorem, given any \( y \) with \... | No |
Corollary 3.6.4. Let \( I \subset \mathbb{R} \) be an interval and \( f : I \rightarrow \mathbb{R} \) be monotone. Then \( f \) has at most countably many discontinuities. | Proof. Let \( E \subset I \) be the set of all discontinuities that are not endpoints of \( I \) . As there are only two endpoints, it is enough to show that \( E \) is countable. Without loss of generality, suppose \( f \) is increasing. We will define an injection \( h : E \rightarrow \mathbb{Q} \) . For each \( c \i... | Yes |
Proposition 3.6.6. If \( I \subset \mathbb{R} \) is an interval and \( f : I \rightarrow \mathbb{R} \) is strictly monotone, then the inverse \( {f}^{-1} : f\left( I\right) \rightarrow I \) is continuous. | Proof. Let us suppose \( f \) is strictly increasing. The proof is almost identical for a strictly decreasing function. Since \( f \) is strictly increasing, so is \( {f}^{-1} \) . That is, if \( f\left( x\right) < f\left( y\right) \), then we must have \( x < y \) and therefore \( {f}^{-1}\left( {f\left( x\right) }\ri... | No |
Proposition 4.1.6. Let \( f : I \rightarrow \mathbb{R} \) be differentiable at \( c \in I \), then it is continuous at \( c \) . | Proof. We know the limits\n\n\[ \mathop{\lim }\limits_{{x \rightarrow c}}\frac{f\left( x\right) - f\left( c\right) }{x - c} = {f}^{\prime }\left( c\right) \;\text{ and }\;\mathop{\lim }\limits_{{x \rightarrow c}}\left( {x - c}\right) = 0 \]\n\nexist. Furthermore,\n\n\[ f\left( x\right) - f\left( c\right) = \left( \frac... | Yes |
Proposition 4.1.7. Let \( I \) be an interval, let \( f : I \rightarrow \mathbb{R} \) and \( g : I \rightarrow \mathbb{R} \) be differentiable at \( c \in I \), and let \( \alpha \in \mathbb{R} \) . (i) Define \( h : I \rightarrow \mathbb{R} \) by \( h\left( x\right) \mathrel{\text{:=}} {\alpha f}\left( x\right) \) . T... | Proof. First, let \( h\left( x\right) \mathrel{\text{:=}} {\alpha f}\left( x\right) \) . For \( x \in I, x \neq c \) we have \[ \frac{h\left( x\right) - h\left( c\right) }{x - c} = \frac{{\alpha f}\left( x\right) - {\alpha f}\left( c\right) }{x - c} = \alpha \frac{f\left( x\right) - f\left( c\right) }{x - c}. \] The li... | Yes |
Proposition 4.1.10 (Chain Rule). Let \( {I}_{1},{I}_{2} \) be intervals, let \( g : {I}_{1} \rightarrow {I}_{2} \) be differentiable at \( c \in {I}_{1} \) , and \( f : {I}_{2} \rightarrow \mathbb{R} \) be differentiable at \( g\left( c\right) \) . If \( h : {I}_{1} \rightarrow \mathbb{R} \) is defined by\n\n\[ h\left(... | Proof. Let \( d \mathrel{\text{:=}} g\left( c\right) \) . Define \( u : {I}_{2} \rightarrow \mathbb{R} \) and \( v : {I}_{1} \rightarrow \mathbb{R} \) by\n\n\[ u\left( y\right) \mathrel{\text{:=}} \left\{ \begin{array}{ll} \frac{f\left( y\right) - f\left( d\right) }{y - d} & \text{ if }y \neq d, \\ {f}^{\prime }\left( ... | Yes |
Lemma 4.2.2. Let \( f : \left\lbrack {a, b}\right\rbrack \rightarrow \mathbb{R} \) be a function differentiable at \( c \in \left( {a, b}\right) \), and \( f \) has a relative minimum or a relative maximum at \( c \) . Then \( {f}^{\prime }\left( c\right) = 0 \) . | Proof. We prove the statement for a maximum. For a minimum the statement follows by considering the function \( - f \) .\n\nLet \( c \) be a relative maximum of \( f \) . In particular, as long as \( \left| {x - c}\right| < \delta \) we have \( f\left( x\right) - f\left( c\right) \leq 0 \) . Then we look at the differe... | Yes |
Theorem 4.2.3 (Rolle). Let \( f : \left\lbrack {a, b}\right\rbrack \rightarrow \mathbb{R} \) be continuous function differentiable on \( \left( {a, b}\right) \) such that \( f\left( a\right) = f\left( b\right) \) . Then there exists a \( c \in \left( {a, b}\right) \) such that \( {f}^{\prime }\left( c\right) = 0 \) . | Proof. As \( f \) is continuous on \( \left\lbrack {a, b}\right\rbrack \) it attains an absolute minimum and an absolute maximum in \( \left\lbrack {a, b}\right\rbrack \) . We wish to apply Lemma 4.2.2 and so we need a minimum or maximum at some \( c \in \left( {a, b}\right) \) . Write \( K \mathrel{\text{:=}} f\left( ... | Yes |
Theorem 4.2.4 (Mean value theorem). Let \( f : \left\lbrack {a, b}\right\rbrack \rightarrow \mathbb{R} \) be a continuous function differentiable on \( \left( {a, b}\right) \) . Then there exists a point \( c \in \left( {a, b}\right) \) such that\n\n\[ f\left( b\right) - f\left( a\right) = {f}^{\prime }\left( c\right) ... | Proof. Define the function \( g : \left\lbrack {a, b}\right\rbrack \rightarrow \mathbb{R} \) by\n\n\[ g\left( x\right) \mathrel{\text{:=}} f\left( x\right) - f\left( b\right) - \frac{f\left( b\right) - f\left( a\right) }{b - a}\left( {x - b}\right) . \]\n\nThe function \( g \) is differentiable on \( \left( {a, b}\righ... | Yes |
Proposition 4.2.6. Let \( I \) be an interval and let \( f : I \rightarrow \mathbb{R} \) be a differentiable function such that \( {f}^{\prime }\left( x\right) = 0 \) for all \( x \in I \) . Then \( f \) is constant. | Proof. Take arbitrary \( x, y \in I \) with \( x < y \) . Then \( f \) restricted to \( \left\lbrack {x, y}\right\rbrack \) satisfies the hypotheses of the mean value theorem. Therefore, there is a \( c \in \left( {x, y}\right) \) such that\n\n\[ f\left( y\right) - f\left( x\right) = {f}^{\prime }\left( c\right) \left(... | Yes |
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