Exam (elaborations) TEST BANK FOR Understanding Analysis 2nd Edition By Stephen Abbott (Instructors' Solution Manual)

1 The Real Numbers 1 1.1 Discussion: The Irrationality of p 2 . . . . . . . . . . . . . . . . . 1 1.2 Some Preliminaries . . . . . . . . . . . . . . . . . . . . . . . . . . 1 1.3 The Axiom of Completeness . . . . . . . . . . . . . . . . . . . . . 6 1.4 Consequences of Completeness . . . . . . . . . . . . . . . . . . . 8 1.5 Cantor’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . 14 2 Sequences and Series 19 2.1 Discussion: Rearrangements of Infinite Series . . . . . . . . . . . 19 2.2 The Limit of a Sequence . . . . . . . . . . . . . . . . . . . . . . . 19 2.3 The Algebraic and Order Limit Theorems . . . . . . . . . . . . . 21 2.4 The Monotone Convergence Theorem and a First Look at Infinite Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25 2.5 Subsequences and the Bolzano–Weierstrass Theorem . . . . . . . 29 2.6 The Cauchy Criterion . . . . . . . . . . . . . . . . . . . . . . . . 31 2.7 Properties of Infinite Series . . . . . . . . . . . . . . . . . . . . . 33 2.8 Double Summations and Products of Infinite Series . . . . . . . . 39 3 Basic Topology of R 45 3.1 Discussion: The Cantor Set . . . . . . . . . . . . . . . . . . . . . 45 3.2 Open and Closed Sets . . . . . . . . . . . . . . . . . . . . . . . . 45 3.3 Compact Sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49 3.4 Perfect Sets and Connected Sets . . . . . . . . . . . . . . . . . . 51 3.5 Baire’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . 55 4 Functional Limits and Continuity 57 4.1 Discussion: Examples of Dirichlet and Thomae . . . . . . . . . . 57 4.2 Functional Limits . . . . . . . . . . . . . . . . . . . . . . . . . . . 57 4.3 Combinations of Continuous Functions . . . . . . . . . . . . . . . 61 4.4 Continuous Functions on Compact Sets . . . . . . . . . . . . . . 66 4.5 The Intermediate Value Theorem . . . . . . . . . . . . . . . . . . 70 4.6 Sets of Discontinuity . . . . . . . . . . . . . . . . . . . . . . . . . 72 vii viii Contents 5 The Derivative 75 5.1 Discussion: Are Derivatives Continuous? . . . . . . . . . . . . . . 75 5.2 Derivatives and the Intermediate Value Property . . . . . . . . . 75 5.3 The Mean Value Theorem . . . . . . . . . . . . . . . . . . . . . . 79 5.4 A Continuous Nowhere-Differentiable Function . . . . . . . . . . 84 6 Sequences and Series of Functions 89 6.1 Discussion: Branching Processes . . . . . . . . . . . . . . . . . . 89 6.2 Uniform Convergence of a Sequence of Functions . . . . . . . . . 89 6.3 Uniform Convergence and Differentiation . . . . . . . . . . . . . 97 6.4 Series of Functions . . . . . . . . . . . . . . . . . . . . . . . . . . 99 6.5 Power Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 102 6.6 Taylor Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 105 7 The Riemann Integral 111 7.1 Discussion: How Should Integration be Defined? . . . . . . . . . 111 7.2 The Definition of the Riemann Integral . . . . . . . . . . . . . . . 111 7.3 Integrating Functions with Discontinuities . . . . . . . . . . . . . 114 7.4 Properties of the Integral . . . . . . . . . . . . . . . . . . . . . . 117 7.5 The Fundamental Theorem of Calculus . . . . . . . . . . . . . . . 120 7.6 Lebesgue’s Criterion for Riemann Integrability . . . . . . . . . . 123 8 Additional Topics 129 8.1 The Generalized Riemann Integral . . . . . . . . . . . . . . . . . 129 8.2 Metric Spaces and the Baire Category Theorem . . . . . . . . . . 133 8.3 Fourier Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 141 8.4 A Construction of R From Q . . . . . . . . . . . . . . . . . . . . 149 Chapter 1 The Real Numbers 1.1 Discussion: The Irrationality of p 2 1.2 Some Preliminaries Exercise 1.2.1. (a) Assume, for contradiction, that there exist integers p and q satisfying (1) µ p q ¶2 = 3: Let us also assume that p and q have no common factor. Now, equation (1) implies (2) p2 = 3q2: From this, we can see that p2 is a multiple of 3 and hence p must also be a multiple of 3. This allows us to write p = 3r, where r is an integer. After substituting 3r for p in equation (2), we get (3r)2 = 3q2, which can be simplified to 3r2 = q2. This implies q2 is a multiple of 3 and hence q is also a multiple of 3. Thus we have shown p and q have a common factor, namely 3, when they were originally assumed to have no common factor. A similar argument will work for p 6 as well because we get p2 = 6q2 which implies p is a multiple of 2 and 3. After making the necessary substitutions, we can conclude q is a multiple of 6, and therefore p 6 must be irrational. (b) In this case, the fact that p2 is a multiple of 4 does not imply p is also a multiple of 4. Thus, our proof breaks down at this point. Exercise 1.2.2. (a) False, as seen in Example 1.2.2. (b) True. This will follow from upcoming results about compactness in Chapter 3. (c) False. Consider sets A = f1; 2; 3g;B = f3; 6; 7g and C = f5g. Note that A (B [ C) = f3g is not equal to (A B) [ C = f3; 5g. 1 2 Chapter 1. The Real Numbers (d) True. (e) True. Exercise 1.2.3. (a) If x 2 (A B)c then x =2 (A B). But this implies x =2 A or x =2 B. From this we know x 2 Ac or x 2 Bc. Thus, x 2 Ac [ Bc by the definition of union. (b) To show Ac [ Bc µ (A B)c, let x 2 Ac [ Bc and show x 2 (A B)c. So, if x 2 Ac [ Bc then x 2 Ac or x 2 Bc. From this, we know that x =2 A or x =2 B, which implies x =2 (A B). This means x 2 (A B)c, which is precisely what we wanted to show. (c) In order to prove (A [ B)c = Ac Bc we have to show, (1) (A [ B)c µ Ac Bc and; (2) Ac Bc µ (A [ B)c: To demonstrate part (1) take x 2 (A [ B)c and show that x 2 (Ac Bc). So, if x 2 (A [ B)c then x =2 (A [ B). From this, we know that x =2A and x =2 B which implies x 2 Ac and x 2 Bc. This means x 2 (Ac Bc). Similarly, part (2) can be shown by taking x 2 (Ac Bc) and showing that x 2 (A[B)c. So, if x 2 (Ac Bc) then x 2 Ac and x 2 Bc. From this, we know that x =2 A and x =2 B which implies x =2 (A [ B). This means x 2 (A [ B)c. Since we have shown inclusion both ways, we conclude that (A[B)c = Ac Bc. Exercise 1.2.4. (a)When a and b have the same sign, consider the following two cases: (i) If a ¸ 0 and b ¸ 0 then we have a + b > 0 which implies ja + bj = a + b. Furthermore, because jaj = a and jbj = b, we have jaj+jbj = a+b. This implies, ja + bj = jaj + jbj, which satisfies the triangle inequality. (ii) If a · 0 and b · 0 then we have a+b · 0 which implies ja+bj = ¡a¡b. Furthermore, since we know jaj = ¡a and jbj = ¡b we have jaj + jbj = ¡a ¡ b. This implies, ja + bj = jaj + jbj, which satisfies the triangle inequality. (b)If a ¸ 0, b < 0, and a + b ¸ 0 then we have ja + bj = a + b = a ¡ (¡b) = jaj ¡ jbj < jaj + jbj. This implies ja + bj · jaj + jbj as desired. Exercise 1.2.5. (a) Observe that ja ¡ bj = ja + (¡b)j · jaj + j ¡ bj = jaj + jbj which implies ja ¡ bj · jaj + jbj. (b) First note that jaj = ja ¡ b + bj · ja ¡ bj + jbj. Taking jbj to the left side of the inequality we get jaj¡jbj · ja¡bj. Reversing the roles of a and b in the previous argument gives jbj ¡ jaj · jb ¡ aj, and because ja ¡ bj = jb ¡ aj the result follows. Exercise 1.2.6. (a) f(A) = [0; 4] and f(B) = [1; 16]. In this case, f(A B) = f(A) f(B) = [1; 4] and f(A [ B) = f(A) [ f(B) = [0; 16]. (b) Take A = [0; 2] and B = [¡2; 0] and note that f(A B) = f0g but f(A) f(B) = [0; 4]. 1.2. Some Preliminaries 3 (c) We have to show y 2 g(A B) implies y 2 g(A) g(B). If y 2 g(A B) then there exists an x 2 AB with g(x) = y. But this means x 2 A and x 2 B and hence g(x) 2 g(A) and g(x) 2 g(B). Therefore, g(x) = y 2 g(A) g(B). (d) Our claim is g(A [ B) = g(A) [ g(B). In order to prove it, we have to show, (1) g(A [ B) µ g(A) [ g(B) and; (2) g(A) [ g(B) µ g(A [ B): To demonstrate part (1), we let y 2 g(A [ B) and show y 2 g(A) [ g(B). If y 2 g(A [ B) then there exists x 2 A [ B with g(x) = y. But this means x 2 A or x 2 B, and hence g(x) 2 g(A) or g(x) 2 g(B). Therefore, g(x) = y 2 g(A) [ g(B). To demonstrate the reverse inclusion, we let y 2 g(A) [ g(B) and show y 2 g(A [ B). If y 2 g(A) [ g(B) then y 2 g(A) or y 2 g(B). This means we have an x 2 A or x 2 B such that g(x) = y. This implies, x 2 A [ B, and hence g(x) 2 g(A[B). Since we have shown parts (1) and (2), we can conclude g(A [ B) = g(A) [ g(B). Exercise 1.2.7. (a) f¡1(A) = [¡2; 2] and f¡1(B) = [¡1; 1]. In this case, f¡1(AB) = f¡1(A)f¡1(B) = [¡1; 1] and f¡1(A[B) = f¡1(A)[f¡1(B) = [¡2; 2]. (b) In order to prove g¡1(A B) = g¡1(A) g¡1(B), we have to show, (1) g¡1(A B) µ g¡1(A) g¡1(B) and; (2) g¡1(A) g¡1(B) µ g¡1(A B): To demonstrate part (1), we let x 2 g¡1(AB) and show x 2 g¡1(A)g¡1(B). So, if x 2 g¡1(A B) then g(x) 2 (A B). But this means g(x) 2 A and g(x) 2 B, and hence g(x) 2 A B. This implies, x 2 g¡1(A) g¡1(B). To demonstrate the reverse inclusion, we let x 2 g¡1(A) g¡1(B) and show x 2 g¡1(A B). So, if x 2 g¡1(A) g¡1(B) then x 2 g¡1(A) and x 2 g¡1(B). This implies g(x) 2 A and g(x) 2 B, and hence g(x) 2 A B. This means, x 2 g¡1(A B). Similarly, in order to prove g¡1(A[B) = g¡1(A)[g¡1(B), we have to show, (1) g¡1(A [ B) µ g¡1(A) [ g¡1(B) and; (2) g¡1(A) [ g¡1(B) µ g¡1(A [ B): To demonstrate part (1), we let x 2 g¡1(A[B) and show x 2 g¡1(A)[g¡1(B). So, if x 2 g¡1(A [ B) then g(x) 2 (A [ B). But this means g(x) 2 A or g(x) 2 B, which implies x 2 g¡1(A) or x 2 g¡1(B). From this we know x 2 g¡1(A) [ g¡1(B). 4 Chapter 1. The Real Numbers To demonstrate the reverse inclusion, we let x 2 g¡1(A) [ g¡1(B) and show x 2 g¡1(A [ B). So, if x 2 g¡1(A) g¡1(B) then x 2 g¡1(A) or x 2 g¡1(B). This implies g(x) 2 A or g(x) 2 B, and hence g(x) 2 A [ B. This means, x 2 g¡1(A [ B). Exercise 1.2.8. (a) There exist two real numbers a and b satisfying a < b such that for all n 2 N we have a + 1=n ¸ b. (b) There exist two distinct rational numbers with the property that every number in between them is irrational. (c) There exists a natural number n where p n is rational but not a natural number. (d) There exists a real number x such that n · x for all n 2 N. Exercise 1.2.9. (a) We will use induction to prove xn · 2, for every n 2 N. For n = 1, we can easily see x1 = 1 · 2. Now, we want to show that if we have xn · 2, then it follows that xn+1 · 2. Starting from the induction hypothesis xn · 2, we multiply across the inequality by 1/2 and add 1 to get 1 2xn + 1 · 1 2 2 + 1 = 2; which is precisely the the desired conclusion xn+1 · 2. By induction, the claim is proved for all n 2 N. Exercise 1.2.10. (a) For n = 1, we can easily see y1 = 1 < 4, and this proves the base case. Now, we want to show that if we have yn < 4, then it follows that yn+1 < 4. Starting from the induction hypothesis yn < 4, we can multiply across the inequality by 3/4 and add 1 to get 3 4yn + 1 < 3 4 4 + 1 = 4; which is the the desired conclusion yn+1 · 4. By induction, the claim is proved for all n 2 N. (b) For n = 1, we can easily see y1 = 1 < 7=4 = y2, proving the base case. Now, we want to show that if we have yn · yn+1, then it follows that yn+1 · yn+2. Starting from the induction hypothesis yn · yn+1, we can multiply across the inequality by 3/4 and add 1 to get 3 4yn + 1 < 3 4yn+1 + 1 which is the the desired conclusion yn+1 · yn+1. By induction, the claim is proved for all n 2 N. 1.2. Some Preliminaries 5 Exercise 1.2.11. We will use induction, this time starting with n = 0, to prove the claim. When n = 0 then A = ;. For this case, the set A has just the empty set as its only subset. Since 20 = 1, the claim is true in this case. Now we have to show that if sets of size n have 2n different subsets, then it follows that sets of size n + 1 have 2n+1 different subsets. Given a set A of size n + 1, first remove an arbitrary element a 2 A. The set Anfag has n elements, and we can use the induction hypothesis to say that there are exactly 2n subsets of Anfag. Said another way, there are precisely 2n subsets of A that do not contain the particular element a. By adding the element a to each of these we will produce 2n new subsets of A. Since every subset of A either contains a or does not contain a, we can be sure that we have listed them all. Thus, the total number of subsets of A is given by 2n (for the subsets without a) plus 2n (for the subsets that do contain a), and 2n + 2n = 2n+1. By induction, the claim is proved for all n 2 N. Exercise 1.2.12. (a) From Exercise 1.2.3 we know (A1 [A2)c = Ac 1 Ac 2 which proves the base case. Now we want to show that if we have (A1 [ A2 [ ¢ ¢ ¢ [ An)c = Ac 1 Ac 2 ¢ ¢ ¢ Ac n, then it follows that (A1 [ A2 [ ¢ ¢ ¢ [ An+1)c = Ac 1 Ac 2 ¢ ¢ ¢ Ac n+1: Since the union of sets obey the associative law, (A1 [ A2 [ ¢ ¢ ¢ [ An+1)c = ((A1 [ A2 [ ¢ ¢ ¢ [ An) [ An+1)c which is equal to (A1 [ A2 [ ¢ ¢ ¢ [ An)c Ac n+1: Now from our induction hypothesis we know that (A1 [ A2 [ ¢ ¢ ¢ [ An)c = Ac 1 Ac 2 ¢ ¢ ¢ Ac n which implies that (A1 [ A2 [ ¢ ¢ ¢ [ An)c Ac n+1 = Ac 1 Ac 2 ¢ ¢ ¢ Ac n Ac n+1: By induction, the claim is proved for all n 2 N. (b) The point here is to distinguish between asserting that a statement is true for all values of n 2 N and asserting that it is true in the infinite case. Induction cannot be used when we have an infinite number of sets. It is used to prove facts that hold true for each value of n 2 N. For instance, in Exercise 1.2.2, we could use induction to show that Tn k=1 Ak is infinite for all choices of n 2 N, but notice that this conclusion is not true for T1 k=1 An. (c) In order to prove ( S1 n=1 An)c = T1 n=1 Ac n we have to show, (1) Ã 1[ n=1 An !c µ 1 n=1 Ac n and; 6 Chapter 1. The Real Numbers (2) 1 n=1 Ac n µ Ã 1[ n=1 An !c : To demonstrate part (1), we let x 2 ( S1 n=1 An)c and show x 2 T1 n=1 Ac n. So, if x 2 ( S1 n=1 An)c then x =2 An for all n 2 N. This implies x is in the complement of each An and by the definition of intersection x 2 T1 n=1 Ac n. To demonstrate the reverse inclusion, we let x 2 T1 n=1 Ac n and show x 2 ( S1 n=1 An)c. So, if x 2 T1 n=1 Ac n then x 2 Ac n for all n 2 N which means x =2 An for all n 2 N. This implies x =2 ( S1 n=1 An) and we can now conclude x 2 ( S1 n=1 An)c. 1.3 The Axiom of Completeness Exercise 1.3.1. (a) For any z 2 Z5 the additive inverse is y = 5 ¡ z. (b) For z = 1 the additive inverse is x = 1, for z = 2 it is x = 3, for z = 3 it is x = 2,and for z = 4 it is x = 4. (c) For any z 2 Z4 the additive inverse is y = 4 ¡ z. However, the multiplicative inverse of 2 does not exist. In general, additive inverses exist in Zn for all values of n. Multiplicative inverses exist for prime values of n only. Exercise 1.3.2. (a) A real number i is the greatest upper bound, or the infimum, for a set A µ R if it meets the following two criteria: (i) i is a lower bound for A; i.e., i · a for all a 2 A, and (ii) if l is any lower bound for A, then l · i. (b) Lemma: Assume i 2 R is a lower bound for a set A µ R. Then, i = inf A if and only if, for every choice of ² > 0, there exists an element a 2 A satisfying i + ² > a. (i) To prove this in the forward direction, assume i = inf A and consider i + ², where ² > 0 has been arbitrarily chosen. Because i + ² > i, statement (ii) implies i + ² is not a lower bound for A. Since this is the case, there must be some element a 2 A for which i + ² > a because otherwise i + ² would be a lower bound. (ii) For the backward direction, assume i is a lower bound with the property that no matter how ² > 0 is chosen, i+² is no longer a lower bound for A. This implies that if l is any number greater than i then l is no longer a lower bound for A. Because any number greater than i cannot be a lower bound, it follows that if l is some other lower bound for A, then l · i. This completes the proof of the lemma. Exercise 1.3.3. (a) Because A is bounded below, B is not empty. Also, for all a 2 A and b 2 B, we have b · a. The first thing this tells us is that B is bounded above and thus ® = supB exists by the Axiom of Completeness. It remains to show that ® = inf A. The second thing we see is that every element 1.3.