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Solutions Manual for:

Understanding Analysis, Second Edition

Stephen Abbott

Middlebury College

June 25, 2015

Author's note

What began as a desire to sketch out a simple answer key for the problems inUnderstanding Analysisinevitably evolved into something a bit more ambi- tious. As I was generating solutions for the nearly 200 odd-numbered exercises in the text, I found myself adding regular commentary on common pitfalls and strategies that frequently arise. My sense is that this manual should be a use- ful supplement to instructors teaching a course or to individuals engaged in an independent study. As with the textbook itself, I tried to write with the in- troductory student rmly in mind. In my teaching of analysis, I have come to understand the strong correlation between how students learn analysis and how they write it. A nal goal I have for these notes is to illustrate by example how the form and grammar of a written argument are intimately connected to the clarity of a proof and, ultimately, to its validity. The decision to include only the odd-numbered exercises was a compromise between those who view access to the solutions as integral to their educational needs, and those who strongly prefer that no solutions be available because of the potential for misuse. The total number of exercises was signicantly increased for the second edition, and almost every even-numbered problem (in the regular sections of the text) is one that did not appear in the rst edition. My hope is that this arrangement will provide ample resources to meet the distinct needs of these different audiences. I would like to thank former students Carrick Detweiller, Katherine Ott, Yared Gurmu, and Yuqiu Jiang for their considerable help with a preliminary draft. I would also like to thank the readers ofUnderstanding Analysisfor the many comments I have received about the text. Especially appreciated are the constructive suggestions as well as the pointers to errors, and I welcome more of the same.

Middlebury, VermontStephen Abbott

May 2015

v viAuthor's note

Contents

Author's notev

1 The Real Numbers1

1.1 Discussion: The Irrationality of

p

2 . . . . . . . . . . . . . . . . . 1

1.2 Some Preliminaries . . . . . . . . . . . . . . . . . . . . . . . . . . 1

1.3 The Axiom of Completeness . . . . . . . . . . . . . . . . . . . . . 4

1.4 Consequences of Completeness . . . . . . . . . . . . . . . . . . . 6

1.5 Cardinality . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7

1.6 Cantor's Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . 11

2 Sequences and Series15

2.1 Discussion: Rearrangements of Innite Series . . . . . . . . . . . 15

2.2 The Limit of a Sequence . . . . . . . . . . . . . . . . . . . . . . . 15

2.3 The Algebraic and Order Limit Theorems . . . . . . . . . . . . . 16

2.4 The Monotone Convergence Theorem and a First Look at

Innite Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20

2.5 Subsequences and the Bolzano{Weierstrass Theorem . . . . . . . 24

2.6 The Cauchy Criterion . . . . . . . . . . . . . . . . . . . . . . . . 26

2.7 Properties of Innite Series . . . . . . . . . . . . . . . . . . . . . 28

2.8 Double Summations and Products of Innite Series . . . . . . . . 31

3 Basic Topology of R35

3.1 Discussion: The Cantor Set . . . . . . . . . . . . . . . . . . . . . 35

3.2 Open and Closed Sets . . . . . . . . . . . . . . . . . . . . . . . . 35

3.3 Compact Sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38

3.4 Perfect Sets and Connected Sets . . . . . . . . . . . . . . . . . . 41

3.5 Baire's Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . 43

4 Functional Limits and Continuity45

4.1 Discussion: Examples of Dirichlet and Thomae . . . . . . . . . . 45

4.2 Functional Limits . . . . . . . . . . . . . . . . . . . . . . . . . . . 45

4.3 Continuous Functions . . . . . . . . . . . . . . . . . . . . . . . . 48

4.4 Continuous Functions on Compact Sets . . . . . . . . . . . . . . 52

4.5 The Intermediate Value Theorem . . . . . . . . . . . . . . . . . . 55

vii viiiContents

4.6 Sets of Discontinuity . . . . . . . . . . . . . . . . . . . . . . . . . 56

5 The Derivative59

5.1 Discussion: Are Derivatives Continuous? . . . . . . . . . . . . . . 59

5.2 Derivatives and the Intermediate Value Property . . . . . . . . . 59

5.3 The Mean Value Theorems . . . . . . . . . . . . . . . . . . . . . 63

5.4 A Continuous Nowhere-Differentiable Function . . . . . . . . . . 65

6 Sequences and Series of Functions69

6.1 Discussion: The Power of Power Series . . . . . . . . . . . . . . . 69

6.2 Uniform Convergence of a Sequence of Functions . . . . . . . . . 69

6.3 Uniform Convergence and Differentiation . . . . . . . . . . . . . 74

6.4 Series of Functions . . . . . . . . . . . . . . . . . . . . . . . . . . 77

6.5 Power Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 79

6.6 Taylor Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 81

6.7 The Weierstrass Approximation Theorem . . . . . . . . . . . . . 84

7 The Riemann Integral87

7.1 Discussion: How Should Integration be Dened? . . . . . . . . . 87

7.2 The Denition of the Riemann Integral . . . . . . . . . . . . . . . 87

7.3 Integrating Functions with Discontinuities . . . . . . . . . . . . . 90

7.4 Properties of the Integral . . . . . . . . . . . . . . . . . . . . . . 93

7.5 The Fundamental Theorem of Calculus . . . . . . . . . . . . . . . 95

7.6 Lebesgue's Criterion for Riemann Integrability . . . . . . . . . . 98

8 Additional Topics103

8.1 The Generalized Riemann Integral . . . . . . . . . . . . . . . . . 103

8.2 Metric Spaces and the Baire Category Theorem . . . . . . . . . . 105

8.3 Euler's Sum . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 109

8.4 Inventing the Factorial Function . . . . . . . . . . . . . . . . . . 113

8.5 Fourier Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 120

8.6 A Construction ofRFromQ. . . . . . . . . . . . . . . . . . . . 123

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 integerspand qsatisfying (1) (p q 2 = 3: Let us also assume thatpandqhave no common factor. Now, equation (1) implies (2)p2= 3q2: From this, we can see thatp2is a multiple of 3 and hencepmust also be a multiple of 3. This allows us to writep= 3r, whereris an integer. After substituting 3rforpin equation (2), we get (3r)2= 3q2, which can be simplied to 3r2=q2. This impliesq2is a multiple of 3 and henceqis also a multiple of

3. Thus we have shownpandqhave 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 getp2= 6q2which

impliespis a multiple of 2 and 3. After making the necessary substitutions, we can concludeqis a multiple of 6, and thereforep

6 must be irrational.

(b) In this case, the fact thatp2is a multiple of 4 does not implypis also a multiple of 4. Thus, our proof breaks down at this point.

Exercise 1.2.2.

Exercise 1.2.3.

(a) False, as seen in Example 1.2.2. (b) True. This will follow from upcoming results about compactness in

Chapter 3.

1

2Chapter 1. The Real Numbers

(c) False. Consider setsA=f1;2;3g;B=f3;6;7gandC=f5g. Note that

A\(B[C) =f3gis not equal to (A\B)[C=f3;5g.

(d) True. (e) True.

Exercise 1.2.4.

Exercise 1.2.5.

(a) Ifx2(A\B)cthenx =2(A\B). But this impliesx =2A orx =2B. From this we knowx2Acorx2Bc. Thus,x2Ac[Bcby the denition of union. (b) To showAc[Bc(A\B)c, letx2Ac[Bcand showx2(A\B)c. So, ifx2Ac[Bcthenx2Acorx2Bc. From this, we know thatx =2Aor x =2B, which impliesx =2(A\B). This meansx2(A\B)c, which is precisely what we wanted to show. (c) In order to prove (A[B)c=Ac\Bcwe have to show, (1)(A[B)cAc\Bcand; (2)Ac\Bc(A[B)c: To demonstrate part (1) takex2(A[B)cand show thatx2(Ac\Bc). So, ifx2(A[B)cthenx =2(A[B). From this, we know thatx =2A andx =2B which impliesx2Acandx2Bc. This meansx2(Ac\Bc). Similarly, part (2) can be shown by takingx2(Ac\Bc) and showing that x2(A[B)c. So, ifx2(Ac\Bc) thenx2Acandx2Bc. From this, we know thatx =2Aandx =2Bwhich impliesx =2(A[B). This meansx2(A[B)c. Since we have shown inclusion both ways, we conclude that (A[B)c=Ac\Bc.

Exercise 1.2.6.

Exercise 1.2.7.

(a)f(A) = [0;4] andf(B) = [1;16]. In this case,f(A\B) = f(A)\f(B) = [1;4] andf(A[B) =f(A)[f(B) = [0;16]. (b) TakeA= [0;2] andB= [2;0] and note thatf(A\B) =f0gbut f(A)\f(B) = [0;4]. (c) We have to showy2g(A\B) impliesy2g(A)\g(B). Ify2g(A\B) then there exists anx2A\Bwithg(x) =y. But this meansx2Aandx2B and henceg(x)2g(A) andg(x)2g(B). Therefore,g(x) =y2g(A)\g(B). (d) Our claim isg(A[B) =g(A)[g(B). In order to prove it, we have to show, (1)g(A[B)g(A)[g(B) and; (2)g(A)[g(B)g(A[B): To demonstrate part (1), we lety2g(A[B) and showy2g(A)[g(B). If y2g(A[B) then there existsx2A[Bwithg(x) =y. But this means

1.2. Some Preliminaries3

x2Aorx2B, and henceg(x)2g(A) org(x)2g(B). Therefore,g(x) =y2 g(A)[g(B). To demonstrate the reverse inclusion, we lety2g(A)[g(B) and show y2g(A[B). Ify2g(A)[g(B) theny2g(A) ory2g(B). This means we have anx2Aorx2Bsuch thatg(x) =y. This implies,x2A[B, and henceg(x)2g(A[B). Since we have shown parts (1) and (2), we can conclude g(A[B) =g(A)[g(B).

Exercise 1.2.8.

Exercise 1.2.9.

(a)f1(A) = [2;2] andf1(B) = [1;1]. In this case, f

1(A\B) =f1(A)\f1(B) = [1;1] andf1(A[B) =f1(A)[f1(B) =

[2;2]. (b) In order to proveg1(A\B) =g1(A)\g1(B), we have to show, (1)g1(A\B)g1(A)\g1(B) and; (2)g1(A)\g1(B)g1(A\B): To demonstrate part (1), we letx2g1(A\B) and showx2g1(A)\g1(B). So, ifx2g1(A\B) theng(x)2(A\B). But this meansg(x)2Aand g(x)2B, and henceg(x)2A\B. This implies,x2g1(A)\g1(B). To demonstrate the reverse inclusion, we letx2g1(A)\g1(B) and show x2g1(A\B). So, ifx2g1(A)\g1(B) thenx2g1(A) andx2g1(B). This impliesg(x)2Aandg(x)2B, and henceg(x)2A\B. This means, x2g1(A\B). Similarly, in order to proveg1(A[B) =g1(A)[g1(B), we have to show, (1)g1(A[B)g1(A)[g1(B) and; (2)g1(A)[g1(B)g1(A[B): To demonstrate part (1), we letx2g1(A[B) and showx2g1(A)[g1(B). So, ifx2g1(A[B) theng(x)2(A[B). But this meansg(x)2Aor g(x)2B, which impliesx2g1(A) orx2g1(B). From this we know x2g1(A)[g1(B). To demonstrate the reverse inclusion, we letx2g1(A)[g1(B) and show x2g1(A[B). So, ifx2g1(A)\g1(B) thenx2g1(A) orx2g1(B). This impliesg(x)2Aorg(x)2B, and henceg(x)2A[B. This means, x2g1(A[B).

Exercise 1.2.10.

Exercise 1.2.11.

(a) There exist two real numbersaandbsatisfyinga < b such that for alln2Nwe havea+ 1=nb. (b) For all real numbersx >0, there existsn2Nsatisfyingx1=n. (c) There exist two distinct rational numbers with the property that every number in between them is irrational.

4Chapter 1. The Real Numbers

Exercise 1.2.12.

Exercise 1.2.13.

(a) From Exercise 1.2.5 we know (A1[A2)c=Ac1\Ac2which proves the base case. Now we want to show that if we have (A1[A2[ [An)c=Ac1\Ac2\ \Acn, then it follows that (A1[A2[ [An+1)c=Ac1\Ac2\ \Acn+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\Acn+1:

Now from our induction hypothesis we know that

(A1[A2[ [An)c=Ac1\Ac2\ \Acn which implies that (A1[A2[ [An)c\Acn+1=Ac1\Ac2\ \Acn\Acn+1:

By induction, the claim is proved for alln2N.

(b) Example 1.2.2 illustrates this phenomenon. (c) In order to prove ( ∪1 n=1An)c=∩1 n=1Acnwe have to show, (1)

1∪

n=1A n) c

1∩

n=1A cnand; (2)

1∩

n=1A cn(

1∪

n=1A n) c

To demonstrate part (1), we letx2(∪1

n=1An)cand showx2∩1 n=1Acn. So, if x2(∪1 n=1An)cthenx =2Anfor alln2N. This impliesxis in the complement of eachAnand by the denition of intersectionx2∩1 n=1Acn. To demonstrate the reverse inclusion, we letx2∩1 n=1Acnand showx2 (∪1 n=1An)c. So, ifx2∩1 n=1Acnthenx2Acnfor alln2Nwhich means x =2Anfor alln2N. This impliesx =2(∪1 n=1An) and we can now conclude x2(∪1 n=1An)c.

1.3 The Axiom of Completeness

Exercise 1.3.1.

(a) A real numberiis the greatest lower bound, or the inmum, for a setARif it meets the following two criteria:

1.3. The Axiom of Completeness5

(i) iis a lower bound forA; i.e.,iafor alla2A, and (ii) iflis any lower bound forA, thenli. (b) Lemma: Assumei2Ris a lower bound for a setAR. Then,i= infA if and only if, for every choice ofϵ >0, there exists an elementa2Asatisfying i+ϵ > a. (i) To prove this in the forward direction, assumei= infAand consider i+ϵ, whereϵ >0 has been arbitrarily chosen. Becausei+ϵ > i, statement (ii) impliesi+ϵis not a lower bound forA. Since this is the case, there must be some elementa2Afor whichi+ϵ > abecause otherwisei+ϵwould be a lower bound. (ii) For the backward direction, assumeiis a lower bound with the property that no matter howϵ >0 is chosen,i+ϵis no longer a lower bound forA. This implies that iflis any number greater thanithenlis no longer a lower bound forA. Because any number greater thanicannot be a lower bound, it follows that iflis some other lower bound forA, thenli. This completes the proof of the lemma.

Exercise 1.3.2.

Exercise 1.3.3.

(a) BecauseAis bounded below,Bis not empty. Also, for alla2Aandb2B, we haveba. The rst thing this tells us is thatBis bounded above and thus= supBexists by the Axiom of Completeness. It remains to show that= infA. The second thing we see is that every element ofAis an upper bound forB. By part (ii) of the denition of supremum,a for alla2Aand we conclude thatis a lower bound forA. Is it the greatest lower bound? Sure it is. Iflis an arbitrary lower bound forAthenl2B, and part (i) of the denition of supremum impliesl. This completes the proof. (b) We do not need to assume that greatest lower bounds exist as part of the Axiom of Completeness because we now have a proof that they exist. By demonstrating that the inmum of a setAis always equal to the supremum of a different set, we can use the existence of least upper bounds to assert the existence of greatest lower bounds. Another way to achieve the same goal is to consider the setA=fa:a2 Ag. IfAis bounded below it follows thatAis bounded above and it is not too hard to prove infA= sup(A).

Exercise 1.3.4.

Exercise 1.3.5.

(a) In the casec= 0,cA=f0gand without too much difficulty we can argue that sup(cA) = 0 =csupA. So let's focus on the case wherec >0: Observe thatcsupAis an upper bound forcA. Now, we have to show ifdis any upper bound forcA, thencsupAd. We knowcadfor alla2A, and thusad=cfor alla2A. This meansd=cis an upper bound forA, and by Denition 1.3.2 supAd=c. But this impliescsupAc(d=c) =d, which is precisely what we wanted to show.

6Chapter 1. The Real Numbers

(b) Assuming the setAis bounded below, we claim sup(cA) =cinfAfor the casec <0. In order to prove our claim we rst showcinfAis an upper bound forcA. Since infAafor alla2A, we multiply both sides of the equation to getcinfAcafor alla2A. This shows thatcinfAis an upper bound for cA. Now, we have to show ifdis any upper bound forcA, thencinfAd. Wequotesdbs_dbs14.pdfusesText_20

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