9780471433316 pdf This page intentionally left blank FFIRS 12152010 10 13 22 Page 1 Introduction to Real Analysis This page intentionally left blank FFIRS 12152010 10 13 22 Page 3 INTRODUCTION TO.9780471433316 pdf This page intentionally left blank FFIRS 12152010 10 13 22 Page 1 Introduction to Real Analysis This page intentionally left blank FFIRS 12152010 10 13 22 Page 3 INTRODUCTION TO.
This page intentionally left blank FFIRS 12/15/2010 10:13:22 Page Introduction to Real Analysis This page intentionally left blank FFIRS 12/15/2010 10:13:22 Page INTRODUCTION TO REAL ANALYSIS Fourth Edition Robert G Bartle Donald R Sherbert University of Illinois, Urbana-Champaign John Wiley & Sons, Inc FFIRS 12/15/2010 10:13:22 Page VP & PUBLISHER PROJECT EDITOR MARKETING MANAGER MEDIA EDITOR PHOTO RESEARCHER PRODUCTION MANAGER ASSISTANT PRODUCTION EDITOR COVER DESIGNER Laurie Rosatone Shannon Corliss Jonathan Cottrell Melissa Edwards Sheena Goldstein Janis Soo Yee Lyn Song Seng Ping Ngieng This book was set in 10/12 Times Roman by Thomson Digital, and printed and bound by Hamilton Printing Company The cover was printed by Hamilton Printing Company Founded in 1807, John Wiley & Sons, Inc has been a valued source of knowledge and understanding for more than 200 years, helping people around the world meet their needs and fulfill their aspirations Our company is built on a foundation of principles that include responsibility to the communities we serve and where we live and work In 2008, we launched a Corporate Citizenship Initiative, a global effort to address the environmental, social, economic, and ethical challenges we face in our business Among the issues we are addressing are carbon impact, paper specifications and procurement, ethical conduct within our business and among our vendors, and community and charitable support For more information, please visit our website: www.wiley.com/go/ citizenship This book is printed on acid-free paper Copyright # 2011, 2000, 1993, 1983 John Wiley & Sons, Inc All rights reserved No part of this publication may be reproduced, stored in a retrieval system or transmitted in any form or by any means, electronic, mechanical, photocopying, recording, scanning or otherwise, except as permitted under Sections 107 or 108 of the 1976 United States Copyright Act, without either the prior written permission of the Publisher, or authorization through payment of the appropriate per-copy fee to the Copyright Clearance Center, Inc 222 Rosewood Drive, Danvers, MA 01923, website www.copyright.com Requests to the Publisher for permission should be addressed to the Permissions Department, John Wiley & Sons, Inc., 111 River Street, Hoboken, NJ 07030-5774, (201)748-6011, fax (201)748-6008, website http://www.wiley.com/go/permissions Evaluation copies are provided to qualified academics and professionals for review purposes only, for use in their courses during the next academic year These copies are licensed and may not be sold or transferred to a third party Upon completion of the review period, please return the evaluation copy to Wiley Return instructions and a free of charge return shipping label are available at www.wiley.com/go/returnlabel Outside of the United States, please contact your local representative Library of Congress Cataloging-in-Publication Data Bartle, Robert Gardner, 1927Introduction to real analysis / Robert G Bartle, Donald R Sherbert – 4th ed p cm Includes index ISBN 978-0-471-43331-6 (hardback) Mathematical analysis Functions of real variables I Sherbert, Donald R., 1935- II Title QA300.B294 2011 515–dc22 2010045251 Printed in the United States of America 10 FDED01 12/08/2010 15:42:42 Page A TRIBUTE This edition is dedicated to the memory of Robert G Bartle, a wonderful friend and colleague of forty years It has been an immense honor and pleasure to be Bob’s coauthor on the previous editions of this book I greatly miss his knowledge, his insights, and especially his humor November 20, 2010 Urbana, Illinois Donald R Sherbert FDED02 12/08/2010 15:43:51 Page To Jan, with thanks and love FPREF 12/09/2010 14:45:10 Page PREFACE The study of real analysis is indispensable for a prospective graduate student of pure or applied mathematics It also has great value for any student who wishes to go beyond the routine manipulations of formulas because it develops the ability to think deductively, analyze mathematical situations and extend ideas to new contexts Mathematics has become valuable in many areas, including economics and management science as well as the physical sciences, engineering, and computer science This book was written to provide an accessible, reasonably paced treatment of the basic concepts and techniques of real analysis for students in these areas While students will find this book challenging, experience has demonstrated that serious students are fully capable of mastering the material The first three editions were very well received and this edition maintains the same spirit and user-friendly approach as earlier editions Every section has been examined Some sections have been revised, new examples and exercises have been added, and a new section on the Darboux approach to the integral has been added to Chapter There is more material than can be covered in a semester and instructors will need to make selections and perhaps use certain topics as honors or extra credit projects To provide some help for students in analyzing proofs of theorems, there is an appendix on ‘‘Logic and Proofs’’ that discusses topics such as implications, negations, contrapositives, and different types of proofs However, it is a more useful experience to learn how to construct proofs by first watching and then doing than by reading about techniques of proof Results and proofs are given at a medium level of generality For instance, continuous functions on closed, bounded intervals are studied in detail, but the proofs can be readily adapted to a more general situation This approach is used to advantage in Chapter 11 where topological concepts are discussed There are a large number of examples to illustrate the concepts, and extensive lists of exercises to challenge students and to aid them in understanding the significance of the theorems Chapter has a brief summary of the notions and notations for sets and functions that will be used A discussion of Mathematical Induction is given, since inductive proofs arise frequently There is also a section on finite, countable and infinite sets This chapter can used to provide some practice in proofs, or covered quickly, or used as background material and returning later as necessary Chapter presents the properties of the real number system The first two sections deal with Algebraic and Order properties, and the crucial Completeness Property is given in Section 2.3 as the Supremum Property Its ramifications are discussed throughout the remainder of the chapter In Chapter 3, a thorough treatment of sequences is given, along with the associated limit concepts The material is of the greatest importance Students find it rather natural though it takes time for them to become accustomed to the use of epsilon A brief introduction to Infinite Series is given in Section 3.7, with more advanced material presented in Chapter vii FPREF 12/09/2010 14:45:10 viii Page PREFACE Chapter on limits of functions and Chapter on continuous functions constitute the heart of the book The discussion of limits and continuity relies heavily on the use of sequences, and the closely parallel approach of these chapters reinforces the understanding of these essential topics The fundamental properties of continuous functions on intervals are discussed in Sections 5.3 and 5.4 The notion of a gauge is introduced in Section 5.5 and used to give alternate proofs of these theorems Monotone functions are discussed in Section 5.6 The basic theory of the derivative is given in the first part of Chapter This material is standard, except a result of Caratheodory is used to give simpler proofs of the Chain Rule and the Inversion Theorem The remainder of the chapter consists of applications of the Mean Value Theorem and may be explored as time permits In Chapter 7, the Riemann integral is defined in Section 7.1 as a limit of Riemann sums This has the advantage that it is consistent with the students’ first exposure to the integral in calculus, and since it is not dependent on order properties, it permits immediate generalization to complex- and vector-values functions that students may encounter in later courses It is also consistent with the generalized Riemann integral that is discussed in Chapter 10 Sections 7.2 and 7.3 develop properties of the integral and establish the Fundamental Theorem of Calculus The new Section 7.4, added in response to requests from a number of instructors, develops the Darboux approach to the integral in terms of upper and lower integrals, and the connection between the two definitions of the integral is established Section 7.5 gives a brief discussion of numerical methods of calculating the integral of continuous functions Sequences of functions and uniform convergence are discussed in the first two sections of Chapter 8, and the basic transcendental functions are put on a firm foundation in Sections 8.3 and 8.4 Chapter completes the discussion of infinite series that was begun in Section 3.7 Chapters and are intrinsically important, and they also show how the material in the earlier chapters can be applied Chapter 10 is a presentation of the generalized Riemann integral (sometimes called the ‘‘Henstock-Kurzweil’’ or the ‘‘gauge’’ integral) It will be new to many readers and they will be amazed that such an apparently minor modification of the definition of the Riemann integral can lead to an integral that is more general than the Lebesgue integral This relatively new approach to integration theory is both accessible and exciting to anyone who has studied the basic Riemann integral Chapter 11 deals with topological concepts Earlier theorems and proofs are extended to a more abstract setting For example, the concept of compactness is given proper emphasis and metric spaces are introduced This chapter will be useful to students continuing on to graduate courses in mathematics There are lengthy lists of exercises, some easy and some challenging, and ‘‘hints’’ to many of them are provided to help students get started or to check their answers More complete solutions of almost every exercise are given in a separate Instructor’s Manual, which is available to teachers upon request to the publisher It is a satisfying experience to see how the mathematical maturity of the students increases as they gradually learn to work comfortably with concepts that initially seemed so mysterious But there is no doubt that a lot of hard work is required on the part of both the students and the teachers Brief biographical sketches of some famous mathematicians are included to enrich the historical perspective of the book Thanks go to Dr Patrick Muldowney for his photograph of Professors Henstock and Kurzweil, and to John Wiley & Sons for obtaining portraits of the other mathematicians BOTH02 12/08/2010 17:1:45 384 Page 384 HINTS FOR SELECTED EXERCISES 20 (b) [n Z n is contained in [n;k J nk and the sum of the lengths of these intervals is 21 (a) (b) (c) (d) The Product Theorem 7.3.16 Rb R b applies Rb We have ầ2t a f g t2 a f ỵ a g2 : LetR t ! in ðbÞ R R b 1=2 b b If a f 6¼ 0; let t ¼ a g2 = a f in bị: X e=2n ẳ e n 22 Note that sgn h is Dirichlet’s function, which is not Riemann integrable Section 7.4 Show that if P is any partition, then L f ; P ị ẳ U f ; P ị ẳ cb aị ẩ ẫ ẩ É If k ! 0, then inf k f xị : x I j ẳ k inf f ðxÞ : x I j : Consider the partition P e :ẳ 0; e=2; ỵ e=2; 2Þ See Exercise 2.4.8 11 If j f xịj M for x ẵa; b and e > 0, let P be a partition such that the total length of the subintervals that contain any of the given points is less than e=M Then U ð f ; P Þ À Lð f ; P Þ < e so that Theorem 7.4.8 applies Also U ð f ; P Þ e, so that U ð f Þ ¼ Section 7.5 Use (4) with n ¼ 4, a ¼ 1, b ¼ 2, h ¼ 1=4 Here 1=4 f 00 ðcÞ 2; so T % 0:697 02 T % 0:782 79 pffiffiffi The index n must satisfy 2=12n2 < 10À6 ; hence n > 1000= % 408:25 S4 % 0:785 39: The index n must satisfy 96=180n4 < 10À6 ; hence n ! 28 12 The integral is equal to the area of one quarter of the unit circle The derivatives of h are unbounded on [0, 1] Since h00 ðxÞ 0, the inequality is T n ðhÞ < p=4 < M n ðhÞ See Exercise 13 Interpret K as an area Show that h00 xị ẳ x2 ị and that 7=2 h4ị xị ẳ 31 þ 4x2 Þð1 À x2 Þ To eight decimal places, p ¼ 3:141 592 65 3=2 14 Approximately 3.653 484 49 15 Approximately 4.821 159 32 16 Approximately 0.835 648 85 17 Approximately 1.851 937 05 18 19 Approximately 1.198 140 23 20 Approximately 0.904 524 24 Section 8.1 Note that f n ðx Þ x=n ! as n ! If x > 0; then j f n ðxÞ À 1j < 1=ðnxÞ: If x > then j f n ðxÞj 1=ðnxÞ ! 0: Àx If x > 0; then < e If x > 0, then < 1: nx x e ẳ x2 ex ịn ! 0, since < eÀx < 10 If x Z, the limit equals If x = Z, the limit equals 11 If x ½0; a, then j f n ðxÞj 14 If x ẵ0; b, then j f n xịj 15 If x ẵa; 1ị, then j f n xịj a=n However, f n nị ẳ 1=2 bn However, f n 21=n ẳ 1=3 1=naị However, f n 1=nị ẳ 12 sin1 > 18 The maximum of f n on ẵ0; 1ị is at x ẳ 1=n, so jj f n jjẵ0; 1ị ẳ 1=neị BOTH02 12/08/2010 17:1:46 Page 385 HINTS FOR SELECTED EXERCISES 385 20 If n is sufficiently large, jj f n jjẵa;1ị ẳ n2 a2 =ena However, jj f n jjẵ0;1ị ẳ 4=e2 23 Let M be a bound for ð f n ðxÞÞ and ðgn ðxÞÞ on A, whence also j f ðxÞj M The Triangle Inequality gives j f n ðxÞgn ðxÞ À f ðxÞgðxÞj M ðj f n xị f xịj ỵ jgn xị gxịjị for x A Section 8.2 The limit function is f xị :ẳ for x < 1; f 1ị :ẳ 1=2, and f xị :ẳ for < x If e > is given, let K be such that if n ! K, then jj f n À f jjI < e=2 Then j f n ðxn Þ À f ðx0 Þj j f n xn ị f xn ịj ỵ j f xn ị f x0 ịj e=2 ỵ j f ðxn Þ À f ðx0 Þj Since f is continuous (by Theorem 8.2.2) and xn ! x0 , then j f ðxn Þ À f ðx0 Þj < e=2 for n ! K 0, so that j f n ðxn Þ À f ðx0 Þj < e for n ! maxfK; K g Here f 0ị ẳ and f xị ẳ for x 0; The convergence is not uniform on [0, 1] Given e :¼ 1, there exists K > such that if n ! K and x A, then j f n ðxÞ À f ðxÞj < 1, so that j f n xịj j f K xịj ỵ for all x A Let M :¼ maxfjj f jjA ; ; jj f KÀ1 jjA ; jj f K jjA ỵ 1g p p f n 1= nị ẳ n=2 10 Here ðgn Þ converges uniformly to the zero function The sequence g0n does not converge uniformly 11 Use the Fundamental Theorem 7.3.1 and Theorem 8.2.4 13 If a > 0, then jj f n jj½a;p 15 Here jjgn jj½0; 1 1=ðnaÞ and Theorem 8.2.4 applies for all n Now apply Theorem 8.2.5 20 Let f n xị :ẳ xn on ẵ0; 1ị Section 8.3 Let A :ẳ x > 0 and let m ! in (5) For the upper estimate on e, take x ¼ and n ¼ to obtain e À 23 < 1=12; so e < 34 Note that if n ! 9, then 2=n ỵ 1ị! <  10À7 <  10À6 Hence e % 2:71828 Evidently En ðxÞ [0, a] Note that ex for x ! To obtain the other inequality, apply Taylors Theorem 6.4.1 to tn =1 ỵ tị tn for t ½0; x ln 1:1 ’ 0:0953 and ln 1:4 % 0:3365 Take n > 19; 999 ln % 0:6931 10 L0 1ị ẳ limẵL1 ỵ 1=nị L1ị=1=nị ẳ lim L1 ỵ 1=nịn ị ẳ Llim1 ỵ 1=nịn ị ẳ Leị ẳ 1: a 11 (c) xyị ẳ EaLxyịị ẳ EaLxị ỵ aLyịị ¼ EðaLðxÞÞ Á EðaLðyÞÞ ¼ xa Á ya a b 12 (b) xa ị ẳ EbLxa ịị ẳ EbaLxịị ¼ xab , and similarly for ðxb Þ 15 Use 8.3.14 and 8.3.9(vii) 17 Indeed, we have loga x ẳ ln xị=ln aị ẳ ẵln xị=ln bị ẵln bị=ln aị if a 6ẳ 1; b 6ẳ Now take a ¼ 10; b ¼ e Section 8.4 If n > 2jxj, then jcos x À C n ðxÞj ð16=15Þjxj2n =ð2nÞ!, so cosð0:2Þ % 0:980 067, cos % 0:549 302 Similarly, sinð0:2Þ % 0:198 669 and sin % 0:841 471 We integrate 8.4.8(x) twice on [0, x] Note that the polynomial on the left has a zero in the interval [1.56,1.57], so 1:56 p=2 BOTH02 12/08/2010 17:1:46 386 Page 386 HINTS FOR SELECTED EXERCISES Exercise 8.4.4 shows that C ðxÞ cosx C ðxÞ for all x R Integrating several times, we get S4 ðxÞ sin x S5 ðxÞ for all x > Show that S4 ð3:05Þ > and S5 ð3:15Þ < (This procedure can be sharpened.) If jxj A and m > n > 2A, then jcm ðxÞ À cn ðxÞj < ð16=15ÞA2n =ð2nÞ!, whence the convergence of ðcn Þ to c is uniform on each interval ẵA; A Dẵcxịị2 sxịị2 ẳ for all x R For uniqueness, argue as in 8.4.4 Let gxị :ẳ f 0ịcxị ỵ f 0ịsxị for x R, so that g00 xị ẳ gxị; g0ị ẳ f 0ị and g0 0ị ẳ f 0ị Therefore hxị :ẳ f xị gxị has the property that h00 xị ẳ hxị for all x R and h0ị ẳ 0; h0 0ị ẳ Thus gxị ẳ f xị for all x R, so that f xị ẳ f 0ịcxị ỵ f 0ịsxị If wxị :ẳ cxị, show that w00 xị ¼ wðxÞ and wð0Þ ¼ 1; w0 ð0Þ ¼ 0, so that wxị ẳ cxị for all x R Therefore c is even Section 9.1 Let sn be the nth partial sum of X an , let tn be the nth partial sum of 1 X jan j, and suppose that an ! for n > P If m > n > P, show that tm À tn ¼ sm À sn Now apply the Cauchy Criterion Take positive terms until the partial sum exceeds 1, then take negative terms until the partial sum is less than 1, then take positive terms until the partial sum exceeds 2, etc Yes If n ! 2, then sn ¼ Àln À ln n ỵ lnn ỵ 1ị Yes We have s2n sn ! na2n ẳ 12 2na2n ị, and s2nỵ1 sn ! 12 2n ỵ 1ịa2nỵ1 Consequently limnan ị ẳ 11 Indeed, if n2 an M for n, then jan j M=n2 pp p1 P n nỵ1ỵ n and note that 13 (a) Rationalize to obtain xn where xn :¼ xn % yn :ẳ 1=2nị Now applyP the Limit Comparison Test 3.7.8 (b) Rationalize and compare with 1=n3=2 P P 14 If an is absolutely convergent, the partial sums of jan j are bounded, say by M Evidently the absolute value of the partial sums P of any subseries of an are also bounded by M Conversely, if every subseries of an is convergent, then the subseries consisting of the strictly positive (and strictly negative) terms are absolutely convergent, whence it follows that P an is absolutely convergent Section 9.2 P 1=n2 Divergent; note that 21=n ! 1 (a) Convergent; compare with (a) (c) Divergent; apply 9.2.1 with bn :¼ 1=n Convergent; use 9.2.4 and note that ðn=ðn þ 1ÞÞn ! 1=e < (a) (c) (e) ðln nÞp < n for large n, by L’Hospital’s Rule Convergent; note that ðln nÞln n > n2 for large n Divergent; apply 9.2.6 or Exercise 3.7.15 (c) (a) Convergent (b) Divergent À Á (c) Divergent À Á (d) Convergent; note that ðln nÞexp Àn1=2 < n exp Àn1=2 < 1=n2 for large n, by L’Hospital’s Rule (e) Divergent (f) Divergent Apply the Integral Test 9.2.6 (a, b) Convergent (c) Divergent If m > n ! K, then jsm sn j (d) Convergent jxnỵ1 j þ Á Á Á þ jxm j < rnþ1 =ð1 À rÞ Now let m ! BOTH02 12/08/2010 17:1:47 Page 387 HINTS FOR SELECTED EXERCISES 387 R1 12 (a) A crude estimate of the remainder is given by s À s4 < xÀ2 dx ¼ 1=5 Similarly s À s10 < 1=11and s À sn < 1=ðn þ 1Þ, so that 999 terms suffice to get s À s999 < 1=1000 (d) If n ! 4, then xnỵ1 =xn 5=8 so (by Exercise 10) js s4 j 5=12 If n ! 10, then xnỵ1 =xn 11=20 so that js À s10 j 10=210 ð11=9Þ < 0:012 If n ¼ 14, then js À s14 j < 0:000 99 Z 1 X pffiffiffi 13 (b) Here < xÀ3=2 dx ¼ 2= n, so js À s10 j < 0:633 and js À sn j < 0:001 nỵ1 n when n > 106 (c) If n ! 4, then js À sn j ð0:694Þxn so that js À s4 j < 0:065 If n ! 10, then js À sn j ð0:628Þxn so that js À s10 j < 0:000 023 14 Note that ðs3n Þ is not bounded 16 Note that, for an integer with n digits, there are ways of picking the first digit and 10 ways of picking each of the other n À digits There is one value of mk from to 9, there is one value from 10 to 19, one from 20 to 29, etc 18 Here limn1 xnỵ1 =xn ịị ẳ c a bị ỵ 1, so the series is convergent if c > a ỵ b and is divergent if c < a ỵ b Section 9.3 (a) (c) Absolutely convergent Divergent (b) (d) Conditionally convergent Conditionally convergent Show by induction that s2 < s4 < s6 < Á Á Á < s5 < s3 < s1 Hence the limit lies between sn and snỵ1 so that js sn j < jsnỵ1 sn j ẳ znỵ1 Use Dirichlets Test with yn ị :ẳ ỵ1; 1; 1; ỵ; ỵ1; 1; 1; Þ Or group the terms in pairs (after the first) and use the Alternating Series Test If f ðxÞ :ẳ ln xịp =xq , then f xị < for x sufficiently large L’Hospital’s Rule shows that the terms in the alternating series approach (a) Convergent (b) Divergent (c) Divergent (d) Divergent 11 Dirichlet’s Test does not apply (directly, at least), since the partial sums of the series generated by ð1; À1; À1; 1; 1; 1; Þ are not bounded 15 (a) Use Abel’s Test with xn :¼ 1=n pffiffiffiffiffi (b) Use the Cauchy Inequality with xn :¼ an ; yn :¼ 1=n, to get P pffiffiffiffiffi P P 1=2 an =n ð an Þ1=2 ð 1=n2 Þ , establishing convergence pffiffiffi À1 (d) Let an :ẳ ẵnln nị2 , which converges by the Integral Test However, bn :ẳ ẵ n ln n , which diverges Section 9.4 (a) Take M n :¼ 1=n2 in the Weierstrass M-Test (c) Since jsin yj jyj, the series converges for all x But it is not uniformly convergent on R If a > 0, the series is uniformly convergent for jxj a (d) If x 1, the series is divergent If < x < 1, the series is convergent It is uniformly convergent on ẵa; 1ị for a > However, it is not uniformly convergent on ð1; 1Þ À Á If r ¼ 1, then the sequence jan j1=n is notbounded Hence if jx0 j > 0, then there are infinitely many k N with jak j1=k > 1=jx0 j so that ak xk0 > Thus the series is not convergent when x0 6¼ Suppose L :ẳ limjan j=janỵ1 jị exists and that < L < It follows from the Ratio Test P that that an xn converges for jxj < L and diverges for jxj > L The Cauchy-Hadamard Theorem implies that L ¼ R BOTH02 12/08/2010 17:1:47 388 Page 388 HINTS FOR SELECTED EXERCISES (a) R ¼ (c) R ¼ 1=e (e) R ¼ À Á Use lim n1=n ¼ (b) R ¼ (d) (f) R ¼ 10 By the Uniqueness Theorem 9.4.13, an ẳ 1ịn an for all n 12 If n N, there exists a polynomial Pn such that f nị xị ẳ e1=x Pn 1=xị for x 6ẳ 13 Let gxị :ẳ for x ! and gxị :ẳ e1=x for x < Show that gnị 0ị ẳ for all n 16 Substitute Ày for x in Exercise 15 and integrate from y ¼ to y ¼ x for jxj < 1, which is justified by Theorem 9.4.11 X Rx 1ịn x2nỵ1 =n!2n ỵ 1ị for x R 19 eÀt dt ¼ n¼0 R p=2 p Á Á Á Á Á ð2n À 1Þ : 20 Apply Exercise 14 and ðsin xÞ2n dx ¼ Á 2 Á Á Á Á Á 2n Section 10.1 (a) Since ti À dðti Þ xiÀ1 and xi (b) Apply (a) to each subinterval (b) ti ỵ dti ị, then xi xi1 2dti ị Consider the tagged partition fẵ0; 1; 1ị; ẵ1; 2; 1ị; ẵ2; 3; 3ị; ẵ3; 4; 3ịg (a) If P_ ẳ fẵxi1 ; xi ; ti ịgniẳ1 and if tk is a tag for both subintervals ½xkÀ1 ; xk and ẵxk ; xkỵ1 , we must have tk ¼ xk We replace these two subintervals by the subinterval ẵxk1 ; xkỵ1 with the tag tk , keeping the d-fineness property (b) No (c) If tk xk1 ; xk ị, then we replace ẵxk1 ; xk by the two intervals ½xkÀ1 ; tk and ½tk ; xk both tagged by tk, keeping the d-fineness property If xkÀ1 xk and if tk is the tag for ½xkÀ1 ; xk , then we cannot have tk > 1, since then tk dtk ị ẳ 12 tk ỵ 1ị > Similarly, we cannot have tk < 1, since then tk ỵ dtk ị ẳ tk ỵ 1ị < Therefore tk ẳ (a) Let dtị :ẳ 12 minfjt À 1j; jt À 2j; jt À 3jg if t 6ẳ 1; 2; and dtị :ẳ for t ẳ 1; 2; (b) Let d2 tị :ẳ minfdtị; d1 tịg, where d is as in part (a) F xị :ẳ 2=3ịx3=2 ỵ 2x1=2 ; F xị :ẳ 2=3ị1 xị3=2 21 xị1=2 ; F xị :ẳ 2=3ịx3=2 ln x 2=3ị for x 0; and F 0ị :ẳ 0; 1=2 F xị :ẳ 2xp ln x 2Þ for x ð0; 1 and F ð0Þ :ẳ 0, F xị :ẳ x2 ỵ Arcsin x, F xị :ẳ Arcsinx 1Þ The partition P_ z need not be de -fine, since the value de ðzÞ may be much smaller than À tagged Á de xj R1 R1 If f were integrable, then f ! sn ẳ 1=2 ỵ 1=3 ỵ ỵ 1=n þ 1Þ (a) (b) (c) (d) (e) (f) 10 WeÀ enumerate the nonzero rational numbers as rk ¼ mk =nk and define de mk =nk ị :ẳ e= nk 2kỵ1 and de xị :ẳ otherwise 12 The function M is not continuous on [À2, 2] 13 L1 is continuous and L01 xị ẳ l xị for x 6¼ 0, so Theorem 10.1.9 applies 15 We have C 01 xị ẳ 3=2ịx1=2 cos1=xị ỵ x1=2 sin1=xị for x > Since the first term in C01 has a continuous extension to [0, 1], it is integrable 16 We have C02 xị ẳ cos1=xị ỵ 1=xị sin1=xị for x > By the analogue of Exercise 7.2.12, the first term belongs to Rẵ0; 17 (a) Take wtị :ẳ tp2 ỵ t so Ew ẳ ; to get (b) Take wtị :ẳ t so Ew ẳ f0g to get 22 ỵ ln 3ị BOTH02 12/08/2010 17:1:48 Page 389 HINTS FOR SELECTED EXERCISES 389 pffiffiffiffiffiffiffiffiffiffi (c) Take wtị :ẳ t so Ew ẳ f1g to get Arctan (d) Take wtị :ẳ Arcsin t so Ew ¼ f1g to get 14 p 19 (a) In fact f xị :ẳ F xị ẳ cosp=xị ỵ p=xị sinp=xị for x > We set f 0ị :ẳ 0, F 0ị :ẳ Note that f is continuous on ð0; 1 (b) F ak ị ẳ and F bk ị ẳ 1ịk =k Apply Theorem 10.1.9 Z n n Z bk X X 1=k (c) If j f j Rà ½0; 1, then jfj j f j for all n N k¼1 k¼1 ak 20 Indeed, sgnð f xịị ẳ 1ịk ẳ mxị on ẵak ; bk so mxị f xị ẳ jmxị f xịj for x ½0; 1 Since the restrictions of m and jmj to every interval [c, 1] for < c < are step functions, they belong to R½c; 1 By Exercise 7.2.11, m and jmj belong to R½0; 1 and Z 1 X X R1 k m ẳ ị =k 2k ỵ ị and 1=k2k ỵ 1ị jmj ẳ 0 kẳ1 kẳ1 21 Indeed, wxị ẳ F0 xị ẳ jcosp=xịj ỵ p=xịsinp=xị Á sgnðcosðp=xÞÞ for x = E by Example 6.1.7(c) Evidently w is not bounded near If x a ẵ ; b , then w x ị ẳ jcosp=xịj ỵ k k Rb p=xịjsinp=xịj so that akk jwj ẳ Fbk ị Fak ị ẳ 1=k; whence jwj = R ẵ0; 22 Here cxị ẳ C0 xị ẳ 2xjcosp=xịj ỵ p sinp=xị sgncosp=xịị for x = f0g [ E1 by Example 6.1.7(b) R b Since c is bounded, Exercise 7.2.11 applies We cannot apply Theorem 7.3.1 to evaluate c since E is not finite, but Theorem 10.1.9 applies and c R½0; 1 Corollary 7.3.15 implies that jcj R½0; 1 23 If p ! 0, then mp f p Mp, where m and M denote the infimum and the supremum of f on Rb Rb Rb Rb M a p If a p ¼ 0, the result is trivial; otherwise, the [a, b], so that m a p a fp conclusion follows from Bolzano’s Intermediate Value Theorem 5.3.7 24 By the Multiplication Theorem 10.1.14, f g Rà ½a; b If g is increasing, then gðaÞf f g Rb Rb Rb Rx Rb gðbÞ f so that gðaÞ a f gðbÞ a f Let K xị :ẳ gaị a f þ gðbÞ x f , so that K is a fg continuous and takes all values between K(b) and K(a) Section 10.2 R1 (a) If Gxị :ẳ 3x1=3 for x ẵ0; then c g ẳ G1ị Gcị ! G1ị ẳ R1 (b) We have c 1=xịdx ẳ ln c, which does not have a limit in R as c ! Rc Here xị1=2 dx ẳ 21 cị1=2 ! as c ! 1À Because of continuity, g1 Rà ½c; 1 for all c ð0; 1ị If vxị :ẳ x1=2 , then jg1 xịj vxị for all x ½0; 1 The ‘‘left version’’ of the preceding exercise implies that g1 Rà ½0; 1 and the above inequality and the Comparison Test 10.2.4 imply that g1 L½0; 1 (a) The function À 1Áand continuous  in Á(0, 1) À à is bounded on [0, 1] (use l’Hospital) ln ln x If x ; , the integrand is (c) If x 0; 12 , the integrand is dominated by 2 À Á dominated by ln 12 lnð1 À xÞ (a) Convergent (b, c) Divergent (d, e) Convergent à (f) Divergent 10 By the Multiplication Theorem 10.1.14, f g R ½a; b Since j f xịgxịj Lẵa; b and jj f gjj Bjj f jj Bj f ðxÞj, then f g 11 (a) Let f xị :ẳ 1ịk 2k =k for x ẵck1 ; ck ị and f 1ị :ẳ 0, where the ck are as in Example 10.2.2(a) Then f ỵ :ẳ maxf f ; 0g = R ẵ0; (b) Use the first formula in the proof of Theorem 10.2.7 Rb 13 (ii) If f xị ẳ gxị for all x ẵa; b, then dist f ; gị ¼ a j f À gj ¼ Rb Rb (iii) dist f ; gị ẳ a j f gj ¼ a jg À f j ¼ distðg; f Þ Rb Rb Rb (iv) distð f ; hÞ ¼ a j f À hj a j f À gj ỵ a jg hj ẳ distf ; gị ỵ distðg; hÞ 16 If ð f n Þ converges to f in L ½a; b, given e > there exists K ðe=2Þ such that if m; n ! K ðe=2Þ then jj f m À f jj < e=2 and jj f n À f jj < e=2 Therefore jj f m À f n jj jj f m f jj ỵ jj f f n jj < e=2 ỵ e=2 ẳ e Thus we may take H eị :ẳ K e=2ị BOTH02 12/08/2010 17:1:49 390 Page 390 HINTS FOR SELECTED EXERCISES 18 If m > n, then jjgm gn jj 1=n ỵ 1=m ! One can take g :¼ sgn 19 No 20 We can take k to be the 0-function Section 10.3 Let b ! maxfa; 1=dð1Þg If P_ is a d -fine partition of [a, b], show that P_ is a d-fine subpartition of ẵa; 1ị Rq If f Lẵa; 1ị, apply the preceding exercise to j f j Conversely, if p j f j < e for q > p ! K ðeÞ, R q Rg Rg Rp Rq then f À f j f j < e so both limg f and limg j f j exist; therefore f ; j f j a a p a R ẵa; 1ị and so f Lẵa; 1ị a If f ; g Lẵa; 1ị, then f ; j f j; g; and jgj belong to R ẵa; 1ị, so Example 10.3.3(a) implies that R1 R1 R1 f ỵ g and j f j ỵ jgj belong to R ẵa; 1ị and that a j f j ỵ jgjị ẳ a j f j ỵ a jgj Since Rg Rg Rg R1 R1 j f ỵ gj j f j ỵ jgj, it follows that a j f ỵ gj a j f j ỵ a jgj a j f j ỵ a jgj, whence jj f ỵ gjj jj f jj ỵ jjgjj Rg Indeed, R1 1=xị dx ẳ ln g, which does not have a limit as g ! Or, use Exercise and the 2p fact that p 1=xịdx ẳ ln > for all p ! Rg If g > 0, then cos x dx ¼ sin g, which does not have a limit as g ! R g Àsx (a) We have e dx ẳ 1=sị1 esg Þ ! 1=s (b) Let GðxÞ :¼ Àð1=sÞeÀsx for x ẵ0; 1ị, so G is continuous R on ½0; 1Þ and GðxÞ ! as x ! By the Fundamental Theorem 10.3.5, we have g ¼ G0ị ẳ 1=s 12 (a) If x ! e, then ln xị=x ! 1=x (b) Integrate by parts on ẵ1; g and then let g ! pffiffiffi 13 (a) jsin xj ! 1= > 1=2 Rand 1=x > 1=n ỵ 1ịp for x np ỵ p=4; np ỵ 3p=4ị g (b) If g > n ỵ 1ịp, then jDj ! 1=4ị1=1 ỵ 1=2 ỵ ỵ 1=n ỵ 1ịị 15 Let u ẳ wxị ẳ x2 Now apply Exercise 14 16 (a) Convergent (f) Convergent 17 (a) (c) 18 (a) (c) (b, c) Divergent (d) Convergent (e) Divergent If f ðxÞ :¼ p sinffiffiffi x, then f = Rà ẵ0; 1ị In Exercise 14, take f xị :ẳ x1=2 sin x and w2 xị :ẳ 1= x Take f xị :ẳ x1=2 sin x, and wxị :ẳ x ỵ 1ị=x Rx f xị :ẳ sin x is in R ẵ0; g , and F xị :ẳ sin t dt ¼ À cos x is bounded on ẵ0; 1ị, and wxị :ẳ 1=x decreases monotonely to Rx F xị :ẳ cos t dt ẳ sin x is bounded on ẵ0; 1ị and wxị :ẳ x1=2 decreases monotonely to 19 Let u ẳ wxị :ẳ x2 20 (a) (c) Rg If g > 0, then eÀx dx ¼ À eÀg ! 1, so ex R ẵ0; 1ị Similarly ejxj ẳ ex Rà ðÀ1; 0 2 eÀx eÀx for jxj ! 1, so ex R ẵ0; 1ị Similarly on ðÀ1; 0 Section 10.4 (a) (c) (a) (c) Converges to at x ¼ 0, to on (0, 1] Not uniform Bounded by Increasing Limit ¼ Converges to on [0, l), to 12 at x ¼ Not uniform Bounded by Increasing Limit ¼ pffiffiffi Converges to x on [0, 1] Uniform Bounded by Increasing Limit ¼ 2=3 Converges to at x ¼ 1, to on (1, 2] Not uniform Bounded by Decreasing Limit ¼ BOTH02 12/08/2010 17:1:50 Page 391 HINTS FOR SELECTED EXERCISES 391 (a) Converges to at x ¼ 0, to on (0, 1] Not uniform Bounded by Decreasing Limit ¼ (c) Converges to Not uniform Bounded by 1/e.pNot ffiffiffiffiffi monotone Limit ¼ (e) Converges to Not uniform Bounded by 1= 2e Not monotone Limit ¼ (a) The Dominated Convergence Theorem applies (b) f k xị ! for x ẵ0; 1Þ, but ð f k ð1ÞÞ is not bounded No obvious dominating function Integrate by parts and use (a) The result shows that the Dominated Convergence Theorem does not apply Suppose that ð f k ðcÞÞ converges for some c ½a; b By the Fundamental Theorem, Rx Rx Rx f k xị f k cị ẳ c f 0k By the Dominated Convergence Theorem, c f 0k ! c g, whence ð f x ðxÞÞ converges for all x ½a; b Note that if f k xị :ẳ 1ịk , then f k xịị does not converge for any x ½a; b Rn Indeed, gxị :ẳ supf f k xị : k N g equals 1=k on (k À 1, k], so that g ẳ ỵ 12 ỵ ỵ 1n : Hence g = R ½0; 1Þ 10 (a) If a > 0, then jðeÀtx sin xị=xj eax for t J a :ẳ a; 1Þ If tk J a and tk ! t0 J a , then the argument in 10.4.6(d) shows that E is continuous at t0 Also, if tk ! 1, then jðeÀtk x sin xÞ=xj eÀx and the Dominated Convergence Theorem implies that Eðtk Þ ! Thus EðtÞ ! as t ! À Á R1 (b) It follows as in 10.4.6(e) thatRE0 t0 ị ẳ À ReÀt0 x sin x dx ¼ À1= t20 þ À1 s s (c) By 10.1.9, Esị Etị ẳ t E tịdt ẳ t t ỵ 1ị dt ẳ Arctan t Arctan s for s; t > But EðsÞ ! and Arctan s ! p=2 as s ! (d) We not know that E is continuous as t ! 0ỵ 12 Fix x I As in 10.4.6(e), if t; t0 ½a; b, there exists tx between t; t0 such that f ðt; xÞÀ f ðt0 ; xị ẳ t t0 ị @f Therefore axị ẵ f ðt; xÞ À f ðt0 ; xÞ=ðt À t0 Þ vðxÞ when @t ðtx ; xÞ t 6¼ t0 Now argue as before and use the Dominated Convergence Theorem 10.4.5 13 (a) If ðsk Þ is a sequence of step functions converging to f a.e., and ðtk Þ is a sequence of step functions converging to g a.e., Theorem 10.4.9(a) and Exercise 2.2.18 imply that ðmaxfsk ; tk gÞ is a sequence of step functions that converges to maxf f ; gg a.e Similarly, for minf f ; gg 14 (a) Since f k M½a; b is bounded, it belongs to Rà ½a; b The Dominated Convergence Theorem implies that f Rà ½a; b The Measurability Theorem 10.4.11 now implies that f M½a; b (b) Since t 7! Arctan t is continuous, Theorem 10.4.9(b) implies that f k :ẳ Arctan gk Mẵa; b Further, j f k ðxÞj 12 p for x ½a; b (c) If gk ! g a.e., it follows from the continuity of Arctan that f k ! f a.e Parts (a, b) imply that f M½a; b and Theorem 10.4.9(b) applied to w ¼ tan implies that g ẳ tan f Mẵa; b 15 (a) Since 1E is bounded, it is in Rà ½a; b if and only if it is in M½a; b (c) 1E0 ẳ 1E (d) 1E[F xị ẳ maxf1E xị; 1F xịg and 1E\F xị ẳ minf1E xị; 1F xịg Further, EnF ẳ E \ F0 (e) If Ek ị is an increasing sequence in M ẵa; b, then 1Ek ị is an increasing sequence in Mẵa; b with 1E xị ẳ lim 1Ek xị, and we can apply Theorem 10.4.9(c) Similarly, ð1Fk Þ is a decreasing sequence in Mẵa; b and 1F xị ẳ lim 1Fk xị (f) Let An :ẳ [nkẳ1 Ek , so that An ị is an increasing sequence in M ẵa; b with [1 n¼1 An ¼ E, so (e) applies Similarly, if Bn :ẳ \nkẳ1 F k , then Bn ị is a decreasing sequence in M ½a; b with \1 nẳ1 Bn ẳ F Rb Rb 16 (a) m;ị ẳ a ¼ and 1E implies mEị ẳ a 1E b a (b) Since 1ẵc; d is a step function, then mẵc; d ị ¼ d À c Rb (c) Since 1E0 ¼ 1E , we have mE0 ị ẳ a 1E ị ẳ b aị mEị (d) Note that 1E[F ỵ 1E\F ẳ 1E ỵ 1F : BOTH02 12/08/2010 392 17:1:50 Page 392 HINTS FOR SELECTED EXERCISES (f) If Ek ị is increasing in M ẵa; b to E, then 1Ek ị is increasing in Mẵa; b to 1E The Monotone Convergence Theorem 10.4.4 applies S (g) If Ck ị is pairwise disjoint and En :ẳ nkẳ1 Ck for n N, then mEn ị ẳ mC1 ị S1 S1 ỵ ỵ mCn Þ Since k¼1 C k ¼ n¼1 En and ðEn Þ is increasing, (f) implies that n X X S m C k ị ẳ mCk ị m kẳ1 C k ẳ limn mEn ị ẳ limn k¼1 n¼1 Section 11.1 If jx À uj < inf fx; xg, then u < x ỵ xị ẳ and u > x x ¼ 0, so that < u < Since the union of two open sets is open, then G1 [ [ Gk [ Gkỵ1 ¼ ðG1 [ Á Á Á [ Gk Þ [ Gkỵ1 is open The complement of N is the union ðÀ1; 1Þ [ ð1; 2Þ [ Á Á Á of open intervals Corollary 2.4.9 implies that every neighborhood of x in Q contains a point not in Q 10 x is a boundary point of A() every neighborhood V of x contains points in A and points in CðaÞ()x is a boundary point of CðaÞ 12 The sets F and C(F) have the same boundary points Therefore F contains all of its boundary points () C ðF Þ does not contain any of its boundary points () C ðF Þ is open 13 x A () x belongs to an open set V A()x is an interior point of A 15 Since AÀ is the intersection of all closed sets containing A, then by 11.1.5(a) it is a closed set containing A Since CðẦ Þ is open, then z CðẦ Þ () z has a neighborhood V e ðzÞ in CðẦ Þ () z is neither an interior point nor a boundary point of A 19 If G 6¼ ; is open and x G, then there exists e > such that V e ðxÞ G, whence it follows that a :¼ x À e is in Ax 21 If ax < y < x then since ax :¼ inf Ax there exists a0 Ax such that ax < a0 ðy; x ða0 ; x G and y G y Therefore 23 If x F and n N, the interval I n in F n containing x has length 1=3n Let yn be an endpoint of I n with yn 6¼ x Then yn F (why?) and yn ! x 24 As in the preceding exercise, take zn to be the midpoint of I n Then zn = F (why?) and zn ! x Section 11.2 Let Gn :¼ ỵ 1=n; 3ị for n N Let Gn :ẳ 1=2n; 2ị for n N If G1 is an open cover of K and G2 is an open cover of K , then G1 [ G2 is an open cover of K1 [ K2 Let K n :ẳ ẵ0; n for n N 10 Since K 6¼ ; is bounded, it follows that inf K exists in R If K n :¼ fk K : k inf K ị ỵ 1=ng, then K n is closed and bounded, hence compact By the preceding exercise \K n 6¼ ;, but if x0 \K n , then x0 K and it is readily seen that x0 ¼ inf K [Alternatively, use Theorem 11.2.6.] 12 Let ; 6¼ K R be compact and let c R If n N, there exists xn K such that supfjc À xj : x K g À 1=n < jc À xn j Now apply the Bolzano-Weierstrass Theorem 15 Let F :¼ fn : n N g and F :ẳ fn ỵ 1=n : n N; n ! 2g Section 11.3 À pffiffiffi pffiffiffiÁ À1 (a) If a < b 0,pthen fp I ị ẳ ;.ffiffiffi Ifpaffiffiffi< < b, then f À1 ðI Þ ¼ À b; b If ffiffi ffi À p f I ị ẳ b; À a [ a; b a < b then BOTH02 12/08/2010 17:1:51 Page 393 HINTS FOR SELECTED EXERCISES 393 f Gị ẳ f ẵ0; eịị ẳ ẵ1; ỵ e2 ị ẳ 0; ỵ e2 ị \ I: Let G :ẳ 1=2; 3=2ị Let F :ẳ ẵ1=2; 1=2 Let f be the Dirichlet Discontinuous Function = First note that if A R and x R, then we have x f À1 ðRnAÞ () f ðxÞ RnA () f ðxÞ A () x = f À1 ðAÞ () x Rnf À1 ðAÞ; therefore, f À1 RnAị ẳ Rn f Aị Now use the fact that a set F R is closed if and only if RnF is open, together with Corollary 11.3.3 Section 11.4 If Pi :ẳ xi ; yi ị for i ¼ 1; 2; 3, then d ðP1 ; P2 ị jx1 x3 jỵjx3 x2 jịỵjy1 y3 jỵ jy3 y2 jị ẳ d P1 ; P3 ị ỵ d P3 ; P2 ị Thus d satisfies the Triangle Inequality Since j f xị gxịj j f xị hxịj ỵ jhxị gxịj d f ; hị ỵ d h; gị for all x ẵ0; 1, it follows that d ð f ; gÞ d f ; hị ỵ d h; gị and d satisfies the Triangle Inequality We have s 6ẳ t if and only if d s; tị ¼ If s 6¼ t, the value of d s; uị ỵ d u; tị is either or depending on whether u equals s or t, or neither Since d Pn ; Pị ẳ supfjxn À xj; jyn À yjg, if d ðPn ; PÞ ! then it follows that both jxn À xj ! and jyn À yj ! 0, whence xn ! x and yn ! y Conversely, if xn ! x and yn ! y, then jxn À xj ! and jyn À yj ! 0, whence d ðPn ; PÞ ! If a sequence ðxn Þ in S converges to x relative to the discrete metric d, then d ðxn ; xÞ ! 0, which implies that xn ¼ x for all sufficiently large n The converse is trivial Show that a set consisting of a single point is open Then it follows that every set is an open set, so that every set is also a closed set (Why?) 10 Let G S2 be open in ðS2 ; d Þ and let x f À1 ðGÞ so that f ðxÞ G Then there exists an e-neighborhood V e ð f ðxÞÞ G Since f is continuous at x, there exists a d-neighborhood Vd(x) such that f (Vd(x)) Ve( f(x)) Since x f À1 ðGÞ is arbitrary, we conclude that f À1 ðGÞ is open in ðS1 ; d Þ The proof of the converse is similar 11 Let G ¼ fGa g be a cover of f ðSÞ R by open sets inÈR It follows É from 11.4.11 that each set f À1 ðGa Þ is open in ðS; d Þ Therefore, the collection f À1 ðGa Þ Éis an open cover of S Since È À1 ðS; d Þ is compact, a finite subcollection f ðGa1 Þ; ; f À1 ðGaN Þ covers S, whence it follows that the sets fGa1 ; ; GaN g must form a finite subcover of G for f (S) Since G was an arbitrary open cover of f (S), we conclude that f (S) is compact This page intentionally left blank BINDEX 12/10/2010 11:57:26 Page 395 INDEX A Abel’s Lemma, 279 Test, 279, 315 Absolute: convergence, 267 ff maximum, 135 minimum, 135 value, 32 Absurdum, see Reductio, 356 Additive function, 116, 134, 156 Additivity Theorem, 213, 294 Algebraic properties of R, 23 ff Almost everywhere, 221 Alternating series, 98, 278 ff And/or, 2, 349 Antiderivative, 216 Antipodal points, 140 Approximate integration, 233 ff., 364 ff Approximation theorems, 145 ff Archimedean Property, 42 Arithmetic Mean, 29, 260 Axiom, 348 B Base, 13, 259 Basepoint, 218, 224 Bernoulli, Johann, 180 Bernoulli’s Inequality, 30, 177 Bessel functions, 176 Biconditional, 351 Bijection, Binary representation, 49 ff Binomial expansion, 287 Bisection method, 137 Bolzano, Bernhard, 124 Bolzano Intermediate Value Theorem, 138 Bolzano-Weierstrass Theorem: for infinite sets, 337 for sequences, 81, 322 Bound: lower, 37 upper, 37 Boundary point, 332 Bounded Convergence Theorem, 251 function, 41, 111, 134, 151 sequence, 63 set, 37 Boundedness Theorem, 135 Bridge, 13 C Canis lupus, 174 Cantor, Georg, 21, 52 set F, 331 Theorem of, 21, 49, 52 Caratheodory’s Theorem, 165 Cartesian Product, Cauchy, A.-L., 54 Condensation Test, 101 Convergence Criterion, 246, 282 Inequality, 225 Mean Value Theorem, 182 Root Test, 271 sequence, 85, 344 Cauchy-Hadamard Theorem, 283 Chain Rule, 166 Change of Variable Theorem, 220, 224, 297 Chartier-Dirichlet Test, 315 Closed interval, 46 set, 327, 345 Closed Set Properties, 328 Closure of a set, 333 Cluster point, 125, 329 Compact set, 334 ff Compactness, Preservation of, 339, 346 Comparison Tests, 98 ff., 304 Complement of a set, Complete metric space, 334 395 BINDEX 12/10/2010 396 11:57:26 Page 396 INDEX Completeness Property of R, 36 ff esp 39 Theorem, 307 Composition of functions, 9, 133 Composition Theorem, 222 Conclusion, 350 Conditional, 350 Conditional convergence, 267 Conjunction, 349 Consistency Theorem, 291 Continuity, 125 ff., 345 absolute, 149 global, 324, 346 uniform, 142 Continuous Extension Theorem, 145 function, 125 ff., 337 ff Inverse Theorem, 156, 340 Contractive sequence, 88 Contradiction, 349 proof by, 356 Contrapositive, 351 proof by, 355 Convergence: absolute, 267 of integrals, 315 ff interval of, 283 in a metric space, 343 pointwise, 241 radius of, 283 of a sequence, 56 of a sequence of functions, 281 of a series, 94 of a series of functions, 281 uniform, 281 Converse, 351 Convex function, 192 ff Cosine function, 263 Countability: of N  N, 18, 358 of Q , 19 of Z, 18 Countable: additivity, 325 set, 17 ff Counter-example, 353 Cover, 333 Curve, space-filling, 368 Cyclops, 76 D D’Alembert’s Ratio Test, 272 Darboux, Gaston, 225 Darboux Intermediate Value Theorem, 178 Integral, 228 ff Decimal representation, 51 periodic, 51 Decreasing function, 153, 174 sequence, 71 DeMorgan’s Laws, 3, 350 Density Theorem, 44 Denumerable set (see also countable set), 17 Derivative, 162 ff higher order, 188 second, 188 Descartes, Rene, 161 Difference: symmetric, 11 of two functions, 111 of two sequences, 61 Differentiable function, 162 uniformly, 180 Differentiation Theorem, 285 Dini, Ulisse, 252 Dini’s Theorem, 252 Direct image, proof, 354 Dirichlet discontinuous function, 127, 207, 209, 221, 291, 321 integral, 311, 320 test, 279 Discontinuity Criterion, 126 Discrete metric, 343 Disjoint sets, Disjunction, 349 Distance, 34, 306 Divergence: of a function, 105, 108 of a sequence, 56, 80, 91 ff Division, in R, 25 Domain of a function, Dominated Convergence Theorem, 318 Double implication, 351 negation, 349 E Element, of a set, Elliptic integral, 287 BINDEX 12/10/2010 11:57:26 Page 397 INDEX Empty set ;, Endpoints of intervals, 46 Equi-integrability, 316 Equivalence, logical, 349 Euler, Leonhard, 76 Euler’s constant, 276 number e, 75, 255 Even function, 171, 216 number, Excluded middle, 349 Existential quantifier 9, 352 Exponential function, 253 ff Exponents, 25 Extension of a function, 144 ff Extremum, absolute, 135 relative, 172, 175, 191 F F (¼ Cantor set), 331 Falsity, 349 Fermat, Pierre de, 161, 198 Fibonacci sequence, 56, 89 Field, 24 d-Fine partition, 149, 289 Finite set, 16 ff First Derivative Test, 175 Fluxions, 161 Fresnel Integral, 314 Function(s), additive, 116, 134, 156 Bessel, 173 bijective, bounded, 41, 111, 134 composition of, 9, 133 continuous, 125 ff., 337 ff convex, 192 ff decreasing, 153, 174 derivative of, 162 difference of, 111 differentiable, 162 direct image of, Dirichlet, 127, 207, 209, 221, 279, 291, 321 discontinuous, 125 domain of, even, 171, 216 exponential, 255 ff gauge, 149 graph of, greatest integer, 129 hyperbolic, 266 image of, increasing, 153, 174 injective, integrable, 201, 290 inverse, 7, 156, 168 inverse cosine, 10 inverse image of, inverse sine, 10 jump of, 155 limit of, 104 ff Lipschitz, 143 logarithm, 257 ff measurable, 320 metric, 342 monotone, 153 multiple of, 111 nondifferentiable, 163, 367 nth root, 157 odd, 171, 216 one-one, onto, oscillation, 361 periodic, 148 piecewise linear, 147 polynomial, 113, 131, 148 power, 159, 258 product of, 111 quotient of, 111 range of, rational, 131 rational power, 159 restriction of, 10 sequence of, 241 ff series of, 281 ff signum, 109, 127 square root, 10, 43 step, 145, 210 sum of, 111 surjective, Thomae’s, 128, 206, 222 Translate, 207 trigonometric, 131, 260 ff values of, Fundamental Theorems of Calculus, 216 ff., 295, 297 G Gallus gallus, 349 Gauge, 149 ff., 252, 289 ff 397 BINDEX 12/10/2010 398 11:57:26 Page 398 INDEX Generalized Riemann integral, 290 ff Geometric Mean, 29, 260 series, 95 Global Continuity Theorem, 338, 346 Graph, Greatest integer function, 129, 224 lower bound (= infimum), 37 H Hadamard-Cauchy Theorem, 283 Hake’s Theorem, 302, 309 Half-closed interval, 46 Half-open interval, 46 Harmonic series, 88, 96, 267 Heine-Borel Theorem, 335 Henstock, Ralph, 289 Higher order derivatives, 188 Horizontal Line Tests, Hyperbolic functions, 266 Hypergeometric series, 277 Hypothesis, 350 induction, 13 I Image, Implication, 350 Improper integrals, 272, 302 ff Increasing function, 153, 174 sequence, 71 Indefinite integral, 218 Indeterminate forms, 181 Indirect proofs, 355 Induction, Mathematical, 12 ff Inequality: Arithmetic-Geometric, 29 Bernoulli, 30, 177 Schwarz, 225 Triangle, 32, 342 Infimum, 37 Infinite limits, 119 series, 94 ff., 267 ff set, 16 ff Injection, Injective function, Integers, Integral: Darboux, 228 ff Dirichlet, 311, 320 elliptic, 287 Fresnel, 314 generalized Riemann, 290 ff improper, 272, 302 ff indefinite, 218 Lebesgue, 289, 304, 362 lower, 227 Riemann, 201 ff Test, for series, 273 upper, 227 Integration by parts, 222, 299 Interchange Theorems: relating to continuity, 248 relating to differentiation, 249 relating to integration, 250, 315 ff relating to sequences, 247 ff relating to series, 282 Interior Extremum Theorem, 172 of a set, 332 point, 332 Intermediate Value Theorems: Bolzano’s, 138 Darboux’s, 178 Intersection of sets, Interval(s), 46 ff characterization of, 47 of convergence, 283 length of, 46 nested, 47 ff partition of, 149, 199 Preservation of, 139 Inverse function, 8, 156, 169 image, Irrational number, 25 Iterated sums, 270 suprema, 46 J Jump, of a function, 155 K K(e)-game, 58 Kuala Lumpur, 349 Kurzweil, Jaroslav, 289 L Lagrange, J.-L., 188 form of remainder, 190 Least upper bound (= supremum), 37 Lebesgue, Henri, 198, 220, 288, 362 Dominated Convergence Theorem, 318 ... Congress Cataloging-in-Publication Data Bartle, Robert Gardner, 192 7Introduction to real analysis / Robert G Bartle, Donald R Sherbert – 4th ed p cm Includes index ISBN 97 8-0 -4 7 1-4 333 1-6 (hardback)... 10:13:22 Page Introduction to Real Analysis This page intentionally left blank FFIRS 12/15/2010 10:13:22 Page INTRODUCTION TO REAL ANALYSIS Fourth Edition Robert G Bartle Donald R Sherbert University... Urbana-Champaign John Wiley & Sons, Inc FFIRS 12/15/2010 10:13:22 Page VP & PUBLISHER PROJECT EDITOR MARKETING MANAGER MEDIA EDITOR PHOTO RESEARCHER PRODUCTION MANAGER ASSISTANT PRODUCTION EDITOR