An Introduction to Analysis SECOND EDITION
JAMES R. KIRKWOOD SWEET BRIAR COLLEGE
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(i) This book is printed on acidfree, recycled paper. Library of Congress CataloginginPublication Data Kirkwood, James R. An introduction to analysis / James R. Kirkwood  2nd ed. p. em. Includes bibliographical references and index. ISBN 0534944221 1. Mathematical analysis I. Title. QA300.K523 1995 9439727 515'.83dc20 CIP Sponsoring Editor: Steve Quigley Editorial Assistant: John Ward Production Editor: Kathleen Wilson Manufacturing Coordinator: Ellen Glisker Marketing Manager: Marianne C. P. Rutter Interior Design: S. London Cover Design: Designworks Interior Illustration: Lisa C. Sparks Typesetting: H. Charlesworth & Co. Ltd. Printing and Binding: Quebecor PrintingMartinsburg Printed and bound in the United States of America. 95 96 97 98 9910 9 8 7 6 5 4 3 2
Contents
Introduction
I'll
1
The Real Number System 4 11 Sets and Functions 4 12 Properties of the Real Numbers as an Ordered Field 13 The Completeness Axiom 25
IHI
Sequences of Real Numbers
36
21 Sequences of Real Numbers 36 22 Subsequences 48 23 The BolzanoWeierstrass Theorem
==
IMMI
52
Topology of the Real Numbers 31 Topology of the Real Numbers
Continuous Functions
60
73
41 Limits and Continuity 73 42 Monotone and Inverse Functions 92
IMI
Differentiation 104 51 The Derivative of a Function 52 Some Mean Value Theorems iii
104 115
60
14
iv
Contents
IMI
Integration
132
61 The Riemann Integral 132 62 Some Properties and Applications of the Riemann Integral 63 The RiemannStieltjes Integral (Optional) 161
Series of Real Numbers
174
72 Operations Involving Series
184
= = 71 Tests for Convergence of Series
174
IM:MI Sequences and Series of Functions 81 Sequences of Functions 200 82 Series of Functions 215 83 Taylor Series 227
IMI
Fourier Series 238 91 Fourier Coefficients 238 92 Representation by Fourier Series
245
Bibliography 266 Appendix: Hints and Answers for Selected Exercises 267 Index 275
200
146
Preface his book provides a mathematically rigorous introduction to analysis of realvalued functions of one variable. The only prerequisite is an elementary calculus sequence. A typical reader would be an average mathematics student for whom this is his or her first experience with an abstract mathematics course. Such students often find themselves floundering and discouraged after the first few weeks of the course, as it is usually impossible to make the transition from primarily computational to primarily theoretical mathematics without some frustration. This text is written in a manner that will help to ease this transition. I have followed the principle that the material should be as clear and intuitive as possible and still maintain mathematical integrity. In some instances this meant sacrificing elegance for clarity. A major goal of this type of course is to teach students to understand mathematical proofs as well as to be able to formulate and write them. In constructing a proof, two processes are involved: ( 1) discovering the ideas necessary for the proof and (2) expressing these ideas in mathematical language. One of the pedagogical features of the text is frequently to supply the ideas of the proof and leave the writeup as an exercise. The text also states why a step in a proof is the reasonable thing to do and which techniques are recurrent. Students often leave a first calculus sequence with little intuition about functions in an abstract setting. Examples, while no substitute for a proof, are a valuable tool in helping to develop intuition and are an important feature of this text. The text uses numerous examples to motivate what is to come, to highlight the need for a particular hypothesis in a theorem, and to make abstract ideas more concrete. Examples can also provide a vivid reminder that what one hopes might be true is not always true. Several of the exercises are devoted to the construction of counterexamples. Most of the exercises range in difficulty from the routine "getting your feet wet" types of problems to the moderately challenging problems. In a few cases optional topics are developed through a series of exercises. (Within the text, optional topics are denoted by a dagger. They are also flagged as optional in the Table of Contents.) I have provided ample material for a onesemester or a twoquarter course. The text begins with a discussion of the axioms of the real number system. The idea of a limit is introduced via sequences, which seems to be the simplest setting
•
v
vi
Preface
in which limits can be rigorously discussed. A sufficient amount of the topology of the real number system is developed to obtain the familiar properties of continuous functions. One of the changes in the second edition is that the property of connectedness is now presented as an optional section, and I give a proof of the Intermediate Value Theorem that does not rely on connectedness. Chapters 16 would make a reasonable onesemester advanced calculus course and Chapters 4, 5 and 6 contain material on continuous functions, differentiable functions and the Riemann integral. Chapter 6 also contains an optional section on the RiemannStieltjes integral. In Chapter 6 the proof of the RiemannLebesgue Theorem is given. This result is often omitted in text of this level. While this proof is substantially more difficult than the other results in the text, the statement of the theorem is not difficult to understand, and it is the fundamental existence theorem for Riemann integration. Having this result available enables the use of more natural and less tedious proofs for several other theorems. If time is limited, you might consider skipping the proof and concentrating on what the theorem says. Chapter 7 covers series of real numbers and can be discussed immediately after Chapter 2 if you are willing to allow the integral test for convergence of series without a formal development of the Riemann integral. Chapter 8 deals with sequences and series of functions, and in Chapter 9 Fourier series is discussed. The chapter on Fourier series is intended to show an application of the earlier work, and the material is developed at a somewhat faster pace. Obviously, the material in a text of this type is not new. All of the theorems and most of the exercises will be familiar to those who have worked in this area. I regret that it is only in rare instances that I am able to give credit to those who may have originated an idea. My intent is for the text to present the material in a way that will make the subject understandable and perhaps exciting to those who are beginning their study of abstract mathematics. There are many people to whom I owe a great deal of thanks for their help in the preparation of this book. I would first of all like to thank the reviewers. Their criticism, comments and suggestions did much to improve the final manuscript. These include: Thomas E. Armstrong University of MarylandBaltimore County J. Marshall Ash De Paul University William Berlinghoff Southern Connecticut State University Sidney Birnbaum California State Polytechnic University Larry E. Bobisud University of Idaho George Ulrich Brauer University of Minnesota James W. Carlson Creighton University Matthew P. Coleman Fairfield University John Firkins Gonzaga University Robert E. Gaines Colorado State University Kenneth B. Hannsgen Virginia Polytechnic Institute and State University Hugo D. Junghenn George Washington University
Preface
vii
Herbert Kamowitz University of Masachusetts I. G. Kastanas California State University, Los Angeles Lawrence C. Moore Duke University Richard C. Roberts University of Maryland Lisa M. Taylor California State University, Sacramento John W. Watson Arkansas Tech University Marshall E. Wick University of Wisconsin I would also like to acknowledge the staff of PWS for their work throughout the development of the text. Special thanks are due to Julie Riddleberger, who typed the manuscript and its many revisions in an exceptionally able manner and always with a cheerful attitude. Finally, I would like to thank my wife, Bessie, for her support. She not only provided invaluable technical help by reading the early drafts of the manuscript but also offered encouragement when it was needed. James Kirkwood
Introduction n most elementary calculus courses, the emphasis is on learning the me=== chanics of calculus rather than on understanding the principles that pro
vide the necessary techniques. Our work will focus on developing the theorems that one uses in calculus from the, axioms of the real numbers. To do this, it is necessary to understand mathematical proofs. Before we begin our study, it is worthwhile to give a very brief introduction to "the rules of the game." In a mathematical system such as the one that we shall develop, there are some objects and rules for manipulating and combining these objects. To develop any sort of theory, we need to have a common agreement about the terms we use and the rules that our system must obey. This agreement is described by a collection of definitions and axioms. There will necessarily be some undefined terms. A prime requirement (and one of the beauties) of mathematics is the lack of ambiguity of its language. It is incumbent upon us then to keep the undefined terms to an absolute minimum. As is common in modern mathematics, we shall take the word set as an undefined term. Axioms and definitions are statements that all parties concerned agree to accept. They do not require proof. (In ancient times axioms were held to be selfevident truths.) An axiom is a statement that cannot be demonstrated in terms of simpler concepts. Axioms can be thought of as the fundamental building blocks of the theory. One characteristic that the axioms and definitions must have is consistency. This means that there is no statement that we can prove to be both true and false from our set of axioms and definitions. The axioms should also be independent. That is, no axiom should be derivable from the others. A third requirement for a set of axioms is that there should be sufficiently many of them to provide a meaningful theory. We want to derive as many results as we can about our system from the axioms and definitions. In other words, we seek to formulate and prove theorems. A theorem is a formula derivable from the axioms and definitions by the rules of inference. The theorems that we shall prove are conditional. They have a hypothesis and a conclusionsometimes called the antecedent and consequent. The hypothesis is a statement that is assumed to be true (the if part). The conclusion is a statement that can be deduced if the conditions of the hypothesis hold (the then part). We want to discuss some of the ways that theorems are stated and how we can prove them. 1
2
Introduction
Suppose a theorem is of the form, "If A, then B." For example: "Iff is a differentiable function, then f is continuous." One way to prove the theorem is directly. This means to show that if A is true, then B must necessarily follow. The statement, "If A, then B" can be alternatively worded, ·~ is sufficient for B." There are also indirect methods of proof. The indirect method of proof that we shall most often use to prove a conditional theorem is to prove the contrapositive of the theorem. The contra positive of "If A, then B" is "If B is not true, then A is not true." The contra positive of our preceding example is, "Iff is a function that is not continuous, then f is not differentiable." The statement, "If B is not true, then A is not true," can be alternatively worded, "B is necessary for A." One of the rules of inference is that a conditional theorem is true if the contra positive is true and vice versa. A second indirect method of proof of a conditional theorem is to show that it is impossible to have both the hypothesis true and the conclusion false. That is, one could prove, "If A, then B" by showing that ·~ and not B" is an impossibility. One example of where we shall use this technique is to prove that a convergent sequence can have only one limit point. In our proof we suppose that there exists a convergent sequence with more than one limit point and show that this is impossible. This method of proof is sometimes called proof by contradiction. Another, sometimes useful, technique of proof is to consider various cases that may occur and to rule out all but one of them. For example, suppose x is a real number about which we know something, and we want to show that x > 0. One of our axioms will be that if x is a real number, then exactly one of the following must be true: (i) x < 0, (ii) x = 0, (iii) x > 0. If we can rule out (i) and (ii), then (iii) must occur. Obviously, this is most useful when there are only a few cases to be considered. If one interchanges the hypothesis and the conclusion of a theorem, the result is called the converse of the theorem. Thus the converse of "If A, then B" is "If B, then A." For example, the converse of "If x equals 4, then x is an integer" is "If x is an integer, then x equals 4." In this case the original statement is true, but the converse is not. If a theorem, "If A, then B," and its converse, "If B, then A," are both true, then A and B are equivalent. To prove a theorem that is worded, ·~ is true if and only if B is true," we must prove two assertions: "If A, then B" and "If B, then A." Sometimes such a theorem is stated, ·~ is necessary and sufficient for B." We shall often use examples to clarify explanations and to help make ideas more concrete. They are also useful in highlighting why a certain hypothesis is necessary in a theorem by showing that the theorem is not true if the hypothesis is omitted. Sometimes a problem will be stated, "Is is true that ...." In such a situation you must prove that the statement is true or find an example where the assertion does not hold. Such an example is called a counterexample. Obviously, showing that something is true for any number of examples does not constitute a proof that a theorem is true. Next, a word about definitions. Definitions are to be interpreted in the biconditional (if and only if) form, even though it has become customary not to state them in that fashion. For example, suppose we were to make the following definition: "A triangle with three sides of equal length is an equilateral triangle."
Introduction
3
We interpret the definition to mean that every triangle with three equal sides is an equilateral triangle and any equilateral triangle has three equal sides. Notice the use of the words every and any in the previous sentence. These words could have been interchanged and the sentence would have exactly the same implications. Other terms that will typically have the same meaning as these two are each and all. Thus, for example, we could equally well have said, "Each triangle with three equal sides is an equilateral triangle." This should be enough to get us started. We shall discuss other points of logic as we encounter them.
The Real Number System he pattern of mathematics is to propose a set of axioms that a collection ===of objects and operations on the objects must satisfy and then to derive as many conclusions as possible about the system under consideration. The system is then the objects and operations together with the axioms. While courses in plane geometry emphasize this pattern, this is often not the case in algebra. Thus one is often adept at algebraic manipulations without realizing why these operations are legitimate. Why, for example, is 3 + 2 = 2 + 3 or 5 x 0 = 0? Throughout the text we shall deal with the real number system. In this first chapter we present the preliminary material that provides the foundation for our work, including the axioms of the real numbers, and some of the consequences of these axioms.
Sets and Functions        S e t s and Their OperatorsPerhaps the two most fundamental concepts in mathematics are sets and functions. In this section we describe some of their properties. The term set is left undefined and is synonymous with collection or aggregate or any of several other terms. A set may contain nothing or it may contain some objects called elements or members. If the set contains nothing, it is called the null set or the empty set and is denoted by 0. One property that a set must have is that it must be well defined. This means that given a set and an object, it is possible to say whether the object is a member of the set. Usually, sets will be denoted with uppercase letters and elements by lowercase letters. To express the idea that "x is an element of the set A," we write XEA,
and to express "x is not an element of the set A," we write
x¢A. 4
Section 11
Sets and Functions
5
It may seem that anything can be a set. This is not the case. Later in this chapter, we shall show that there can be no largest set. Thus there is no set that contains everything. There are other restrictions that a set must satisfy in order to avoid
some paradoxes, but these will not affect our work.t Definition: Let A be a set. The complement of A, denoted Ac, is the set of elements not in A. •
We have just said that there is no largest set, and if Ac is everything not in A, then there is a difficulty. In particular, what should 0c be? This problem is overcome by having a universal set in the background that contains just the objects
in which we are interested. What this universal set actually is depends on the situation under consideration and is usually clear from the context of the discussion. For us, ,the universal set will usually be the real numbers. Thus if A is the set whose elements are the numbers 1 and 2, then Ac is the set whose elements are all real numbers except 1 and 2. Most often we shall describe a set in one of two ways. One is by simply listing the elements in the set. The other is by describing a property that those elements in the set, and only those elements, satisfy. For example, if A is the set made up of the elements 1 and 2, then we could write A={1,2}
or
A={xlx 2 3x+2=0}.
The latter way of describing the elements in a set is called set builder notation. If we write A= {xiP},
it means that A consists of all elements (in the universal set) that satisfy property P. In our example property P is the condition that the number be a solution to the equation x 2  3x + 2 = 0. Definition: Let A and B be sets. We say that A is a subset of B if every element of A is also an element of B. •
If A is a subset of B, then we write
AcB or B::::JA. Definition: Two sets are equal if they contain exactly the same elements. •
That is, a set is determined by its elements and the order of the elements is immaterial. The usual way to prove that two sets A and B are equal is to prove that A c Band B cA. Given two sets, we can construct new sets using these sets as building blocks. We now describe the most common ways of doing this.
t One of the most famous paradoxes arises if we try to construct a set that has itself as an element, that is, a set A with A EA. This paradox is called Russell's paradox after the twentiethcentury philosopher and mathematician, Bertrand Russell.
Chapter 1
6
The Real Number System
Definition: Let A and B be sets. Then (i) The union of
A and B, denoted Au B, is the set of elements that are in
A or B. A and B, dehoted An B, is the set of elements that are in both A and B. (iii) The complement of B relative to A (or A minus B), denoted A\ B, is the set of elements that are in A but not in B. • (ii) The intersection of
Thus
AuB= {xlxEA or xEB} AnB= {xlx=A and xEB} A\B= {xlxEA and x¢B} =AnBc. When we use the word or, we mean it in the inclusive sense. That is, Au B consists of the elements that are in A or B, or both A and B.
Example 11: Let A= {2, 3, 4} and B = {3, 4, 5}. Then AuB = {2, 3, 4, 5}, AnB = {3, 4}, A \B = {2}, B\A = {5}. The ideas of intersection and union are extendable to larger collections of sets. If J represents a set such that Ai is a set for each j E J (such a set J is called an indexing set),
U Ai = {xlx E Ai for some j E J}
then
jE J
n
and
Aj = {xlx E Aj for every j E J}.
jE J
Definition: Let A and B be sets. If An B = 0, then A and B are said to be disjoint. If {Ailj disjoint if
E
J} is a collection of sets, the collection is said to be pairwise if i, j
E
J
and
i # j. •
The last type of set construction that we consider is the Cartesian product.
Definition: Let AI> A 2 , • .• , An be a finite collection of sets. The Cartesian product of A 1 , A 2 , ••• , An, denoted A 1 x A 2 x · · · x An, is defined by A 1 x A 2 x · · · x An= {(a 1 , a2 ,
... ,
a.) Ia; E A;, i = 1, ... , n}. •
Thus the Cartesian product of two sets A and B is the set of all ordered pairs where the first coordinate comes from A and the second coordinate comes from B. We show in Exercise 9, Section 11, that if A and B are nonempty sets, then A x B is different from B x A unless A= B.
Section 11
Sets and Functions
7
Set theory is a mathematical subject in itself, and we shall discuss only those results needed to prove some later theorems. To illustrate a technique that is often used to prove elementary set theoretic properties, we present the following result.
Theorem 11: If A, B, and C are sets, then (a)
A\ (B u C) = (A\ B) n (A\ C).
(b)
A\(BnC)=(A\B)u(A\C).
Proof of (a): To accomplish the proof, we show (i) A\ (B u
C) c (A\ B) n (A\ C), and
(ii) (A\ B) n (A\ C) c A\ (B u C).
To prove (i), we choose xEA\(BuC) arbitrarily and must show xE{A\B)n(A\C). Since xEA\(BuC), it must be that xEA and x¢(BuC). Since x ¢ (Bu C), then x ¢ B and x ¢C. Thus x E (A \B) and x E (A\ C), so x E (A\ B)n(A \C), which proves (i). To prove (ii), we choose x E (A\ B)n(A \C) arbitrarily and must show x E A\ (B u C). Since x E (A\ B), x E A and x ¢ B; and since x E (A\ C), x E A and x ¢C. Thus x ¢ (Bu C) and x E A, so x E A\ (Bu C), which proves (ii). The proof of (b) is left for Exercise 7, Section 11. •
Corollary 11 (De Morgan's Laws): If Band Care sets, then (a) (b)
(BuCY =Bene. (BnCY=Beuc.
Proof of (a): Let A be the universal set in the theorem. Then A\ (B u C) = An (B u C)< = (B u C)
x 2 E
~(f)= ~(f)
{xlx # 2}. We shall show that f is with f(xd = f(x 2 ). Then
xi4 = x~4. x 1  2 x 2 2' or, since xl> x 2 # 2, we can write this as x1 + 2
= x 2 + 2,
so that x 1 = x 2 and f is 11. Next we show that f is not onto. To do this, we must find a number y for which there is no x E ~(f) with f(x) = y. The clue to finding this number y is the fact that f(x) = x + 2 if x # 2 and 2 is not in ~(f). Thus 2 + 2 = 4 is
10
Chapter 1 The Real Number System a reasonable candidate for y. Suppose f(x) = 4. x 2 4 =4 x2 '
Then so
But this equation is true only for x with f(x) = 4.
=
2 and 2 ¢: fliJ(f). Thus there is no x
E fliJ(f)
Definition: Suppose f and g are functions from the real numbers to the real numbers. Then: (i) (ii)
f + g is the function whose domain is f(x) + g(x) for x E fliJ(f) n fliJ(g). f · g is the function whose domain is
fliJ(f)nfliJ(g) and (f + g)(x)= fliJ(f) n fliJ(g) and (f · g)(x) =
f(x) · g(x) for x E fliJ(f) n fliJ(g). (iii)
f jg is the function whose domain is fliJ
(iv)
(~) = fliJ(f)nfliJ(g)n {x E fliJ(g)Jg(x) i= 0}
and
(~) (x) = ~~;
for all x E fliJ(f jg). If IX is a real number, then IX/ is a function whose domain is fliJ(f) and (1X/)(x) =IX· f(x) for x E fliJ(f). •
Notice that (iv) is actually a special case of (ii). There is really no reason to require that our functions f and g must have domain and range in the real numbers, but we do need to be able to add, multiply, and divide f(x) and g(x) for the definition to make sense. There is another way to combine functions that will also be important to us.
Definition: Let A, B, and C be sets and f and g be functions with fliJ(f) c A, ~(f)
c B, fliJ(g) c B with ~(f) c fliJ(g) and ~(g) c C. Then the composition of g with f, denoted go f, is the function from fliJ(f) to C defined by (go f)(x) = g(f(x)). •
Figure 11 illustrates this definition.
gof
Figure 11
Section 11
Sets and Functions
11
Example 15: Let f(x) = x2 + 3 and g(x) = )x 2. Then £d(g) = {xlx): 2} =>~(f)= {xlx): 3} and go f(x) = g(f(x)) = g(x 2 + 3) = )(x 2 + 3) 2 = )x 2 + 1.
Theorem 12: Suppose f: A~ B and g: B ~C. (a) (b)
Iff and g are 11 functions, then go f is a 11 function. Iff and g are onto functions, then go f is an onto function.
Proof: We give the proof of part (a) and leave the proof of part (b) for Exercise 17, Section 11. To prove part (a), we choose x 1 , x 2 E A with x 1 =!= x 2 and must show (go f)(xd =!=(go f)(x 2 ). Now if x 1 =!= x 2 , then f(x 1 ) =!= f(x 2 ), since f is 11. If f(x 1 ) =!= f(x 2 ), then g(f(x 1 )) =!= g(f(x 2 )), since g is 11. Thus if x 1 =!= x 2 , g(f(xd) =!= g(f(x 2 )); that is, (go f)(xd =!=(go f)(xz). • Suppose that f: X~ Y is a 11 function. Then we can define a function f 1 : ~(f)~X by f 1 (y) = x if and only if y = f(x). The reason! needs to be 11 for f  1 to exist as a function can be seen when we consider what would happen if there were x 1 , x 2 EX with f(xd = f(x 2 ) = y. What should we do with f 1 (y)? Should it go to x 1 or x 2 ? This is a fatal predicament and is the reason f must be 11. Also notice that iff and f 1 are as we have said, then f 1 (f(x)) = x for x E £d(f) and f(f  1 (y)) = y for y E £d(f  1 ). Thus f of  1 is the identity function on £d(f 1 ) (that is, (Jo f 1 )(y) = y), and f 1 of is the identity function on £d(f). With this in mind, we make the following definition.
Definition: Let f: X~ Y be a 11 function. Then the function f 1 :~(f)~ X defined by f 1 (y) = x if y = f(x) is called the inverse of the function f. • Example 16: Let f(x) = x 3 + 1. We shall later establish methods to show this is a 11 function. For now, assume that it is. We want to find f 1 . We know f 1 (f(x)) = x; so if we let y = f(x), this condition becomes f 1 (y) = x. Thus if we can solve for x in terms of y, we shall be done. So let y = x 3 + 1 or
Then
x
= ;} y
 1.
f 1 (y)=x=;}yl.
If we visualize a function f as a set of ordered pairs, then f 1 is the function that reverses the coordinates. So if
f
then
=
/ 1
{(x1, Y1), (Xz, Yz), · · · , (xn, Yn)},
= {(yl>
xd, (Yz, Xz), ... , (yn, xn)}.
This gives another way of seeing why f needs to be 11. For f  1 to be a function, all the first coordinates of its ordered pairs must be different. Thus Y; =!= Yi if i =!= j, which means f is 11.
12
Chapter 1
The Real Number System
An important observation: Iff is a 11 function, then f!fi(f) = f!li(f  1 ) and f!fi(f 1) = f!li(f).
The notationf 1 is used another way. Associated with every functionf: X+ Y (whether f is 11 or not) is a mapping from subsets of Yto subsets of X, denoted by f 1 and defined by f 1 (B)
=
{x
E
Xlf(x)
E
B}
for B c Y. Thus f  1 (B) c X for B c Y. The set f  1 (B) is called the inverse image of B under f To say that f(x) E B is equivalent to saying x Ef 1 (B). The use of f  1 to denote two different kinds of functions is unfortunate but universal. However, the way in which f 1 is to be interpreted is usually clear from the context of the problem and rarely leads to confusion.
Example 17: Let X={1,2,3,4}, Y={2,4,5}, and f:X+Y be defined by f(1)=5, f(2) = 4, f(3) = 5, f(4) = 4. Then
f 1 {4, 5} = {1, 2, 3, 4}, f 1 {5} = {1, 3}, f 1 {2}=0. Example 18: Let f:X+ Y withAl> A 2 c X. We show that f(A 1 uA 2 ) cf(A 1 )uf(A 2 ). Let y Ef(A 1 uA 2 ). Then there is an x E A1 uA 2 withf(x) = y. Now x E A1 , or x E A 2 • If x E Al> then y = f(x) Ef(Ad, and if x E A 2 , then y = f(x) Ef(A 2 ). Thus y = f(x) Ef(Ad or y = f(x) Ef(A 2 ), soy Ef(Adu f(A 2 ).
Example 19: Let f:X+Y with B 1 ,B2 cX. We show that f 1 (Bdnf 1 (B 2 )c f 1 (B 1 nB 2 ). Let x Ef 1 (Bdnf 1 (B 2 ). Then x Ef 1 (Bd so f(x) E B1 , and x Ef 1 (B 2 ) so f(x) E B 2 • Thus f(x) E B1 and f(x) E B 2 so f(x) E B1 n B 2 , and therefore x Ef 1 (B 1 nB2).
Exercises 11          1. Find AuB, AnB, A \B, B\A, Ax B, and B x A if A and Bare (a) A= {2, 3, 6}, B = {1, 3, 5}. (b) A={0,1}, B=0. (c) A={1,2,3,4}, B={2,3}.
2. Find !!fi(f) and fY1(f) if f(x) is given by (a) f(x) =
l:l.
(b) f(x)
= x 2 ~ 1.
1
(c) f(x)=   .
)x3
Section 11
Sets and Functions
13
3. Supposef(x)=3x2 +7. Findf 1 (A) if (a) A= {9, 10, 11}. (b) A= {1, 2, 3}. 4. Find (fog)( 1) and (go f)( 1) iff and g are defined by (a) f(x) =sin x, g(x) =In x. (b) f(x) = ~, g(x) = x 3 + 4. (c) f(x) = n, g(x) =tan x. 5. Let f(x) = x and g(x) = 1/x. Is f =gog? 6. Find the domain of fog and go f if f(x) = rx+i. and g(x) = 1/(x 6). 7. (a) Prove that A\ (B n C)= (A\ B) u (A\ C) for any sets A, B, and C. (b) Prove that (B n C)'= Be u CC for any sets A, B, and C. 8. Let {A; Ii E I} be a collection of sets. Show that: (a) (
nA;)c = U (Ai).
iel
(b) (
U A;)c =
n(Ai).
iel
iel
iel
9. Show that A x B = B x A for nonempty sets A and B if and only if A = B. 10. Suppose B = B1 uB 2 . Show that Ax B =(Ax BI)u(A x B 2 ). Is it always true that (a) (A 1 n A 2 ) x (B 1 n B 2 ) = (A 1 x Bd n (A 2 x B 2 )? (b) (A 1 uA 2 ) x (B 1 uB 2 )=(A 1 x BI)u(A 2 x B 2 )? 11. Show that for any sets A, B, and C
(a) AnB cA. (b) AuB:::) A. (c) Au B =A if and only if B cA. (d) AnB =A if and only if A c B. (e) A u(B n C)= (AuB)n(A u C). (/) A n(Bu C)= (A nB)u(A n C). (g) A \ B = A if and only if An B = 0. (h) A\ B = 0 if and only if B :::) A.
Note: One can generalize parts (e) and (f) to show that Au(n B;) = n (AuB;) iel
An(U B;) = U (AnB;)
and
iel
iel
iel
for any collection of sets {Bd i E I}. We shall use this generalization in later chapters. 12. Show that for any sets A and B (a) A= (AnB)u(A \B). (b) Au B =(A\ B) u (B \A) u (An B) and that the sets (A\ B), (B \A), and (An B) are pairwise disjoint. 13. Let f: X+ Y with Al> A 2 c X and Bl> B 2 c Y. Show that: (a) j(A1 u Az) =/(Ad u f(Az). (b) f(AlnAz) c f(Ad nf(Az). (c) r 1 (Bl UBz) = r 1 (BI)ur 1 (Bz). (d) r 1 (BlnBz)=r 1 (Bdnr 1 (Bz). (e) r 1(Y\Bd=X\r 1(BI). (f) Give an example where f(A 1n A 2 ) i' f(Ad nf(A 2 ).
Note: One can generalize parts (a)( d) to arbitrary collections of sets. For example, it can be shown that
r
1(
U B;) = iU f
iE 1
E
1
(B;)
1
for any collections of subsets of Y, {Bd i E I}. We shall use this generalization in later chapters.
14
Chapter 1
The Real Number System
14. Prove that iff: X+ Y and for every pair of sets A, B c X, f(A n B)= f(A) n f(B), then f is a 11 function. 15. Prove that if f: X+ Y is a 11 function, then for any sets A, B c X, f(A n B)= f(A) nf(B). 16. Suppose f: X+ Y and A c X, B c Y. Show that (a) f(f 1 (B)) c B. Give an example where f(r 1 (B)) #B. (b) A c f  1 (/(A)). Give an example where A# f  1 (/(A)). (c) f is a 11 function if and only iff  1 (/(A)) =A for every A c X. 17. 18. 19. 20.
Suppose f: X+ Y and g: Y+ Z are onto functions. Show that go f maps X onto Z. Show that iff and g are 11 functions such thatjo g is defined, then (jo g) 1 = g 1 o f 1. Show that f: X+ Y is 11 and onto if and only if for each set A c X, f(A') = [f(A)]'. Suppose that Ak is a set for each positive integer k. (a) Show that x E n~ 1 (uf=n Ak) if and only if x E Ak for infinitely many sets Ak. (b) Show that x E u:'= 1 (nf=n Ak) if and only if x E Ak for all but finitely many of the sets Ak.
Properties of the Real Numbers As an Ordered Field The underlying structure for the topics that we shall study is the real number system. The properties of the real numbers can be broadly grouped into two classes: algebraic and geometric, or topological. In this section we begin by stating some of the axioms that govern the algebraic properties of the real number system and deriving some familiar consequences.
Axioms of a FieldThe real number system, which will be denoted from now on by of a mathematical structure known as a complete ordered field.
~.
is an example
Definition: A field F is a nonempty set together with two operations and · called addition and multiplication, which satisfy the following axioms:
+
The operations + and · are binary operations; that is, if a, b E F, then a + b and a· b are uniquely determined elements of F. (A2) The operations + and · are associative. That is, for a, b, c E F (A1)
(a+ b)+ c =a+ (b +c)
and (a· b)· c =a· (b ·c).
+ and · are commutative. That is, for a + b = b + a and a · b = b · a.
(A3)
The operations
(A4)
The distributive property holds. That is, for a, b, c E F a· (b +c)= (a· b)+ (a· c).
a, b E F
Section 12 (AS)
Properties of the Real Numbers as an Ordered Field
15
There exist additive and multiplicative identities. That is, there are elements 0 and 1 in F for which 0 + a = a and
1· a = a
for every a E F. (A6) There exists an additive inverse for each a E F. That is, if a
E
F, there is
an element in F denoted a for which
a+ (a)= 0. (A7)
For each a E F for which a=!= 0, there is a multiplicative inverse. That is, if a E F and a =1= 0, there is an element in F denoted a 1 or 1/a for which
a·a 1 =1. • We shall prove a few propositions to show how these axioms can be used to derive some of the other wellknown properties of the real numbers. This is not a complete list, and a thorough study of these properties is better left to a course in abstract algebra. Theorems 13 and 14 provide the key to the proofs of these properties.
Theorem 13: Let F be a field. Then the additive identity and the multiplicative identity are unique. Proof: We show that the additive identity is unique and leave it for Exercise 1, Section 12, to show that the multiplicative identity is unique. The usual way to prove uniqueness of an object is to suppose that there are two elements that possess the property that defines the object and then to show that the elements are equal. Suppose F is a field with two additive identities, call them 0 and 0. Since 0 is an additive identity and since 0 is an additive identity
Since addition is commutative, we have
0 = 0 + 0 = 0 + 0 = 0. So 0 and 0 are the same. Thus there is only one additive identity. •
Theorem 14: Let F be a field. Then the additive inverse and multiplicative inverse of an element are unique. Proof: We do the proof for the multiplicative inverse. Let a E F, a =!= 0. Suppose b and b are multiplicative inverses of a. That is, suppose a · b = b · a = 1 and a · b = b · a = 1. Then
b = b · 1 = b ·(a· b)= (b ·a)· b = 1 · b = b
16
Chapter 1
The Real Number System
where we have used the associative property in the third equality. Thus b = b, and the multiplicative inverse is unique. It is left for Exercise 1, Section 12, to show that the additive inverse is unique. •
Theorem 15: Let F be a field. Then a· 0 = 0 for every a E F. Proof: For aEF, a+O=a. Then a·a=a·(a+O)=a·a+a·O. Adding (a· a) to both sides of the equation gives 0 =a· 0. • Theorem 16: Let F be a field and a, b E F. Then (a)
(b) (c)
a·(b)=(a)·b= (a· b). (a)= a. (a)· (b)= a· b.
Proof: The main tools that we use to construct the proof are Theorem 15 and the fact that additive inverses are unique. (a)
By Theorem 15 and Axiom A4 0= a ·0 = a·(b +(b))= a· b + a·(b).
(b)
(c)
Thus (a· b) and a·(b) are additive inverses of a· b. Since additive inverses are unique, (a· b)= a· (b). Showing that (a· b)= (a)· b is similar and is left as Exercise 2, Section 12. By definition (a) is the additive inverse of(a). But a+(a)=O, so a is also the additive inverse of (a). Using the fact that additive inverses are unique, we conclude a= (a). By Theorem 15 0 =(a)· 0 =(a)· (b +(b))= (a)· b +(a)· (b). Thus (a)·(b) is the additive inverse of (a)· b. By part (a) of this theorem, (a)·b= (a·b), so (a)·(b) and a·b are additive inverses of (a· b). Thus a· b = (a)· (b), since additive inverses are unique. •
Hereafter a· b will usually be written as ab and a+ (b) will be written as ab.
        T h e Order AxiomDefinition: Let F be a field. Then F is an ordered field if it satisfies the additional axiom: (AS) There is a nonempty subset P ofF (called the positive subset) for which
If a, bE P, then a+ b E P (closure under addition). If a, bE P, then abE P (closure under multiplication). (iii) For any a E F exactly one of the following holds: a E P, a E P, or a = 0 (law of trichotomy). • (i)
(ii)
Section 12
Properties of the Real Numbers as an Ordered Field
17,
For us, P will be the set of positive real numbers. We use this axiom to define an order relation on the real numbers.
Definition: Let F be an ordered field, and let P be the positive subset of F. Let a, bE F. We say a< b if b a E P. We say a:::;; b if a< b or a= b. The statements a < b and b > a are equivalent. • Remark: This ordering also induces an ordering on any subset of the real
numbers. Notice that if a,bEF, then either baEP (a 0, then a < 0.
Theorem 17: Let F be an ordered field. For a, b, c E F the following hold: (a) (b)
(c) (d)
(e)
If a < b, then a+ c < b + c. If a< b and b < c, then a< c. (This says the relation < is transitive.) If a < b and c > 0, then ac < be. If a < b and c < 0, then be < ac. If a # 0, then a 2 > 0.
Proof: The main device that is used in proving the theorem is Axiom A8 of the definition of an ordered field. (a)
By definition of the relation (a+ c). The observant reader will protest that we have not shown that a c = (a+ c). This is left as Exercise 2, Section 12. Since a< b and b < c, then (b a) E P and (c b) E P. Since P is closed under addition, (c b)+ (b a)= c +((b)+ b)+ (a)= c a E P.
(c) (d) (e)
Thus a< c. The hypotheses imply that (b a) E P and c E P. Since P is closed under multiplication, (b a)c =(be ac) E P. That is, ac 0 (i.e., a E P) or a< 0 (i.e., 0 a= a E P) by the trichotomy property of P. Since P is closed under multiplication, in the first case we have a· a= a 2 E P. In the second case (a)· (a)= a 2 E P. So in either case, a 2 > 0. •
Definition: An interval of real numbers is a set A containing at least two numbers such that if r, s E A with r < s and if t is a number such that r < t < s, then tEA. •
18
Chapter 1
The Real Number System
A set consisting of a single point is sometimes called a degenerate interval. An interval is one of the following sets: ~
(a,b)={xla 1. if 0 < x < 1.
We leave the proof as Exercise 10, Section 12. •
Section 12
Properties of the Real Numbers as an Ordered Field
19
Theorem 110: Let x andy be positive real numbers with x < y, and let s be a positive rational number. Then x• < y•.
We leave the proof for Exercise 7, Section 12. • Since xa = 1/xa we can make analogous conclusions for negative rational numbers. These theorems are important because later they will enable us to define positive numbers raised to irrational powers. For example, we need them to know what 2..[3 means if we are going to discuss the function f(x) = 2x where x can be any real number.
Mathematical InductionSome of the steps involved in the proofs of Theorems 18 and 19 and some other theorems we shall prove involve a method of proof called proof by induction. We digress for a bit to discuss this technique. We state the principles involved and give some examples of how the technique is used, but do not elaborate on the reasons for its validity. There is more than one form of mathematical induction, but the one that we use is as follows: Induction Principle: Suppose that for each positive integer n there is a statement P(n). Suppose that (i) P( 1) is true. (ii) For any positive integer k, if P(k) is true, then P(k + 1) must be true.
Then P(n) is true for every positive integer n.
Example 110: We show that for each positive integer n
I j=l
j= n(n+ 1)_
2
According to the Induction Principle, we must first check that P( 1) is true. P( 1) is the statement
±j=1(1+1) 2
j=l
which is true. Next we assume that P(k) is true. That is, we assume
t
j=l
j= k(k+ 1)_
2
We must next show that P(k + 1) is true; that is,
kf j= (k+ 1)[(k+ 1)+ 1]_ j=l
2
(1)
20
Chapter 1
The Real Number System
The verification that P(k + 1) is true will almost always use the assumption that P(k) is true. We exhibit that here. Now k+l
I
j=l
k(k + 1)
k
j=
I
j+(k+ 1)=
+(k+ 1)
2
j=l
where the second equality follows from ( 1). Then k(k+ 1)
2
k(k+ 1)+2(k+ 1)
+ (k + 1) =
2
(k+ 1)(k+2)
=
2
(k+ 1)[(k+ 1)+ 1]
2· which shows that P(k + 1) is true. Thus, by the Induction Principle, we can conclude that
I
j= n(n+ 1)
2
j=l
for every positive integer n. As the second example, we prove the Binomial Theorem. To do this, we need to establish some notation. For a positive integer k we define k! (read "k factorial") to be the product of all the positive integers up to and including k. That is, k! = 1 · 2 · 3 · · · k. Also 0! is defined to be 1. For m?: k and m and k nonnegative integers, let
( m) k
m! =
k!(m k)!"
To facilitate the proof of the Binomial Theorem, we first prove the following theorem.
Theorem 111: Form and j positive integers with j
~
m
(~)+C:J=(m;1). Proof: We have ( m) j
+
( m ) m! m! j1 =j!(mj)!+[m(j1)]!(j1)! m!(m+ 1j)
m!j
j!(m j)!(m + 1j) + (m + 1j)!(j 1)!j =m![(m+1j)+j]= (m+1)! =(m+1). • (m + 1  j)!j! (m + 1  j)!j! j
21
Properties of the Real Numbers as an Ordered Field
Section 12
Theorem 112 (Binomial Theorem): Let a and b be real numbers, and let m be a positive integer. Then (a+b)m=
I
(~)aibmj.
j=O
}
Proof: We do the proof by induction. For m = 1 we have
±(~)aib 1 i=( 1 )b+( 1 )a=b+a=(a+W.
j=O
1
}
Assume the theorem is true for m = k. That is, assume that (a+ b)k =
I
j=O
(~) aibk i.
(2)
}
We must show the theorem is true for m = k + 1. That is, we must show that (a+b)k+1=
k+1(k+1) L . aibk+li. j=O
(3)
}
Once this is done, the Induction Principle says that we have accomplished the proof. To prove (3): (a+
b)k+ 1=(a+ b)k(a +b)= (it(~) aibk i) (a+ b)
by (2). Expanding this product gives
I
(~) ai+ 1bki + I (~) aibk+li.
j=O
j=O
}
}
We eventually want to add these two summands. We begin by letting j the first summand. Using this substitution
I
+ 1= t
k )atbk(f1) k£1( t1 = I ( k ) a'bk+1t + ak+1. t 1
in
(~)ai+1bki=
(=1
j=O}
(4)
(=1
In the second summand, we split the first term off so that
I (~) j=O
aibk+1i =
I j= 1
}
(~) aibk+1i + bk+1.
(5)
}
In (4) we rename the index of summation j. Then from (4) and (5), we have (a+b)k+1=ak+1+
I(.;1
k )aibk+1i+
j=1
c
I (~)aibk+1i+bk+1
j=1
J
=ak+1 +J1[ ~ 1) +C)Jaibk+1i +bk+1.
(6)
Chapter 1
22
The Real Number System
By Theorem 111,
ak+1
+
cJ ~
c~ 1}
I (k ~ 1) aibk+1i + bk+1 j= 1
since
+C)=
ak+ 1 =
so (6) can be written =
}
G: ~)
k£1 (k ~ 1) aibk+li, j=O
ak+ 1 b0
and
bk + 1 =
C;
}
1) a0 bk + 1 .
•
      T h e Absolute Value FunctionAn extremely important function on the real numbers is the absolute value function. It is particularly important because it provides a basis for the geometrical or topological properties of the real numbers, and these are the properties with which we shall be primarily concerned.
Definition: For a E
~,the
absolute value of a, denoted lal, is given by
lal = {
a~
a,
if
a,
if a< 0. •
The following theorem gives some of the properties of the absolute value function that we shall often use. ~ the following hold: Ia I ~ 0 with equality if and only if a = 0.
Theorem 113: For a, b E (a)
(h)
lal=lal. lal::::::a::::::lal. labl = lal·lbl. 1/lbl=l1/blifb#O. Ia/b I = Ia 1/1 b I if b # o. Ia I< b if and only if b (g)
Proof: We give the proof of part (h) and leave the proof of the remaining parts as Exercises 13 and 14, Section 12. To prove Ia + bl:::;; Ia I+ lbl, note that by part (c) I a I :::::: a:::::: Ia I and
I b I :::::: b :::::: Ib I
so that by adding the inequalities (which we show is valid in Exercise 4, Section 12), I a I  Ib I :::::: a + b :::::: Ia I + Ib I or so by part (g). •
 (I a I + Ib I) :::::: a + b :::::: Ia I + Ib I
Ia + b I :::::: Ia I + Ib I
Properties of the Real Numbers as an Ordered Field
Section 12
23
The absolute value function and the triangle inequality will be used extensively. Suppose we define a function d: ~ x ~+ ~ by d(a, b)= Ia bl. Then (i) (ii) (iii)
d(a, b);;;::: 0 and d(a, b)= 0 only if a= b, since Ia bl;;;::: 0, and Ia hi= 0 only if a= b. d(a, b)= d(b, a), since d(a, b)= Ia hi= 1(a b)l = lb al = d(b, a). d(a, c)::::;; d(a, b)+ d(b, c), since
d(a,c)=lacl = l(a b)+ (b c) I::::;; Ia hi+ lb cl
(by the triangle inequality)
= d(a, b)+ d(b, c).
Definition: If X is a set and d is a function d: X x X+~ satisfying the preceding properties (i), (ii), and (iii), then d is called a metric on X. • A metric may be thought of as a function that measures the distance between points, and it will often be advantageous to think of 1a  b 1 as the distance between a and b.
Exer cises 12          1. (a) Show that the multiplicative identity in a field is unique. (b) Show that the additive inverse in a field is unique.
2. If F is a field and a, b E F, show that 1'
(a) (a·b)=(a)·b. (b) ab= (a+b).
3. Show that if F is an ordered field and a, b, c E F with c < 0 and a < b, then ac > be. 4. If F is an ordered field and a, b, c, d E F with a < b and c < d, then (a) Show a + c < b + d. (b) Give an example to show that it is not necessarily true that ac < bd. 5. Show that if F is an ordered field and a, b E F with a ~ b and b ~a, then a = b. 6. Show that for any positive integer n ~
(a)
(b)
•2 _
i~/
f. / j=l
=
n(n + 1)(2n + 1) .
6
[n(n + 2
1)]
2
•
(c) For any real number x =11
1 x•+l 1 +x+x2 + · ·· +x"=
1x
I
''
The following exercises refer to the field of real numbers. 7. Show that (a) 1 > 0. (b) If 0 1 and n is a positive integer, then x•+ 1 > x" > 1. If m and n are positive integers with m > n, then xm > x". (c) If x > 0 and n is a positive integer, then 0 < x < (x + 1)". 9. (a) Show that if 0 < x < 1 and n is a positive integer, then x 1'" < 1. If m and n are positive integers with m > n, then x 11m > x 1'"· (b) Show that if x > 1 and n is a positive integer, then x 1'" > 1. If m and n are positive integers with m > n, then x 11m < x 1'"· 10. (a) Show that if 0 < x < 1 and s1 and s2 are positive rational numbers with s1 < s2 , then xs• > Xs 2 • [Hint: Write s1 = ajn, s2 = bjn where a, b, and n are positive integers with a< b.] (b) Show that if x > 1 and s1 and s2 are positive rational numbers with s1 < s2 , then xs• < xsz. (e)
11. If a, b;::, 0, show that J;b ~(a+ b)/2. This says that the arithmetic mean of two nonnegative numbers dominates the geometric mean. / 12 Use induction to show that ( 1 +a)";::, 1 + na for n a positive integer and a>  1. 13. Prove parts (a)(g) of Theorem 113. 14. (a) Showthatlallbl~labl. (b) Show that /lallblj
~ Ia bi.
15. Prove by induction that if a 1, ... , a. are real numbers, then Ia 1 + · · · + a. 1~ la1l + .. · + Ia. I. 16. Suppose a., L, and e are real numbers. Show that Ia. Ll < e if and only if a. E (L e, L+ e).
l'y. Show that the function d is a metric where dx _ lx Yl ( ,y)1+1xyl 18. (a) Let al> ... , a. and bl> ... , b. be real numbers. Prove the Schwarz inequality. That is,
[Hint: For any real numbers IX and {3, I;;= 1(1Xai f3bi) 2 ;::, 0. Expand this and choose and f3 judiciously.] (b) When does equality hold in the Schwarz inequality? IX
19. (a) Use the Schwarz inequality to prove the Minkowski inequality. That is,
for real numbers al> ... , a. and b1, ... , b•. (b) When does equality hold in Minkowski's inequality?
20. Show that if al> a2 , . . . , a.> 0, then occur?
(I:= 1ak)(L:= 1 1/ak);::, n
2•
When does equality
Section 13
The Completeness Axiom
25
I
·21. Show that if ak,bk>O for k=l, ... ,n with I~~ 1 bk=l, then (L~~ 1 akbd 2 ~ I~~ 1 a~bk. 22. Let f(x) = a.x" + a._ 1 x" 1 + · · · + a 1 x + a0 (a. #0). Let A =max{la0 1, la 1 1, ... , Ia.!}, and let B = nAJia.l. Show that f(x) # 0 if lxl >B. L/ 23. Show that if a < b + e for every e > 0, then a ~ b. p
The Completeness Axiom We begin this section by discussing the final axiom of the real numbersthat of completeness. This completeness axiom may be the least familiar of the axioms of the real numbers, but it is of fundamental importance anytime that limits are discussed. It is interesting to notice how this axiom is needed when we discuss the Archimedean property of real numbers, Cauchy sequences, and the existence of the Riemann integral, to name a few examples. There is more than one way to state the completeness axiom of the real numbers. Our approach uses the idea of a bounded set. Definition: Let A be a set of real numbers. If there is a real number b for which x :( b for every x E A, then b is said to be an upper bound for A. A set that has an upper bound is said to be bounded above. If there is a number c such that c :( x for every x E A, then c is said to be a lower bound for A. A set that has a lower bound is said to be bounded below. A set that is bounded above and below is said to be bounded. A set that is not bounded is said to be unbounded. • Definition: Let A be a set of real numbers that is bounded above. The number b is called the least upper bound (or supremum) of the set A (denoted l.u.b. A or sup A) if b is an upper bound of A, and (ii) If c is also an upper bound of A, then b ~ c. • (i)
Note that (ii) is equivalent to saying that if d < b, then d is not an upper bound of A. Definition: Let A be a set of real numbers that is bounded below. The number b is called the greatest lower bound (or irifimum) of the set A (denoted g.l.b. A or inf A) if
b is a lower bound of A, and (ii) If c is a lower bound of A, then c :( b. • (i)
Definition: Let S be an ordered field. Then S is said to be complete if for any nonempty subset A of S that is bounded above, the least upper bound of A is inS. •
26
Chapter 1 The Real Number System The Axiom of Completeness: The final axiom of the real numbers is: (A9)
The real numbers are complete. •
The definition of completeness could have been stated in terms of lower bounds. The definition we gave is equivalent to the following, as we show in Exercise 7, Section 13: LetS be an ordered field. Then Sis complete if for any nonempty subset A of S that is bounded below, the greatest lower bound of A is inS. An example of an ordered field that is not complete is the rational numbers. For example, if we let
then A is bounded above, but it has no least upper bound in the rational numbers.
Theorem 114: If the least upper bound and greatest lower bound of a set of real numbers exist, they are unique. Proof: We show that the least upper bound of a set is unique. Let A be a set of real numbers that is bounded above. Suppose rx and fi. are both least upper bounds of A. Thus rx and fi. are both upper bounds of A. Since rx is a least upper bound and fi. is an upper bound, then
Likewise, since fi. is a least upper bound and rx is an upper bound, then
Thus fi. = rx, so the least upper bound is unique. We leave the proof of the uniqueness of the greatest lower bound as Exercise 5, Section 1 3. • Using the axiom of completeness and uniqueness of the least upper bound, we can define what it means to raise a positive number to an irrational power.
Definition: Let x > 1 be a positive real number and r a positive irrational number. Then the number x' is the least upper bound of the set A where A= {xPip is a positive rational number less than r}. • If 0 < x < 1 and r is a positive irrational number, then x' is defined to be 1/y' where y = 1/x > 1.
Since xr = 1/x', this also gives a way of defining a positive number raised to a negative irrational exponent. Theorem 115 gives an alternative way of defining the least upper bound and greatest lower bound. This theorem will provide the characterization of these numbers that we shall use most often.
Section 13
The Completeness Axiom
27
Theorem 115: (a)
(b)
A number a is the least upper bound of a set of real numbers A if and only if (i) a is an upper bound of A, and (ii) Given any e > 0, there is a number x(B) in A for which x(B) >a e. A number {3 is the greatest lower bound of a set of real numbers A if and only if (i) {3 is a lower bound of A, and (ii) Given any B > 0, there is a number x(s) in A for which x(B) < {3 +B.
Before proving Theorem 115, it is worthwhile to give a detailed discussion of condition (ii) of part (a), since this will be a recurrent theme throughout the text. One thinks of B as being a very small positive number. Suppose a= l.u.b. A. What (ii) is saying is that if first we choose B, then we can find a number x(B) in A that is larger than a B. This means a  B is not an upper bound for A. Thus for any B > 0, a  e is not an upper bound for A. The order of doing things here is crucial. First, e is chosen, and then x(B) is found. The number x(B) in A will often depend on e as the notation is supposed to emphasize.
Proof: We give the proof of part (a) and leave the proof of part (b) as Exercise 3, Section 13. First, suppose a= l.u.b. A. We show that (i) and (ii) must hold. By definition, (i) holds. To show that (ii) holds, let B > 0 be given. Then a B 0,
=IX B.
By (ii) there is a number x(B) E A with x(B) >a B = fi,
so fi is not an upper bound for A. •
Example 111: The purpose of this example is to clarify the comments preceding the proof of Theorem 115. Let
A={!,~'~'~' ... }= {n: 11 n ~ E
}·
28
Chapter 1 The Real Number System We claim 1 is the least upper bound of A. Clearly, n
  < 1 for n E n+1
~.
so 1 is an upper bound for A. Now choose e = .01 > 0. Then we claim there is an element in A larger than 1  e = .99. In fact, 999/1000 = .999 > .99 and 999/1000 is in A. Suppose that we had chosen e = .0001. Then 1 e = .9999, and the number we chose before, 999/1000, no longer works. We have to choose another number in A, say 99,999/100,000 = .99999, to exceed 1 e. The point is that in this example no number in A exceeds 1  e for every e > 0. However, it seems plausible (we need Theorem 118 to actually prove it) that if we choose e > 0 first, then we can find an element of A that exceeds 1  e. Example 111 shows that the least upper bound of a set need not be an element of the set. On the other hand, it might be. If we modify our example by adding 1 to the set, so that if B =Au {1} = {1, !, ~, ~, ... },then 1 is the least upper bound of B and 1 E B. Notice that if the set A is a finite set, then l.u.b. A is just the largest number in the set. As we have seen, if A is infinite, then l.u.b. A may or may not be in A. The following result may provide some additional insight into the concept of the least upper bound of a set.
Theorem 116: Let a= l.u.b. A, and suppose a¢: A. Then for any e > 0, the interval (a e, a) contains an infinite number of points of A. Proof: Choose e > 0. By definition of l.u.b. A and since a ¢: A, the interval B, a) contains at least one point of A. Suppose that interval contains only a finite number of points of A (call them x 1 , . . . , xn), and suppose they occur as in Figure 12. Since there are only a finite number of these points, we can select the largest, which in this case is xn. Now xn is larger than any other number in A, but is smaller than a. Thus a is not the least upper bound of A.
(a
ue
Xz
x3
Figure 1  2   
XII
•
We proved this result by proving the contrapositive of the proposition. That is, we assumed the conclusion was false and showed the hypothesis could not hold. This is a common method of proof.
Definition: Let A be a set of real numbers, and suppose c is a real number. Then cA is the set of real numbers given by cA
=
{cxtx E A}. •
Section 13
The Completeness Axiom
29
Example 112: Let A= { 1, 2, 4}. Then 3A = { 3, 6, 12} and 2A = {2, 4, 8}. Notice that sup A = 4, infA= 1,
sup 3A = 12 = 3 sup A, inf3A= 3=3infA,
inf ( 2A) =  8 = 2 sup A,
and
sup (2A) = 2 = 2 inf A. This is an example of an important principle, which is given by Theorem 117.
Theorem 117: Let A be a bounded set of real numbers, and suppose cis a real number. Then (a)
(b)
If c > 0 (i) l.u.b. (cA) = c l.u.b. A. (ii) g.l.b. (cA) = c g.l.b. A. If c < 0 (i) l.u.b. (cA) = c g.l.b. A. (ii) g.l.b. (cA) = c l.u.b. A.
Proof: We give the proof of part (ii) of(b) and leave the proof of the remaining parts as an exercise. Let A be a set of real numbers with x E A, x :( IX. Thus, since c < 0,
IX=
l.u.b. A and suppose c < 0. For any
ex~ CIX.
Therefore ciX :(ex for any x E A, so ciX is a lower bound for cA. Next we show ciX is the greatest lower bound. Let e > 0 be given. We must show that there is a number ex( e) E cA with ex(e)< CIX +e. Now e/( c)> 0, and since IX= sup A, there is a number x(e) E A with x(e) >IX e/( c). Then cx(e)
0, then there is a positive integer n such that na > b. Proof: Let A = {ka Ik is a positive integer}. We shall show that A is not bounded above. If A is not bounded above, then b is not an upper bound. Thus there is some positive integer n for which na >b. To show that A is not bounded above, we suppose that A is bounded above and show that this is impossible.
30
Chapter 1
The Real Number System
If A is bounded above, then by the completeness property, A has a least upper bound, which we call rx. Since a > 0, there is a number in A, call it Na, such that
rxa 1 so that
1 N"
c>
"
Clearly, N depends one, and we emphasize this by writing N(s) for N. Now
N(a) 1 N(a) E A
and
1
B
0. Use the Binomial Theorem to find a number b such that (z +b)"< y. This shows that z was not l.u.b. A. A similar argument works if one assumes y < z".] B = {xlx" > y and x > 0}.
(c) Let
Show that B is nonempty and bounded below. (d) Show that l.u.b. A= g.!. b. B.
/ 10. Let A be an uncountable set and B a countable set. Show that A\ B is uncountable.
11. (a) Prove that the rational numbers are countable. (b) Prove that the irrational numbers are uncountable. 12. Show that the set of open intervals with rational endpoints is countable. 13.J(a) Let A and B be countable sets. Show that the Cartesian product A x B is countable. (b) Let Al> ... , A. be countable sets. Show that
A1 x · · · x A.= {(al> ... , a.)lai E Ai}
is countable. 14. Show that the set of polynomials with rational coefficients is countable. 15. Construct a 11 onto function from {0, 1, 2, ... } to {1, 2, 3, ... }. Prove that the function
you have constructed is 11 and onto.
Section 13
The Completeness Axiom
35
/16. Show that the function f: 7l.> f:::l given by f(n) = {
2n + 1,
if n)!: 0
2n,
if n < 0
is 11 and onto. 17. Let a be greater than 1 and r a positive irrational number. Let A= {axlx is a positive rational number< r}
and
B
= {axlx is a positive rational number> r}.
Show that: (a) The set A is bounded above, and the set B is bounded below. (b) If sEA and t E B, then s < t. (c) What assumptions, if any, must be made to show l.u.b. A= g.!. b. B? 18. (a) Show that if a> 1 and x and y are real numbers with x < y, then ax< aY. Recall that we have proven this in the case that x and y are rational numbers. \)(b) Show that if 0 aY.
Sequences of Real Numbers
Sequences of Real Numbers One of the great advantages of calculus is that it enables us to solve problems of a dynamic nature, that is, problems in which a change in the variables occurs. The technique that calculus uses to deal with this type of problem is the limiting process. In this chapter we introduce sequences of real numbers. Sequences provide perhaps the simplest setting for the rigorous study of limits, and sequences will also be indispensable in studying more complex topics.
Definition: A sequence of real numbers is a function from the positive integers into the real numbers. • The functional concept is not the most convenient way to visualize a sequence, and we shall establish a more intuitive viewpoint. If f is the function in the definition, then the range off is the set {f(l ),f(2), ... }.
The numbers f( 1),f(2 ), ... are called the terms of the sequence and f(n) is called the nth term of the sequence. The domain of a sequence is always the positive integers. Therefore if we list only the range or terms of the sequence in their natural order of appearance, then the sequence will be completely described. It is customary to further simplify the notation and write fn for f(n). So now our sequence is written as a subscripted variable within braces {!1 ,f2 , . . . } or more often as {f.}. Thus {x.} will represent a sequence whose first term is x 1 , whose second term is x 2 , and so on. Sequences are often viewed as an infinite string of numbers. Two sequences are equal if and only if they are equal term by term. That is, not only must the numbers in the sequences be the same, but they must also appear in the same order. Thus the sequence {1, 2, 3, 4, 5, ... } is not equal to the sequence {2, 1, 3, 4, 5, ... } even though they consist of the same numbers. 36
Section 21
Sequences of Real Numbers
37
Often a sequence is represented by a function within braces, which describes the nth term of the sequence. For example, we might represent the sequence {1,!,l,;\, ... } as {1/n}, since the nth term of the sequence is equal to 1/n.
Example 21: The sequence 1, 1, 1, . . . or {1, 1, 1, ... } may be represented as {1} or {1}:= 1 where the indices are used to emphasize that the object under consideration is a sequence and not merely a set. (Remember, a set is defined by the elements within the set and is independent of the order in which the elements appear, whereas a sequence is determined by the elements and the order in which they appear.)
Example 22: The sequence 2, 4, 6, 8, ... or {2, 4, 6, 8, ... } may be represented as {2n} or {2n}:=l· Definition: We say that the sequence of real numbers {x.} converges to the number L if, for any t: > 0, there is a positive integer N(t:) such that if n is a positive integer larger than N(a), then lx. Ll N(a), then Ia. Ll 0 be given, and then Find how large n must be to ensure that
Thus we need l1/n 2 1 1/a, so n > 1//"e. In this case, we could take N(a) to be any positive integer as large as 1//"e. So if n > N(a) > 1//"e, then 1(2 1/n2 )  21 < B, and we have proved that {2 ( 1/n2 )} + 2 as our intuition suggested.
Example 25: Let {x.} be the sequence {1, 2, 1, 4, 1, 6, 1, 8, ... }. This sequence diverges for the following reason: Suppose that we choose B =!.Then no matter what values we take for Land N(a), there are numbers in the sequence beyond the N(a) term such that lx.LI >!.
Section 21
Sequences of Real Numbers
39
There are two ways in which a sequence can diverge that will be of special interest to us. We now define these.
Definition: A sequence of real numbers {x.} is said to diverge to infinity if, given any number M, there is a positive integer N(M) such that if n > N(M), then x. > M. In this case we write lim x. = oo or {x.}+ oo. • One should think of Mas being a large positive number. The number N(M) tells us beyond which term we need to go so that all the terms after the N(M)th term exceed M. Figure 22 illustrates this idea.
•
• M
•
• • •

• • •
•
• •
N(M)
Figure 2  2   
Definition: A sequence of real numbers {x.} is said to diverge to negative infinity if, given any number K, there is a positive integer N(K) such that if n > N(K), then x. < K. In this case we write lim x. =  oo or {x.}+ oo. • In this definition one should think of K as being a very large negative number.
Example 26: Let {x.} = {n 2 }. We show that {x.} diverges to oo. Choose a positive number M. If we let N(M) be the smallest integer larger than (e.g., if M = 101, then N(M) = 11), and if n > N(M), then
.JM
x. = n2 > N(M) 2 > (.fMf = M.
Example 27: Let {x.} = {1, 2, 1, 3, 1, 4, ... }. Then {x.} diverges because there is no number L for which all the terms beyond some point may be made arbitrarily close
40
Chapter 2
Sequences of Real Numbers
to L. The sequence does not diverge to oo because there is no point in the sequence beyond which the terms are arbitrarily large. (The 1's are always there.) We now prove some theorems that describe the behavior of convergent sequences.
Theorem 21: A sequence of real numbers can converge to at most one number. • Writing mathematical proofs involves two processes: ( 1) formulating the ideas of the proof and (2) writing these ideas in mathematical language appealing only to the definitions and previously proven results. Here we give the ideas involved in the proof and leave it for Exercise 5, Section 21, to express these ideas in mathematical language.
Proof: We shall suppose that there is a sequence, {xn}, that converges to two different numbers, L and M, and show this is impossible. Picture L and M on the number line in Figure 23. Draw an interval about L and an interval about M that do not intersect each other, as shown in Figure 24. In writing your proof, you should specify how large these intervals are. M
L
Figure 2  3   
L
L+E
M
M+E
Figure 2  4   If {xn} converges to L, then all the terms beyond some point, say N 1 , in the sequence must be in (L e, L + e). If { xn} converges to M, then all the terms in the sequence beyond some point, say N2 , must lie in (Me, M +e). Then if n is larger than both N1 and N2 , Xn must be in both (L e, L+ e) and (Me, M +e). But this is impossible, since the intervals do not intersect. •
Theorem 22: The sequence of real numbers {an} converges to L if and only if for every e > 0 all but a finite number of terms of {an} lie in the interval (Le,L+e). Proof: Suppose that the sequence {an} converges to L. Let e > 0 be given. By definition, there is a number N(e) such that if n > N(e), then ian Li N(e), then an E (L e, L+ e) so that ian Li N2 (e), then
Then ifn>max{N1 (e),N2 (e)}, both conditions hold, so
lanbn abl:::; lanllbn bl + lbllan al N(e), then i(anfbn) (ajb)i 0 with (1/lbnl) < M for every n. Since {an} +a, there is a number N 1 (e) such that if n > N1 (e), then ian ai < ej(2M); and since {bn} +b, there is a number N2 (e) such that if n > N2 (e), then
eibi ibn bi < 2Mial + 1 So if N(e) =max {N1 (e), N2 (e)}, and if n > N(e), then
IG:) (~)I~
l:nllan ai +
~~~~~lb bni < Mlan ai +~~~~ibn bi
Me IaiM ( elbl ) an+ d for every n, the sequence is strictly monotone increasing (decreasing). • Any of these types of sequences is called a monotone sequence.
44
Chapter 2
Sequences of Real Numbers
Monotone sequences are nice because the converse of Theorem 23 is true. That is, bounded monotone sequences converge, as we show in Theorem 26. This is not true for arbitrary bounded sequences, as the sequence {( 1)"} shows.
Theorem 26: A bounded monotone sequence converges. Proof: We give the proof only for bounded monotone increasing sequences. Let {a.} be such a sequence. To show that the sequence converges, we need a candidate for the number to which it converges. There is really only onethe least upper bound of the set of terms of {a.}. Call this number L. By definition of least upper bound, L has two properties: (i) (ii)
a. ~ L for every n. For any given e > 0, there is a positive integer N(e) for which aN(e) > L e.
Since {a.} is monotone increasing, if n > N(e), then a.;;?: aN(e) > L e. But Thus L e N(e); that is,
a.~
L.
la.LI N(e). • Corollary 26: (a)
(b)
A monotone increasing sequence either converges or diverges to oo. A monotone decreasing sequence either converges or diverges to  oo. •
Example 28: Let {x.} be the sequence of numbers defined by x 1 =sin 1 x 2 =larger of {sin 1, sin 2}
xk =largest of {sin 1, sin 2, ... , sink}
where the angle is given in radian measure. Then {x.} is monotone increasing and bounded above by 1, since sin x ~ 1. Thus by Theorem 26, {x.} converges. Notice that we have not exhibited the number to which {x.} converges.
Example 29: Later we shall define the natural logarithm function, but you are probably already familiar with its importance and the frequency with which its base, the number e, occurs in mathematics. We shall not use sequences to define e, but one way that it can be defined is by
e = lim ( 1 + n+ 00
~)". n
Section 21
Sequences of Real Numbers
45
Here we shall show that the sequence {xn} = {(1 + 1/n)n} is an increasing sequence. To show the sequence is increasing, we use the Binomial Theorem. We have
n n(n1)···(nk+1)(1)k =I n k=o k! n 1 ( 11) ( 12) ··· ( 1 k1) = I.(1)
k=o k.
and
Xn +1 = :
t: ~ (
n
1) ( 1  n
Now,
n
~ 1) ( 1 
n
n
!
n 1 ~I,, k= 1 k.
1) · · · ( 1  : :
~).
j j 1>1n+1 n
so that the kth term for Xn+ 1 is larger than the kth term for xn. In addition, Xn +1 has one additional term so that Xn +1 > Xn. Later we shall show that this sequence is bounded above so that {(1 + 1/n)n} is a convergent sequence. The next theorem is a special case of a more general result that we shall prove later.
Theorem 27: Let An= [an, bn] be a sequence of intervals such that An=> An+ 1 for n = 1, 2, .... Suppose that limnco (bn an)= 0. Then there is a real number p for which
n An= {p}. 00
n=1
That is, there is exactly one point common to every An. • We leave the proof for Exercise 12, Section 21.
Definition: A sequence of sets {An} such that An=> An+ 1 is called a nested sequence of sets. • In Chapter 1 we defined the least upper bound and greatest lower bound of a set of real numbers. Theorem 28 gives a fact that will be important later.
Theorem 28: Let A be a nonempty set of real numbers that is bounded above. Then there is a sequence of numbers {xn} such that (i) (ii)
Xn E A, n = 1, 2, .... lim Xn = l.u.b. A.
An analogous result holds for g.l.b. A.
46
Chapter 2
Sequences of Real Numbers
Proof: Let a= l.u.b. A. If a E A, then let x. =a for every n. If a$ A, then, by definition of l.u.b. A, for each positive integer n there is a number x. E A with 1 n
a N(e), then
Ia. ami< e.
•
In Exercises 13 and 14, Section 23, we prove the following very important result.
Theorem 29: A sequence converges if and only if it is a Cauchy sequence. • This theorem turns out to be surprisingly powerful and we shall use it extensively. Part of its power comes from the fact that, to show a sequence converges using this theorem, we do not need to have a candidate for the number to which the sequence converges.
Exercises 21          1. Each of the following sequences converges: (a) Find the number to which the sequence converges. (b) If L is the number to which the sequence converges, find how large n must be so
that Ia. Ll < .01. 0 find N(e) such that Ia. Ll < e if n > N(e). (i) {a.}= {3 1/(4n)}. (iii) {a.}= {(n2 + 1)jn 2 }. (ii) {a.}= {(2n + 3)/(3n 1)}. (iv) {a.}= {1 + 1/2"}. 2. Construct a sequence {a.} that converges to 0, but a. =1= 0 for any n. 3. Assume that n = 3.1415926535 ... is irrational. Describe how you would construct a sequence {a.} that has all of the following properties: (i) Each term of {a.} is rational, (ii) {a.} is monotone increasing, and (iii) {a.} converges to n. This shows that the rational numbers are not complete. 4. Let r be a rational number. (a) Construct a strictly increasing sequence of rational numbers that converges to r. (b) Construct a strictly increasing sequence of irrational numbers that converges to r. (You may use the fact that n is irrational.) (c) Fore>
Section 21
Sequences of Real Numbers
47
5. Prove Theorem 21. 6. Give an example of a bounded sequence that does not converge. 7. Prove part (b) of Theorem 24. 8. (a) Suppose that {an} +a and that an~ 0 for every n. Show that a~ 0. [Hint: Suppose that a< 0 and take e =  a/2. Show that {an} cannot converge to a.] (b) Show that if {an} + L and an :( K for every n, then L :( K. (c) Suppose that {an} and {bn} are sequences with an:( bn for every n, and suppose 'that {an} +a and {bn} +b. Show that a:( b. (d) If {an} and {bn} are sequences with 0 :(an:( bn for every nand {bn} +0, show that {an} +0. 9. (a) Suppose that {an} and {bn} are sequences with {an}+ L. Show that if {an bn} +0, then {bn} +L. (b) Show that if {an}, {bn}, and {en} are sequences with an :( bn :( en for every n and if {an} +Land {en} +L, then {bn} +L. 10. Suppose that {bn} + b and b f= 0 and bn f= 0 for any n. Show that there is a number K > 0 such that Ibn I> K for every n. You may assume bn > 0 to simplify the writeup. 11. (a) If{an}+a,showthat{lanl}+lal. (b) Find an example of a sequence {an} such that { Ian I} + 1 but {an} does not converge. (c) If {lanl} +0, show that {an} +0.
12. Let In be an interval [an, bnJ. Suppose that {In} is a sequence of nested intervals; that is, In 2 In+! for each n so that an:( an+l and bn ~ bn+l for each n.
Show that {an} and {bn} are convergent sequences. If {an} +a and {bn} +b, show that a:( b. If {bn an} +0, show a= b. If {In} is a nested sequence of closed bounded intervals whose lengths go to 0, show that n;:"= 1 In consists of exactly one point. (e) Is the result of part (d) true if the intervals are not required to be closed? (f) Is the result of part (d) true if the intervals are not required to be bounded? 13. Complete the proof of Theorem 26 by showing that a bounded monotone decreasing sequence converges. 14. Give an example of two divergent sequences whose sum converges. 15. (a) Prove that the sequence {(1n2 )/n} diverges to oo. (b) Prove that {n 2  4n + 7} diverges to oo. 16. Let {an} and {bn} be sequences that diverge to oo. (a) Show that {anbn} diverges to oo. (b) Show that {an+ bn} diverges to oo. 17. Give examples of sequences {an} and {bn} that diverge to oo for which (a) {an bn} diverges to 00. (b) {anbn} converges to 5. 18. Prove that if {an} is a sequence of nonnegative numbers and {an} +a, then (a) (b) (c) (d)
{./a,. }..... ;a.
19. Show that the sequence {an} converges where (a) an= 1· 3 · 5 · · · (2n1)/(2·4· 6 · · · (2n)). (b) a1 = .1, a2 = .12, ... , an= .12 ... n. 20. (a) Suppose that {an} is bounded and that {bn} +0. Prove that {anbn} converges to 0. (b) Find an example where {bn} + 0 but {anbn} does not converge to 0. 21. Show that a monotone increasing sequence that does not converge must diverge to oo.
48
Chapter 2
Sequences of Real Numbers
22. (a) Suppose {a.} +a. Let {b.} be the sequence defined by b = al + ... +a. n • n
Show that {b.} +a. (b) Let {a.}= {(1)"}. Show that {a.} diverges, but the sequence {b.} defined by
+an b = a+··· 1 n n converges. 23. (a) Show that if n;?:: 3, then {n 11"} is a monotone decreasing sequence. (b) Show that lim n11" = 1. 24. Suppose that {x.} is a sequence of positive numbers and lim(x:: 1 ) =L. (a) Show that if L > 1, then lim x. = oo, and if L < 1, lim x. = 0. (b)
Construct a sequence of positive numbers {x.} such that lim Xn+l = 1
x. and the sequence {x.} diverges. (c) Show that lim nk/rx" = 0 for rx > 1 and k a positive integer. 25. Give a condition equivalent to the statement, "The sequence {x.} does not converge to L," that does not use negatives.
Subsequences Definition: Let {a.} be a sequence of real numbers, and let n 1 < n1
... in the definition.
Definition: Let {a.} be a sequence of real numbers. We say that the number L is a subsequential limit of {a.} if there is a subsequence of {a.} that converges to L. • We shall adopt the convention that oo or  oo are subsequential limit points of a sequence if there are subsequences that diverge to those values. We shall also adopt the convention that oo is the least upper bound of a set that is not bounded above and  oo is the greatest lower bound of a set that is not bounded below. With these conventions, every sequence has a subsequential limit point and every set has a least upper bound and a greatest lower bound. Whenever a letter such as Lis used to denote a subsequential limit point, it will be understood to represent a real number. We shall often refer to a subsequential limit of a sequence as merely a limit point of the sequence.
Theorem 210: A sequence {a.} converges to L if and only if every subsequence of {a.} converges to L. Proof: We first show that if {a.} converges to L, then any subsequence of {a.} converges to L. Let {a.J be a subsequence of {a.}, and let e > 0 be given. Since {a.} converges to L, there is a number N(e) such that if n > N(e), then Ia. Ll K(e), then la.k Ll N(e), then a.k is a term of {a.} past the N(e) term, and so la.k Ll 0, the interval (L e, L + e) contains infinitely many terms of {a.}.
50
Chapter 2
Sequences of Real Numbers
Proof: First suppose that L is a subsequential limit point of {an}. Then there is a subsequence {ank} of {an} that converges to L. Let e > 0 be given. Then there is a number K(e) such that if k > K(e), then lank Ll < e; that is, ankE(Le,L+e), and so (Le,L+e) contains infinitely many terms of {an}. Conversely, suppose that for any e > 0 the interval (L e, L+ e) contains infinitely many terms of {an}. We shall show that Lis a subsequential limit point of {an} by constructing a subsequence that converges to L. The construction proceeds inductively. Step 1. Let e = 1. There is some term of {an} in (L1, L+ 1), say an,· Step 2. Let e = !. There are infinitely many terms of {an} in (L !, L+ !), so we can choose one beyond the n1 term. Call it anz· That is, it is possible to choose n1 > n1 with an 2 E (L !, L + !). Induction Step: Assume that for n = k (a positive integer) it is possible to find a term of {an} beyond the nk _ 1 term in (L 1/k, L + 1/k). Choose one of these terms, and let it be ank· Let n = k + 1. Letting e = 1/(k + 1) in the hypothesis, there are infinitely many terms of {an} in (L1/(k+1), L+ 1/(k + 1)). Choose one beyond the nkth term, and let it be a"k+t ·
Thus we have shown by induction that it is possible to extract a subsequence {ank} of {an} such that ank E (L1/k, L+ 1/k) fork= 1, 2, .... We show that {anJ converges to L. Let e > 0 be given. By the Archimedean Principle, there is a positive integer K(e) with 1/K(e) K(e), then
ank E ( Lso that {ank}
+ L.
i' D L+
c
(L e, L+ e),
•
Corollary 211: Let {an} be a sequence of real numbers. Then Lis a subsequential limit of {an} if and only if given any e > 0 and any positive integer N, there is a positive integer n(e, N) > N for which Ian(e,N) Ll 1 and define a sequence {a.} by a 1 = 1 and a.+ 1 = k( 1 + a.)/(k +a.). (a) Show that {a.} converges. (b) Find lim a•.
14. Let {a.} be a sequence such that {a 2. } converges to Land {azn+d converges to L. Show that {a.} converges to L. 15. The Fibonacci sequence {a.} is defined by a0 = 1, a 1 = 1, and a.= a._ 1 + a.z for n;?:: 2. Let the sequence {f.} be defined by f.= a.!a._ 1 for n;?:: 1. Assuming that {f.} converges (which it does), find limf•. 16. Use the results of Exercises 9 and 10 of Section 12 and Exercises 4 and 5 of this section to show that for a > 0 and a given e > 0, there is a () > 0 such that if IxI < o, then 1ax 11 0, the interval (x s, x +e) contains infinitely many points of A. • Notice that there is no requirement that a limit point of a set be an element of the set. For example, 0 is a limit point of the set {t, t, !, ... }. Recall that [a, b] is the set {xI a :::::; x :::::; b}. In the proof of Theorem 212, when we say we divide the interval [a, b] into two intervals of equal length, these intervals will be understood to be [a, (a+ b)/2] and [(a+ b)/2, b].
Theorem 212 (BolzanoWeierstrass Theorem): Every bounded infinite set of real numbers has at least one limit point. Proof: Let A be a bounded set of real numbers. Since A is bounded, there is a positive number M such that A c [  M, M]. Divide [ M, M] into two closed intervals of equal length, [ M, OJ and [0, M]. At least one of these intervals contains an infinite number of points of A. Choose one of the intervals that contains an infinite number of points of Acall it A1 . Notice that the length of A1 isM= 2Mj2. Divide A1 into two closed intervals of equal length. Choose one of the subintervals that contains infinitely many points of A, and call it A 2 • The length of A 2 is 2M/2 2 • See Figure 25.
Section 23
The BolzanoWeierstrass Theorem
53
A3A2A~•
Figure 2  5   Continue the process inductively so that for each positive integer k, Ak is a closed interval oflength 2M/2k = M/2k1, and Ak contains infinitely many points of A. Notice that An ::::JAn+ 1 , and if An= [an, bn], then M
lim(bn an)= lim 2n_ 1
=
0.
Thus, by Theorem 27, n;;'= 1 An has exactly one pointcall it p. We claim that p is a limit point of A. Let e > 0 be given. We shall show that (p e, p +e) contains infinitely many points of A by showing that (p e, p +e) ::::JAN for some positive integer N.
M
p
p
M p+
p+E
2NJ
2N I
Figure 2  6   
pE
Choose N so that 2M/2N = M/2N 1 0. But this occurs if and only if (i) or (ii) occurs. •
54
Chapter 2
Sequences of Real Numbers
Example 210: Let {an} = {0, 1, 0, 1, ... }. Then the set {0, 1} contains all of the elements of the terms of the sequence {an}. Since 0 and 1 each appear infinitely often, they are each subsequential limit points for the sequence {an}. They are not limit points of the set made up of the terms of {an}, since this set is {0, 1} and as a finite set it has no limit points.
Example 211:
t, .. .}.
t, .. .}
Let {an}= {1, !, 1, ~' 1, Here {1, !, ~' is the set of elements that makes up the terms of the sequence {an}. The number 1 appears infinitely often in the sequence, and 0 is the only limit point of the set {1,!, ~' Thus 0 and 1 are exactly the subsequential limit points of the sequence.
t, ... }.
The next theorem uses the BolzanoWeierstrass Theorem to prove an important characteristic of sequences.
:/ Theorem 214: Every bounded sequence has a convergent subsequence. Proof: Let
{an} be a bounded sequence. There are two possibilities:
Case 1: There is a number L for which an infinite number of terms of {an}
assume the value L. Suppose these are the n1 , n2 , •.• terms of the sequence with n1 < n2 < · · ·. Then the subsequence {an 1 , an 2 , •• • } converges to L. Case 2: The terms of the sequence take on an infinite number of distinct values. (Case 1 does not exclude this possibility, but if Case 1 does not occur, then Case 2 must occur.) By the BolzanoWeierstrass Theorem, there is a number p that has the property that (p a, p +a) contains an infinite number of terms of {an} for any a > 0. By Theorem 211 there is a subsequence of {an} that converges to p. •
Corollary 214: A bounded sequence that does not converge has more than one subsequential limit point. The proof is left for Exercise 17, Section 23. • Notice that if {an} is a sequence for which a~ an~ b for every nand if Lis a subsequential limit point of {an}, then, by Theorem 25, a~ L~ b. We next examine a characteristic of unbounded sequences.
Theorem 215: sequence that is unbounded above has a subsequence that diverges to oo. (b) A sequence that is unbounded below has a subsequence that diverges to  oo. (a) A
Proof of (a): Let {an} be a sequence that is unbounded above. We shall inductively construct a subsequence of {an} that diverges to oo.
Section 23
The BolzanoWeierstrass Theorem
Let P(k) be the statement that there is a positive integer ank
nk
>
nk_ 1
55 with
> k. We shall verify that P(k) is true for each positive integer k. Step 1: Since {an} is unbounded above, there is a term, say an 1 with an 1 > 1. Thus P( 1) is true. Step 2: There are infinitely many terms of {an} that exceed 2. Why are there infinitely many? Because if there were only finitely many, then we could choose the largest number in this finite set to be an upper bound of {an}. The reason we need an infinite number of terms is that now we can choose a term an 2 > 2 with n2 > n1 . Induction Step: Now assume P(k) holds, that is, that we have found ank > k with nk > nk_ 1 . Then, as in Step 2, there are infinitely many terms of {an} larger than k + 1, so we can choose a term ank+l > k + 1 with nk+ 1 > nk. Thus P(k + 1) is true. By induction P(k) is true for all k. Thus there is a sequence of positive integers n1 < n2 < n3 < ...
with
ank
> k. This subsequence {ank} diverges to oo. •
Example 212: Let {an} be the sequence {1,!,2,i,3,~,4, ... }. Then the subsequence {1, 2, 3, 4, ... } diverges to oo. Theorem 216 just puts some of our earlier results together to provide yet another way of characterizing convergent sequences. The proof amounts to citing these earlier results and is a useful exercise.
Theorem 216: A sequence {an} converges if and only if it is bounded and has exactly one subsequential limit point. • ~
Since we have adopted the convention that oo and  oo are subsequential limit points for the sequence {an}, if there are subsequences that diverge to those values, then every sequence has at least one subsequential limit point. Using the fact that  oo < L < oo for any real number L, we can make the following definition.
Definition: Let {an} be a sequence of real numbers. Then lim sup an = lim an is the least upper bound of the set of subsequential limit points of {an}, and lim inf an = lim an is the greatest lower bound of the set of subsequential limit points of {an}. • Remark: We shall show in Exercise 16, Section 23 that the supremum of a set of limit points of a sequence is a limit point of the sequence, as is the infimum. Assuming this, we shall now refer to lim an and lim an as the largest and smallest respectively of the limit points of the sequence {an}.
56
Chapter 2
Sequences of Real Numbers
Notice that lim an is not necessarily the largest value of {an}, but the largest of the limit points of {an}. In fact, if {an}
=
{1, !, !, i,
· · ·},
then lim an = 0 even though every term of the sequence is larger than 0. We next characterize lim an and lim an for a bounded sequence {an}.
Theorem 217: Let
{an} be a bounded sequence of real numbers. Then:
(a) lim an = L if and only if, for any a > 0, there are infinitely many terms of {an} in (L B, L +a) but only finitely many terms of {an} with an> L +B. (b) lim an = K if and only if, for any a > 0, there are infinitely many terms of {an} in (K
B,
K +a) but only finitely many terms of {an} with an< K
B.
Proof of (a): Suppose lim an= L. Since Lis a limit point of {an}, we know by Theorem 211 that there are infinitely many terms of {an} in (La, L +a). Thus we need to show that there are only finitely many terms of {an} that exceed L + a for any positive number a. We do the proof by contradiction; that is, we shall suppose that for some a > 0 there are an infinite number of terms of {an} that exceed L + B. We shall then show that there is a subsequential limit of {an} larger than L. This will contradict the hypothesis that L = lim an is the largest of the limit points of {an}. Suppose there is an a> 0 for which there are an infinite number of terms of {an} that exceed L + B. Since {an} is bounded above, say by M, there are infinitely many terms of {an} between L + B and M. From those terms of {an} that exceed L +a, we construct a subsequence {ank}. (We leave the argument of how this may be done until Exercise 5, Section 23.) Now {ank} is a sequence whose terms lie between L+ a and M. Thus {ank} has a convergent subsequence, and since each term of {ank} is at least as large as L + a, this limit is at least as large as L + B. Thus we have shown the existence of a limit point of {an} larger than L, contradicting the hypothesis. Conversely, suppose {an} is a bounded sequence such that for any a > 0 the interval (La, L + a) contains infinitely many terms of {an}, but only finitely many terms of {an} exceed L +B. Since (La, L +a) contains infinitely many terms of {an} for any B > 0, L is a limit point of {an}. Suppo~ M > L. We shall show that M is not a limit point of {an}. This will mean L = lim an. Let a= (M L)/2, (see Figure 27). Then (M a, M +a) contains only finitely many terms of {an} (since every number in (M a, M +a) exceeds L+ a); soMis not a limit point of {an}. • The proof of part (b) is similar and is left for Exercise 7, Section 23.
L
M
Figure 2  7   
M+E
Section 23
The BolzanoWeierstrass Theorem
57
_Sorollary 217: A bounded sequence {an} of numbers converges if and only if lim an= lim an. •
Theorem 218: Let {an} and {bn} be bounded sequences. Then (a) (b)
lim(an + bn) ~lim an+ lim bn. lim an+ lim bn ~ lim(an + bn).
Remark: We first observe that equality does not always hold. If
{an}= {(1t}
then but
and
{bn} = {(1t+ 1 },
lim an= lim bn = 1, an + bn =
Q. for every n.
Thus, lim(an + bn) = 0. (Contrast this with Exercise 8 of Section 13.)
Proof: We give only the proof of the first inequality and leave the proof of the second inequality until Exercise 8, Section 23. Let K =lim an and L= lim bn, and let e > 0 be given. To show that lim(an + bn) ~lim an+ lim bn = K + L, it is enough to show that there are at most finitely many terms of {an + bn} that exceed K + L + e, by Theorem 217. By Theorem 217, there are at most finitely many terms of {an} that exceed K + e/2. Call those that do exceed K + e/2 the nl> ... , n, terms. At most finitely many terms of {bn} exceed L + e/2. Call those that do exceed L + e/2 the m1 , •.. , m. terms. Now if n =f. n1 , ... , n, m1 , . . . , m., then an + bn < ( K +
~) + ( L + ~) = K + L + e. •
Many of our applications of sequences will deal with functions. We close this chapter with such an application that will be used quite often when we discuss integration theory.
Definition: We say that a function f is bounded if the range of f is a bounded set. • Iff is bounded, we denote l.u.b. !J£(f) by sup f and g.l.b. !J£(f) by inff
:».
Theorem 219: Let f and
g be bounded functions with the same domain
Then (a)
(b)
sup(f +g)~ supf +sup g. infj + inf g ~ inf(f +g).
(Compare this with Theorem 218.)
Proof: We prove part (a). The proof of part (b) is similar. We then give an example to show that the equality need not hold. To prove part (a), we know
58
Chapter 2
Sequences of Real Numbers
by Theorem 28 that we can choose a sequence {Yn} such that Yn n = 1, 2, ... and lim Yn For each Yn
E ~(f +g),
there is an Xn Yn
But
=
f(xn)
= E
sup(f +g).
£2(! +g) such that
(f + g)(xn) = f(xn) ~
E ~(f +g),
+ g(xn).
supf and g(xn) ~sup g
for all n. Thus sup{f +g)= lim Yn
~
supf +sup g. •
Example 213: Let f(x)
=
{
0,
0 ~X~ 1
1,
1 0, (L e, L+ e) contains infinitely many points of A. 13. The purpose of this exercise is to prove that every Cauchy sequence is a convergent
Section 23
14.
15.
16.
17.
The BolzanoWeierstrass Theorem
59
sequence. Let {a.} be a Cauchy sequence. (a) Show that {a.} is bounded. (b) Show that there is at least one subsequential limit point for {a.}. (c) Prove there is no more than one subsequential limit point of {a.}. (d) Show that {a.} converges. Show that a convergent sequence is a Cauchy sequence. Let A be an uncountable set of numbers. (a) Show that A has a (finite) limit point. (b) Is it possible for A to have at most a finite number of limit points? Show that the supremum (l.u.b.) of a set of limit points of a sequence {a.} is a limit point of {a.}. Likewise, show that the infimum of a set of limit points of a sequence is a limit point of the sequence. Show that a bounded sequence that does not converge has more than one subsequential limit point.
Topology of the Real Numbers ne of our primary objectives will be to derive some of the properties of continuous realvalued functions on the real numbers. Even though we have not yet defined what a continuous function is, it is likely that you have some idea of the importance of continuous functions from elementary calculus. The study of sets and continuous functions is an area of mathematics called topology. In this chapter we define some of the topological properties that sets can have, and we determine which sets of real numbers have these properties. Later we shall prove some results that describe how continuous functions behave on sets that have these properties. For example, one of the topological properties a set may possess is that of compactness. As a consequence of this property, a continuous function on a closed bounded interval must attain its maximum and minimum values on the interval.
•
Topology of the Real Numbers        O p e n and Closed S e t s       The topological properties with which we shall be concerned are defined in terms of open sets. The property of being open is not an inherent quality of a set, but in the setting in which we shall work, an open set is defined as follows:
Definition: A set U of real numbers is said to be open if, for each x E U, there is a number J(x) > 0 such that (x J(x), x + J(x)) c U • In the definition the number J(x) depends on what x is. The idea is similar to some others we have encountered in that to show a set U is open, we first choose x E U and then show there is a number J(x) (which may be different for different values of x) such that (x J(x), x + J(x)) c U. Now that we have made the point that J(x) is a number that depends on x, we shall frequently use the less cumbersome notation of 0 such that (x b, x +b) c A, then x is said to be an interior point of A. The set consisting of all the interior points of A is called the interior of A and is denoted int(A).
64
Chapter 3
Topology of the Real Numbers
If, for every 13 > 0, the interval (x 13, x + 13) contains a point in A and a point not in A, then x is said to be a boundary point of A. The set of all boundary points of A is called the boundary of A and is denoted b(A). (c) If, for every 13 > 0, the interval (x 13, x + 13) contains a point of A distinct from x, then x is said to be a limit point (or cluster point or accumulation point) of A. (d) A point x E A is said to be an isolated point of A if there is a 13 > 0 such that (x 13, x + 13)nA = {x}. • (b)
The definition of a limit point is equivalent to the one given in Chapter 2 (as was proved in Exercise 12, Section 23).
Example 32: Let A= (0, 1]. Then one can verify that the interior points are (0, 1), the boundary points are {0, 1}, and the limit points are [0, 1].
Example 33: Let A= rational numbers. There are no interior points. Every real number is a boundary point and a limit point.
Example 34: Let A = the integers. There are no interior points. The boundary points are A, and there are no limit points. Each point of A is an isolated point. Theorem 35 gave a fairly tangible characterization of open sets. Unfortunately, no such characterization exists for closed sets, but we shall be able to determine some of their properties. It is often true that the simplest way to prove that a set is closed is to prove that its complement is open.
Theorem 36: A set is closed if and only if it contains all of its boundary points.
Proof: Let A be a closed set, and let x be a boundary point of A. We shall show that x must be in A by supposing that x ¢ A and showing this is impossible. Suppose x ¢A. Then x E Ac and, since A is closed, Ac is open. Thus there is a 13 > 0 such that (x 13, x + 13) c Ac. This is an interval about x containing no points of A, so x cannot be a boundary point of A. Thus a closed set must contain all of its boundary points. Conversely, suppose that A contains all of its boundary points. We shall show that A is closed by showing Ac is open. Pick x E Ac. Then x is not a boundary point of A and x ¢A, so there must be some 13 > 0 such that (x 13, x + 13) contains no point of A. (If x E Ac and if, for every 13 > 0, there was a point of A m (x 13, x + 13), then x would be a boundary point of A.) Thus Ac is open. •
Section 31
65
Topology of the Real Numbers
We make two observations that will be verified in Exercise 15, Section 31. First, if x is a limit point of A and x ¢. A, then x is a boundary point of A. Likewise, if x is a boundary point of A and x ¢. A, then x is a limit point of A. This does not imply that boundary points and limit points are the same, and some of the earlier examples show they are not. The second observation is that if x is a boundary point of A, then xis a boundary point of Ac. These observations make the proofs of the following corollaries almost immediate.
Corollary 36: (a)
(b)
/
A set is closed if and only if it contains all of its limit points. A set is open if and only if it contains none of its boundary points. •
Definition: Let A be a set of real numbers. The closure of A, denoted the set consisting of A and its limit points. •
A, is
By our earlier observations, A could also be defined to be the set consisting of A and its boundary points. Thus in our previous examples, the closure of (0, 1] is [0, 1], the closure of the rational numbers is the real numbers, and the closure of the integers is the integers.
Theorem 37: Let A be a set of real numbers. Then A is a closed set. Proof: We again show that a set is closed by showing that its complement is open. Let x E (A)'. We must show that some interval about x contains no points of A. Now x ¢.A and xis not a boundary point of A, so there is a b > 0 such that (x b, x +b) contains no points of A. We aren't quite done, because we need to show also that some interval about x contains no boundary points of A. It is possible to show that (x b, x +b) contains no boundary points of A, but it is a little easier to argue that (x b/2, x + b/2) contains no boundary points of A. If y E (x b/2, x + b/2), then
(y ~, y+ ~)
c
(x b, x +b).
But (x b, x +b) contains no points of A, so y is not a boundary point of A. •
Compactness Definition: Let A be a set. The collection of sets {I a} a Ed is said to be a cover of A if A c uaEdia. If each set Ia is open, the collection is said to be an open cover of A. If the number of sets in the collection d is finite, then the collection is said to be a finite cover. • Example 35: A = (0, 1), and let In = ( 1/n, 1  1/n) for each integer n larger than 2. Then each set In is open and A c Un>Zin. (Why?)
66
Chapter 3
Topology of the Real Numbers
Example 36: Let A be the set of real numbers, and let our collection of open sets be the collection of all intervals centered at rational numbers of rational radius (i.e., {(x (), x + b)lx is a rational number and (j a positive rational number}). This is a countable collection of open sets and forms a cover of A.
Definition: A set A is said to be compact if every open cover of A has a finite subcover. • Note: An open cover of A, Ua}aE""' has a finite subcover if there is a finite number of sets in the collection that covers A.
Compact sets will be important in studying the properties of continuous functions as we mentioned at the beginning of the section. We want to find conditions that characterize compact sets. We first observe that neither of the sets in Examples 35 and 36 is compact. In Example 35, if
is a finite subcollection of the open cover and N = max {n1 ,
... ,
nk},
then 1/(2N) is not in any of the sets in the finite subcover. In Example 36, if {(xl (jl, X1
+ bd, · · ·, (xn (in, Xn + (jn)}
is a finite subcollection, and if x0 =max{x 1 , . . . ,xn}
and
b0 =max{b 1 , ... ,bn},
then x 0 + 2() 0 is a real number that is not in the finite subcollection. Our task of characterizing compact sets is somewhat involved. We begin by proving a more general form of a result we derived in Theorem 27 concerning nested closed intervals.
Theorem 38: Let {A 1 , A2 , . . . } be a countable collection ofnonempty, closed bounded sets of real numbers such that A; :::J Aj if i ~j. Then nA; # 0. Proof: Since each set A; is nonempty, we can select a number a; E A;. Since A1 :::J A; for every i and A1 is bounded, then {a 1 , a2 , ••. } is a bounded sequence of real numbers. As we showed in Chapter 2, this means there is a subsequence {a;J that converges to some number p. We shall show that pEA; for every i. Let e > 0 be given. Then a;k E
(p  e, p + e)
for k sufficiently large.
But by the nested property of the sets A;, this means (p e, p +e) contains a point of every A;. Thus either pEA; or p is a limit point of A; for every A;. But each A;
Section 31
Topology of the Real Numbers
67
is closed and must therefore contain all of its limit points. Thus p E n A; so
nAd0. • Corollary 38: Let {A 1 , A2 , .•. } be a countable collection of closed bounded sets of real numbers such that A;::::JAj if i 0 such that (a b, a+ b) c U (since U is open). But then (a b, a+ b)n V, #
which contradicts U n V = 0. Second, a¢. V because a¢. V,, and if a E V, there must be an interval V.. in the collection of disjoint intervals whose union is V such that a E V... But then V, n V.. # 0 (why?), which contradicts the assumption that V, and V.. are disjoint. Thus a ¢. U and a ¢. V. Now a< a< b, and if a, bE A and A is an interval, then a EA. This contradicts the assumption that A c U u V. Thus A must be connected. • One can use the property of connectedness to prove the Intermediate Value Theorem for continuous functions. This theorem states that iff is a continuous function on [a, b] and x 1 and x 2 E [a, b] with f(x 1 ) 12 , .•• }, there is an A; with I; c A;. Show that B = u~ 1 A;. (a) Give an example of a set A such that A is not connected, but Au {1} is connected. (b) Show that in part (a) 1 must be a boundary point of A. Prove that the following two conditions are equivalent: (i) Every infinite set of points in A has a limit point in A. (ii) If {x.} is a sequence of points in A, there is a subsequence {x•• } that converges to some point in A. Show that a closed subset of a compact set is compact. (a) Show that if A is compact and B is closed, then An B is compact. (b) If A is compact and B is bounded, must An B be compact? Justify your answer. Show that a compact set contains its supremum and infimum. Show that a set A is open relative to V if, given any x 0 E A, there is a b > 0 such that
(x 0

b, x 0 + b)n VcA.
Continuous Functions
Limits and Continuity In this section we define continuous functions and determine some of their properties. First we define what it means for a function to be continuous at a point and then what it means for a function to be continuous on a set. We shall find that there are two ways in which a function can be continuous on a set. Throughout this chapter (and the rest of the text), unless otherwise stated, f will be a function from a subset of the real numbers into the real numbers and f0(f) will denote the domain of f.
        L i m i t of a FunctionThe property of continuity is intimately connected with the idea of the limit of a function, and this will be our starting point.
Definition: Let f be a function with domain ::0(!), and suppose that x 0 is a limit point of ::0(!). The limit off as x approaches x 0 is L if, given any 8 < 0, there is a 6(8) > 0 such that if 0 < lx x0 1< 6(8) and x E ::0(/), then lf(x) Ll < 8. In this case we write limx__,xJ(x) = L. • The idea is that f(x) may be made arbitrarily close to L by making x sufficiently close to x 0 • The numbers 8 and 6(8) should be thought of as very small numbers. Again the order of selection of the two numbers is important. First 8 is chosen, and then 6(8) is found. Usually, 6(8) will depend on 8. Two other points are important: First, x 0 need not be in f0(f) but only a limit point of ::0(!). Thus f(x 0 ) need not be defined. Second, even if x 0 E f0(f), it is not required that f(x 0 ) = limx__,xJ(x). In fact, the value off at x 0 is completely irrelevant in view of the condition 0 < Ix  x 0 1. Figure 41 presents these ideas from a graphical viewpoint. 73
74
Chapter 4
L+e
Continuous Functions
I
I
1
I
~~
L
I
Le
~1
I I I I
•
I I
Figure 4  1   
In
the proof of Theorem 41, we shall want to suppose that L. This is equivalent to saying that there is some positive number c: 0 such that, for any (j > 0, there is a number x for which limx~xJ(x) =1
X E
with lf(x) Ll
(x 0  b,
Xo
+b)() f0(f),
X
=f. Xo
~ B0 •
Example 41: Let f(x) = (2x 2  8)/(x 2). We shall show that limx~d(x) = 8. Notice that 2 ¢ f0(f). Let c: > 0 be given. We must find a b(c:) > 0 such that 2x 2  8 I 1   8 x2
1, then I~ 01 x0 /2, which will be true if lx x0 1< lxol/2. Then IXo  xI 2 IXo  xI < =lxxol· lxllxol lxo/211xol lx61 Now if lx x 0 1< Blx61/2, then 2 2 Blx61 lx6llx xol < lx61. 2
=B.
Thus if b(B) =min {Bix61/2, lxol/2} and if lx x 0 1< b(B), then
1 ~_I_I 0 such that if lx x 0 1< b and x, x 0 E (0, 1), then
I ~]_~ X
0 and x 0 E A, there is a number J(x 0 , e) such that if x E :!iJ(f) and lx x 0 1< b(x0 , e), then lf(x) f(x 0 )1 0.
Since f is continuous at x 0 , there is a b > 0 such that if Ix  x 0 I < b and x E :!iJ(f), then lf(x)f(x 0 )l 0 be given. Then (f(x 0 )  e, f(x 0 ) +e) is an open set. The hypotheses state that f 1 (f(x 0 )  e,f(x 0 ) +e)= {x E :!iJ(f)lf(x) E (f(x 0 )  e,f(x 0 ) +e)} is an open set relative to :!iJ(f). Now x 0 Ej  1 (f(x 0 )  e,f(x 0 ) +e)
84
Chapter 4
Continuous Functions
so there is an interval (x 0 
~. x 0
+ ~)
such that
(x 0  ~. x 0 + ~)n~(f) cf 1 (f(x 0 )  s,f(x 0 ) +e) for some~> 0. But this says that if lx x 0 1 0, then (a)
f 1 ((a, b))= [0, b2 ) (c)
which is open relative to ~(f), since [0, b2 ) = (b2 , b2 )n [0, oo ). For a~ 0, f 1 ((a, b))= (a 2 , b2 ). Thus if A= u(a;, b;), then
f 1 (A) =
uf 1 ((a;, b;))
is an open set. We now have three ways to characterize a continuous function. We summarize these in a theorem.
Theorem 46: Let f be a function with domain ~(f). The following are equivalent to the condition that f is continuous on ~(f). (a)
Given any B > 0 and x 0 E ~(f), there is a number ~(B, x 0 ) > 0 such that if lx x0 1< ~(B, x0) and x E ~(f), then lf(x) f(x 0 )1 0, then there is
a~>
0 such that f(x) > 0 if x
E
(c ~. c + ~)n~(f). •
Corollary 48: Iff is a continuous function at x = c and if f(c) < 0, then a~> 0 such that f(x) < 0 if x E (c ~. c + ~)n~(f). •
there is
We prove a stronger version of Theorem 48 and Corollary 48 in Exercise 12, Section 41.
Theorem 49: Iff is a continuous function on [a, b] and if f(a) and f(b) are of opposite signs, then there is a number c E (a, b) for which f(c) = 0. Proof: We do the proof in the case f(a) > 0 and f(b) < 0. The proof of the other case is nearly identical. Let A= {tlf(x) > 0 if x E [a, t]}. Now A =1 0 since a E A, and b is an upper bound of A. Let c = l.u.b. A. Note that a< c < b, because f is continuous and f(a) > 0 andf(b) < 0 so there is a 8 > 0 such thatf(x) > 0 on [a, a+ 8) andf(x) < 0 on (b  8, b] by Theorem 48 and its corollary. We claim f(c) = 0. Suppose this is not the case. If f(c) < 0, then there is a~> 0 such that f(x) < 0 if x E (c ~. c + ~). But then c (~/2) would be an upper bound of A, contradicting c = l.u.b. A. If f(c) > 0, then there is a ~ > 0 such that f(x) > 0 if x E (c ~. c + ~). If this is the case, then f(x) > 0 on [a, c (~/2)] and f(x) > 0 on (c ~. c + ~) so that f(x) > 0 on [a, c + (~/2)]. This contradicts that c is an upper bound of A. • Corollary 49 (The Intermediate Value Theorem): Iff is a continuous function on [a, b] and a is a number between f(a) and f(b), then there is a number c E (a, b) where f(c) =a. Proof: For definiteness, assume f(a) > f(b). Then f(b) 0 and g(b) = f(b) a< 0. Thus by Theorem 49, there is a number c E (a, b) for which g(c) = f(c) a= 0; that is, f(c) =a. • Definition: Let f be a function defined on an interval I. We say that f has the intermediate value property if whenever x 1 , x 2 E I with f(xd =1 f(x 2 ), then for any number a between f(xd and f(x 2 ) there is ar.. x 3 between x 1 and x2 with j(x3) =a. • It may seem reasonable to conjecture that a function that satisfies the intermediate value property must be continuous. However, this is not the case, as the function
f(x)= { shows. (See Exercise 6, Section 41.)
sin(l/x),
x=f0
0,
x=O
Section 41
Limits and Continuity
87
f c
a
b
Figure46
Uniform ContinuityAs a final point in this section, we return to an earlier example in which we considered the function f(x) = 1/x. We showed that this function was continuous on (0, 1), but as the point x 0 became closer to 0, the number x 0 eX~} ~(e)=min { 2'2
became smaller even with B fixed. Furthermore, we showed that no value of ~(e) can work for every x0 E (0, 1), even with B fixed. In this example suppose we had restricted our set to be [!, 1) instead of (0, 1). Then
. {x2'2 eX~} . {1 e} ~mm 4'8
~(e)=mm
(taking x 0 = !, the smallest it can be). Now we do have a situation where it is possible to get a ~(e) that works for every x in the set, once B is chosen. This difference of behavior turns out to be very important. One use of it occurs in our study of integration.
Definition: Let f be a function from a set A to the real numbers. We say that f is uniformly continuous. on A if, given B > 0, there is a ~(e)> 0 such that if x, yEA and lx Yl 0 be given. Since f is continuous on A, given x E A, there is a number b(x) such that if yEA and lx yi < b(x), then lf(x) f(y)i < ej2. Now if it were possible to choose ;5 = inf{.5(x)lx E A} so that ;5 > 0, then we would be done. However, we are taking the infimum of an infinite set of positive numbers, which may be 0. If we could somehow reduce this to a finite set of positive numbers, then we could then select the minimum of this set, which would be positive. This is what compactness allows us to do. For each x E A, b(x)) ( x .5(x) 2 ,x+ 2
Figure 4  7   
Section 41
Limits and Continuity
89
is an open set containing x, so
is an open cover of A. Since A is compact, some finite subcollection, say
covers A. Let
. {~12'''''2~n} .
~=mm Now~>
0, since this set is finite. We claim that if x' and x" then lf(x') f(x")l 1 if x is irrational or x = 0 m n
if x =  in lowest terms.
Use sequences to show that f is not continuous at the nonzero rational numbers. 4. Prove parts (a) and (c) of Theorem 43. 5. Suppose that f, g, and h are functions such that f(x),:::; g(x),:::; h(x) for every x except possibly x =a. Show that if limf(x) =lim h(x) = L, then limx~.g(x) = L. 6. Show that the function f(x)
={
sin(1/x),
x *0
0,
x=O
satisfies the intermediate value property on [0, 1]. 7. (a) Show that x" is continuous for any positive integer n. (b) Prove that any polynomial f(x) = a.x" + a._ 1 x" 1 + · · · + a 0 is continuous. 8. Give examples of the following. (a) A function on an interval [0, 1] that does not assume its supremum. (b) A continuous function on [ 1, oo) that does not assume its infimum. (c) A continuous function on (0, 1) that does not assume its supremum or infimum. 9. Show that any function whose domain is the integers is uniformly continuous. 10. Show that f(x)
={
*0
x sin( 1/x),
x
0,
x=O
is continuous at x = 0. 11. (a) Show that if the function f is continuous, then If I is continuous. (b) Give an example where If I is continuous, but f is not continuous.
12. Suppose that f is a continuous function whose domain is the real numbers. (a) If f(a) > 0, show there is a D> 0 such that f(x)): !/(a) if x E (aD, a+ D). (b) If f(a) < 0, show there is a D> 0 such that f(x),:::; !/(a) if x E (aD, a+ D). 13. (a) Show that f(x) =sin x is a continuous function. You may use the fact that Isin xl,:::; lxl. (b) Show that f(x) =cos x is a continuous function. (c) Show that f(x) =tan x is continuous for x E (  n/2, n/2 ).
Section 41
Limits and Continuity
91
14. Show that f(x) = 3x 3 +sin x 1 has a root (i.e., a value of x where f(x) = 0) between 1 and 1.
15. (a) Let f: [0, 1]> [0, 1] be continuous. Show there is a x E [0, 1] where f(x) = x. (b) Show there is a value of x for which cos x = x. (c) Suppose f is a continuous function on [0, 2] with f(O) = f(2). Show there is an x E [0, 1] where f(x) = f(x + 1).
16. Show that if a> 0 and n is a positive integer, there is a number b for which b" =a. 17. (a) Show that a function f is continuous if and only if for each closed set A, f 1 (A) is closed relative to £»(!). (b) Show that iff is a continuous function, then {xlf(x) =a} is a closed set relative
to :?.&(f) for any real number a. 18. (a) Suppose f and g are continuous functions on ~ with f(x) = g(x) except possibly at xi> ... , x •. Show that f(x;) = g(x;) for i = 1, ... , n. (b) Show that iff and g are continuous functions on~ with f(x) = g(x) for any rational number x, then f(x) = g(x) for all x E ~. 19. Suppose f is continuous on [a, b] with f(a) = f(b). Let
M=Lu.b.{f(x)lxE[a,b]} and m =g.!. b. {f(x)lx E [a, b]}.
Show that if m < c < M, then there exist xi> x 2 E [a, b], x 1 f=. x 2 , where f(xd = f(x 2 ) =c. 20. Suppose f is a continuous function on ~. (a) Show that g(x) = f(x +c) is continuous on ~. (b) Show that h(x) = f(ax +b) is continuous on ~. 21. Show that f(x) =ax is a continuous function for any a> 0. 22. Suppose f is uniformly continuous on (a, b) and is continuous at x =a and x =b. Show that f is uniformly continuous on [a, b]. 23. Suppose f is continuous on (a, b) and that the function F defined by
limx~af(x)
f(x), F(x) = { ~i~f(x),
if x
E
and
limx~bf(x)
exist. Show that
(a, b)
if x =a
limf(x), if x = b x~b
is uniformly continuous on [a, b]. 24. (a) Show that iff is uniformly continuous on a bounded interval J, then f is bounded on I. (b) Give an example of a continuous function on (0, 1) that is not bounded. (c) Give an example of a bounded continuous function on (0, 1) that is not uniformly continuous. 25. (a) Show that f(x) = x is uniformly continuous on ( oo, oo ). (b) Show that f(x) = x 2 is not uniformly continuous on ( oo, oo ). Thus the product
of uniformly continuous functions is not always uniformly continuous. (c) Show that the sum of two uniformly continuous functions is uniformly continuous. (d) Show that the product of two uniformly continuous functions on a bounded interval
is uniformly continuous.
92
Chapter 4 Continuous Functions
26. Show that iff is a continuous function on [a, b], then f([a, b]) is either a point or a closed, bounded interval. 27. Show that iff is uniformly continuous on (a, b) and {x.} c (a, b) is a Cauchy sequence,
then {f(x.)} is a Cauchy sequence. 28. (a) Give an example of a function f that is continuous on (0, 1) and a Cauchy sequence {x.} c (0, 1) for which {f(x.)} is not a Cauchy sequence. (b) Show that iff is a function that is continuous but not uniformly continuous on (a, b), then there is a Cauchy sequence {x.} c (a, b) for which {f(x.)} is not a Cauchy sequence. (c) Is the result in part (b) necessarily true if the interval is (a, ro )? 29. (This problem uses the notion of connectedness.) (a) Show that iff is a continuous function and A is a connected set, then f(A) is a connected set. (b) Use the result of part (a) to prove the Intermediate Value Theorem.
30. Suppose f and g are continuous functions on [a, b]. Define (f v g)(x) =max {f(x), g(x)}
(fA g)(x) =min {f(x), g(x)}. Show that f v g and fAg are continuous on [a, b]. 31. A function f is said to be convex on [a, b] if for x1o x 2 E [a, b] and 0 ~A.~ 1,
(a) Give an example of a convex function. (b) Show that a convex function is continuous.
Monotone and Inverse Functions In this section we continue our examination of limits and continuous functions. We begin by defining the limit of a function when either the independent variable or the function becomes infinite. We then define the left and right limit of a function, which leads to the idea of left and right continuity. The idea of onesided continuity is important for several reasons, and we shall use it to help classify the ways in which a function may be discontinuous. We then investigate increasing and decreasing functions and find there is only one type of discontinuity that such functions can have. Finally, we use this information to examine the properties of the inverse of a continuous 11 function.
Limits Involving InfinityIn Section 41 we defined what it means to say, "The limit of the function f as x approaches x 0 is L." In that definition x 0 and L were assumed to be real numbers. We now extend the idea of the limit of a function to the cases where either x 0 or L may be infinite.
Section 42
Monotone and Inverse Functions
93
Definition: Let f be a function with domain !0(!), and suppose that x0 is a limit point of !:0(!). We say the limit off as x approaches x 0 is oo if, given any number M, there is a number ~(M) > 0 such that if 0 < lx x0 1< ~(M) and x E !0(!), then f(x) > M. • In this case we write limx~xof(x) = oo. In a similar manner we can define what it means for the limit off as x approaches x 0 to be  oo. It is a useful exercise to formulate the definition. The number M in the definition should be thought of as a very large positive number and ~(M) as a very small number. Figure 48 illustrates this idea. The order of selection of the two numbers is that first M is chosen and then ~(M) is found; ~(M) will depend on M. The relationship between M and ~(M) is that as M becomes larger, ~(M) becomes smaller.
Example 411: Let f(x) = 1/(x 1f Then !0(!) = {x lx =f. 1}. We shall show limx~d(x) = oo. Choose any number M > 0. (If M ~ 0, any number works for ~(M).) If 0 < lx 11 < 1//M, then f(x) = (x 1)2 > M
which is what we needed to show. Thus we could take ~(M) = 1//M.
Definition: Let f be a function with domain !0(!) such that !:0(!) n (M, oo) =f.
for any number M.
Then we say the limit off as x goes to oo is L provided that, given any e > 0, there
M
Figure48
94
Chapter 4 Continuous Functions
is a number N(B) such that if x > N(B) and x
E ~(f),
then lf(x) Ll N(B) no matter how large N(B) is.
f as
Example 412: We show that x 2 1 lim  2 1 = 1.
x~oo X
+
Let B> 0 be given. We need to find a number N(B) such that if x > N(B), then
+ 1 1 I )(2/B) 1. (We assume B< 2, so that (2/B) 1 > 0.) Thus we may take N(B) = )(2/B) 1.
Definition: Let f be a function with domain ~(f) such that ~(f) n (M, oo) # 0 for any number M. We say the limit off as x approaches oo is oo provided that, given any number K, there is a number N(K) such that if
N(E)
Figure 4  9   
Section 42 x > N(K) and x
E
Monotone and Inverse Functions
95
!!/J(f), then f(x) > K. •
In this case we write limx_.oof(x) = oo. Likewise, we could define lim f(x) =  oo, x+oo
lim f(x) = oo,
and
x+oo
lim f(x) =  oo. x+oo
In the definition, the numbers K and N(K) should both be thought of as very large positive numbers. First K is chosen; then N(K) is found and will depend onK.
Example 413: We show that limx_.oo[; = oo. Choose K > 0. We must find a number N(K) such that if x > N(K), then[;> K. But if x > K 2 , then[;> K, so we could take N(K) = K 2 •
Right and LeftHand LimitsNext we discuss the notions of right and lefthand limits, which may be familiar topics from elementary calculus. Here, we look at the behavior of a function as x approaches a given point specifically from the right or the left. As was mentioned in the introduction, we shall use these properties to help classify the types of discontinuities that a function may have.
Definition: Let f be a function with domain !!/J(f). Let x0 be a limit point of !!/J(f) n [x 0 , oo ). We say the limit off as x approaches x 0 from the right is L if, given any e > 0, there is a b(e) > 0 such that if 0 < x x 0 < b(e) and x E !!/J(f), then lf(x) Ll 0, there is a b(e) > 0 such that if 0 < x 0  x < b(e) and x E !!/J(f), then lf(x) Ll 0, there is a c5(e) > 0 such that if 0:;;;; x x 0 < c5(e) (0:;;;; x 0  x < c5(s)) and x E fiiJ(f), then lf(x) f(xo)l f(x 2 )), f is strictly monotone increasing (decreasing). A function that is either monotone increasing or decreasing is said to be monotone. •
Monotone functions have some nice properties that arbitrary functions do not necessarily possess. Your understanding of some of these properties may become more intuitive if we examine the graph of such a function. (Later we shall prove, in theorems or exercises, that our ideas were correct.)
Figure 411
Suppose f is an increasing function as shown in Figure 411. It seems reasonable that either the graph has no breaks (is continuous) or, if a break does occur, then it must be similar to what happens at x 0 in Figure 411. In this case, limf(x) xjx 0
and
limf(x) x~x 0
98
Chapter 4
Continuous Functions
both exist and are rx and /3, respectively. Also, rx < f3 and f(x) does not assume any value between rx and f3 except f(x 0 ). Thus, iff is not continuous at some value, then it will not have the intermediate value property. It also appears that a strictly monotone function must be 11.
Theorem 413: Let f be a monotone function with domain an open interval (a, b). Then limxtxof(x) and limx~x 0 f(x) exist for each x 0 E (a, b). Proof: We do the proof in the case that f is an increasing function. Choose x 0 E (a, b). We shall show limxtxof(x) exists. Since f is increasing, if a< x < x 0 , then f(x) ~ f(x 0 ). Thus f(x 0 ) is an upper bound for the set A= {f(x)la < x < x 0 }.
Since A is bounded above, it has a least upper bound rx. We shall show limxtxof(x) = rx. (Notice rx ~ f(x 0 ).) Let 8 > 0 be given. By definition of least upper bound, there is an x 1 with a< x 1 < x 0 such that rx 8 < f(xd ~ rx. Since f is increasing, if x 1 < x < x 0 , then f(xd ~ f(x) ~ rx, so that rx 8 < f(x) ~ rx. Thus if x 1 < x < x 0 , lf(x) rxl < 8. Take b(8)=x0 x 1 • If0x 0 } = limf(x). X~Xo
Thus iff is an increasing function, there is a discontinuity at x 0 if and only if f3 > rx. If this is the case, there is a rational number y(x 0 ) between rx and f3. Notice also that iff is an increasing function with discontinuities at x 0 and x 1 with x 0 < xl> then /Lg.l.b.{f(x)lx>x 0 }
~l.u.b.{f(x)lx 1,
then f has left and righthand derivatives at x = 1, but is not differentiable at x = 1. 5. Here we derive the formulas for the derivatives of the trigonometric functions. (a) The area of a sector of a circle of radius r sub tended by an angle e is given by !r 2 e, and the length of the arc subtended is given by re, where e is measured in radians. Consider a circle of radius 1 with central angle e. Use Figure 52 to argue that
! sin e cos e < !e < ! tan e. (b) Show that
lim eo
1 cos e
e
= 0.
(c) Show that
1 cos e lim::
(d) Use the fact that limeo cos
e= 1 to show
e2
sine lim= 1. eo e (e) Use the formula sin(()(+ {3) =sin
I)( cos f3 +sin f3 cos I)( to show that if f(x) = sin(x), then f'(x) =cos x. (/) Use the fact that cos x = sin(n/2 x) to show that if f(x) =cos x, then f'(x) = sin x. (g) Derive the formulas for the other trigonometric functions. 6. Show that d/dx sin x = (n/180) cos x if x is measured in degrees. 7. Show that iff is differentiable on (a, b), and f'(c) > 0, where c E (a, b) and iff' is continuous at x = c, then there is an interval about c on which f is increasing.
114
Chapter 5
Differentiation
(1, tan ll)
Figure 5  2   8. (a) Suppose lf(x +h) f(x)l ~ Kh" for some constant K and IX> 0. Show that continuous. (b) Suppose lf(x +h) f(x)l ~ Kh" for some constant K and IX> 1. Show that differentiable and f'(x) = 0. (c) Suppose lf(h) f(O)I ~h. Must f be differentiable at x = 0? Hint: Consider f(x) = {
!x,
if x is rational
x,
If x IS urattonal.
.
. .
f
is
f
is
.
9. Verify that if
x _ {x + 2x 2 sin( 1/x), f( )  0,
if x # 0 'f = 0,
1 X
then f'(x) = 1 if x = 0. Also show that f'(x) takes on positive and negative values arbitrarily close to 0 and that f(O) < f(x) if x < !. 10. Show that
f(x)
={
x 2 sin( 1/x),
x# 0
0,
x=O
is differentiable, but the derivative is not continuous at x 11. For what values of IX is the function x•, { f(x) = 0, differentiable at x 12. Show that
= 0.
if x is rational . uratton . . a! I'f x Is
= 0? Show that 0 is the only point at which f can be differentiable.
f(x)
={
x sin(1/x),
if x # 0
0,
if
is continuous but not differentiable at x
= 0.
X=
Section 52
Some Mean Value Theorems
115
13. Suppose f and g have nth order derivatives on (a, b). Let h = f ·g. Show that, for c E (a, b), h(n)(c) =
I (nkl\,(k)(c)g f(a). Then there is some c E (a, b) where f(c) = M and thusfhas a relative maximum at x = c so f'(c) = 0. (A similar argument works if m < f(a).) • In the exercises we construct examples that show Rolle's Theorem fails if any of the hypotheses is omitted.
Theorem 58 (Mean Value Theorem): Suppose f is continuous on [a, b] and differentiable on (a, b). Then there is a number c E (a, b) for which f'(c) = f(bi
=~(a).
Proof: First notice that [f(b) f(a)]/(b a) is the slope of the line joining the points (a, f(a)) and (b, f(b)). We are to prove there is a point c at which the slope of the tangent line is parallel to this line. (See Figure 53.)
Figure 53Let g(x) = f(x) [f(bi =~(a)] x. The graph of
[ f(b) f(a)J x ba
is a straight line passing through the origin parallel to the line joining (a, f(a)) and (b,f(b)), so it appears that g(a) = g(b). (See Figure 54.) We show that this is the case. Now
Section 52
Some Mean Value Theorems
117
[ f(b)  f(a)]x ba
a
b
Figure 5  4   g(a) = f(a) _ [f(bi
=~(a)] a= bf(ai =;(b)
g(b) = f(b) _ [f(b) f(a)J b = bf(a) af(b). ba ba
and
Also g is continuous on [a, b] and differentiable on (a, b), since f satisfies these hypotheses. Thus g(x) satisfies the hypotheses of Rolle's Theorem, so there is a point c E (a, b) where g'(c) = 0. g'(c) = f'(c) [f(b) f(a)J = 0 ba
But
f'(c) = f(bi
so
=~(a).
•
Corollary 5B(a): Suppose f is continuous on [a, b] and differentiable on (a, b). If f'(x) on (a, b), then f is constant on [a, b].
=
Proof: Choose c E (a, b]. We shall show that f(c) = f(a). By the Mean Value Theorem, there is a point rx E (a, c) with f(c) f(a) = f'(rx)(c a).
But f'(rx)
=
0, so f(c)
=
f(a). •
Corollary 5B(b): Suppose f and g are continuous on [a, b] and differentiable on (a, b). If f'(x) = g'(x) on (a, b), then f and g differ by a constant.
Proof: Let h(x) = f(x) g(x). Then h satisfies the hypotheses of Corollary 58(a), so h(x) is constant. •
118
Chapter 5
Differentiation
Example 55:
Suppose f and g are differentiable functions with f(a) = g(a) and f'(x) < g'(x) for x >a. We shall show that f(b) < g(b) for b >a. To do this, define h(x) = g(x) f(x). Then h(a) = 0 and h'(x) > 0 for x >a. Thus h(b) h(a) = h'(c)(b a) for some c E (a, b).
Now h'(c) > 0, (b a)> 0, so h(b) > h(a) = 0. But h(b) = g(b) f(b). Example 56:
We show that ln(1 + x) < x for x > 0. Let f(x) = ln(1 + x), g(x) = x. Then f(O)=g(O)=O,
1 f'(x)= 1 +x'
g'(x)= 1
so f'(x) < g'(x) for x > 0. Thus, by Example 55, f(x)=ln(1+x)O.
Example 57:
Iff is differentiable at x = c and g is differentiable on an interval containing f(c +h) for h sufficiently small, and if g' is continuous at f(c), then we can use the Mean Value Theorem to give an easy proof of the Chain Rule. For h sufficiently small
g(f(c +h)) g(f(c)) _ , [f(c +h) f(c)J h g (8 ) h
for some 8 between f(c +h) and f(c) by the Mean Value Theorem. Now f is differentiable at c, so . f(c +h) f(c) =f'() 1lm h c. h~o
Also f is continuous at c, so limh~of(c +h)= f(c), and thus limh~o 8 = f(c) (since 8 is between f(c +h) and f(c)). Now g' is continuous at f(c), so lim g'(8) = g'(f(c)). h~o
Thus
g'(f(c))f'(c) =lim g'(8) lim f(c + h~o
h~ f(c)
h~o
=lim g'(8) [f(c +h) f(c)J h~o h =lim g(f(c +h)) g(f(c)) =(go f)'( c). h~o h
Section 52
Some Mean Value Theorems
119
Theorem 59 (Generalized Mean Value Theorem): Suppose f and g are functions that are continuous on [a, b] and differentiable on (a, b). Then there is a point c E (a, b) where [f(b) f(a)]g'(c)
=
[g(b) g(a)]f'(c).
Remarks: Notice that this reduces to the Mean Value Theorem in the case g(x) = x. The geometric interpretation of the theorem is as follows: If we describe a curve parametrically in the xy plane by x=f(t), y=g(t) for t E [a, b], then there is a point on the curve where the slope of the line tangent to the curve is parallel to the line joining the points (f(a), g(a)) and (f(b), g(b)). This curve is not necessarily the graph of a function. Proof: We want to define a function h(x) that satisfies the hypotheses of Rolle's Theorem so that the conclusion of Rolle's Theorem will provide a proof of this theorem. To this end, define h(x) = f(x) o:g(x)
where ex is a constant, which will be chosen to force h(a) = h(b). Now h(a) = f(a) o:g(a)
and
h(b) = f(b) o:g(b).
So if h(a) = h(b), then f(a) o:g(a) = f(b) o:g(b)
or f(b) f(a) = o:(g(b) g(a)).
Now if g(b) = g(a) and f(b) # f(a), then no such ex exists. However, if g(b) # g(a), then we let f(b) f(a) ex= g(b) _ g(a)
and so
h(x) = f(x)
[f(b) f(a)J g(b) _ g(a) g(x).
Now h(x) satisfies the hypotheses of Rolle's Theorem, so there is a number c E (a, b) where h'(c) = f'(c) [f(b) f(a)J g'(c) = 0 g(b) g(a)
f' (c) [g(b) g(a)] =
or
g' (c) [f(b) f(a)].
Now suppose g(b) = g(a). Then g satisfies the hypotheses of Rolle's Theorem, so there is a point c E (a, b) where g'(c) = 0. Then f'(c)[g(b) g(a)]
=
g'(c)[f(b) f(a)]
because both sides of the equation are 0. •
F
Example 58: 4x3

7x 2 and g(x) = x 4  5 on [0, 2]. Then f(O) = 0, f(2) = 4, g(O) = 5, f'(x)
=
12x 2  14x,
g(2) = 11,
and g'(x) = 4x3 .
Chapter 5
120
Differentiation
Theorem 59 guarantees a value of x between 0 and 2 where (12x 2 14x)[ll(5)] =4x 3 (40).
This value is 6
Jii.
Taylor's TheoremWe next discuss a result known as Taylor's Theorem that gives another application of the Mean Value Theorem and will preview a later topic, Taylor series. Suppose that f is a function with derivatives f', f", ... ,J 0 and g(x) > 0 if x E (a, a+ b).) Thus f(xd
g(xd
=
f'(xo) [ 1 g(xz)/g(xd] g'(xo) [1 f(xz)/ f(xl)] ·
(2)
Now hold x 2 fixed, and let x 1 vary. Since lim g(xd = limf(xd = oo x 1 !a
x 1 !a
then lim 1  g(x 2 )/g(xd 1  f(xz)/ f(xd
So
=
1.
(3)
x,!a
1  g(x 2 )/g(xd
Now let
h(xd = 1 f(xz)lf(xd·
(Remember x 2 is fixed.) Now by (3) there is a b2 > 0 such that if 0 < x 1  a< b2 (i.e., x 1 E (a, a+ b2 )), then lh(xd11 < 2(1LI +(e/3))' Now by (2), f(xl) LI l g(xd
=
lf'(xo) ·h(xt)LI g'(xo)
::::;; If'(xo). h(xd f'(xo) I + If'(xo) L I g' (xo) g' (xo) g' (xo) =
lf'(xo)llh(xd11+ lf'(xo) LI g' (xo) g' (xo)
c, then f has a relative maximum at x = c. (ii) Show that if f'(x) < 0 for x < c and f'(x) > 0 for x > c, then f has a relative minimum at x =c. (b) Find the relative extrema of f(x) = x 213 (8 x) 2 on [ 10, 10], and classify them as maxima or minima.
130
Chapter 5
Differentiation
5. (a) Show that Isin x sin Yl ~ lx Yl· Note that this implies Isin xl ~ lxl(b) Use part (a) to show cos x;;:: 1 x 2 /2 for x > 0. (c) Use part (b) to show sin x;;:: x x 3/6 for x > 0. (d) Use part (c) to show cos x ~ 1 x 2 /2 + x 4 j24 for x > 0. 6. (a) Suppose f"(x) exists and is continuous on (a, b) and c E (a, b). Show that , . f(c+h)2f(c)+ f(ch) hm h2 =!(c).
h0
(b) Let f(x) = xlxl. Show that
. f(O +h) 2/(0) + f(O h) hm h2 =0
h0
7.
8.
9.
10.
but f"(O) does not exist. (a) Suppose f is a function whose derivative exists for every x, and suppose that f has n distinct roots. Show that f' has at least n 1 distinct roots. (b) Is it possible that f' has more roots than f? Find a bound on the maximum difference of sin x and x x 3 /6 + x 5/120 on [0, 1]. [Hint: In Taylor's Theorem, let a= 0 and x = 1. What is the maximum value that [(j(c))/(n + 1)!](x a)"+ 1 can take on?] Use Taylor's Theorem to find accurate to .0001. Use L'H6pital's Rule and any formulas from elementary calculus you wish to evaluate the following limits:
J8
(a)
lim
cos x  1 + x 2 /2  x 4 j24
x+0
(b)
lim (ezx x)1fx. xo
X
6
(c) lim x 2 fx.
(e)
lim xnf2
(!: 2
x) tan x.
ax+b 11. Let f(x) =   d (ad be =I= 0). ex+ (a) Show f is a 11 function and find f 1. (b) For what values of a, b, c, and d (if any) does f 1 =
f
12. Letf(x)=x 7 +x 5 +x 3 +x. Find (f 1)'(4). 13. For n/2 ~ x ~ n/2, sin x is a 11 function. Define f(x) = sin x on [ n/2, n/2]. Show that
(r1)'(x)= _1__
.J1 x
2
14. Suppose one knew that iff is 11 and differentiable with f' (x) =I= 0 on (a, b), then f  1 is differentiable on its domain. Use the Chain Rule to find a formula for (f 1)'(f(x0 )). 15. Suppose f is differentiable on (a, b) and c E (a, b). (a) Show that
r
~~
f(c +h) f(c h)= f'( ) 2h c.
Section 52
Some Mean Value Theorems
131
(b) Show that
f(x) f(c) = f'(c)(x c)+ g(x)(x c)
where lim g(x) = 0. x~c
16. Suppose f' exists on (a, b) and c E (a, b). Show that there is a sequence {x.} c (a, b), x. # c where {f'(x.)}+ f'(c). 17. Suppose that f' exists and is increasing on (0, co) and that f is continuous on [0, co) with f(O) = 0. Show that g(x) = [f(x)]/x is increasing on (0, co). 18. Here we show that the derivative of a function must satisfy the intermediate value property. Suppose f is differentiable on an open interval containing [a, b] and f'(a) < f'(b). Let fJ be between f'(a) and f'(b). We shall show that there is a c E (a, b) where f'(c) =e. (a) Define a function g on [a, b] by
g(x)= {
f(x) f(a), xa f'(a),
x#a x=a.
Show that g(x) attains every value between f'(a) and [f(b) f(a)]/(b a). (b) Define a function h on [a, b] by
h(x)= {
f(b) f(x), bx
x# b
f'(b),
x=b.
Show that h(x) attains every value between f'(b) and [f(b) f(a)]/(b a). (c) Let y = [f(b) f(a)]/(b a). Our proof will assume f'(a) < y < f'(b), but the idea of the proof is valid for any relationship for y, f'(a), and f'(b). Choose fJE(f'(a),f'(b)). Show that if fJE[y,f'(b)], there is a number x 0 E[a,b] where h(xo) =e. (d) Show that if f) E [f'(a), y], there is a number Xo E [a, b] where g(xo) =e. (e) Use the Mean Value Theorem to show there is a number c E [x 0 , b] where f'(c) = h(xo) = fJ or a number C E [a, Xo] where f'(c) = g(xo) = fJ. 19. Show that if f(x) is a function whose derivative f'(x) is monotonic, then f'(x) is continuous.
·Integration n this chapter we study the theory of integration. In the first section we define the Riemann integral of a bounded function defined on a closed, bounded interval. We shall see that the integral of such a function does not always exist. One of the main results of the first section is the determination of a condition that characterizes Riemann integrable functions. That is, we shall find a condition such that a function is Riemann integrable if and only if it satisfies this condition. In Section 62 we derive some of the properties of the Riemann integral and prove some of the computational devices of elementary calculus, including the Fundamental Theorem of Calculus and the integration by parts formula. There are many kinds of integrals. The Riemann integral was historically the first one invented and is perhaps the simplest. In Section 63 we study a more general integral, the RiemannStieltjes integral. We say that this is a more general integral because the Riemann integral is a special case of the RiemannStieltjes integral. Also, there are applications where the RiemannStieltjes integral provides an acceptable tool and the Riemann integral is inappropriate.
•
~The Riemann Integral Throughout this chapter we shall assume that f is a bounded function whose domain includes the interval [a, b] unless otherwise stated. We begin our study of integration with some results about sets of numbers.
Theorem 61: If A and B are nonempty bounded sets of real numbers with A cB, then
sup A
~
sup B
and
inf A
The proof is left for Exercise 1, Section 61. • 132
~
inf B.
Section 61
The Riemann Integral
133
Theorem 62: (a) (b)
Suppose A and B are nonempty sets of real numbers such that if x E A and y E B, then x ::::;: y. Then sup A and inf B are finite, and sup A ::::;: inf B. Suppose that A and B are as in part (a). Then sup A= inf B if and only if, for any e > 0, there exist an x(e) E A and y(e) E B such that y(e) x(e) 0, inf B  sup A < e since inf B;;::, sup A by part (a).
Let e > 0 be given. Suppose there are numbers x(e) E A and y(e) E B such that y(e) x(e) sup A e/2, and there is a number y(e) E B with inf B::::;: y(e) < inf B + ej2. Then, if a= inf B =sup A, we have
so 0 < y(e) x(e) 0 such that if 0 < h < J, then sup { lf(x') f(x")ll x', x" E (x h, x +h) n E&(f)}
0 such that
+ b)nE&(f)} < 8. Since f is continuous at x, there is a b > 0 such that if Ix  y I < b and then lf(x) f(y)l < 8/2. Thus if x', x" E (x J, x + J)nE&(f), then sup { lf(x') f(x")ll x', x" E (x b, x
lf(x') f(x")l
~
lf(x') f(x)l
8
y E £&(!),
8
+ lf(x) f(x")l < 2 + 2 = 8.
•
We now state and prove the RiemannLebesgue Theorem. The proof of the RiemannLebesgue Theorem is substantially more difficult than any we have encountered thus far. However, one does not have to understand the proof of a theorem in order to understand the statement of the theorem and to be able to apply it. Keep this fact in mind if the following proof seems overwhelming.
142
Chapter 6
Integration
Theorem 611 (RiemannLebesgue Theorem): A bounded function f defined on [a, b] is Riemann integrable on [a, b] if and only if the set of discontinuities off on [a, b] has measure zero. Proof: Suppose f is a bounded function defined on [a, b] that is continuous except on a set of measure zero. We show that f is Riemann integrable. Let M =sup {f(x)lx E [a, b]}
m = inf {f(x)lx E [a, b]}.
and
(Assume M m > 0, which is the case unless f is constant.) Let
A= {x E [a, b] If is not continuous at x}. Recall that x
E
A if and only if osc(f; x) > 0. Thus, for any s > 0, the set A.= {x E [a, b] I osc(f; x) ~ s}
is a subset of A. Since A has measure zero, so does A •. Let e > 0 be given. There is a countable collection of open intervals {I 1 , 12 , such that 00
U I;
A,I 0 and any real number r. (i)
Property (i) is immediate by the definition of In x. We exhibit two ways to prove properties (ii), (iii), and (iv). The first method is to appeal directly to the definition of In x. We use this approach to prove (ii). Now lnab= i
ab dt
=
1
t
ia dt
+
1
t
iab dt a
. t
In the integral J:b(dt/t), we make the change of variables u = t/a. rab dt =
Then
and so
Ja ln(ab) =
jab dt =
J1
t
t
lb du J1 u
ja dt + jb du =In a+ In b.
J1
t
J1
u
We shall prove (iv) using the following ideas. Suppose f(x) and g(x) are differentiable functions on an interval (c, d) and f'(x) = g'(x) for all x E (c, d). Then f(x) = g(x) + k on (c, d) for some constant k. If there is a point x 0 E (c, d) at which we know f(x 0 ) and g(x 0 ), then we can solve for k; that is, k = f(x 0 )  g(x 0 ).
Section 62
Some Properties and Applications of the Riemann Integral
157
Let f(x) = ln(x') and g(x) = r In x for x > 0. Then r 1 f'(x) = rx' 1 =
x'
and g'(x) =
X
r
.t X
Thus f(x)=g(x)+k. At x 0 =1, f(x 0 )=ln(1')=0 and g(x 0 )=rln(1)=0, so k = 0. Hence f(x) = g(x). We leave the proof of (iii) for Exercise 15, Section 62. Next we obtain an estimate for Inn for n, a positive integer. We do this by estimating J~ (dx/x) by upper and lower Riemann sums. Fix a positive integer n, and let P = {1, 2, 3, ... , n} be a partition of [1, n]. Now, on the interval [k, k + 1], 1 1 1 k+1""x""k
~~
so if f(x) = 1/x, then 1 1 §(f; P) = 2 + 3 + ...
f" dx
1
+ ~ ~ ]1
1
1
~ ~ 1 + 2 + 3 + ...
1
_
+ n 1 = S(f; P).
We shall show that ! + l + · · · + 1/n can be made indefinitely large by taking n sufficiently large. Notice that
! + l > 2(~) =! (! + i) + (~ + t +! + ~) > 2(~) + 4(i) = 2(!) and one can show 1 2
Thus
n 1 2
1 2" 1
1 2"
+ ... +   > (n 1)1 2
1 3
1 2"  1 ""
 0, there is a step function g. defined on [a, b] such that
(b) Let
lg.(x) f(x)l < e for every x
If f ~1
and
g.
E
[a, b]
0 be given. There is a partition P of [a, b] such that S(f; g; P)  §(f; g; P)
3ln n so that
Jn
exp(f;r) > exp(3ln n) = n3 and In n < n. Thus
and the series
L 1/n
2
converges by the pseries test.
186
Chapter 7
Series of Real Numbers
Exercises 71          1. Determine the convergence or divergence of a series whose nth term is given by
n4 (a)
3n 2
+ 4n 1'
(d)
1 n(ln n)(ln(ln n))'
(g)
rn+l ,fn.
n! n"
(c)
n! e"
(/) ~·
(b).
n~
10.
. (n!)z (2n)! ·
n~2.
(In nin"'
1 (k) (In n)..fio'
(J)
n+
4fn.
n! e
(e).
(h)
1
1 (i) (In n)2 n'
n~2.
n~2.
2. (a) Show that I;'= 2 1/(ln k)" diverges for any value of n. (b) Show that I;'= 2 (ln k)"/k 1 +• converges for any value of n and any e > 0. 3. Show that the series I ak converges if and only if, given any e > 0, there is a number N (which depends on e) such that
Ik~n I< ak
for any m > n > N.
e
4. This is a somewhat different type of comparison test. Suppose that I a. and I b. are series of positive terms and that there is a positive number c such that a. ~ cb. for every n. Show that if I b. converges, then I a. converges, and if I a. diverges, then I b. diverges. 5. Use Exercise 4 to show that if I a. and I b. are series of positive numbers and lim a./b.= 1, then I a. converges if and only if I b. converges. [Hint: for n large enough, ~ ~·
J;,;n•
rx
Section 71
Tests for Convergence of Series
187
12. Suppose that I ak is a convergent series whose terms are positive. Let {S.} be the sequence of partial sums for I ak. Let S =lim s. and let Pk = ak+ljak. (a) Suppose there is a number N for which Pk;:;:. Pk+ 1 if k;:;:. Nand PN < 1. Show that S SN:::; aN+d(1 PN). (b) Suppose there is a number N for which Pk:::; pk+ 1 if k:::; N and that there is a number p < 1 for which p =limp •. Show that S SN:::; aN+d(1 p).
13. Suppose {a.} and {b.} are series of positive terms. (a) Suppose a.+l/a.:::; bn+l!b. and I b. converges. Show that I a. converges. (b) Show that if a.+l!a.:::; b.+l/b. and I a. diverges, then I b. diverges. 14. (a) Use the Binomial Theorem to show that if k is a positive integer, then, for n sufficiently large, (1 k/n):::; (11/n)k. (b) Suppose k is a positive integer larger than 1, and I a. is a series of positive terms with
k)
an+1 ( 1. :::; a. n
Show that I a. converges. [Hint: Use Exercise 13 to compare a.+l!a. with ( 1/nk+ 1 )/( 1/nk).] (c) Show that the series whose nth term is
(2)( 4) · · · (2n)
(5)(7) · · · (2n
+ 5)
converges. 15. Later we shall show that I:'=o 1/k! =e. For now, assume this is true. Let s. = I~=o 1/k!, so that lim s. =e. We want to estimate e s•. (a) Show that I:'= 1 1/(n + 1 )k = 1/n. (b) Show that I:'=n+ 1 1/k! < 1/(n!n) so that e s. < 1/(n!n). (c) How large must n be to ensure I;'= 1 1/k! approximates e to within 10 10 ? 16. Suppose I a. is a series of positive terms with lim(a.+ da.) = L. Show that lim(a.) 11" = L. This may be done by the following steps: (a) First suppose L > 0. Then choose e > 0 with 0 < e < L. Then there is an N such that if n > N, L e < (b) Show that (L etaN < aN+k < (L (c) Now
an+1
 < L a.
+e.
+ e)kaN. aN (Let 1 or p > 1.
Theorem 711: Let I an be a conditionally convergent series. Choose the positive terms of the series I an, and form a series I b. from these terms. Choose the negative terms from the series I a., and form a series I en from these terms. Then I b. diverges to oo and I en diverges to  oo. Idea of the Proof: We first argue that at least one of the series I bn or en must diverge. If I bn converges to L and I en converges to  K, then it is not hard to show that the sequence of partial sums for I Ian I is bounded by L + K, so the series I an would then be absolutely convergent. Now suppose that exactly one of the series Ibn or I en diverges. We shall argue that the series I an must diverge. Suppose that I bn diverges to oo and I c. converges to  K. Then one can show that the series L an must diverge to oo, so that I an was not conditionally convergent. •
I
Rearrangement and Regrouping of SeriesOne often would like to do certain operations with series that are valid arithmetic operations, such as grouping the terms. The problem is that the usual rules of arithmetic do not always apply to series. For example, if we have a finite collection of numbers that we want to add, say 1 + 2 + 3 + · · · + 10, then we can group the terms any way we like by inserting parentheses, say (1 + 2) + (3 + 4) + .. · + (9 + 10), and get the same answer. Consider, however, if the series
11+11+11+"' is grouped as
1 + (1 + 1) + (1 + 1) + ... = 1,
then we get a different answer than if the terms are grouped as
( 1  1) + ( 1  1) + ... = 0. Thus, one cannot blindly apply the associative law to series. The same holds for the commutative law. With a finite number of terms, we are free to move the numbers around any way we want, and the answer will always be the same. For example, 1+2+3=2+1+3=3+1+2=~
With series, we shall see this is not always the case. That is, one is not always free to rearrange the terms of the series. By a rearrangement of a series I an, we mean a new series Ibn whose terms consist of the same numbers as those of I a., but which appear in a different order. Thus if
an=1+!+i+i+!+ ... and
Section 72
then Ibn is a rearrangement of definition.
I
Operations Involving Series
191
an. More precisely, we give the following
Definition: Let f be a 11 onto function from the positive integers to the positive integers. Let I an be a series, and define a new series I bn by bn = af(n)
for n, a positive integer. Then Ibn is a rearrangement of I an. • In our preceding example, /(1) = 2, /(2) = 1, and f(n) = n, otherwise. The next theorem shows that by rearranging a series, it may be possible to get a series that behaves radically different from the original series.
Theorem 712: Let I an be a conditionally convergent series. Then given any number L (which may be finite or infinite), it is possible to construct a rearrangement of I an that converges to L. Proof: We demonstrate the proof in the case that L is positive and finite. Choose L, and let B > 0 be given. Let bk denote the kth nonnegative term of the series I an and ck denote the kth negative term of the series I an. Since I an converges, there is a number N such that Ian I < B if n?:: N. Notice that this implies Ibk I < B and Ick I < B if k?:: N. By Theorem 711 Ibn diverges to 00 and I en diverges to  00. We inductively construct a rearrangement of L an that converges to L. First add b1 + · · · + bn 1 so that
+ , . , + bnl > L b1 + · · · + bn 1 :s:; L. b1
but
1
That is, we choose the first numbers in the rearrangement by adding the first positive terms of I an until L is exceeded. Next we add the first negative terms of I a" until our sum falls below L. That lS,
+ ... + bnl + c1 + ... + Cnz < L b1 + ... + bnl + c1 + ... + cnz 1 ?:: L. b1
but
Now we add the next positive terms until we just exceed L, and so on. We claim this rearrangement of I an converges to L. Suppose that {Sn} is the sequence of partial sums for the rearranged series. Suppose n is large enough so that we have included at least N of the b;'s and at least N of the c;'s. Then
so lim Sn
=
L and the rearranged series converges to L. •
We now know that some caution is in order when trying to apply the rules of arithmetic to series. A reasonable question is: When can we regroup or rearrange series without affecting the convergence of the original series?
192
Chapter 7
Series of Real Numbers
First we further investigate the effect of inserting parentheses into a series. Suppose that we begin with a series a 1 + a2 + · · · and then add parentheses, grouping some of the terms. The new series looks like
(al + ... +anl)+(anl+l + ... +anJ+ .. ·. To sum the new series, we first sum the terms within parentheses and then add the parenthetical terms. In effect, we have made a new series L bn where
... + anl hz = anl + 1 + + anz
bl
=
al +
We say that the series L bn is obtained from the series L an by inserting parentheses. We want to know when is L an= L bn. We know by our earlier example that this is not always the case.
Theorem 713: Suppose L an converges to L and L bn is obtained from L an by inserting parentheses. Then L bn also converges to L. That is, inserting parentheses has no effect on a convergent series. Proof: Let {Sn} be the sequence of partial sums associated with L an. The hypotheses imply that lim Sn = L. Let {T,} be the sequence of partial sums associated with L bn. That is, Tl = bl = al
+ ... + anl = Snl
Tz = bl + hz = al +
+ anz = snz
where n1 < n2 < · · · < nk < ···.That is, {T,} is a subsequence of {Sn}. But a subsequence of a convergent sequence must converge to the same number as the original sequence. Thus
Ibn= lim
r, =lim sn = L.
•
For a rearrangement of a series to exhibit the same convergence as the original series, we need more restrictive hypotheses, as Theorem 712 shows. What is needed is absolute convergence.
Theorem 714: Let L an be an absolutely convergent series. Let La~ be the series whose terms are defined by a'= { n
an, 0,
Section 72 and
I
Operations Involving Series
193
a~ be the series whose terms are defined by
a"= {a", n 0,
if an< 0 if an;?! 0.
Then I a~ and I a~ are absolutely convergent series and The proof is left for Exercise 4, Section 72. •
I
a" =
L a~ + I
a~.
Theorem 715: Let I a" be an absolutely convergent series, and let I a~ and I a~ be defined as in Theorem 714. Suppose I a~= P and I a~= M. (Note that M ~ 0.) Then any rearrangement of I a" converges to P + M. Proof: Let I b" be a rearrangement of ,I a", and letfbe the 11 onto function that defines the arrangement, that is, b" = af(n)· In Exercise 2(b), Section 72 we show that I b" converges absolutely and thus I b" converges. Let B > 0 be given. We shall show that
Since
I
a~ =
P and
I
a~ =
M, there is a number N 1 such that if n > N 1 , then
Iktl a~ pI < ~
(1)
and a number N 2 such that if n > N 2 , then (2) Since
I
b" converges absolutely, there is a number N3 such that if n;?! N3 , (3)
Choose N > N3 such that {b1 ,
... ,
bN} contains
{ a 1, ... , aN1 } u { a 1, ... , aN2
}
= { a 1, ... ,
aN}
where N =max {Nl> N2 }. (It is left for Exercise 5, Section 72, to show that this is always possible.) Now
ILbn(P+M)I~ I Jl bn(P+M)I + ln=~+l bnl =I( ,I' bn P) +(,I" bn M)l +I n=~+l bn I
I'
where b" indicates we sum over the b;s that are nonnegative and whose subb" is the sum over b;s that are negative and scripts are smaller than N, and
I"
Chapter 7
194
Series of Real Numbers
whose subscripts are smaller than N. Now Nl
L' bn ~ L
n=1
a~
IL' bn PI < 3 and B
so that Thus
(Why?)
and ~"
B
I L. bn M I < }"
I(L' bn P) + (L" bn M)l +I
n=~+ 1 bn I 00
~IL'bnPI+IL"bnMI+
c 3
c 3
L lbnl n=N+1
c 3
0 for every n, the series called alternating series. •
L (1 )"an and L (1 )" + 1 an are
That is, an alternating series is one in which each term has a different sign from the preceding term.
Theorem 716 (Alternating Series Test): Let L (l)n+ 1an be an alternating series such that an+ 1 > 0 for every n. lim an = 0.
(i) an~ (ii)
Then
L (l)n +1an and L (1 )nan converge.
L(
l)n +1an. Let {Sn} be the sequence of partial sums associated with
Proof: We do the proof for the series
L (1 )" + 1 an. Then
s2k = (a1 a2) + (a3 a4) + ... + (a2k1 a2k).
Since the numbers an are decreasing, each parenthetical term is nonnegative so that S2 k ~ 0. Also {S2n} is an increasing sequence since the parenthetical terms are nonnegative. But
since the parenthetical terms are all positive. Thus {S2 n} is a bounded monotone increasing sequence that must converge. Suppose that the limit is L. Now
Section 72 lim S2 k+ 1 = lim S2 k
so that
Operations Involving Series
195
+ lim a2 k+ 1 = lim S2 k + 0 = L
since lim an= 0. Thus {Sn} converges to L. (See Exercise 14, Section 22.) • The proof of the theorem also gives part (a) the following result.
Corollary 716: If L(l)n+ 1 an is an alternating series that satisfies the hypotheses of Theorem 716 and converges to L, then L 0 and x
E
E, there is a number N(t:, x) such that
lfn(x) f(x)l < e if n > N(t:, x). Then the function f is said to be the pointwise limit of the sequence of functions Un}. • It is a crucial point (and one that is often misunderstood initially) that in consideration of the pointwise limit, the number N(t:, x) depends on both the number e and the point x. Just as in the case of convergence of sequences of numbers where no number N will necessarily work for every e, no number N will necessarily work for every x E E, even with e fixed. The following example may help to clarify this.
Example 85: Let fn(x) = xn for x E (0, 1). Then Un(x)} converges to 0 for every x E (0, 1). Table 81 has the column headings of some values of x between 0 and 1. The row headings are for certain e values. The entries in the table are the minimum values of n for the values of e and x in that particular row and column to make lxnl 0, there is a number N(e) (which
Section 81
Sequences of Functions
205
depends on e, but not on x) such that if n > N(e), then lfn(x) f(x)l < e for every x
E
E.
Ifthe function f is the uniform limit of the sequence of functions Un}, then Un} is said to converge uniformly to f. •
Remark: The crucial difference between the pointwise limit and the uniform limit is that in the uniform limit we are able to choose a number N(e) that works for every x E E. This is not necessarily possible in the case of pointwise convergence. Figure 82 illustrates this. In the uniform limit, if we draw a belt extending an amount eon either side of the function f, then for n > N(e), fn(x) must be within this belt for every x.
Figure 8  2   We leave the proof of the following theorem for Exercise 1, Section 81. It provides one of the most convenient ways to show that a sequence of functions converges uniformly.
Theorem 81: Suppose that Un} converges pointwise to f on a set E. Let Mn = sup lfn(x) f(x)l. xEE
Then {fn} converges uniformly to f onE if and only if lim Mn = 0. •
Example 86: Let fn(x) = xn for x E [0, !J. We claim that Un} converges uniformly to 0. To prove this, note that
Chapter 8
206
for every then
X E
[0,
Sequences and Series of Functions
n So given e > 0, if 1/2n
lln el/ln 2 fore< 1),
X E
[0,
n
This same method will work to show that {xn} converges uniformly to 0 on [0, 1 et] for any et with 0 < et < 1. However, {xn} does not converge uniformly to 0 on [0, 1).
Theorem 82: Let f be the uniform limit of a sequence of continuous functions Un}. (That is, each functionfn is continuous for n = 1, 2, .... ) Thenfis continuous. Remark: Thus, while the pointwise limit of a sequence of continuous functions may not be continuous, the uniform limit of a sequence of continuous functions, if it exists, is continuous.
Proof: Suppose each function fn is defined on E, a subset of the real numbers. Choose x 0 E E, and let t: > 0 be given. We need to show that there is a (j > 0 such that if x E E and lx x 0 1< b, then lf(x) f(x 0 )1 0, which depends on both the point x 0 and the function fN 1 , such that
(7) if lx x 0 1 N. •
Example 87: Let f.(x)
={
1/n, 0:::::; x:::::; n 0,
x>n.
Then {f.(x)} converges to 0 uniformly on [0, oo ), but
f"
fn (x) dx =
f G)
dx = 1 # 0 =
f"
0 dx.
Thus, as stated earlier, the interval of integration must be finite for Theorem 83 to apply. Notice that Example 84 shows a uniformly convergent sequence of differentiable functions for which lim.~ oo (f~(x )) does not exist. Thus even stronger conditions than uniform convergence are needed to ensure that differentiability conditions carry through to the limiting function. Before addressing that problem, we give a condition that ensures the uniform convergence of a sequence of functions that is very useful.
Theorem 84 (Cauchy Criterion for Uniform Convergence): Let {f.} be a sequence of functions defined onE, a subset of the real numbers. Then {f.} converges uniformly to a function f on E if and only if, given e > 0, there is a number N(e) (which depends on e but not on x) such that lf.(x) fm(x)l < e for every x
E
E whenever m, n > N(e).
Remark: Such a sequence of functions is said to be uniformly Cauchy on E.
Section 81
Sequences of Functions
209
Proof: First suppose that Un} converges uniformly to f on E. Let e > 0 be given. Then there is a number N such that lfn(x) f(x)l < t:/2 for every x E E if n > N. Thus, if both m and n exceed N, then lfn(x) fm(x)l ~ lfn(x) f(x)l
e
e
+ lf(x) fm(x)l < 2 + 2 =e.
Conversely, let e > 0 be given. The hypotheses say that there is a number N such that lfn(x) fm(x)l < t:/2 for every x E E if n, m > N. Thus, for each x E E, Un(x)} is a Cauchy sequence of real numbers, so we may define a function f(x) onE by f(x) = limn~oofn(x) for x E E. Clearly Un} converges pointwise to f onE; the problem is to show that the convergence is uniform. Now
lfn(x) f(x)l
~
lfn(x) fm(x)l
+ lfm(x) f(x)l
for any nand m. Note that as long as n, m > N, lfn(x) fm(x)l < t:/2 for any x E E. Now for the crucial step: For a given x there is a number Nx that depends on both e and x such that if m > Nx, then lfm(x) f(x)l < t:/2, since Un} converges to f pointwise. Thus, by letting m vary with x, we have shown that
lfn(x) f(x)l ~ lfn(x) fm(x)l if n > N and x on E. •
E
e
e
+ lfm(x) f(x)l < 2 + 2 =
e
E. Since N is independent of x, Un} converges uniformly to
f
As we noted earlier, in Example 84 with the sequence of functions
fn(x)
sin nx =
J;z ,
the functions converge uniformly to 0 but lim f~(x) # 0. Thus, to get the desired result for differentiation, we need more than uniform convergence. The result that tells us when the derivative of the limit of a sequence of functions is the limit of the derivatives of the functions is considerably more involved. To derive our result, we shall use the following theorem.
Theorem 85: Let Un} be a sequence of functions that converges uniformly to
f
on [a, b]\ {x0 } where x 0 exists. Then
E
[a, b]. Suppose that, for each n = 1, 2, ... ,
limx~xofn(x)
lim (lim fn(x)) X1>Xo
n1oo
=
lim (lim fn(x)). n1oo
x1x0
Proof: Let Ctn = limx~xJn(x). We first show that {etn} is a Cauchy (and thus a convergent) sequence. Let e > 0 be given. Since Un} converges uniformly to f on [a,b]\ {x0 }, by Theorem 84 Un} is uniformly Cauchy on [a,b]\ {x0 }. Thus there is a number N such that, if n, m > N, then lfn(x) fm(x)l < Now
e
3
letn etml ~ letn fn(x)l
for every x
E
[a, b]\ {x0 }.
+ lfn(x) fm(x)l + lfm(x) etml·
(9)
Chapter 8 Sequences and Series of Functions
210
Choose n and m larger than N so that Equation (9) holds. Since lim f.(x) =a. for n = 1, 2, ... xx0
then, for each positive integer k, there is a number c5k > 0 such that lctk fk(x)l < e/3 if 0 < lx x 0 1< c5k
0 N 2 , then let. ct 0 1< e/3. Let N =max {N1 , N2 }. Then
Also, there is a c5N > 0 such that lfN(x) ctNI
1. Therefore we may conclude d  (x') = rx' 1
dx
on ( 1, oo) for any irrational number r. The case for x between 0 and 1 is left for Exercise 7, Section 81.
Exercises 81          1. Suppose Un(x)} converges to f(x) pointwise on a set E. Let
Mn =sup lfn(x) f(x)l. xEE
Show that Un} converges to f uniformly onE if and only if limn~oo M" = 0. 2. Show that if {f"(x)} converges uniformly on (a, b) and if {f"(x)} converges at x =a and x = b, then {f"(x)} converges uniformly on [a, b]. 3. Show that if fn(x) = x" on [0, 1], then lim
r r fn(x) dx =
limfn(x) dx.
Thus uniform convergence is not a necessary condition for convergence of the integrals. 4. Give an example of a sequence of functions that is not continuous at any point, but which converges uniformly to a continuous function.
Chapter 8 Sequences and Series of Functions
214
5. Show that the following sequences of functions converge uniformly to 0 on the given sets: (a) { si::x} on [a, oo) where a> 0.
(c)
L:
(b) {xenx} on
(d) {
nx} on (0, 1).
ln(1
[0, oo ).
+ nx)} on
n
[0, M].
6. Let f.(x) = xenx + ((n + 1)/n) sin x. (a) Find a function f(x) such that {f.(x)} converges to f(x) on (b) Show that the convergence in part (a) is uniform.
[0, oo ).
7. Complete the work to show that d d)x')=rx' 1 on (0, oo)
by showing (a) {x'· 1 } is uniformly bounded on [1, M] (see Exercise 12). (b) 11 x''•1 may be made uniformly small on [1, M] by making r r. sufficiently close to 0. (c) Show that for x E (0, 1) d dx (x') = rx'1. [Hint: For x E (0, 1), x = 1/y for some y E (1, oo ).] 8. (a) Show that the sequence of functions f.(x) = x/n converges pointwise but not uni
formly to 0 on [0, oo ). (b) Show that the sequence of functions f.(x) = xjn converges to 0 uniformly on the
9. 10.
11.
12.
interval [0, M] for any number M. Show that for any a E (0, 1), the sequence of functions f.(x) = x" converges uniformly to 0 on [0, 1 a] but does not converge uniformly on [0, 1). (a) Prove that if {f.(x)} converges uniformly to the function f(x) on each of the sets A1 , A 2 , ..• , A., then it converges uniformly on u7 A;. (b) If {f.(x)} converges uniformly to f(x) on AI> A 2 , ••• , must it do so on u'[' A;? (a) Show that if {f.} converges uniformly to f, and if each function f. is bounded, then f is bounded. (b) If {f.} converges pointwise to f and if each fn is bounded, must f be bounded? (c) Let {f.} be a sequence of bounded functions that converges uniformly to f on an interval [a, b]. Suppose that for some e' > 0 and some N, lfN(x) f(x)l < e'/2 for all x E [a, b]. Let M(f) =sup {f(x)lx E [a, b]} and M(fN) =sup {fN(x)lx E [a, b]}. Show that IM(f) M(fN)I < e'. (d) In the proof of Theorem 83, we claimed that since lf(x) fN(x)l < e/(6(b a)) for all x E [a, b], then IM;(f) M;(fN)I < e/(3(b a)). Use part (c) to justify this claim. (e) (This exercise provides a stronger result than part (c).) Suppose that {f.} is a sequence of bounded functions that converges uniformly to a function f on [a, b]. Suppose that for some e' > 0 and some N, lfN(x) f(x)l < e'/2 for all x E [a, b]. If M(f) and M(fN) are as in part (c), show that IM(f) M(fN)I ~ e'/2. (a) A sequence of functions {f.(x)} is said to be uniformly bounded on a set E if there is a number M such that lf.(x)l ~ M for every n = 1, 2, ... and every x E E. Show that a uniformly convergent sequence of bounded functions is uniformly bounded. (b) Is the result in part (a) true if "uniformly convergent" is replaced with "pointwise convergent"?
Section 82
Series of Functions
215
13. (a) If {f.} and {g.} are sequences of functions that converge uniformly to the functions f and g, respectively, on a set E, show that {f.+ g.} converges uniformly to f + g on E. (b) Find an example where {f.} converges uniformly to f and {g.} converges uniformly tog, but {f.g.} does not converge uniformly to fg. (c) Show that if {f.(x)} converges to 0 uniformly on a set E and if g(x) is bounded on E, then {g(x)f.(x)} converges uniformly to 0 on E. (d) Suppose that {f.} converges uniformly to f and {g.} converges uniformly to g on a set E, and that each fn and g. is bounded on E. Show that {f.g.} converges uniformly to fg on E. 14. (a) Give an example of a sequence of functions {f.} that satisfies all of the following conditions: (i) Each function f. is continuous on [0, 1]. (ii) The sequence {f.} converges to a function f on [0, 1]. (iii) There is a sequence {x.} c [0, 1] with {x.} converging to x, but {f.(x.)} does not converge to f(x). (b) Show that if we require the sequence of functions in part (a) to be uniformly convergent, then {f.(x.)} must converge to f(x). 15. (a) Suppose {f.} is a sequence of monotone increasing functions that converges to a function f on [a, b]. Show that f is monotone increasing. (b) Show that if {f.} is a monotone increasing sequence of functions that converges to a continuous function f on [a, b], then the convergence is uniform. 16. Let {f.(x)} be the sequence of functions defined by
nenx
f.(x)
Note that f.(x)
~
= {0
if X
~0
if X< 0.
0. Show that:
(a) eoo.f.(x) dx = 1. (b) If g is a function that is continuous at x
!~~
J:
= 0, then
f.(x)g(x) dx = g(O).
(Such a sequence of functions is called an approximate identity.)
Series of Functions Convergence of Series of FunctionsIn this section we want to form a series of functions. We begin with a sequence of functions {f.}, with each function defined on a set E. Then, for each x E E and each positive integer n, define
Thus, for each x E E, {S.(x)} is the sequence of partial sums for the series of numbers l:f.(x). Suppose that lim.~ co s.(x) exists for each x E E. Then we can
216
Chapter 8
Sequences and Series of Functions
define a function f on E by f(x) = lim Sn(x).
Definition: With the preceding notation, we say that the series of functions
L fn converges to f on E and write
J= "iJn· If {S"} converges uniformly to f on E, then we say the series uniformly to f on E. •
L fn
converges
As with series of numbers, we shall refer to the sequence {Sn} as the sequence of partial sums for the series L fn Therefore, the question of convergence, or uniform convergence of a series of functions, is exactly the same as the question of convergence or uniform convergence of the associated sequence of partial sums. Thus it should not be surprising when we appeal to the results of Section 81 to prove many of our results about series, similar to our use of the results of Chapter 2 on sequences of numbers to prove the theorems about series in Chapter 7. As we stated earlier, series of functions have important applications in the physical and social sciences as well as mathematics. In these applications one often wants to perform an operation such as integration or differentiation on the series by applying the operation to each term of the series. As we know from our work with sequences, this is not always legitimate, and usually uniform convergence (and possibly more) is needed to ensure that this procedure is valid. We now give some conditions that guarantee uniform convergence of a series of functions.
Theorem 87 (Cauchy Criterion for Uniform Convergence of a Series): Let Un} be a sequence of functions defined on a set E. The series L fn converges uniformly on E if and only if, given e > 0, there is a number N(e) such that if m > n > N(e), then
for every x
E
E.
Remark: Again notice that in uniform convergence N(e) depends only on e and not on the choice of x E E, and that the inequality must hold for every XEE.
Proof: We apply the Cauchy criterion for uniform convergence to the sequence of partial sums {Sn} for Lfn· Now Sn(x)=fl(x)+ · · · + fn(x) Sm(x)=fl(x)+ · · · + fn(x)+ fn+l(x)+ · · · + fm(x)
since m > n. Thus
Section 82
Series of Functions
217
and the sequence of partial sums {Sn(x)} converges uniformly onE if and only if, given 8 > 0, there is a number N(8) such that if m > n > N(8), then ISm(x) Sn(x)l =
Ii=t+t /;(x) I
n,
Ii=t+t
/;(x)
I~ i=t+t 1/;(x)l ~ i=t+t M;
for every x
E
E.
By the Cauchy criterion for convergence of real numbers, given 8 > 0, there is a number N(8), such that if m > n > N(8), then
i=n+l
Thus for every xEE, if m>n>N(8). So by Theorem 87 the series uniformly on E. •
Lfn
converges
Example 89: Let fn(x) = n2 enx on [ 1, oo ). We show that L fn converges uniformly. Since enx is a decreasing function, lfn(x)l ~ n2 en on [1, oo]. The series L n2 en converges. (Why?) So by the Weierstrass Mtest, the series L n2 enx converges uniformly on [1, oo ). In fact, using an almost identical argument, one can show that the series L nkenx converges uniformly on [a, oo) for any a> 0 and any number k. We now state some corollaries, which give conditions on series that ensure certain operations on series are legitimate.
218
Chapter 8 Sequences and Series of Functions
Corollary 82: If Un} is a sequence of continuous functions defined on a set E and if L fn converges uniformly to a function f on E, then f is continuous on E. Proof: Let SN(x) = f 1 (x) + · · · + fN(x). Each function .t;(x) is continuous, so SN(x) is continuous, since it is the sum of a finite number of continuous functions. Thus by Theorem 82
is continuous, since this is the uniform limit of a sequence of continuous functions. •
Corollary 83: Let Un} be a sequence of Riemann integrable functions on a finite interval [a, b]. If Lfn converges uniformly to f on [a, b], then J!f(x) dx exists and
Proof: Again, let SN(x) = / 1 (x) + · · · + fN(x). Since each function };(x) is integrable on [a, b], we have that SN is integrable on [a, b]. Thus {Sn} is a sequence of integrable functions that converges uniformly to Lfn, so by Theorem 83 L
i bfn(x) dx = a
Nib fn(x) dx =
lim L
N+oo 1
= lim Noo
a
ib LNfn(x) dx
lim N+oo
a
rb SN(x) dx = rb (lim
Ja
Ja
1
SN(x)) dx
NHXJ
=
rb L fn(x) dx.
Ja
The step lim
rb SN(x) dx = rb (lim
Noo Ja
Ja
SN(x)) dx
N+oo
is where uniform convergence is used. •
Corollary 85: Let Un} be a sequence of functions that are differentiable on an open interval containing [a, b] such that (i)
Lfn(x0 ) converges for some x 0 E [a, b], and converges uniformly to a function h(x) on [a, b].
(ii) Lf~(x)
Then (a) (b)
Lfn(x) converges uniformly to a function f(x) on [a, b], and Lf~(x) = h(x) = f'(x) for x E (a, b).
The proof to this corollary is similar to the previous two. •
Section 82
Series of Functions
219
Power SeriesDefinition: Let {an}:'=o be a sequence of real numbers. An expression of the form 00
a0 + a 1 (x c)+ a 2 (x c) 2 + · · · =
L an(x c)" 0
is called a power series in x  c. The numbers an are the coefficients for the power series. • To simplify the notation somewhat, we shall first consider power series with c=O. For a given set of coefficients, we want to answer the following questions: (i)
For what values of x does the power series converge? For values of x where the power series converges, define 00
f(x) = a0 + a1 x + a2 x 2 + · · · =
L anxn.
n=O Then: (ii) (iii) (iv)
When is f(x) continuous? When is L anxn dx = L an xn dx? When is f'(x) = nanxnl?
J
L
J
To answer these questions, we shall apply our earlier theorems with fn(x)= anxn. Notice that fn(x) = anxn is infinitely differentiable and continuous and is integrable over any finite interval. Also notice that a power series converges if x = 0. Beyond that, there may be values of x for which the power series converges and other values of x for which it diverges. For example, the power series 1 + x + x 2 + x3 + · · · is one with an = 1 for every n. It is also a geometric series for a given x with r = x. Thus the series converges if Ixi< 1 and diverges if Ixi~ 1. The next theorem characterizes the behavior of power series.
Theorem 89: Let
L anxn be a power series, and let A= lim(lanl 11n)
and
1 R = ~·
(where R = 0 if A= oo and R = oo if A= 0). Then the power series (a) (b)
(c)
L anxn
Converges absolutely if Ixi< R. Diverges if Ixi> R. Assumes 0 < R < oo. Then the power series anxn converges uniformly on [ R + e, R e] for any e > 0. If R = oo, the series converges uniformly on any compact interval.
L
220
Chapter 8 Sequences and Series of Functions
Proof: To prove parts (a) and (b) for 0 ~ =
R.
If A.= 0, we note that lxiA. < 1 for all values of x, and if A.= oo, the series diverges if lxl > 0. This proves parts (a) and (b). To prove part (c) in the case R < oo, let e > 0 be given, and let p = R e/2. Consider Figure 83. f.
f.
p=R
p= R+z
R
2
Rf.
R+f.
R
Figure 8  3   
Now and
1
1
R< Re/2.
Thus, there is a number N such that if n > N, then 1 p Thus for n > N,
if lxl (O)
x2
L  n!x n = 1 2! +x44!  x66!+ ... oo
n=O
oo
(1tx2n
= n=O I
(2n)'. .
We are not saying, although it will turn out to be the case, that (1tx2n (2 )' n=O n· oo
cos X=
L
but only that the series on the right is the Maclaurin series for cos x.
Example 815: Find the Taylor series for f(x) = ln x about c = 1. f(x)=lnx
/(1)=0
a0 =0
f'(x)=_!_
/'(1)=1
a 1 =1
X
1
f"(x)=x2 f"'(x)
=
_3_ x3
/"(1) = 1
/'"(1) = 2
2 1 a 3  3! 3 a4
=
(2)(3) 1 4!  4
(l)n1
a= n
n
230
Chapter 8
Sequences and Series of Functions
So the Taylor series for f(x) = ln x about c = 1 is (x 1)
(x1? 2
+
(x1)3 3
 .. · =
oo (l)nl(xl)n
L n=l
n
·
We again emphasize that we are not saying (although it will turn out to be the case if lx 11 < 1) that lnx=
oo (1t+ 1 (x1t
L n=l
n
.
One may wonder if we are not being overly cautious by not immediately writing
f (nl(c)
oo
f(x) =
L  n.1
(x ct
n=O
for functions with infinitely many derivatives at x =c. However, the function e1/xz, x#O f(x)= { 0, x=O has the property J a. We have k n1 j(a) f(x)= k~O ~(xa) +Rn_ 1 (x).
r
r
r
Let m = min{f(t)lt E [a, x]}. Then m
(xt)" 1 dt~
(xt)" 1 j 0 or a= 0. If a> 0, write I _ I I(J;...Ja)(J;..+.Ja)l Iva. val. 1 1 (v a.+ v a) 19. (a) Show that {a.} is a decreasing sequence of positive numbers.
20. (a) Show there is a number M for which Ia. b. 01 :( Mlb. 01. 22. (a) Choose such that if n > then Ia. a I < e/3. Choose K such that if k > K, then
N
N,
Ia1 +; ; : aN I
1, and so L > 1. 4. (a) If a11" < 1 for some n, then (a 11")" < 1 but (a 11")" =a. a1fn 4 (b)   = a1/(n(n+1)] > 1 a1fn+1 . 5. (a) Let b = 1/a. Then b > 1. Now apply Exercise 4. 7. (a) Show that {a.} is an increasing sequence that is bounded above by 2. 8. (b) a 1 = 1, a 2 = 1, a 3 = 2, a4 = 1, a5 = 2, a6 = 3, a 7 = 1, ... 8. (c) Enumerate the rational numbers. 9. (a) (1 + 1/2n) 3 " = [(1 + 1/2n)2"P2 . Now (1 + 1/2n)2 " is a subsequence of (1 + 1/n)". 10. Show that lam+n a. I< b"(bm + bm 1 + · · · + 1). Then show
1 + ...
+ bm =
1 bm+ 1 1b
1 < . 1b
11. Show that lan+ 1  a. I= lao a 11/2" 1. In fact, the series converges to a0 +~la 1  a 0 1if a 1 > a0 and a 1 +~la 1  a0 1if a 1 < a0 .
270
Hints and Answers for Selected Exercises
Section 23, page 5 8       5. (b) Let a., be the ith term in the sequence {a.} that exceeds L+ e. 12. Suppose some interval (L e, L+ e) contains only infinitely many points, say x 1 , Let ex=! min { IL xd Ii = 1, ... , N}, and consider (Lex, L+ ex).
... ,
xN.
13. (a) Show there is an N such that if m > N, then lam aNI< 1, so that Iami< IaNI+ 1. The set {la 1 1, ... , laNI, laNI + 11} is a finite set and is bounded. 13. (c) Suppose that L and M are subsequential limit points of {a.}. Let 6 = ~IL MI. Show that if ak E (L e,L+ e) and am E (M e,M +e), then lak ami> 6.
15. (a) Let A. = [n, n + 1] n A for n, an integer. If each A. is countable, then u. E z A.= A is countable. Thus some interval [n, n + 1] must contain an uncountable number of points. 16. Let A =sup {ex Iex is a limit point of {a"}}. Suppose {b.} is a sequence of limit points of {a.} converging to A (so each bk is a limit point of {a.}). For every k, choose a number from {a.}, say a"k' such that a.k E (bk 1/k, bk + 1/k) and ni < nk if j < k. Show that {a•.} converges to A.
Section 31, page 71        3. (a) Any interval of the form [0, a) or (a, b) or (a, oo) where a> 0. 4. If x + ex E A, then there is an open interval about x +ex that is contained in A. This will contradict the definition of ex.
5. If z E I x n I Y, then I xu I Y is an open interval containing x. Since I x is the maximal open interval, Ix=Ixuiy. 7. (b) No. 9. (b) Let A= rational numbers, B =irrational numbers. 11. (a) Neither open nor closed. Limit points= IRI., int(A) = 0. A= IRI.. Boundary of A= IRI.. 14. Enumerate the rational numbers {rl> r 2 , ••• }. Let E.= {(r. r;, r. + r;) Ii = 1, 2, ... }.
Section 41, page 8 9        3. Let x 0 be a nonzero rational number, and let {x.} be a sequence of irrational numbers converging to x 0 . Consider {f(x.)} and f(x 0 ). 6. One only need show that if x 0 E (0, 1], then, for any y 1 between 0 and f(x 0 ), there is an x 1 E (0, x 0 ) with f(xd = y 1 . This is sufficient because f is continuous on (0, 1].
> 0, take b = ~. 11. (a) Use llf(x)llf(y)l k lf(x) f(y)l. 12. Take 6 = l!f(a)l. There is a b > 0 such that 9. For any
6
lf(a) f(x)l < 6 if x
E
(a b, a+ b).
15. (a) Consider g(x) = f(x) x. What about g(O) and g( 1)? 15. (c) Consider g(x) = f(x) f(x + 1). 19. Iff is not constant, either f(a) i= M or f(a) i= m. Suppose f(a) i= M, and suppose c is between f(a) and M. There is an x 0 E (a, b) where f(x 0 ) = M. Consider f on [a, x 0 ] and [x 0 , b].
Hints and Answers for Selected Exercises
271
25. (b) fx 2  l l = fx yffx + yf. Suppose fore=!, there is a o> 0 such that if fx yf < o, then fx 2 ll (x). 15. Show that f(cxn) _ f'(c) I~ I fln II f(fln) f(c) _ f'(c) I If(f3n)fln  cxn fln  cxn fln  c +1 1  f3nc llf(cxn)f(c) f'(c)l. fln  CXn CXn C
C
Section 52, page 129       5. (b) Let f(x) =cos x, g(x) = 1 x 2 j2. Then f(O) = g(O) and f'(x) > g'(x). 7. (b) Consider f(x) = x 2 + 1. 16. Use the Mean Value Theorem. 17. Show g'(x) > 0.
272
Hints and Answers for Selected Exercises
Section
6t page 1 4 5       
5. (a) (1/n) L~=J(k/n) is a Riemann sum for f(x) on the interval [0, 1] with partition {0, 1/n, 2/n, ... , njn}.
5. (b) Use part (a) with f(x) = x. 8. Show that A~ is open. Do this by showing that if t < s and osc(f; x 0 ) = t, then there is an interval about x 0 where osc(f; x) < s if x is in the interval about x 0 • 11. (b) The function!· g is continuous at x 0 unless either for g fails to be continuous at x 0 . 12. (b) Show that the lower Riemann sum off for any partition is 0. 13. Use the fact that §(f; P,) ~ S(f; P,) ~ S(f; P,).
Section 62, page 159       1. (c) No.
2. (a) Writer f=
;t
1
r f
2. (b) For given e > 0, there is a b > 0 such that if lx Yi < b, then lf(x) f(y)l 0, there is a point x 0 and an interval about x 0 where lf(x)l >Lefor every x in the interval. Now show that, for n sufficiently large, II f lin> L e. 14. (a) No.
18. e2 20. (a) Define
](x) =
{
f(x),
if X#
C
limf(x),
if X=
C.
x~c
Then J is continuous, and F'(c) =](c)= lim](x) = limf(x). x+c
x+c
Section 63, page 172       +2 5. (a) f(x)=g(x)h(x) where 2. (a) 2e 2
h(x)={x 2 , 0,
xE[1,0] XE(0,1],
g(x) = {
0, 2
x,
X E [
1, OJ
XE(0,1].
Hints and Answers for Selected Exercises
273
Section 71, page 1 8 6       3. Let {Sn} be the sequence of partial sums for L an. Then {Sn} converges if and only if it is a Cauchy sequence. 8. Show that the series nk/An converges.
L
13. (a) Show that an~ bn(adbd. 14. (c) Use Exercise 14(b).
Section 81, page 2 1 3       6. (a) f(x) =sin x 8. (a) Suppose there is anN such that lx/nl < 1 if n > N for every x. Show this is impossible by choosing x = N + 1.
9. To show uniform convergence on [0, 1 ex], show that ( 1 ex)"< e if n is sufficiently large. To show nonuniform convergence on [0, 1), note that if e =! and if N is any positive integer, xN >!for x > {!) 1/N E (0, 1). 10. (b) Consider fn(x) = xjn, A;= [i, i + 1]. 13. (b) Let fn(x) = gn(x) = x + 1/n on [0, oo ). 14. (a) Let fn(x) = xn on [0, 1]. Choose Xn = 11/n. 15. (a) Suppose there are x, y E [a, b] with x < y where f(x) > f(y). Show this is impossible by taking e = (~)[f(x) f(y)]. 15. (b) f is uniformly continuous. Given e > 0, choose N such that if lx yi ~ 1/N, then lf(x) f(y)i < e/2. Let X;= a+ i/N. Thenfn(x) converges to f(x) at each X;. Choose N; such that if n > N;, then lfn(x;) f(x;)l < e/2. Show that if n > N =max {N;}, then lfn(x) f(x)l