﻿ Introduction to Mathematical Analysis I - 3rd Edition

# Introduction to Mathematical Analysis I - 3rd Edition

Introduction to Mathematical Analysis I 3rd edition Beatriz Lafferriere Gerardo Lafferriere Mau Nam Nguyen Portland State University Introduction to Mathematical Analysis I MTH 311

Contents 1 TOOLS FOR ANALYSIS. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7 1.1 Basic Concepts of Set Theory 7 1.2 Functions 11 1.3 The Natural Numbers and Mathematical Induction 15 1.4 Ordered Field Axioms 19 1.5 The Completeness Axiom for the Real Numbers 23 1.6 Applications of the Completeness Axiom 27 2 SEQUENCES . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33 2.1 Convergence 33 2.2 Limit Theorems 41 2.3 Monotone Sequences 45 2.4 The Bolzano-Weierstrass Theorem 51 2.5 Limit Superior and Limit Inferior 54 3 LIMITS AND CONTINUITY . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61 3.1 Limits of Functions 61 3.2 Limit Theorems 67 3.3 Continuity 74 3.4 Properties of Continuous Functions 78

3.5 Uniform Continuity 83 4 DIFFERENTIATION. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 89 4.1 Definition and Basic Properties of the Derivative 89 4.2 The Mean Value Theorem 95 4.3 Some Applications of the Mean Value Theorem 102 4.4 L’Hôpital’s Rule 104 4.5 Taylor’s Theorem 111 5 ADDITIONAL TOPICS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 115 5.1 Topology of the Real Line 115 5.2 Continuity and Compactness 122 5.3 Limit Superior and Limit Inferior of Functions 125 5.4 Lower Semicontinuity and Upper Semicontinuity 131 5.5 Convex Functions and Derivatives 136 5.6 Nondifferentiable Convex Functions and Subdifferentials 141 References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 151 Solutions and Hints for Selected Exercises . . . . . . . . . . . . . . . . . . . . . . 153

Preface Our goal in this set of lecture notes is to provide students with a strong foundation in mathematical analysis. Such a foundation is crucial for future study of deeper topics of analysis. Students should be familiar with most of the concepts presented here after completing the calculus sequence. However, these concepts will be reinforced through rigorous proofs. The lecture notes contain topics of real analysis usually covered in a 10-week course: the completeness axiom, sequences and convergence, continuity, and differentiation. In addition, the notes include many carefully selected exercises of various levels of difficulty. Hints and solutions to selected exercises are available in the back of the book. For each section, there is at least one exercise with hints or fully solved. For those exercises, besides the solutions, there are explanations about the process itself and examples of more general problems where the same technique may be used. Exercises with solutions are indicated by a ▶and those with hints are indicated by a ▷. The last chapter contains additional topics. These include topological properties of the real line, generalizations of the extreme value theorem and more contemporary topics that expand on the notions of continuity or optimization. Lower and upper semicontinuity, differentiation of convex functions, and generalized differentiation of non-differentiable convex functions can be used as optional mathematical projects. Finally, to make it easier for students to navigate the text, the electronic version of these notes contains many hyperlinks that students can click on to go to a definition, theorem, example, or exercise at a different place in the notes. These hyperlinks can be easily recognized because the text or number is on a different color and the mouse pointer changes shape when going over them. Changes in the Third Edition This third edition includes a number of improvements based on recommendations from students and colleagues and on our own experience teaching the course over the last several years. We reorganized the narrative in multiple sections. More significantly, the first four chapters now offer a streamlined presentation of the main topics without going into more abstract topological properties of the real line. While various definitions such as limits and continuity are defined in some

6 generality for functions defined on arbitrary subsets of R, the main results are stated and proved for functions defined on intervals. We leave the notions of open sets, closed sets and compact sets to Chapter 5. We also moved to this chapter various results about continuous functions defined on compact sets. A second important feature of this edition is the addition of more detailed explanations on various techniques for proving limits (of sequences and of functions) directly from the definition. These explanations are intended to make the text more welcoming for students who are tackling various proofs in analysis for the first time. Finally, we corrected a number of typos that have stubbornly persisted in the second edition. We have used these notes multiple times to teach the one-quarter course Introduction to Mathematical Analysis I at Portland State University. We are currently completing the second volume that will include Riemann integration and series as well as additional special topics for exploration. Acknowledgements We would like to thank our colleagues in the Fariborz Maseeh Department of Mathematics and Statistics for their constructive feedback and thoughtful insights. We also want to acknowledge the many students in our courses who offered suggestions. Finally, we give special thanks to Karen Bjork, Head of Digital Initiatives, Cataloging, & eAccess at the Portland State University Library, for her continuing support for this project. The creation and update of this textbook were supported in part by a PSU faculty enhancement grant and by an OER grant.

Basic Concepts of Set Theory Functions The Natural Numbers and Mathematical Induction Ordered Field Axioms The Completeness Axiom for the Real Numbers Applications of the Completeness Axiom 1. TOOLS FOR ANALYSIS This chapter discusses various mathematical concepts and constructions which are central to the study of many fundamental results in analysis. Generalities are kept to a minimum in order to move quickly to the heart of analysis: the structure of the real number system and the notion of limit. The reader should consult the bibliographical references for more details. 1.1 Basic Concepts of Set Theory Intuitively, a set is a collection of objects with certain properties. The objects in a set are called the elements or members of the set. We usually use uppercase letters to denote sets and lowercase letters to denote elements of sets. If ais an element of a set A, we write a∈A. If ais not an element of a set A, we write a̸∈A. To specify a set, we can list all of its elements, if possible, or we can use a defining rule. For instance, to specify the fact that a set Acontains four elements a,b,c,d, we write A={a,b,c,d}. To describe the set E containing all even integers, we write E={x : x =2k for some integer k}. We say that a set Ais a subset of a set Bif every element of Ais also an element of B, and write A⊂Bor B⊃A. Two sets are equal if they contain the same elements. If Aand Bare equal, we write A=B. The following result is straightforward and very convenient for proving equality between sets. Theorem 1.1.1 Two sets AandBare equal if and only if A⊂BandB⊂A. If A⊂BandAdoes not equal B, we say that Ais a proper subset of B, and write A⊊B.

8 1.1 Basic Concepts of Set Theory The set 0/ ={x : x̸ =x} is called the empty set. This set clearly has no elements. Using Theorem 1.1.1, it is easy to show that all sets with no elements are equal. Thus, we refer tothe empty set. Throughout this book, we will discuss several sets of numbers which should be familiar to the reader: • N={1,2,3, . . .}, the set of natural numbers or positive integers. • Z={0,1,−1,2,−2, . . .}, the set of integers (that is, the natural numbers together with zero and the negative of each natural number). • Q={m/n: m,n∈Z,n̸=0}, the set of rational numbers. • R, the set of real numbers. • Intervals. For a,b∈R, a≤b, we define: [a,b] ={x ∈R: a≤x ≤b}. (a,b) ={x ∈R: a<x <b}. [a,b) ={x ∈R: a≤x <b}. (a,b] ={x ∈R: a<x ≤b}. We call intervals of the form[a,b] closed intervals and intervals of the form(a,b) open intervals. Moreover, we will use the symbols ∞and−∞in the following definitions: [a,∞) ={x ∈R: a≤x}. (−∞,b] ={x ∈R: x ≤b}. (a,∞) ={x ∈R: a<x}. (−∞,b) ={x ∈R: x <b}. We will say more about the symbols ∞and−∞in Section 1.5. Since the real numbers are central to the study of analysis, we will discuss them in great detail in Sections 1.4, 1.5, and 1.6. For two sets Aand B, the union, intersection, difference, andsymmetric difference of Aand Bare given respectively by: A∪B={x : x ∈Aor x ∈B}. A∩B={x : x ∈Aandx ∈B}. A\B={x : x ∈Aandx / ∈B}. A∆B= (A\B)∪(B\A). If A∩B=0/ , we say that AandBare disjoint. The difference of Aand Bis also called the complement of Bin A. If X is a universal set, that is, a set containing all the objects under consideration, then the complement of AinX is denoted simply by Ac.

9 Theorem 1.1.2 Let A,B, andCbe subsets of a universal set X. Then the following hold: (i) A∪Ac =X. (ii) A∩Ac =0/ . (iii) (Ac)c =A. (iv) (Distributive law) A∩(B∪C) = (A∩B)∪(A∩C). (v) (Distributive law) A∪(B∩C) = (A∪B)∩(A∪C). (vi) (DeMorgan’s law) A\(B∪C) = (A\B)∩(A\C). (vii) (DeMorgan’s law) A\(B∩C) = (A\B)∪(A\C). (viii) A\B=A∩Bc. Proof: We prove some of the results and leave the rest for the exercises. (i) Clearly, A∪Ac ⊂X since both Aand Ac are subsets of X. Now let x ∈X. Then either x is an element of Aor it is not an element of A. In the first case, x ∈Aand, so, x ∈A∪Ac. In the second case, x ∈Ac and, so, x ∈A∪Ac. Thus, X ⊂A∪Ac. Applying Theorem (1.1.1) it follows that A∪Ac =X. (ii) No element x can be simultaneously inAand not in A. Thus, A∩Ac =0/ . (iv) Let x ∈A∩(B∪C). Then x ∈A and x ∈B∪C. Therefore, x ∈B or x ∈C. In the first case, since x is also in Awe get x ∈A∩Band, hence, x ∈(A∩B)∪(A∩C). In the second case, x ∈A∩Cand, hence, x ∈(A∩B)∪(A∩C). Thus, in all cases, x ∈(A∩B)∪(A∩C). This shows A∩(B∪C) ⊂(A∩B)∪(A∩C). Now we prove the other inclusion. Let x ∈(A∩B)∪(A∩C). Then x ∈A∩B or x ∈A∩C. In either case, x ∈A. In the first case, x ∈B and, hence, x ∈B∪C. It follows in this case that x∈A∩(B∪C). In the second case, x∈Cand, hence, x∈B∪C. Again, we conclude x∈A∩(B∪C). Therefore, (A∩B)∪(A∩C) ⊂A∩(B∪C). Applying Theorem (1.1.1) the equality follows. □ A set whose elements are sets is often called a collection/family of sets and is often denoted by script letters such as A or B. Let I be a nonempty set such that to eachi ∈I corresponds a set Ai. Then the family of all sets Ai as i ranges over I is denoted by {Ai : i ∈I}. Such a family of sets is called anindexed family and the set I is called the index set. Consider the indexed family of sets {Ai : i ∈I}. The union and intersection of this family as i ranges over I is defined respectively by [ i∈I Ai ={x : x ∈Ai for some i ∈I} and \ i∈I Ai ={x : x ∈Ai for every i ∈I}. ■ Example 1.1.1 The following examples illustrate the notation. (a) Let the index set be I =Nand for each n∈Nwe have An = [−n,n]. Then [ n∈N An =R and \ n∈N An = [−1,1].

10 1.1 Basic Concepts of Set Theory (b) Here we let the index set be J = (0,1] and for each s ∈J we have As = (−s,s). Then [ s∈J As = (−1,1) and \ s∈J As ={0}. The proofs of the following properties are similar to those in Theorem 1.1.2. We include the proof of part (i) and leave the rest as an exercise. Theorem 1.1.3 Let {Ai : i ∈I}be an indexed family of subsets of a universal set X and let Bbe a subset of X. Then the following hold: (i) B∪ Ti∈I Ai =Ti∈I(B∪Ai). (ii) B∩ Si∈I Ai =Si∈I(B∩Ai). (iii) B\ Ti∈I Ai =Si∈I(B\Ai). (iv) B\ Si∈I Ai =Ti∈I(B\Ai). (v) Ti∈I Ai c =Si∈I Ac i . (vi) Si∈I Ai c =Ti∈I Ac i . Proof of (i): Let x ∈B∪ Ti∈I Ai . Thenx ∈Bor x ∈Ti∈I Ai. If x ∈B, thenx ∈B∪Ai for all i ∈I and, thus, x ∈Ti∈I(B∪Ai). If x ∈Ti∈I Ai, then x ∈Ai for all i ∈I. Therefore, x ∈B∪Ai for all i ∈I and, hence, x ∈Ti∈I(B∪Ai). We have thus showed B∪ Ti∈I Ai ⊂Ti∈I(B∪Ai). Now let x ∈Ti∈I(B∪Ai). Then x ∈B∪Ai for all i ∈I. If x ∈B, then x ∈B∪ Ti∈I Ai . If x̸ ∈B, then we must have that x∈Ai for all i ∈I. Therefore, x∈Ti∈I Ai and, hence, x∈B∪ Ti∈I Ai . This proves the other inclusion and, so, the equality. □ We want to consider pairs of objects in which the order matters. Given objects aand b, we will denote by (a,b) the ordered pair where ais the first element and bis the second element. The main characteristic of ordered pairs is that (a,b) = (c,d) if and only if a=c and b=d. Thus, the ordered pair (0,1) represents a different object than the pair (1,0) (while the set {0,1}is the same as the set {1,0})1. Given two sets AandB, the Cartesian product of AandBis the set defined by A×B:={(a,b) : a∈Aandb∈B}. ■ Example 1.1.2 If A={1,2}andB={−2,0,1}, then A×B={(1,−2),(1,0),(1,1),(2,−2),(2,0),(2,1)}. ■ Example 1.1.3 If A and B are the intervals [−1,2] and [0,7] respectively, then A×B is the rectangle [−1,2] ×[0,7] ={(x,y): −1≤x ≤2, 0≤y ≤7}. We will make use of cartesian products in the next section when we discuss functions. 1For a precise definition of ordered pair in terms of sets see [Lay13]

11 Exercises 1.1.1 Prove the remaining items in Theorem 1.1.2. 1.1.2 ▶Let Y andZ be subsets of X. Prove that (X\Y)∩Z=Z\(Y∩Z). 1.1.3 Prove the remaining items in Theorem 1.1.3. 1.1.4 Let A, B, C, and Dbe sets. Prove the following. (a) (A∩B)×C= (A×C)∩(B×C). (b) (A∪B)×C= (A×C)∪(B×C). (c) (A×B)∩(C×D) = (A∩C)×(B∩D). 1.1.5 Let A⊂X and B⊂Y. Determine if the following equalities are true and justify your answer: (a) (X×Y)\(A×B) = (X\A)×(Y\B). (b) (X×Y)\(A×B) = [(X\A)×Y] ∪[X×(Y\B)]. 1.2 Functions Definition 1.2.1 Let X and Y be sets. Afunction from X into Y is a subset f ⊂X×Y with the following properties: (i) For all x ∈X there is y ∈Y such that (x,y) ∈ f . (ii) If (x,y) ∈ f and(x,z) ∈ f , then y =z. The set X is called the domainof f , the set Y is called the codomainof f , and we write f : X →Y. The range of f is the subset of Y defined by{y ∈Y : there is x ∈X such that (x,y) ∈ f }. It follows from the definition that, for eachx ∈X, there is exactly one element y ∈Y such that (x,y) ∈ f . We will write y = f (x). If x ∈X, the element f (x) is called the value of f at x or the image of x under f . Note that, in this definition, a function is a collection of ordered pairs and, thus, corresponds to the geometric interpretation of the graph of a function given in calculus. In fact, we will refer indistinctly to the function f or to the graph of f . Both refer to the set {(x, f (x)) : x ∈X}. Let f : X →Y and g: X →Y be two functions. Then the two functions are equal if they are equal as subsets of X×Y. It is easy to see that f equals gif and only if f (x) =g(x) for all x ∈X. Note that it is implicit in the definition that two equal functions must have the same domain and codomain. Let f : X →Y be a function and let Abe a subset of X. The restriction of f on A, denoted by f|A, is a new function fromAintoY given by f|A(a) =f (a) for all a∈A.

12 1.2 Functions Definition 1.2.2 Let f : X →Y be a function. (i) We say f is surjective (or maps X onto Y) if for every element y ∈Y, there exists an element x ∈X such that f (x) =y. (ii) We say f is injective (or one-to-one) if whenever x1,x2 ∈X, x1̸ =x2, then f (x1)̸ = f (x2). Equivalently, f is one-to-one if for all x1,x2 ∈X with f (x1) =f (x2), it follows that x1 =x2. (iii) If f is both surjective and injective, we say f is bijective (or a one-to-one correspondence). Remark 1.2.1 When the function f is a bijection, for anyy∈Y, there exists a unique element x ∈X such that f (x) =y. This element x is then denoted by f −1(y). In this way, we already built a function fromY toX called the inverse of f . Theorem 1.2.1 Let f : X →Y. If there are two functions g: Y →X and h: Y →X such that g(f (x)) =x for every x ∈X and f (h(y)) =y for every y ∈Y, then f is bijective and g=h=f −1. Proof: First we prove that f is surjective. Let y ∈Y and set x =h(y). Then, from the assumption on h, we have f (x) =f (h(y)) =y. This shows that f is surjective. Next we prove that f is injective. Let x1,x2 ∈X be such that f (x1) = f (x2). Then x1 = g(f (x1)) =g(f (x2)) =x2. Thus, f is injective. We have shown that for each y ∈Y, there is a unique x ∈X, which we denote f −1(y), such that f (x) =y. Since for such a y, g(y) =g(f (x)) =x, we obtaing(y) =f −1(y). Since f (h(y)) =y, we also conclude that h(y) =x =f −1(y). □ ■Example 1.2.1 Consider the function f : (1,2] →[3,4) given by f (x) =4−(x−1) 2. We show that f is bijective. First let x1,x2 ∈(1,2] be such that f (x1) =f (x2). That is, 4−(x1−1) 2 =4−(x 2−1) 2. Then (x1 −1) 2 = (x 2 −1) 2. Since bothx 1 >1 and x2 >1, we conclude that x1 −1=x2 −1 and, so, x1 =x2. This proves f is injective. Next let y∈[3,4). We need to findx∈(1,2] such that f (x) =y. For that, we set up 4−(x−1)2 = y and solve for x. We get, x =√4−y+1. Note that since y <4, 4−y has a square root. Also note that since 3≤y <4, we have 1≥4−y >0 and, hence, 2≥ √4−y+1>1. Therefore, x ∈(1,2]. This proves f is surjective. Definition 1.2.3 Let f : X →Y be a function and let Abe a subset of X. Then the image of A under f is given by f (A) ={f (a) : a∈A}. It follows from the definition that f (A) ={b∈Y : b= f (a) for some a∈A}. Moreover, f is surjective if and only if f (X) =Y. Definition 1.2.4 Let f : X →Y be a function and let Bbe a subset of Y. Then the preimage of B under f is given by f −1(B) ={x ∈X : f (x) ∈B}. Remark 1.2.2 Note that, despite the notation, the definition of preimage does not require the function to have an inverse. It does not even require the function to be injective. The examples below illustrate these concepts.

13 ■ Example 1.2.2 Let f : R→Rbe given by f (x) =3x−1. Let A= [0,2) andB={1,−4,5}. Then f (A) = [−1,5) and f −1(B) ={2 3,−1,2}. ■ Example 1.2.3 Let f : R→Rbe given by f (x) =−x+7. Let A= [0,2) and B= (−∞,3]. Then f (A) = (5,7] and f −1(B) = [4,∞). ■ Example 1.2.4 Let f : R→Rbe given by f (x) =x 2. Let A= (−1,2) and B= [1,4). Then f (A) = [0,4) and f −1(B) = (−2,−1] ∪[1,2). Theorem 1.2.2 Let f : X →Y be a function, let Abe a subset of X, and let Bbe a subset of Y. The following hold: (i) A⊂f −1(f (A)). (ii) f (f −1(B)) ⊂B. Proof: We prove (i) and leave (ii) as an exercise. Let x ∈A. By the definition of image, f (x) ∈ f (A). Now, by the definition of preimage, x ∈ f −1(f (A)). □ Theorem 1.2.3 Let f : X →Y be a function, let A,B⊂X, and let C,D⊂Y. The following hold: (i) If C⊂D, then f −1(C) ⊂f −1(D). (ii) f −1(D\C) =f −1(D)\ f −1(C). (iii) If A⊂B, then f (A) ⊂f (B). (iv) f (A\B) ⊃f (A)\ f (B). Proof: We prove (ii) and leave the other parts as an exercise. We show first f −1(D\C) ⊂f −1(D)\ f −1(C). Let x ∈ f −1(D\C). Then, from the definition of inverse image, we get f (x)∈D\C. Thus, f (x)∈Dand f (x)̸∈C. Hencex∈f −1(D) andx̸∈f −1(C). We conclude that x ∈ f −1(D)\ f −1(C). Next we prove f −1(D)\ f −1(C) ⊂f −1(D\C). Let x∈ f −1(D)\ f −1(C). Thus, x∈ f −1(D) and x̸∈ f −1(C). Therefore, f (x) ∈Dand f (x)̸ ∈C. This means f (x) ∈D\Cand, so, x∈ f −1(D\C). □ Theorem 1.2.4 Let f : X →Y be a function, let {Aα}α∈I be an indexed family of subsets of X, and let {Bβ}β∈J be an indexed family of subsets of Y. The following hold: (i) f (Sα∈I Aα) =Sα∈I f (Aα). (ii) f (Tα∈I Aα) ⊂Tα∈I f (Aα). (iii) f −1(Sβ∈J Bβ) =Sβ∈J f − 1(Bβ). (iv) f −1(Tβ∈J Bβ) =Tβ∈J f − 1(Bβ). Proof: We prove (i) and leave the other parts as an exercise. First we show f (Sα∈I Aα) ⊂Sα∈I f (Aα). Let y ∈ f (Sα∈I Aα). From the definition of image of a set, there is x ∈Sα∈I Aα such that y =f (x). From the definition of union of a family of sets, there is α0 ∈I such that x ∈Aα0. Therefore, y =f (x) ∈ f (Aα0) and, so, y ∈Sα ∈I f (Aα).

14 1.2 Functions We now prove that Sα∈I f (Aα) ⊂f (Sα∈I Aα). Let y∈Sα∈I f (Aα). From the definition of union of a family of sets, there is α0 ∈I such that y ∈ f (Aα0). From the definition of image of a set, there is x ∈Aα0 such that y =f (x). We have x ∈Aα0 ⊂Sα∈I Aα. Therefore, y =f (x) ∈ f (Sα∈I Aα). By Theorem 1.1.1 the result follows. □ Definition 1.2.5 Let f : X →Y and g: Y →Z be two functions. Then the composition function g◦ f of f and g is the function g◦ f : X →Z given by (g◦ f )(x) =g(f (x)) for all x ∈X. Theorem 1.2.5 Let f : X →Y andg: Y →Z be two functions and let B⊂Z. The following hold: (i) (g◦ f )−1(B) =f −1(g−1(B)). (ii) If f andgare injective, then g◦ f is injective. (iii) If f andgare surjective, theng◦ f is surjective. (iv) If g◦ f is injective, then f is injective. (v) If g◦ f is surjective, then gis surjective. Proof: We prove (iv) and leave the other parts as an exercise. Suppose g◦ f is injective and let x1,x2 ∈X be such that f (x1) = f (x2). Then (g◦ f )(x1) = g(f (x1)) =g(f (x2)) = (g◦ f )(x2). Since g◦ f is injective, it follows that x1 =x2. We conclude that f is injective. □ Definition 1.2.6 Asequence of elements of a set Ais a function with domain Nand codomain A. We discuss sequences in detail in Chapter 2. Definition 1.2.7 We say that set Ais finite if it is empty or if there exists a natural number nand a one-to-one correspondence f : A→ {1,2, . . . ,n}. A set is infinite if it is not finite. We leave it as an exercise to prove that the union of two finite sets is finite. It is also easy to show, by contradiction, that Nis infinite. Exercises 1.2.1 ▶Let f : X →Y be a function. Prove that: (a) If f is one-to-one, then A=f −1(f (A)) for every subset Aof X. (b) If f is onto, then f (f −1(B)) =Bfor every subset Bof Y. 1.2.2 Let f : R→Rbe given by f (x) =x2−3 and let A= [−2,1) andB= (−1,6). Find f (A) and f −1(B). 1.2.3 Prove that each of the following functions is bijective. (a) f : (−∞,3] →[−2,∞) given by f (x) =|x−3| −2. (b) g: (1,2) →(3,∞) given by g(x) =3/(x−1). 1.2.4 Prove that if f : X →Y is injective, then the following hold: (a) f (A∩B) =f (A)∩f (B) for A,B⊂X.

15 (b) f (A\B) =f (A)\ f (B) for A,B⊂X. 1.2.5 Prove part (ii) of Theorem 1.2.2. 1.2.6 Prove the remaining parts of Theorem 1.2.3. 1.2.7 Prove the remaining parts of Theorem 1.2.4. 1.2.8 Prove the remaining parts of Theorem 1.2.5. 1.2.9 Prove that the union of two finite sets is finite. Hint: it is easier to show when the sets are disjoint. 1.3 The Natural Numbers and Mathematical Induction We will assume familiarity with the set Nof natural numbers, with the usual arithmetic operations of addition and multiplication onN, and with the notion of what it means for one natural number to be less than another. In addition, we will also assume the following property of the natural numbers. Well-Ordering Property of the Natural Numbers: If Ais a nonempty subset of N, then there exists an element ℓ ∈Asuch that ℓ ≤x for all x ∈A. To paraphrase the previous property, every nonempty subset of positive integers has a smallest element. The principle of mathematical induction is a useful tool for proving facts about sequences. Theorem 1.3.1 — Principle of Mathematical Induction. For each natural number n∈N, suppose that P(n) denotes a proposition which is either true or false. Let A={n∈N: P(n) is true}. Suppose the following conditions hold: (i) 1∈A. (ii) For eachk ∈N, if k ∈A, then k+1∈A. Then A=N. Proof: Suppose conditions (i) and (ii) hold. Assume, by way of contradiction, that A̸=N. Set B=N\A, that is, B={n ∈N: P(n) is false}. Then B is a nonempty subset of N. By the WellOrdering Property of the natural numbers, there exists a smallest element ℓ ∈B. By condition (i), 1̸∈B. Hence, ℓ ≥2. It follows that k =ℓ−1 is a natural number. Since k < ℓ, thenk̸ ∈Band, hence, we have that P(k) is true. By condition (ii), we obtain that P(k+1) is true. But k+1=ℓ, andP(ℓ) is false, since ℓ ∈B. This is a contradiction, so the conclusion follows. □ The principle says that, given a list of propositions P(n), one for each n∈N, if P(1) is true and, moreover, P(k+1) is true whenever P(k) is true, then all propositions are true. We will refer to this principle as mathematical induction or simply induction. Condition (i) above is called the base case and condition (ii) the inductive step. When proving (ii), the statement P(k) is called the inductive hypothesis. ■ Example 1.3.1 Prove using induction that 1+2+· · ·+n= n(n+1) 2 for all n∈N.

16 1.3 The Natural Numbers and Mathematical Induction The statement P(n) is the equality 1+2+· · · +n=n(n+1) 2 for n∈N. Now the base case says that 1=1(1+1) 2 , which is clearly true. Suppose P(k) is true for some k ∈N. That is, suppose that 1+2+· · · +k =k(k+1) 2 for some k ∈N(this is the inductive hypothesis). Now we have 1+2+· · ·+k+(k+1) = k(k+1) 2 +(k+1) = k(k+1)+2(k+1) 2 = (k+1)(k+2) 2 . This shows that P(k+1) is true. We have now proved conditions (i) and (ii) of Theorem 1.3.1. Therefore, by the principle of mathematical induction we conclude that 1+2+· · ·+n= n(n+1) 2 for all n∈N. ■ Example 1.3.2 Prove using induction that 7 n −2n is divisible by 5 for all n∈N. The statement P(n) is 7n−2n is divisible by 5 for n∈N. For n=1, we have 7−2=5, which is clearly a multiple of 5. Therefore the base case is true. Suppose now that P(k) is true for some k∈N. That is, there is an integer j such that 7k −2k =5j. It follows that 7k =2k +5j. Now, substituting this expression below, we have 7k+1−2k+1 =7·7k−2·2k =7(2k+5j)−2·2k =7·2k−2·2k+7·5j =2k(7−2)+5·7j =5(2k+7j). Hence 7k+1 −2k+1 is a multiple of 5. This completes the proof of the inductive step. We conclude by induction that 7n −2n is divisible by 5 for all n∈N. ■ Example 1.3.3 Prove using induction that n+1≤2n for all n∈N. The statement P(n) is the inequality n+1≤2n for n∈N. For n=1, we have 1+1=2≤21, so the base case is true. Suppose next that P(k) is true for some that k ∈N. That is, k+1≤2k for some k ∈N. Then (k+1)+1≤2k +1. Since 2k is a positive integer, we also have 1≤2k. Therefore, (k+1)+1≤2k +1≤2k +2k =2· 2k =2k+1. This completes the proof of the inductive step. We conclude by the principle of mathematical induction that n+1≤2n for all n∈N. The following result is known as the Generalized Principle of Mathematical Induction. It simply states that we can start the induction process at any integer n0, and then we obtain the truth of all statements P(n) for n≥n0. Theorem 1.3.2 — Generalized Principle of Mathematical Induction. Let n0 ∈N and for each natural n ≥n0, suppose that P(n) denotes a proposition which is either true or false. Let A={n∈N: P(n) is true}. Suppose the following two conditions hold: (i) n0 ∈A. (ii) For eachk ∈N, k ≥n0, if k ∈A, then k+1∈A.

17 Then {k ∈N: k ≥n0} ⊂A. Proof: Suppose conditions (i) and (ii) hold. Assume, by way of contradiction, that the conclusion is false. That is, {k ∈N: k ≥n0}̸ ⊂A. Set B={n ∈N: n ≥n0 andP(n) is false}. Then B is a nonempty subset of N. By the Well-Ordering Property of the natural numbers, there exists a smallest element ℓ ∈B. By condition (i), n0̸ ∈B. Hence, ℓ ≥n0 +1. It follows that k =ℓ−1≥n0. Since k < ℓ, k̸ ∈Band, so, we have that P(k) is true. By condition (ii), we obtain that P(k+1) is true. But k+1=ℓ, and P(ℓ) is false, since ℓ ∈B. This is a contradiction, so the conclusion follows. □ ■ Example 1.3.4 Prove by induction that 3n<2 n for all n≥4. Let P(n) be the statement 3n<2n for n∈N, n≥4. P(n) is true for n=4 since 12<16. Suppose next that P(k) is true for some k ∈N, k ≥4. That is, 3k <2k for some k ∈N, k ≥4. Now, 3(k+1) =3k+3<2k +3<2k +2k =2k+1, where the second inequality follows since k ≥4 and, so, 2k ≥16>3. This shows that P(k+1) is true. Thus, by the Generalized Principle of Induction, 3n<2n for all n≥4. Next we present another variant of the induction principle which makes it easier to prove the inductive step. Despite its name, this principle is equivalent to the standard one. Theorem 1.3.3 — Principle of Strong Induction. For each natural n∈N, suppose that P(n) denotes a proposition which is either true or false. Let A={n∈N: P(n) is true}. Suppose the following two conditions hold: (i) 1∈A. (ii) For eachk ∈N, if 1,2, . . . ,k ∈A, then k+1∈A. Then A=N. Remark 1.3.1 Note that the inductive step above says that, in order to prove P(k+1) is true, we may assume not only that P(k) is true, but also that P(1), P(2),. . . ,P(k−1) are true. There is also a generalized version of this theorem where the condition (i) is replaced by n0 ∈A, for some integer n0 >1. We illustrate this theorem with the following example. ■ Example 1.3.5 Prove by induction that every positive integer greater than 1 is either a prime number or a product of prime numbers. Let P(n) be the statement nis either a prime number or a product of prime numbers, for n∈N, n≥2. Clearly, the statement is true for n=2. Suppose that P(2), P(3),. . . ,P(k) are true for some k ∈N, k ≥2. If k+1 is prime, then P(k+1) is true and the result follows. Otherwise, there are positive integers p,q>1 such that k+1=pq. Since p,q≤k, by the inductive assumption applied to both pandqwe can find prime numbers r1,. . . ,rℓ ands1,. . . ,smsuch that p=r1· · ·rℓ andq=s1· · ·sm (note that ℓ andmmay both equal 1). But then k+1=r1· · ·rℓs1· · ·sm. Thus, the statement holds true for k+1. The conclusion now follows by the generalized principle of strong induction.

18 1.3 The Natural Numbers and Mathematical Induction Exercises 1.3.1 Prove the following using Mathematical Induction. (a) 12 +22 +· · ·+n2 =n(n+1)(2n+1) 6 for all n∈N. (b) 13 +23 +· · ·+n3 =n 2(n+1)2 4 for all n∈N. (c) 1+3+· · ·+(2n−1) =n2 for all n∈N. 1.3.2 Prove the following using Mathematical Induction. (a) 9n −5n is divisible by 4 for all n∈N. (b) 7n −1 is divisible by 3 for all n∈N. (c) 32n −1 is divisible by 8 for all n∈N. (d) xn −yn is divisible by x−y for all n∈Nwhere x,y ∈Z, x̸ =y. 1.3.3 Prove the following using Mathematical Induction. (a) 1+3n≤4n for all n∈N. (b) 1+2n≤2n for all n∈N, n≥3. (c) n2 ≤3n for all n∈N. (d) n3 ≤3n for all n∈N. (Hint: Check the cases n=1 andn=2 directly and then use induction for n≥3.) 1.3.4 Given a real number a̸=1, prove that 1+a+a2 +· · ·+an = 1−an+1 1−a for all n∈N. 1.3.5 ▶The Fibonacci sequence is defined by a1 =a2 =1 and an+2 =an+1 +an for n≥1. Prove that an = 1 √5h 1+√5 2 n − 1− √5 2 ni. 1.3.6 Let a≥ −1. Prove by induction that (1+a)n ≥1+na for all n∈N. 1.3.7 ▷Let a,b∈Randn∈N. Use Mathematical Induction to prove the binomial theorem (a+b)n = n ∑ k=0 n k akbn−k, where n k = n! k!(n−k)! .

19 1.4 Ordered Field Axioms In this section and the next we present an axiomatic description of the set of real numbers. That is, we will assume that there exists a set, denoted byR, satisfying the ordered field axioms, stated below, together with the completeness axiom, presented in the next section. In this way we identify the basic properties that characterize the real numbers. After listing the ordered field axioms we derive from them additional familiar properties of the real numbers. We conclude the section with the definition of absolute value of a real number and with several results about it that will be used often later in the text. We assume the existence of a set R(the set of real numbers) and two operations +and· (addition and multiplication) assigning to each pair of real numbers x,y, unique real numbers x+y andx· y and satisfying the following properties: (A1) (x+y)+z =x+(y+z) for all x,y,z ∈R. (A2) x+y =y+x for all x,y ∈R. (A3) There exists a unique element 0∈Rsuch that x+0=x for all x ∈R. (A4) For eachx ∈R, there exists a unique element −x ∈Rsuch that x+(−x) =0. (M1) (x· y)· z =x· (y· z) for all x,y,z ∈R. (M2) x· y =y· x for all x,y ∈R. (M3) There exists a unique element 1∈Rsuch that 1̸=0 and x· 1=x for all x ∈R. (M4) For each x ∈R\ {0}, there exists a unique element x−1 ∈Rsuch that x· (x−1) =1. (We also write 1/x instead of x−1.) (D1) x· (y+z) =x· y+x· z for all x,y,z ∈R. We often write xy instead of x· y. In addition to the algebraic axioms above, there is a relation <on Rthat satisfies the order axioms below: (O1) For all x,y ∈R, exactly one of the three relations holds: x =y, y <x, or x <y. (O2) For all x,y,z ∈R, if x <y andy <z, then x <z. (O3) For all x,y,z ∈R, if x <y, then x+z <y+z. (O4) For all x,y,z ∈R, if x <y and 0<z, then xz <yz. We will use the notation x ≤y to mean x <y or x =y. We may also use the notation x >y to represent y <x and the notationx ≥y to mean x >y or x =y. A set Ftogether with two operations +and · and a relation<satifying the 13 axioms above is called anordered field. Thus, the real numbers are an example of an ordered field. Another example of an ordered field is the set of rational numbers Qwith the familiar operations and order. The integers Zdo not form a field since for an integer mother than 1 or −1, its reciprocal 1/mis not an integer and, thus, axiom (M4) above does not hold. In particular, the set of positive integers Ndoes not form a field either. As mentioned above the real numbers Rwill be defined as the ordered field which satisfies one additional property described in the next section: the completeness axiom. From these axioms, many familiar properties of Rcan be derived. Some examples are given in the next proposition. The proof illustrates how the given axioms are used at each step of the derivation.

20 1.4 Ordered Field Axioms Proposition 1.4.1 For x,y,z ∈R, the following hold: (i) If x+y =x+z, then y =z. (ii) −(−x) =x. (iii) If x̸ =0 and xy =xz, then y =z. (iv) If x̸ =0, then 1/(1/x) =x. (v) 0x =0=x0. (vi) −x = (−1)x. (vii) x(−z) = (−x)z =−(xz). (viii) If x >0, then−x <0; if x <0, then−x >0. (ix) If x <y andz <0, thenxz >yz. (x) 0<1. Proof: (i) Suppose x+y =x+z. Adding −x (which exists by axiom (A4)) to both sides, we have (−x)+(x+y) = (−x)+(x+z). Then axiom (A1) gives [(−x)+x] +y = [(−x)+x] +z. Thus, again by axiom (A4), 0+y =0+z and, by axiom (A3), y =z. (ii) Since (−x)+x =0, we have (by uniqueness in axiom (A4)) −(−x) =x. The proofs of (iii) and (iv) are similar. (v) Using axiom (D1) we have 0x= (0+0)x=0x+0x. Adding−(0x) to both sides (axiom (A4)) and using axioms (A1) and (A3), we get 0=−(0x)+0x =−(0x)+(0x+0x) = (−(0x)+0x)+0x =0+0x =0x. That 0x =x0 follows from axiom (M2). (vi) Using axioms (M3) and (D1) we get x+ (−1)x =1x+ (−1)x = (1+ (−1))x. From axiom (A4) we get 1+(−1) =0 and from part (v) we get x+(−1)x =0x =0. From the uniqueness in axiom (A4) we get (−1)x =−x as desired. (vii) Using axioms (D1) and (A3) and part (v) we have xz+x(−z) =x(z+(−z)) =x0=0. Thus, using axiom (A4) we get that x(−z) =−(xz). The other equality follows similarly. (viii) Fromx >0, using axioms (O3) and (A3) we have x+(−x) >0+(−x) =−x. Thus, using axiom (A4), we get 0>−x. The other case follows in a similar way. (ix) Since z <0, by part (viii), −z >0. Then, by axiom (O4), x(−z) <y(−z). Combining this with part (vii) we get −xz <−yz. Addingxz+yz to both sides and using axioms (A1), (O3), (A2), and (A3) we get yz = (−xz+xz)+yz =−xz+(xz+yz) <−yz+(xz+yz) =−yz+(yz+xz) = (−yz+yz)+xz =xz.

21 (x) Axiom (M3) gives that 1̸=0. Suppose, by way of contradiction, that 1<0. Then by part (ix), 1· 1 >0· 1. Since 1· 1 =1, by axiom (M3) and 0· 1 =0 by part (v), we get 1 >0 which is a contradiction. It follows that 1>0. □ Note that we can assume that the set of all natural numbers is a subset of R(and of any ordered field, in fact) by identifying the 1 in Nwith the 1 in axiom (M3) above, the number 2 with 1+1, 3 with 1+1+1, etc. Furthermore, since 0<1 (from part (x) of the previous proposition), axiom (O3) gives, 1<2<3, etc. (in particular all these numbers are distinct). In a similar way, we can include ZandQas subsets. We say that a real number x is irrational if x ∈R\Q, that is, if it is not rational. Definition 1.4.1 Given x ∈R, define the absolute value of x by |x| =( x, if x ≥0; −x, if x <0. Figure 1.1: The absolute value function. The following properties of absolute value follow directly from the definition. Proposition 1.4.2 Let x,y,M∈Rand suppose M>0. The following properties hold: (i) |x| ≥0. (ii) | −x| =|x|. (iii) |xy| =|x||y|. (iv) |x| <Mif and only if −M<x <M. (The same holds if <is replaced with≤.) Proof: We prove (iv) and leave the other parts as an exercise. Suppose |x| <M. We consider two cases, x ≥0 and x <0. Suppose first x ≥0. Then |x| =x and since M>0, −M<0. Hence, −M<0≤x =|x| <M. Now suppose x <0. Then |x| =−x. Therefore, −x <Mand, so x >−M. It follows that −M<x <0<M.

22 1.4 Ordered Field Axioms For the converse, suppose−M<x<M. Again, we consider two cases. If x≥0, then|x| =x<M as desired. Next suppose x <0. Now, −M<x implies M>−x. Then |x| =−x <M. The result follows. □ Note that as a consequence of part (iv) above, since |x| ≤ |x| we get −|x| ≤x ≤ |x|. The next theorem will play an important role in the study of limits. Theorem 1.4.3 — Triangle Inequality. Given x,y ∈R, |x+y| ≤ |x| +|y|. Proof: From the observation above, we have −|x| ≤x ≤ |x| and − |y| ≤y ≤ |y|. Adding up the inequalities gives −|x| − |y| ≤x+y ≤ |x| +|y|. Since −|x| − |y| =−(|x| +|y|), the conclusion follows from Proposition 1.4.2 (iv). □ Corollary 1.4.4 For any x,y ∈R, ||x| − |y|| ≤ |x−y|. Remark 1.4.1 The absolute value has a geometric interpretation when considering the numbers in an ordered field as points on a line. The number |x| denotes the distance from the number x to 0. More generally, the number d(x,y) =|x−y| is the distance between the points x andy. It follows easily from Proposition 1.4.2 that d(x,y) ≥0, and d(x,y) =0 if and only if x =y. Moreover, the triangle inequality implies that d(x,y) ≤d(x,z)+d(z,y) for all real numbers x,y,z. Exercises 1.4.1 Prove that nis an even integer if and only if n2 is an even integer. (Hint: prove the “if” part by contraposition, that is, prove that if nis odd, then n2 is odd.) 1.4.2 Prove parts (iii) and (iv) of Proposition 1.4.1 1.4.3 Let x ∈R. Prove that (a) if 0<x <1, thenx2 <x. (b) if x >1, thenx <x2. 1.4.4 Let x,y,z,w∈R. Suppose 0<x <y and 0<z <w. Prove that xz <yw. 1.4.5 Let x,y ∈R. Prove the following. (a) xy ≤1 2(x 2 +y2). (b) If x ≥0 and y ≥0, then√xy ≤1 2(x+y).

23 (c) If n>0, thenxy ≤1 2(nx 2 +1 ny 2). 1.4.6 Prove parts (i), (ii), and (iii) of Proposition 1.4.2. 1.4.7 ▶Prove Corollary 1.4.4. 1.4.8 Given two real numbers x andy, prove that max{x,y}= x+y+|x−y| 2 and min{x,y}= x+y− |x−y| 2 . 1.4.9 Let x,y,M∈R. Prove the following: (a) |x|2 =x2. (b) |x| <Mif and only if x <Mand−x <M. (c) |x+y| =|x| +|y| if and only if xy ≥0. 1.4.10 Let x,y,z ∈R. Prove the following statements. (a) If 0≤x <ε for all ε >0, thenx =0. (b) The following are equivalent: (i) y ≤z. (ii) y <z+ε for all ε >0. 1.5 The Completeness Axiom for the Real Numbers There are many examples of ordered fields. However, we are interested in the field of real numbers. There is an additional axiom that will distinguish this ordered field from all others. In order to introduce our last axiom for the real numbers, we first need some definitions. Definition 1.5.1 Let Abe a subset of R. A number Mis called anupper bound of Aif x ≤Mfor all x ∈A. If Ahas an upper bound, thenAis said to be bounded above. Similarly, a number Lis a lower bound of Aif L≤x for all x ∈A, Ais said to be bounded belowif it has a lower bound. We also say that Ais bounded if it is both bounded above and bounded below. It follows that a set Ais bounded if and only if there exist M∈Rsuch that |x| ≤Mfor all x ∈A (see Exercise 1.5.1). The following concept plays a central role in the study of the real numbers. Definition 1.5.2 — Supremum of a set. Let Abe a nonempty set that is bounded above. We call a number α a least upper bound or supremumof A, if the following conditions hold: (1) x ≤α for all x ∈A(that is, α is an upper bound of A). (2) If Mis an upper bound of A, then α≤M(this means α is smallest among all upper bounds). It can be shown that if Ahas a supremum, then it has only one (see Exercise 1.5.2). In this case, we denote such a number by supA.

24 1.5 The Completeness Axiom for the Real Numbers ■ Example 1.5.1 (a) sup[0,3) =sup[0,3] =3. First consider the set [0,3] ={x ∈R: 0≤x ≤3}. We will show that 3 satisfies conditions (1) and (2) in the definition of supremum. By the definition of the given set, we see that for all x ∈[0,3], x ≤3. Thus 3 is an upper bound. This verifies condition (1). To verify condition (2) suppose that M is an upper bound of [0,3]. Since 3 ∈[0,3], we get 3 ≤M. Therefore condition (2) holds. It follows that 3 is indeed the supremum of [0,3]. Consider next the set [0,3) ={x∈R: 0≤x<3}. We again verify that 3 satisfies conditions (1) and (2) in the definition of supremum. Condition (1) follows as before since 3 is an upper bound of [0,3). For condition (2), because 3 is not in the set we cannot proceed as before. Suppose that Mis an upper bound of [0,3) and assume, by way of contradiction, that 3>M. Since Mis an upper bound of [0,3), we have that M>0. Set x=M+3 2 . Then 0<M= M+M 2 < M+3 2 < 3+3 2 =3 or M<x <3. This is a contradiction since Mis an upper bound of [0,3) and x ∈[0,3). We conclude that 3≤Mand, hence, 3 is the supremum of [0,3). (b) sup{3,5,7,8,10}=10. Clearly 10 is an upper bound of the set. Moreover, any upper boundMmust satisfy 10≤Mas 10 is an element of the set. Thus 10 is the supremum. (c) sup (−1)n n : n∈N = 1 2 . Note that if n∈Nis even, then n≥2 and (−1)n n = 1 n ≤ 1 2 . If n∈Nis odd, then (−1)n n =− 1 n <0< 1 2 . This shows that (−1)n/n≤1 2 for all n∈N. Hence 1/2 an upper bound of the set. Also 1/2 is an element of the set, it follows as in the previous example that 1/2 is the supremum. (d) sup{x2 : −2<x <1, x ∈R}=4. Set A={x2 : −2 <x <1, x ∈R}. If y ∈A, then y =x2 for some x ∈(−2,1) and, hence, |x| <2. Therefore, y =x2 =|x|2 <4. Thus, 4 is an upper bound of A. Suppose Mis an upper bound of Abut M<4. Choose a number y ∈Rsuch that M<y <4 and 0<y. Set x =− √y. Then−2<x <0<1 and, so, y =x2 ∈A. However, y >Mwhich contradicts the fact that Mis an upper bound. Thus 4≤M. This proves that 4=supA. Remark 1.5.1 Let Abe a nonempty subset of R. If there is an element aM∈Asuch that aM≥x for all x ∈A, we say that aM is the maximum of Aand write aM=maxA. If there is an element am∈A such that am≤x for all x ∈A, we say that am is the minimum of Aand write am=minA. It is clear that if Ahas a maximum element then it is bounded above since maxAis an upper bound. Also, if an upper bound belongs to the set then it is the maximum of the set.

25 It should also be noted that a set may have neither a maximum nor a minimum element. Consider for example the set A= (0,1). If a∈(0,1), then 0<a<1 and therefore there are elements b1 and b2 in(0,1) such that 0<b1 <a<b2 <1. This shows that ais neither a maximum nor a minimum element of A. The following proposition is convenient in working with suprema. Proposition 1.5.1 Let Abe a nonempty subset of Rthat is bounded above. Then α=supAif and only if (1’) x ≤α for all x ∈A, (2’) For any ε >0, there exists a∈Asuch that α−ε <a. Proof: Suppose first that α=supA. Then clearly (1’) holds (since this is identical to condition (1) in the definition of supremum). Now let ε >0. Since α−ε <α, condition (2) in the definition of supremum implies that α−ε is not an upper bound of A. Therefore, there must exist an element a of Asuch that α−ε <aas desired. Conversely, suppose conditions (1’) and (2’) hold. Then all we need to show is that condition (2) in the definition of supremum holds. Let Mbe an upper bound of A and assume, by way of contradiction, that M<α. Set ε =α−M. Note that ε >0. By condition (2’), there is a ∈A such that a>α−ε =α−(α−M) =M. This contradicts the fact that Mis an upper bound. The conclusion now follows. □ The following is an axiom of the real numbers and is called the completeness axiom. The Completeness Axiom. Every nonempty subset Aof Rthat is bounded above has a least upper bound. That is, supAexists and is a real number. This axiom distinguishes the real numbers from all other ordered fields and it is crucial in the proofs of the central theorems of analysis. There is a corresponding definition for the infimum of a set. Definition 1.5.3 Let Abe a nonempty subset of Rthat is bounded below. We call a number β a greatest lower bound or infimumof A, denoted by β =infA, if the following conditions hold: (1) x ≥β for all x ∈A(that is, β is a lower bound of A). (2) If Lis a lower bound of A, then β ≥L(this means β is largest among all lower bounds). Using the completeness axiom, we can prove that if a nonempty set is bounded below, then its infimum exists (see Exercise 1.5.5). ■ Example 1.5.2 (a) inf(0,3] =inf[0,3] =0. (b) inf{3,5,7,8,10}=3. (c) inf (−1)n n : n∈N =−1. (d) inf{1+ 1 n : n∈N}=1. (e) inf{x2 : −2<x <1,x ∈R}=0.

26 1.5 The Completeness Axiom for the Real Numbers The following proposition is useful when dealing with infima and its proof is completely analogous to that of Proposition 1.5.1. Proposition 1.5.2 Let Abe a nonempty subset of Rthat is bounded below. Thenβ =infAif and only if (1’) x ≥β for all x ∈A, (2’) For any ε >0, there exists a∈Asuch that a<β+ε. The following is a basic property of suprema. Additional ones are described in the exercises. Theorem 1.5.3 Let A and B be nonempty sets and A⊂B. Suppose B is bounded above. Then supA≤supB. Proof: Let Mbe an upper bound for B, then for x ∈B, x ≤M. In particular, it is also true that x ≤Mfor x ∈Asince A⊂B. Thus, Ais also bounded above. Now, since supBis an upper bound for B, it is also an upper bound for A. Then, by the second condition in the definition of supremum, supA≤supBas desired. □ It will be convenient for the study of limits of sequences and functions to introduce two additional symbols. Definition 1.5.4 The extended real number systemconsists of the real fieldRand the two symbols ∞and−∞. We preserve the original order inRand define −∞<x <∞for every x ∈R The extended real number system does not form an ordered field, but it is customary to make the following conventions: (a) If x is a real number, thenx+∞=∞, x+(−∞) =−∞. (b) If x >0, thenx· ∞=∞, x· (−∞) =−∞. (c) If x <0, thenx· ∞=−∞, x· (−∞) =∞. (d) ∞+∞=∞, −∞+(−∞) =−∞, ∞· ∞= (−∞)· (−∞) =∞, and (−∞)· ∞=∞· (−∞) =−∞. We denote the extended real number set by R. The expressions 0· ∞, ∞+(−∞), and(−∞)+∞ are left undefined. The set Rwith the above conventions will be convenient when describing results about limits in later chapters. Definition 1.5.5 If A̸=0/ is not bounded above inR, we will write supA=∞. If Ais not bounded below in R, we will write infA=−∞. With this definition, every nonempty subset of Rhas a supremum and an infimum in R. To complete the picture we adopt the following conventions for the empty set: sup 0/ =−∞ and inf 0/ =∞. Exercises 1.5.1 Prove that a subset Aof Ris bounded if and only if there is M∈Rsuch that |x| ≤Mfor all x ∈A.

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