Basic Analysis. Introduction to Real Analysis

Transcription

1 Bsic Anlysis Introduction to Rel Anlysis by Jiří Lebl My 29, 2013

2 2 Typeset in LATEX. Copyright c Jiří Lebl This work is licensed under the Cretive Commons Attribution-Noncommercil-Shre Alike 3.0 United Sttes License. To view copy of this license, visit licenses/by-nc-s/3.0/us/ or send letter to Cretive Commons, 171 Second Street, Suite 300, Sn Frncisco, Cliforni, 94105, USA. You cn use, print, duplicte, shre these notes s much s you wnt. You cn bse your own notes on these nd reuse prts if you keep the license the sme. If you pln to use these commercilly (sell them for more thn just duplicting cost), then you need to contct me nd we will work something out. If you re printing course pck for your students, then it is fine if the dupliction service is chrging fee for printing nd selling the printed copy. I consider tht duplicting cost. During the writing of these notes, the uthor ws in prt supported by NSF grnt DMS See for more informtion (including contct informtion).

3 Contents Introduction About this book About nlysis Bsic set theory Rel Numbers Bsic properties The set of rel numbers Absolute vlue Intervls nd the size of R Deciml representtion of the rels Sequences nd Series Sequences nd limits Fcts bout limits of sequences Limit superior, limit inferior, nd Bolzno-Weierstrss Cuchy sequences Series More on series Continuous Functions Limits of functions Continuous functions Min-mx nd intermedite vlue theorems Uniform continuity Limits t infinity Monotone functions nd continuity The Derivtive The derivtive Men vlue theorem

4 4 CONTENTS 4.3 Tylor s theorem Inverse function theorem The Riemnn Integrl The Riemnn integrl Properties of the integrl Fundmentl theorem of clculus The logrithm nd the exponentil Improper integrls Sequences of Functions Pointwise nd uniform convergence Interchnge of limits Picrd s theorem Metric Spces Metric spces Open nd closed sets Sequences nd convergence Completeness nd compctness Continuous functions Fixed point theorem nd Picrd s theorem gin Further Reding 237 Index 239

5 Introduction 0.1 About this book This book is one semester course in bsic nlysis. It strted its life s my lecture notes for teching Mth 444 t the University of Illinois t Urbn-Chmpign (UIUC) in Fll semester Lter I dded the metric spce chpter to tech Mth 521 t University of Wisconsin Mdison (UW). A prerequisite for the course should be bsic proof course, for exmple using [H], [F], or the unfortuntely rther pricey [DW]. It should be possible to use the book for both bsic course for students who do not necessrily wish to go to grdute school (such s UIUC 444), but lso s more dvnced one-semester course tht lso covers topics such s metric spces (such s UW 521). Here re my suggestions for wht to cover in semester course. For slower course such s UIUC 444: 0.3, , , , , , For more rigorous course covering metric spces tht runs quite bit fster (such s UW 521): 0.3, , , , , , , It should lso be possible to run fster course without metric spces covering ll sections of chpters 0 through 6. The pproximte number of lectures given in the section notes through chpter 6 re very rough estimte nd were designed for the slower course. The book normlly used for the clss t UIUC is Brtle nd Sherbert, Introduction to Rel Anlysis third edition [BS]. The structure of the beginning of the book somewht follows the stndrd syllbus of UIUC Mth 444 nd therefore hs some similrities with [BS]. A mjor difference is tht we define the Riemnn integrl using Drboux sums nd not tgged prtitions. The Drboux pproch is fr more pproprite for course of this level. Our pproch llows us to fit within semester nd still spend some extr time on the interchnge of limits nd end with Picrd s theorem on the existence nd uniqueness of solutions of ordinry differentil equtions. This theorem is wonderful exmple tht uses mny results proved in the book. For more dvnced students, mteril my be covered fster so tht we rrive t metric spces nd prove Picrd s theorem using the fixed point theorem s is usul. Other excellent books exist. My fvorite is Rudin s excellent Principles of Mthemticl Anlysis [R2] or s it is commonly nd lovingly clled bby Rudin (to distinguish it from his other 5

6 6 INTRODUCTION gret nlysis textbook). I hve tken lot of inspirtion nd ides from Rudin. However, Rudin is bit more dvnced nd mbitious thn this present course. For those tht wish to continue mthemtics, Rudin is fine investment. An inexpensive nd somewht simpler lterntive to Rudin is Rosenlicht s Introduction to Anlysis [R1]. There is lso the freely downlodble Introduction to Rel Anlysis by Willim Trench [T]. A note bout the style of some of the proofs: Mny proofs trditionlly done by contrdiction, I prefer to do by direct proof or by contrpositive. While the book does include proofs by contrdiction, I only do so when the contrpositive sttement seemed too wkwrd, or when contrdiction follows rther quickly. In my opinion, contrdiction is more likely to get beginning students into trouble, s we re tlking bout objects tht do not exist. I try to void unnecessry formlism where it is unhelpful. Furthermore, the proofs nd the lnguge get slightly less forml s we progress through the book, s more nd more detils re left out to void clutter. As generl rule, I use := insted of = to define n object rther thn to simply show equlity. I use this symbol rther more liberlly thn is usul for emphsis. I use it even when the context is locl, tht is, I my simply define function f (x) := x 2 for single exercise or exmple. Finlly, I would like to cknowledge Jn Mříková, Glen Pugh, Pul Vojt, Frnk Betrous, nd Sönmez Şhutoğlu for teching with the book nd giving me useful feedbck. Frnk Betrous wrote the University of Pittsburgh version extensions, which served s inspirtion for mny of the recent dditions. I would lso like to thnk Dn Stonehm, Jeremy Sutter, Eliy Gwett, Dniel Alrcon, Steve Hoerning, Yi Zhng, Nicole Cviris, n nonymous reder, nd in generl ll the students in my clsses for suggestions nd finding errors nd typos.

7 0.2. ABOUT ANALYSIS About nlysis Anlysis is the brnch of mthemtics tht dels with inequlities nd limits. The present course dels with the most bsic concepts in nlysis. The gol of the course is to cquint the reder with rigorous proofs in nlysis nd lso to set firm foundtion for clculus of one vrible. Clculus hs prepred you, the student, for using mthemtics without telling you why wht you hve lerned is true. To use, or tech, mthemtics effectively, you cnnot simply know wht is true, you must know why it is true. This course shows you why clculus is true. It is here to give you good understnding of the concept of limit, the derivtive, nd the integrl. Let us use n nlogy. An uto mechnic tht hs lerned to chnge the oil, fix broken hedlights, nd chrge the bttery, will only be ble to do those simple tsks. He will be unble to work independently to dignose nd fix problems. A high school techer tht does not understnd the definition of the Riemnn integrl or the derivtive my not be ble to properly nswer ll the student s questions. To this dy I remember severl nonsensicl sttements I herd from my clculus techer in high school, who simply did not understnd the concept of the limit, though he could do ll problems in clculus. We strt with discussion of the rel number system, most importntly its completeness property, which is the bsis for ll tht comes fter. We then discuss the simplest form of limit, the limit of sequence. Afterwrds, we study functions of one vrible, continuity, nd the derivtive. Next, we define the Riemnn integrl nd prove the fundmentl theorem of clculus. We discuss sequences of functions nd the interchnge of limits. Finlly, we give n introduction to metric spces. Let us give the most importnt difference between nlysis nd lgebr. In lgebr, we prove equlities directly; we prove tht n object, number perhps, is equl to nother object. In nlysis, we usully prove inequlities. To illustrte the point, consider the following sttement. Let x be rel number. If 0 x < ε is true for ll rel numbers ε > 0, then x = 0. This sttement is the generl ide of wht we do in nlysis. If we wish to show tht x = 0, we show tht 0 x < ε for ll positive ε. The term rel nlysis is little bit of misnomer. I prefer to use simply nlysis. The other type of nlysis, complex nlysis, relly builds up on the present mteril, rther thn being distinct. Furthermore, more dvnced course on rel nlysis would tlk bout complex numbers often. I suspect the nomenclture is historicl bggge. Let us get on with the show...

8 8 INTRODUCTION 0.3 Bsic set theory Note: 1 3 lectures (some mteril cn be skipped or covered lightly) Before we strt tlking bout nlysis we need to fix some lnguge. Modern nlysis uses the lnguge of sets, nd therefore tht is where we strt. We tlk bout sets in rther informl wy, using the so-clled nïve set theory. Do not worry, tht is wht mjority of mthemticins use, nd it is hrd to get into trouble. We ssume tht the reder hs seen bsic set theory nd hs hd course in bsic proof writing. This section should be thought of s refresher Sets Definition A set is collection of objects clled elements or members. A set with no objects is clled the empty set nd is denoted by /0 (or sometimes by {}). Think of set s club with certin membership. For exmple, the students who ply chess re members of the chess club. However, do not tke the nlogy too fr. A set is only defined by the members tht form the set; two sets tht hve the sme members re the sme set. Most of the time we will consider sets of numbers. For exmple, the set S := {0,1,2} is the set contining the three elements 0, 1, nd 2. We write 1 S to denote tht the number 1 belongs to the set S. Tht is, 1 is member of S. Similrly we write 7 / S to denote tht the number 7 is not in S. Tht is, 7 is not member of S. The elements of ll sets under considertion come from some set we cll the universe. For simplicity, we often consider the universe to be the set tht contins only the elements we re interested in. The universe is generlly understood from context nd is not explicitly mentioned. In this course, our universe will most often be the set of rel numbers. While the elements of set re often numbers, other object, such s other sets, cn be elements of set. A set my contin some of the sme elements s nother set. For exmple, T := {0,2} contins the numbers 0 nd 2. In this cse ll elements of T lso belong to S. We write T S. More formlly we hve the following definition. The term modern refers to lte 19th century up to the present.

9 0.3. BASIC SET THEORY 9 Definition (i) A set A is subset of set B if x A implies tht x B, nd we write A B. Tht is, ll members of A re lso members of B. (ii) Two sets A nd B re equl if A B nd B A. We write A = B. Tht is, A nd B contin exctly the sme elements. If it is not true tht A nd B re equl, then we write A B. (iii) A set A is proper subset of B if A B nd A B. We write A B. When A = B, we consider A nd B to just be two nmes for the sme exct set. For exmple, for S nd T defined bove we hve T S, but T S. So T is proper subset of S. At this juncture, we lso mention the set building nottion, {x A : P(x)}. This nottion refers to subset of the set A contining ll elements of A tht stisfy the property P(x). The nottion is sometimes bbrevited (A is not mentioned) when understood from context. Furthermore, x A is sometimes replced with formul to mke the nottion esier to red. Exmple 0.3.3: The following re sets including the stndrd nottions. (i) The set of nturl numbers, N := {1, 2, 3,...}. (ii) The set of integers, Z := {0, 1,1, 2,2,...}. (iii) The set of rtionl numbers, Q := { m n : m,n Z nd n 0}. (iv) The set of even nturl numbers, {2m : m N}. (v) The set of rel numbers, R. Note tht N Z Q R. There re mny opertions we wnt to do with sets. Definition (i) A union of two sets A nd B is defined s A B := {x : x A or x B}. (ii) An intersection of two sets A nd B is defined s A B := {x : x A nd x B}. (iii) A complement of B reltive to A (or set-theoretic difference of A nd B) is defined s A \ B := {x : x A nd x / B}.

10 10 INTRODUCTION (iv) We sy complement of B nd write B c if A is understood from context. The set A is either the entire universe or is the obvious set contining B. (v) We sy tht sets A nd B re disjoint if A B = /0. The nottion B c my be little vgue t this point. If the set B is subset of the rel numbers R, then B c mens R \ B. If B is nturlly subset of the nturl numbers, then B c is N \ B. If mbiguity would ever rise, we will use the set difference nottion A \ B. A B A B A B A B A B B A \ B B c Figure 1: Venn digrms of set opertions. We illustrte the opertions on the Venn digrms in Figure 1. Let us now estblish one of most bsic theorems bout sets nd logic. Theorem (DeMorgn). Let A,B,C be sets. Then or, more generlly, (B C) c = B c C c, (B C) c = B c C c, A \ (B C) = (A \ B) (A \C), A \ (B C) = (A \ B) (A \C).

11 0.3. BASIC SET THEORY 11 Proof. The first sttement is proved by the second sttement if we ssume tht set A is our universe. Let us prove A \ (B C) = (A \ B) (A \C). Remember the definition of equlity of sets. First, we must show tht if x A \ (B C), then x (A \ B) (A \C). Second, we must lso show tht if x (A \ B) (A \C), then x A \ (B C). So let us ssume tht x A \ (B C). Then x is in A, but not in B nor C. Hence x is in A nd not in B, tht is, x A \ B. Similrly x A \C. Thus x (A \ B) (A \C). On the other hnd suppose tht x (A \ B) (A \C). In prticulr x (A \ B) nd so x A nd x / B. Also s x (A \C), then x / C. Hence x A \ (B C). The proof of the other equlity is left s n exercise. We will lso need to intersect or union severl sets t once. If there re only finitely mny, then we simply pply the union or intersection opertion severl times. However, suppose tht we hve n infinite collection of sets ( set of sets) {A 1,A 2,A 3,...}. We define A n := {x : x A n for some n N}, n=1 A n := {x : x A n for ll n N}. n=1 We cn lso hve sets indexed by two integers. For exmple, we cn hve the set of sets {A 1,1,A 1,2,A 2,1,A 1,3,A 2,2,A 3,1,...}. Then we write ( A n,m = A n,m ). n=1 m=1 n=1 m=1 And similrly with intersections. It is not hrd to see tht we cn tke the unions in ny order. However, switching unions nd intersections is not generlly permitted without proof. For exmple: {k N : mk < n} = /0 = /0. n=1 m=1 n=1 However, {k N : mk < n} = N = N. m=1 n=1 m= Induction A common method of proof is the principle of induction. We strt with the set of nturl numbers N = {1,2,3,...}. The nturl ordering on N (tht is, 1 < 2 < 3 < 4 < ) hs wonderful property.

12 12 INTRODUCTION The nturl numbers N ordered in the nturl wy possess the well ordering property. We tke this property s n xiom. Well ordering property of N. Every nonempty subset of N hs lest (smllest) element. By S N hving lest element, we men tht there exist n x S, such tht for every y S, we hve x y. The principle of induction is the following theorem, which is equivlent to the well ordering property of the nturl numbers. Theorem (Principle of induction). Let P(n) be sttement depending on nturl number n. Suppose tht (i) (bsis sttement) P(1) is true, (ii) (induction step) if P(n) is true, then P(n + 1) is true. Then P(n) is true for ll n N. Proof. Suppose tht S is the set of nturl numbers m for which P(m) is not true. Suppose tht S is nonempty. Then S hs lest element by the well ordering property. Let us cll m the lest element of S. We know tht 1 / S by ssumption. Therefore m > 1 nd m 1 is nturl number s well. Since m ws the lest element of S, we know tht P(m 1) is true. But by the induction step we see tht P(m 1 + 1) = P(m) is true, contrdicting the sttement tht m S. Therefore S is empty nd P(n) is true for ll n N. Sometimes it is convenient to strt t different number thn 1, but ll tht chnges is the lbeling. The ssumption tht P(n) is true in if P(n) is true, then P(n + 1) is true is usully clled the induction hypothesis. Exmple 0.3.7: Let us prove tht for ll n N we hve 2 n 1 n!. We let P(n) be the sttement tht 2 n 1 n! is true. By plugging in n = 1, we see tht P(1) is true. Suppose tht P(n) is true. Tht is, suppose tht 2 n 1 n! holds. Multiply both sides by 2 to obtin 2 n 2(n!). As 2 (n + 1) when n N, we hve 2(n!) (n + 1)(n!) = (n + 1)!. Tht is, 2 n 2(n!) (n + 1)!, nd hence P(n + 1) is true. By the principle of induction, we see tht P(n) is true for ll n, nd hence 2 n 1 n! is true for ll n N.

13 0.3. BASIC SET THEORY 13 Exmple 0.3.8: We clim tht for ll c 1, we hve tht 1 + c + c c n = 1 cn+1 1 c. Proof: It is esy to check tht the eqution holds with n = 1. Suppose tht it is true for n. Then 1 + c + c c n + c n+1 = (1 + c + c c n ) + c n+1 = 1 cn+1 1 c + c n+1 = 1 cn+1 + (1 c)c n+1 1 c = 1 cn+2 1 c. There is n equivlent principle clled strong induction. The proof tht strong induction is equivlent to induction is left s n exercise. Theorem (Principle of strong induction). Let P(n) be sttement depending on nturl number n. Suppose tht (i) (bsis sttement) P(1) is true, (ii) (induction step) if P(k) is true for ll k = 1,2,...,n, then P(n + 1) is true. Then P(n) is true for ll n N Functions Informlly, set-theoretic function f tking set A to set B is mpping tht to ech x A ssigns unique y B. We write f : A B. For exmple, we define function f : S T tking S = {0,1,2} to T = {0,2} by ssigning f (0) := 2, f (1) := 2, nd f (2) := 0. Tht is, function f : A B is blck box, into which we stick n element of A nd the function spits out n element of B. Sometimes f is clled mpping nd we sy tht f mps A to B. Often, functions re defined by some sort of formul, however, you should relly think of function s just very big tble of vlues. The subtle issue here is tht single function cn hve severl different formuls, ll giving the sme function. Also, for mny functions, there is no formul tht expresses its vlues. To define function rigorously first let us define the Crtesin product. Definition Let A nd B be sets. Then the Crtesin product is the set of tuples defined s follows. A B := {(x,y) : x A,y B}.

14 14 INTRODUCTION For exmple, the set [0,1] [0,1] is set in the plne bounded by squre with vertices (0,0), (0,1), (1,0), nd (1,1). When A nd B re the sme set we sometimes use superscript 2 to denote such product. For exmple [0,1] 2 = [0,1] [0,1], or R 2 = R R (the Crtesin plne). Definition A function f : A B is subset f of A B such tht for ech x A, there is unique (x, y) f. We then write f (x) = y. Sometimes the set f is clled the grph of the function rther thn the function itself. The set A is clled the domin of f (nd sometimes confusingly denoted D( f )). The set is clled the rnge of f. R( f ) := {y B : there exists n x such tht f (x) = y } Note tht R( f ) cn possibly be proper subset of B, while the domin of f is lwys equl to A. We usully ssume tht the domin of f is nonempty. Exmple : From clculus, you re most fmilir with functions tking rel numbers to rel numbers. However, you hve seen some other types of functions s well. For exmple the derivtive is function mpping the set of differentible functions to the set of ll functions. Another exmple is the Lplce trnsform, which lso tkes functions to functions. Yet nother exmple is the function tht tkes continuous function g defined on the intervl [0,1] nd returns the number 10 g(x)dx. Definition Let f : A B be function, nd C A. Define the imge (or direct imge) of C s f (C) := { f (x) B : x C}. Let D B. Define the inverse imge s f 1 (D) := {x A : f (x) D}. Exmple : Define the function f : R R by f (x) := sin(πx). Then f ([0, 1/2]) = [0,1], f 1 ({0}) = Z, etc.... Proposition Let f : A B. Let C, D be subsets of B. Then f 1 (C D) = f 1 (C) f 1 (D), f 1 (C D) = f 1 (C) f 1 (D), f 1 (C c ) = ( f 1 (C) ) c. Red the lst line s f 1 (B \C) = A \ f 1 (C). Proof. Let us strt with the union. Suppose tht x f 1 (C D). Tht mens tht x mps to C or D. Thus f 1 (C D) f 1 (C) f 1 (D). Conversely if x f 1 (C), then x f 1 (C D). Similrly for x f 1 (D). Hence f 1 (C D) f 1 (C) f 1 (D), nd we re hve equlity. The rest of the proof is left s n exercise.

15 0.3. BASIC SET THEORY 15 The proposition does not hold for direct imges. We do hve the following weker result. Proposition Let f : A B. Let C, D be subsets of A. Then The proof is left s n exercise. f (C D) = f (C) f (D), f (C D) f (C) f (D). Definition Let f : A B be function. The function f is sid to be injective or one-to-one if f (x 1 ) = f (x 2 ) implies x 1 = x 2. In other words, f 1 ({y}) is empty or consists of single element for ll y B. We cll such n f n injection. The function f is sid to be surjective or onto if f (A) = B. We cll such n f surjection. A function f tht is both n injection nd surjection is sid to be bijective, nd we sy f is bijection. When f : A B is bijection, then f 1 ({y}) is lwys unique element of A, nd we cn consider f 1 s function f 1 : B A. In this cse, we cll f 1 the inverse function of f. For exmple, for the bijection f (x) := x 3 we hve f 1 (x) = 3 x. A finl piece of nottion for functions tht we need is the composition of functions. Definition Let f : A B, g: B C. The function g f : A C is defined s (g f )(x) := g ( f (x) ) Crdinlity A subtle issue in set theory nd one generting considerble mount of confusion mong students is tht of crdinlity, or size of sets. The concept of crdinlity is importnt in modern mthemtics in generl nd in nlysis in prticulr. In this section, we will see the first relly unexpected theorem. Definition Let A nd B be sets. We sy A nd B hve the sme crdinlity when there exists bijection f : A B. We denote by A the equivlence clss of ll sets with the sme crdinlity s A nd we simply cll A the crdinlity of A. Note tht A hs the sme crdinlity s the empty set if nd only if A itself is the empty set. We then write A := 0. Definition Suppose tht A hs the sme crdinlity s {1,2,3,...,n} for some n N. We then write A := n, nd we sy tht A is finite. When A is the empty set, we lso cll A finite. We sy tht A is infinite or of infinite crdinlity if A is not finite.

16 16 INTRODUCTION Tht the nottion A = n is justified we leve s n exercise. Tht is, for ech nonempty finite set A, there exists unique nturl number n such tht there exists bijection from A to {1,2,3,...,n}. We cn lso order sets by size. Definition We write A B if there exists n injection from A to B. We write A = B if A nd B hve the sme crdinlity. We write A < B if A B, but A nd B do not hve the sme crdinlity. We stte without proof tht A = B hve the sme crdinlity if nd only if A B nd B A. This is the so-clled Cntor-Bernstein-Schroeder theorem. Furthermore, if A nd B re ny two sets, we cn lwys write A B or B A. The issues surrounding this lst sttement re very subtle. As we do not require either of these two sttements, we omit proofs. The truly interesting cses of crdinlity re infinite sets. We strt with the following definition. Definition If A = N, then A is sid to be countbly infinite. If A is finite or countbly infinite, then we sy A is countble. If A is not countble, then A is sid to be uncountble. The crdinlity of N is usully denoted s ℵ 0 (red s leph-nught). Exmple : The set of even nturl numbers hs the sme crdinlity s N. Proof: Given n even nturl number, write it s 2n for some n N. Then crete bijection tking 2n to n. In fct, let us mention without proof the following chrcteriztion of infinite sets: A set is infinite if nd only if it is in one-to-one correspondence with proper subset of itself. Exmple : N N is countbly infinite set. Proof: Arrnge the elements of N N s follows (1,1), (1,2), (2,1), (1,3), (2,2), (3,1),.... Tht is, lwys write down first ll the elements whose two entries sum to k, then write down ll the elements whose entries sum to k + 1 nd so on. Then define bijection with N by letting 1 go to (1,1), 2 go to (1,2) nd so on. Exmple : The set of rtionl numbers is countble. Proof: (informl) Follow the sme procedure s in the previous exmple, writing 1/1, 1/2, 2/1, etc.... However, leve out ny frction (such s 2/2) tht hs lredy ppered. For completeness we mention the following sttement. If A B nd B is countble, then A is countble. Similrly if A is uncountble, then B is uncountble. As we will not need this sttement in the sequel, nd s the proof requires the Cntor-Bernstein-Schroeder theorem mentioned bove, we will not give it here. We give the first truly striking result. First, we need nottion for the set of ll subsets of set. Definition If A is set, we define the power set of A, denoted by P(A), to be the set of ll subsets of A. For the fns of the TV show Futurm, there is movie theter in one episode clled n ℵ 0 -plex.

17 0.3. BASIC SET THEORY 17 For exmple, if A := {1, 2}, then P(A) = {/0,{1},{2},{1, 2}}. Note tht for finite set A of crdinlity n, the crdinlity of P(A) is 2 n. This fct is left s n exercise. Tht is, the crdinlity of P(A) is strictly lrger thn the crdinlity of A, t lest for finite sets. Wht is n unexpected nd striking fct is tht this sttement is still true for infinite sets. Theorem (Cntor ). A < P(A). In prticulr, there exists no surjection from A onto P(A). Proof. There of course exists n injection f : A P(A). For ny x A, define f (x) := {x}. Therefore A P(A). To finish the proof, we hve to show tht no function f : A P(A) is surjection. Suppose tht f : A P(A) is function. So for x A, f (x) is subset of A. Define the set B := {x A : x / f (x)}. We clim tht B is not in the rnge of f nd hence f is not surjection. Suppose tht there exists n x 0 such tht f (x 0 ) = B. Either x 0 B or x 0 / B. If x 0 B, then x 0 / f (x 0 ) = B, which is contrdiction. If x 0 / B, then x 0 f (x 0 ) = B, which is gin contrdiction. Thus such n x 0 does not exist. Therefore, B is not in the rnge of f, nd f is not surjection. As f ws n rbitrry function, no surjection exists. One prticulr consequence of this theorem is tht there do exist uncountble sets, s P(N) must be uncountble. This fct is relted to the fct tht the set of rel numbers (which we study in the next chpter) is uncountble. The existence of uncountble sets my seem unintuitive, nd the theorem cused quite controversy t the time it ws nnounced. The theorem not only sys tht uncountble sets exist, but tht there in fct exist progressively lrger nd lrger infinite sets N, P(N), P(P(N)), P(P(P(N))), etc Exercises Exercise 0.3.1: Show A \ (B C) = (A \ B) (A \C). Exercise 0.3.2: Prove tht the principle of strong induction is equivlent to the stndrd induction. Exercise 0.3.3: Finish the proof of Proposition Exercise 0.3.4: ) Prove Proposition b) Find n exmple for which equlity of sets in f (C D) f (C) f (D) fils. Tht is, find n f, A, B, C, nd D such tht f (C D) is proper subset of f (C) f (D). Exercise (Tricky): Prove tht if A is finite, then there exists unique number n such tht there exists bijection between A nd {1,2,3,...,n}. In other words, the nottion A := n is justified. Hint: Show tht if n > m, then there is no injection from {1,2,3,...,n} to {1,2,3,...,m}. Nmed fter the Germn mthemticin Georg Ferdinnd Ludwig Philipp Cntor ( ).

18 18 INTRODUCTION Exercise 0.3.6: Prove ) A (B C) = (A B) (A C) b) A (B C) = (A B) (A C) Exercise 0.3.7: Let A B denote the symmetric difference, tht is, the set of ll elements tht belong to either A or B, but not to both A nd B. ) Drw Venn digrm for A B. b) Show A B = (A \ B) (B \ A). c) Show A B = (A B) \ (A B). Exercise 0.3.8: For ech n N, let A n := {(n + 1)k : k N}. ) Find A 1 A 2. b) Find n=1 A n. c) Find n=1 A n. Exercise 0.3.9: Determine P(S) (the power set) for ech of the following: ) S = /0, b) S = {1}, c) S = {1,2}, d) S = {1,2,3,4}. Exercise : Let f : A B nd g: B C be functions. ) Prove tht if g f is injective, then f is injective. b) Prove tht if g f is surjective, then g is surjective. c) Find n explicit exmple where g f is bijective, but neither f nor g re bijective. Exercise : Prove tht n < 2 n by induction. Exercise : Show tht for finite set A of crdinlity n, the crdinlity of P(A) is 2 n. Exercise : Prove n(n+1) = n n+1 for ll n N. Exercise : Prove n 3 = ( ) 2 n(n+1) 2 for ll n N. Exercise : Prove tht n 3 + 5n is divisible by 6 for ll n N. Exercise : Find the smllest n N such tht 2(n + 5) 2 < n 3 nd cll it n 0. Show tht 2(n + 5) 2 < n 3 for ll n n 0. Exercise : Find ll n N such tht n 2 < 2 n.

19 0.3. BASIC SET THEORY 19 Exercise : Finish the proof tht the principle of induction is equivlent to the well ordering property of N. Tht is, prove the well ordering property for N using the principle of induction. Exercise : Give n exmple of countble collection of finite sets A 1,A 2,..., whose union is not finite set. Exercise : Give n exmple of countble collection of infinite sets A 1,A 2,..., with A j A k being infinite for ll j nd k, such tht j=1 A j is nonempty nd finite.

20 20 INTRODUCTION

21 Chpter 1 Rel Numbers 1.1 Bsic properties Note: 1.5 lectures The min object we work with in nlysis is the set of rel numbers. As this set is so fundmentl, often much time is spent on formlly constructing the set of rel numbers. However, we tke n esier pproch here nd just ssume tht set with the correct properties exists. We need to strt with some bsic definitions. Definition An ordered set is set A, together with reltion < such tht (i) For ny x,y A, exctly one of x < y, x = y, or y < x holds. (ii) If x < y nd y < z, then x < z. For exmple, the rtionl numbers Q re n ordered set by letting x < y if nd only if y x is positive rtionl number. Similrly, N nd Z re lso ordered sets. We write x y if x < y or x = y. We define > nd in the obvious wy. Definition Let E A, where A is n ordered set. (i) If there exists b A such tht x b for ll x E, then we sy E is bounded bove nd b is n upper bound of E. (ii) If there exists b A such tht x b for ll x E, then we sy E is bounded below nd b is lower bound of E. (iii) If there exists n upper bound b 0 of E such tht whenever b is ny upper bound for E we hve b 0 b, then b 0 is clled the lest upper bound or the supremum of E. We write sup E := b 0. 21

22 22 CHAPTER 1. REAL NUMBERS (iv) Similrly, if there exists lower bound b 0 of E such tht whenever b is ny lower bound for E we hve b 0 b, then b 0 is clled the gretest lower bound or the infimum of E. We write inf E := b 0. When set E is both bounded bove nd bounded below, we sy simply tht E is bounded. Note tht supremum or infimum for E (even if they exist) need not be in E. For exmple the set {x Q : x < 1} hs lest upper bound of 1, but 1 is not in the set itself. Definition An ordered set A hs the lest-upper-bound property if every nonempty subset E A tht is bounded bove hs lest upper bound, tht is sup E exists in A. The lest-upper-bound property is sometimes clled the completeness property or the Dedekind completeness property. Exmple 1.1.4: For exmple Q does not hve the lest-upper-bound property. The set {x Q : x 2 < 2} does not hve supremum. The obvious supremum 2 is not rtionl. Suppose tht x 2 = 2 for some x Q. Write x = m/n in lowest terms. So (m/n) 2 = 2 or m 2 = 2n 2. Hence m 2 is divisible by 2 nd so m is divisible by 2. We write m = 2k nd so we hve (2k) 2 = 2n 2. We divide by 2 nd note tht 2k 2 = n 2 nd hence n is divisible by 2. But tht is contrdiction s we sid m/n ws in lowest terms. Tht Q does not hve the lest-upper-bound property is one of the most importnt resons why we work with R in nlysis. The set Q is just fine for lgebrists. But nlysts require the lest-upper-bound property to do ny work. We lso require our rel numbers to hve mny lgebric properties. In prticulr, we require tht they re field. Definition A set F is clled field if it hs two opertions defined on it, ddition x + y nd multipliction xy, nd if it stisfies the following xioms. (A1) If x F nd y F, then x + y F. (A2) (commuttivity of ddition) If x + y = y + x for ll x,y F. (A3) (ssocitivity of ddition) If (x + y) + z = x + (y + z) for ll x,y,z F. (A4) There exists n element 0 F such tht 0 + x = x for ll x F. (A5) For every element x F there exists n element x F such tht x + ( x) = 0. (M1) If x F nd y F, then xy F. (M2) (commuttivity of multipliction) If xy = yx for ll x,y F. (M3) (ssocitivity of multipliction) If (xy)z = x(yz) for ll x,y,z F. (M4) There exists n element 1 (nd 1 0) such tht 1x = x for ll x F. (M5) For every x F such tht x 0 there exists n element 1/x F such tht x(1/x) = 1. (D) (distributive lw) x(y + z) = xy + xz for ll x,y,z F.

23 1.1. BASIC PROPERTIES 23 Exmple 1.1.6: The set Q of rtionl numbers is field. On the other hnd Z is not field, s it does not contin multiplictive inverses. We will ssume the bsic fcts bout fields tht cn esily be proved from the xioms. For exmple, 0x = 0 cn be esily proved by noting tht xx = (0 + x)x = 0x + xx, using (A4), (D), nd (M2). Then using (A5) on xx we obtin 0 = 0x. Definition A field F is sid to be n ordered field if F is lso n ordered set such tht: (i) For x,y,z F, x < y implies x + z < y + z. (ii) For x,y F, x > 0 nd y > 0 implies xy > 0. If x > 0, we sy x is positive. If x < 0, we sy x is negtive. We lso sy x is nonnegtive if x 0, nd x is nonpositive if x 0. Proposition Let F be n ordered field nd x,y,z F. Then: (i) If x > 0, then x < 0 (nd vice-vers). (ii) If x > 0 nd y < z, then xy < xz. (iii) If x < 0 nd y < z, then xy > xz. (iv) If x 0, then x 2 > 0. (v) If 0 < x < y, then 0 < 1/y < 1/x. Note tht (iv) implies in prticulr tht 1 > 0. Proof. Let us prove (i). The inequlity x > 0 implies by item (i) of definition of ordered field tht x + ( x) > 0 + ( x). Now pply the lgebric properties of fields to obtin 0 > x. The vice-vers follows by similr clcultion. For (ii), first notice tht y < z implies 0 < z y by pplying item (i) of the definition of ordered fields. Now pply item (ii) of the definition of ordered fields to obtin 0 < x(z y). By lgebric properties we get 0 < xz xy, nd gin pplying item (i) of the definition we obtin xy < xz. Prt (iii) is left s n exercise. To prove prt (iv) first suppose tht x > 0. Then by item (ii) of the definition of ordered fields we obtin tht x 2 > 0 (use y = x). If x < 0, we use prt (iii) of this proposition. Plug in y = x nd z = 0. Finlly to prove prt (v), notice tht 1/x cnnot be equl to zero (why?). Suppose 1/x < 0, then 1/x > 0 by (i). Then pply prt (ii) (s x > 0) to obtin x( 1/x) > 0x or 1 > 0, which contrdicts 1 > 0 by using prt (i) gin. Hence 1/x > 0. Similrly 1/y > 0. Thus (1/x)(1/y) > 0 by definition of ordered field nd by prt (ii) (1/x)(1/y)x < (1/x)(1/y)y. By lgebric properties we get 1/y < 1/x. Product of two positive numbers (elements of n ordered field) is positive. However, it is not true tht if the product is positive, then ech of the two fctors must be positive.

24 24 CHAPTER 1. REAL NUMBERS Proposition Let x,y F where F is n ordered field. Suppose tht xy > 0. Then either both x nd y re positive, or both re negtive. Proof. It is cler tht both possibilities cn hppen. If either x nd y re zero, then xy is zero nd hence not positive. Hence we cn ssume tht x nd y re nonzero, nd we simply need to show tht if they hve opposite signs, then xy < 0. Without loss of generlity suppose tht x > 0 nd y < 0. Multiply y < 0 by x to get xy < 0x = 0. The result follows by contrpositive Exercises Exercise 1.1.1: Prove prt (iii) of Proposition Exercise 1.1.2: Let S be n ordered set. Let A S be nonempty finite subset. Then A is bounded. Furthermore, inf A exists nd is in A nd sup A exists nd is in A. Hint: Use induction. Exercise 1.1.3: Let x,y F, where F is n ordered field. Suppose tht 0 < x < y. Show tht x 2 < y 2. Exercise 1.1.4: Let S be n ordered set. Let B S be bounded (bove nd below). Let A B be nonempty subset. Suppose tht ll the inf s nd sup s exist. Show tht inf B inf A sup A sup B. Exercise 1.1.5: Let S be n ordered set. Let A S nd suppose tht b is n upper bound for A. Suppose tht b A. Show tht b = sup A. Exercise 1.1.6: Let S be n ordered set. Let A S be nonempty subset tht is bounded bove. Suppose tht sup A exists nd tht sup A / A. Show tht A contins countbly infinite subset. In prticulr, A is infinite. Exercise 1.1.7: Find (nonstndrd) ordering of the set of nturl numbers N such tht there exists nonempty proper subset A N nd such tht sup A exists in N, but sup A / A. Exercise 1.1.8: Let F = {0,1,2}. ) Prove tht there is exctly one wy to define ddition nd multipliction so tht F is field if 0 nd 1 hve their usul mening of (A4) nd (M4). b) Show tht F cnnot be n ordered field. Exercise 1.1.9: Let S be n ordered set nd A is nonempty subset such tht sup A exists. Suppose tht there is B A such tht whenever x A there is y B such tht x y. Show tht sup B exists nd sup B = sup A.

25 1.2. THE SET OF REAL NUMBERS The set of rel numbers Note: 2 lectures, the extended rel numbers re optionl The set of rel numbers We finlly get to the rel number system. Insted of constructing the rel number set from the rtionl numbers, we simply stte their existence s theorem without proof. Notice tht Q is n ordered field. Theorem There exists unique ordered field R with the lest-upper-bound property such tht Q R. Note tht lso N Q. As we hve seen, 1 > 0. By induction (exercise) we cn prove tht n > 0 for ll n N. Similrly we cn esily verify ll the sttements we know bout rtionl numbers nd their nturl ordering. Let us prove one of the most bsic but useful results bout the rel numbers. The following proposition is essentilly how n nlyst proves tht number is zero. Proposition If x R is such tht x 0 nd x ε for ll ε R where ε > 0, then x = 0. Proof. If x > 0, then 0 < x/2 < x (why?). Tking ε = x/2 obtins contrdiction. Thus x = 0. A relted simple fct is tht ny time we hve two rel numbers < b, then there is nother rel number c such tht < c < b. Just tke for exmple c = +b 2 (why?). In fct, there re infinitely mny rel numbers between nd b. The most useful property of R for nlysts is not just tht it is n ordered field, but tht it hs the lest-upper-bound property. Essentilly we wnt Q, but we lso wnt to tke suprem (nd infim) willy-nilly. So wht we do is to throw in enough numbers to obtin R. We hve lredy seen tht R must contin elements tht re not in Q becuse of the lest-upperbound property. We hve seen tht there is no rtionl squre root of two. The set {x Q : x 2 < 2} implies the existence of the rel number 2 tht is not rtionl, lthough this fct requires bit of work. Exmple 1.2.3: Clim: There exists unique positive rel number r such tht r 2 = 2. We denote r by 2. Proof. Tke the set A := {x R : x 2 < 2}. First if x 2 < 2, then x < 2. To see this fct, note tht x 2 implies x 2 4 (use Proposition 1.1.8, we will not explicitly mention its use from now on), hence ny number x such tht x 2 is not in A. Thus A is bounded bove. On the other hnd, 1 A, so A is nonempty. Uniqueness is up to isomorphism, but we wish to void excessive use of lgebr. For us, it is simply enough to ssume tht set of rel numbers exists. See Rudin [R2] for the construction nd more detils.

26 26 CHAPTER 1. REAL NUMBERS Let us define r := sup A. We will show tht r 2 = 2 by showing tht r 2 2 nd r 2 2. This is the wy nlysts show equlity, by showing two inequlities. We lredy know tht r 1 > 0. In the following, it my seem tht we re pulling certin expressions out of ht. When writing proof such s this we would of course come up with the expressions only fter plying round with wht we wish to prove. The order in which we write the proof is not necessrily the order in which we come up with the proof. Let us first show tht r 2 2. Tke positive number s such tht s 2 < 2. We wish to find n h > 0 such tht (s + h) 2 < 2. As 2 s 2 > 0, we hve 2s+1 2 s2 > 0. We choose n h R such tht 0 < h < 2s+1 2 s2. Furthermore, we ssume tht h < 1. (s + h) 2 s 2 = h(2s + h) ( ) < h(2s + 1) since h < 1 ( < 2 s 2 since h < 2 ) s2. 2s + 1 Therefore, (s + h) 2 < 2. Hence s + h A, but s h > 0 we hve s + h > s. So s < r = sup A. As s ws n rbitrry positive number such tht s 2 < 2, it follows tht r 2 2. Now tke positive number s such tht s 2 > 2. We wish to find n h > 0 such tht (s h) 2 > 2. As s 2 2 > 0 we hve s2 2 2s > 0. We choose n h R such tht 0 < h < s2 2 2s nd h < s. s 2 (s h) 2 = 2sh h 2 < 2sh < s 2 2 ( ) since h < s2 2. 2s By subtrcting s 2 from both sides nd multiplying by 1, we find (s h) 2 > 2. Therefore s h / A. Furthermore, if x s h, then x 2 (s h) 2 > 2 (s x > 0 nd s h > 0) nd so x / A. Thus s h is n upper bound for A. However, s h < s, or in other words s > r = sup A. Thus r 2 2. Together, r 2 2 nd r 2 2 imply r 2 = 2. The existence prt is finished. We still need to hndle uniqueness. Suppose tht s R such tht s 2 = 2 nd s > 0. Thus s 2 = r 2. However, if 0 < s < r, then s 2 < r 2. Similrly 0 < r < s implies r 2 < s 2. Hence s = r. The number 2 / Q. The set R \ Q is clled the set of irrtionl numbers. We hve just seen tht R \ Q is nonempty. Not only is it nonempty, we will see lter tht is it very lrge indeed. Using the sme technique s bove, we cn show tht positive rel number x 1/n exists for ll n N nd ll x > 0. Tht is, for ech x > 0, there exists unique positive rel number r such tht r n = x. The proof is left s n exercise.

27 1.2. THE SET OF REAL NUMBERS Archimeden property As we hve seen, there re plenty of rel numbers in ny intervl. But there re lso infinitely mny rtionl numbers in ny intervl. The following is one of the fundmentl fcts bout the rel numbers. The two prts of the next theorem re ctully equivlent, even though it my not seem like tht t first sight. Theorem (i) (Archimeden property) If x,y R nd x > 0, then there exists n n N such tht nx > y. (ii) (Q is dense in R) If x,y R nd x < y, then there exists n r Q such tht x < r < y. Proof. Let us prove (i). We divide through by x nd then (i) sys tht for ny rel number t := y/x, we cn find nturl number n such tht n > t. In other words, (i) sys tht N R is not bounded bove. Suppose for contrdiction tht N is bounded bove. Let b := supn. The number b 1 cnnot possibly be n upper bound for N s it is strictly less thn b (the supremum). Thus there exists n m N such tht m > b 1. We dd one to obtin m + 1 > b, which contrdicts b being n upper bound. Let us tckle (ii). First ssume tht x 0. Note tht y x > 0. By (i), there exists n n N such tht n(y x) > 1. Also by (i) the set A := {k N : k > nx} is nonempty. By the well ordering property of N, A hs lest element m. As m A, then m > nx. As m is the lest element of A, m 1 / A. If m > 1, then m 1 N, but m 1 / A nd so m 1 nx. If m = 1, then m 1 = 0, nd m 1 nx still holds s x 0. In other words, m 1 nx < m. We divide through by n to get x < m/n. On the other hnd from n(y x) > 1 we obtin ny > 1 + nx. As nx m 1 we get tht 1 + nx m nd hence ny > m nd therefore y > m/n. Putting everything together we obtin x < m/n < y. So let r = m/n. Now ssume tht x < 0. If y > 0, then we just tke r = 0. If y < 0, then note tht 0 < y < x nd find rtionl q such tht y < q < x. Then tke r = q. Let us stte nd prove simple but useful corollry of the Archimeden property. Corollry inf{1/n : n N} = 0. Proof. Let A := {1/n : n N}. Obviously A is not empty. Furthermore, 1/n > 0 nd so 0 is lower bound, nd b := inf A exists. As 0 is lower bound, then b 0. Now tke n rbitrry > 0. By the Archimeden property there exists n n such tht n > 1, or in other words > 1/n A. Therefore cnnot be lower bound for A. Hence b = 0.

28 28 CHAPTER 1. REAL NUMBERS Using supremum nd infimum We wnt to mke sure tht suprem nd infim re comptible with lgebric opertions. For set A R nd number x define x + A := {x + y R : y A}, xa := {xy R : y A}. Proposition Let A R be bounded nd nonempty. (i) If x R, then sup(x + A) = x + sup A. (ii) If x R, then inf(x + A) = x + inf A. (iii) If x > 0, then sup(xa) = x(sup A). (iv) If x > 0, then inf(xa) = x(inf A). (v) If x < 0, then sup(xa) = x(inf A). (vi) If x < 0, then inf(xa) = x(sup A). Do note tht multiplying set by negtive number switches supremum for n infimum nd vice-vers. The proposition lso implies tht if A is nonempty nd bounded then xa nd x + A re nonempty nd bounded. Proof. Let us only prove the first sttement. The rest re left s exercises. Suppose tht b is n upper bound for A. Tht is, y b for ll y A. Then x + y x + b for ll y A, nd so x + b is n upper bound for x + A. In prticulr, if b = sup A, then sup(x + A) x + b = x + sup A. The other direction is similr. If b is n upper bound for x + A, then x + y b for ll y A nd so y b x for ll y A. So b x is n upper bound for A. If b = sup(x + A), then And the result follows. sup A b x = sup(x + A) x. Sometimes we need to pply supremum or infimum twice. Here is n exmple. Proposition Let A,B R be nonempty sets such tht x y whenever x A nd y B. Then A is bounded bove, B is bounded below, nd sup A inf B. Proof. Any x A is lower bound for B. Therefore x inf B for ll x A, so inf B is n upper bound for A. Hence, sup A inf B.

29 1.2. THE SET OF REAL NUMBERS 29 We hve to be creful bout strict inequlities nd tking suprem nd infim. Note tht x < y whenever x A nd y B still only implies sup A inf B, nd not strict inequlity. This is n importnt subtle point tht comes up often. For exmple, tke A := {0} nd tke B := {1/n : n N}. Then 0 < 1/n for ll n N. However, sup A = 0 nd inf B = 0. The proof of the following often used elementry fct is left to the reder. A similr sttement holds for infim. Proposition If S R is nonempty bounded set, then for every ε > 0 there exists x S such tht sup S ε < x sup S. To mke using suprem nd infim even esier, we my wnt to write sup A nd inf A without worrying bout A being bounded nd nonempty. We mke the following nturl definitions. Definition Let A R be set. (i) If A is empty, then sup A :=. (ii) If A is not bounded bove, then sup A :=. (iii) If A is empty, then inf A :=. (iv) If A is not bounded below, then inf A :=. For convenience, nd re sometimes treted s if they were numbers, except we do not llow rbitrry rithmetic with them. We mke R := R {, } into n ordered set by letting < nd < x nd x < for ll x R. The set R is clled the set of extended rel numbers. It is possible to define some rithmetic on R. Most opertions re extended in n obvious wy, but we must leve, 0 (± ), nd ± ± undefined. We refrin from using this rithmetic so s it leds to esy mistkes s R is not field. Now we cn tke suprem nd infim without fer of emptiness or unboundedness. In this book we mostly void using R outside of exercises, nd leve such generliztions to the interested reder Mxim nd minim By Exercise we know tht finite set of numbers lwys hs supremum or n infimum tht is contined in the set itself. In this cse we usully do not use the words supremum or infimum. When we hve set A of rel numbers bounded bove, such tht sup A A, then we cn use the word mximum nd the nottion mxa to denote the supremum. Similrly for infimum; when set A is bounded below nd inf A A, then we cn use the word minimum nd the nottion min A. For exmple, mx{1,2.4,π,100} = 100, min{1,2.4,π,100} = 1. While writing sup nd inf my be techniclly correct in this sitution, mx nd min re generlly used to emphsize tht the supremum or infimum is in the set itself.

30 30 CHAPTER 1. REAL NUMBERS Exercises Exercise 1.2.1: Prove tht if t > 0 (t R), then there exists n n N such tht 1 n 2 < t. Exercise 1.2.2: Prove tht if t 0 (t R), then there exists n n N such tht n 1 t < n. Exercise 1.2.3: Finish the proof of Proposition Exercise 1.2.4: Let x,y R. Suppose tht x 2 + y 2 = 0. Prove tht x = 0 nd y = 0. Exercise 1.2.5: Show tht 3 is irrtionl. Exercise 1.2.6: Let n N. Show tht either n is either n integer or it is irrtionl. Exercise 1.2.7: Prove the rithmetic-geometric men inequlity. Tht is, for two positive rel numbers x,y we hve xy x + y 2. Furthermore, equlity occurs if nd only if x = y. Exercise 1.2.8: Show tht for ny two rel numbers such tht x < y, we hve n irrtionl number s such tht x < s < y. Hint: Apply the density of Q to x nd y. 2 2 Exercise 1.2.9: Let A nd B be two nonempty bounded sets of rel numbers. Let C := { + b : A,b B}. Show tht C is bounded set nd tht sup C = sup A + sup B nd inf C = inf A + inf B. Exercise : Let A nd B be two nonempty bounded sets of nonnegtive rel numbers. Define the set C := {b : A,b B}. Show tht C is bounded set nd tht sup C = (sup A)(sup B) nd inf C = (inf A)(inf B). Exercise (Hrd): Given x > 0 nd n N, show tht there exists unique positive rel number r such tht x = r n. Usully r is denoted by x 1/n.