Basic Analysis I. Introduction to Real Analysis, Volume I. May 7, 2018 (version 5.0)

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1 Bsic Anlysis I Introduction to Rel Anlysis, Volume I by Jiří Lebl My 7, 2018 (version 5.0)

2 2 Typeset in LATEX. Copyright c Jiří Lebl This work is dul licensed under the Cretive Commons Attribution-Noncommercil-Shre Alike 4.0 Interntionl License nd the Cretive Commons Attribution-Shre Alike 4.0 Interntionl License. To view copy of these licenses, visit by-nc-s/4.0/ or or send letter to Cretive Commons PO Box 1866, Mountin View, CA 94042, USA. You cn use, print, duplicte, shre this book s much s you wnt. You cn bse your own notes on it nd reuse prts if you keep the license the sme. You cn ssume the license is either the CC-BY-NC-SA or CC-BY-SA, whichever is comptible with wht you wish to do, your derivtive works must use t lest one of the licenses. During the writing of these notes, the uthor ws in prt supported by NSF grnts DMS nd DMS The dte is the min identifier of version. The mjor version / edition number is rised only if there hve been substntil chnges. Edition number strted t 4, tht is, version 4.0, s it ws not kept trck of before. 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 nd bounded functions 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 Tylor s theorem Inverse function theorem

4 4 CONTENTS 5 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 271 Index 273 List of Nottion 279

5 Introduction 0.1 About this book This first volume is one semester course in bsic nlysis. Together with the second volume it is yer-long course. 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). Volume II ws dded to tech Mth 4143/4153 t Oklhom Stte University (OSU). A prerequisite for these courses is usully bsic proof course, using for exmple [H], [F], or [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 first few chpters of the book cn be used in n introductory proofs course s is for exmple done t Iow Stte University Mth 201, where this book is used in conjunction with Hmmck s Book of Proof [H]. With volume II one cn run yer-long course tht lso covers multivrible topics. It my mke sense in this cse to cover most of the first volume in the first semester while leving metric spces for the beginning of the second semester. 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 course such s UIUC 444 within semester nd still spend some 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.

6 6 INTRODUCTION 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 gret nlysis textbook, big Rudin). I took 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, Sönmez Şhutoğlu, Jim Brndt, Kenji Kozi, Arthur Busch, Anton Petrunin, Mrk Meilstrup, Hrold P. Bos, nd Atill Yılmz for teching with the book nd giving me lots of useful feedbck. Frnk Betrous wrote the University of Pittsburgh version extensions, which served s inspirtion for mny more recent dditions. I would lso like to thnk Dn Stonehm, Jeremy Sutter, Eliy Gwett, Dniel Pimentel-Alrcón, Steve Hoerning, Yi Zhng, Nicole Cviris, Kristopher Lee, Boyue Bi, Hnnh Lund, Trevor Mnnell, Mitchel Meyer, Gregory Beuregrd, Chse Medors, Andres Ginnopoulos, Nick Nelsen, Ru Wng, Trevor Fncher, 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 (nd severl vribles if volume II is lso considered). Clculus hs prepred you, the student, for using mthemtics without telling you why wht you 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 students 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 the problems in in the textbook. 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, nd we prove those inequlities by estimting. To illustrte the point, consider the following sttement. Let x be rel number. If 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. Suppose next we relly wish to prove the equlity x = 0. In nlysis, we prove two inequlities: x 0 nd x 0. To prove the inequlity x 0, we prove x < ε for ll positive ε. To prove the inequlity x 0, we prove 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, covered lightly, or left s reding) 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. The reder hs hopefully seen the very bsics of set theory nd proof writing before, nd this section should be quick 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. By :=, we men we re defining wht S is, rther thn just showing equlity. We write 1 S to denote tht the number 1 belongs to the set S. Tht is, 1 is member of S. At times we wnt to sy tht two elements re in set S, so we write 1,2 S s shorthnd for 1 S nd 2 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 objects, such s other sets, cn be elements of set. A set my lso 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. See Figure 1 for digrm. The term modern refers to lte 19th century up to the present.

9 0.3. BASIC SET THEORY 9 S T Figure 1: A digrm of the exmple sets S nd its subset T. Definition (i) A set A is subset of set B if x A implies x B, nd we write A B. Tht is, ll members of A re lso members of B. At times we write B A to men the sme thing. (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. For exmple, for S nd T defined bove T S, but T S. So T is proper subset of S. If A = B, then A nd B re simply two nmes for the sme exct set. Let us 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). For exmple, using S = {0,1,2} s bove, {x S : x 2} is the set {0,1}. 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}.

10 10 INTRODUCTION (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}. (iv) We sy complement of B nd write B c insted of A \ B if the set A is either the entire universe or is the obvious set contining B, nd is understood from context. (v) We sy 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 cn rise, we use the set difference nottion A \ B. A B A B A B A B A B B A \ B B c Figure 2: Venn digrms of set opertions, the result of the opertion is shded. We illustrte the opertions on the Venn digrms in Figure 2. 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 the 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 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 x (A \ B) (A \C). In prticulr, x (A \ B), 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. The result bove we clled Theorem, while most results we cll Proposition, nd few we cll Lemm ( result leding to nother result) or Corollry ( quick consequence of the preceding result). Do not red too much into the nming. Some of it is trditionl, some of it is stylistic choice. It is not necessrily true tht Theorem is lwys more importnt thn Proposition or Lemm. 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 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 the order of unions nd intersections is not generlly permitted without proof. For exmple: However, {k N : mk < n} = /0 = /0. n=1 m=1 n=1 {k N : mk < n} = N = N. m=1 n=1 m=1 Sometimes, the index set is not the nturl numbers. In this cse we need more generl nottion. Suppose I is some set nd for ech λ I, we hve set A λ. Then we define A λ := {x : x A λ for some λ I}, λ I A λ := {x : x A λ for ll λ I}. λ I

12 12 INTRODUCTION Induction When sttement includes n rbitrry nturl number, common method of proof is the principle of induction. We strt with the set of nturl numbers N = {1,2,3,...}, nd we give them their nturl ordering, tht is, 1 < 2 < 3 < 4 <. By S N hving lest element, we men tht there exists n x S, such tht for every y S, we hve x y. The nturl numbers N ordered in the nturl wy possess the so-clled well ordering property. We tke this property s n xiom; we simply ssume it is true. Well ordering property of N. Every nonempty subset of N hs lest (smllest) element. 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 S is the set of nturl numbers m for which P(m) is not true. Suppose S is nonempty. Then S hs lest element by the well ordering property. Let us cll m the lest element of S. We know 1 / S by ssumption. Therefore m > 1 nd m 1 is nturl number s well. Since m is 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, 2 n 1 n! (recll n! = 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 P(n) is true. Tht is, suppose 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, 1 + c + c c n = 1 cn+1 1 c. Proof: It is esy to check tht the eqution holds with n = 1. Suppose 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. Sometimes, it is esier to use in the inductive step tht P(k) is true for ll k = 1,2,...,n, not just for k = n. This principle is clled strong induction nd is equivlent to the norml induction bove. The proof tht equivlence 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 or mp, nd we sy 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 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. The Crtesin product is the set of tuples defined s A B := {(x,y) : x A,y B}. 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).

14 14 INTRODUCTION 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 } It is possible tht the rnge R( f ) is 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 sw 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 1 0 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 of D s f 1 (D) := { x A : f (x) D }. 1 f f ({1,2,3,4}) = {b,c,d} f ({1,2,4}) = {b,d} 2 b f ({1}) = {b} 3 c f 1 ({,b,c}) = {1,3,4} f 1 ({}) = /0 4 d f 1 ({b}) = {1,4} Figure 3: Exmple of direct nd inverse imges for the function f : {1,2,3,4} {,b,c,d} defined by f (1) := b, f (2) := d, f (3) := c, f (4) := b. 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.

15 0.3. BASIC SET THEORY 15 Red the lst line of the proposition s f 1 (B \C) = A \ f 1 (C). Proof. Let us strt with the union. Suppose x f 1 (C D). Tht mens 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 hve equlity. The rest of the proof is left s n exercise. 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, for ll y B the set f 1 ({y}) is empty or consists of single element. 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 : R R defined by 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) ) Reltions nd equivlence clsses We often compre two objects in some wy. We sy 1 < 2 for nturl numbers, or 1/2 = 2/4 for rtionl numbers, or {,c} {,b,c} for sets. These re exmples of reltions. Definition Given set A, binry reltion on A is subset R A A, which re those pirs where the reltion is sid to hold. Insted of (,b) R, we write R b. Exmple : Tke A := {1, 2, 3}. First, let the reltion be <. Then the corresponding set of pirs is { (1,2),(1,3),(2,3) }. So 1 < 2 holds s (1,2) is in the set, but 3 < 1 does not hold s (3,1) is not in the set. Similrly, the reltion = is defined by the pirs { (1,1),(2,2),(3,3) }. Any subset of A A is reltion. Let us define the reltion vi { (1,2),(2,1),(2,3),(3,1) }, then 1 2 nd 3 1 re true, but 1 3 is not.

16 16 INTRODUCTION Definition Let R be reltion on set A. Then R is sid to be (i) reflexive if R for ll A. (ii) symmetric if R b implies br. (iii) trnsitive if R b nd br c implies R c. If R is reflexive, symmetric, nd trnsitive, then it is sid to be n equivlence reltion. Exmple : Let A := {1,2,3} s bove. The reltion < is trnsitive, but neither reflexive nor symmetric. The reltion defined by { (1,1),(1,2),(1,3),(2,2),(2,3),(3,3) } is reflexive nd trnsitive, but not symmetric. Finlly, reltion defined by { (1,1),(1,2),(2,1),(2,2),(3,3) } is n equivlence reltion. Equivlence reltions re useful in tht they divide set into sets of equivlent elements. Definition Let A be set nd R n equivlence reltion. An equivlence clss of A, often denoted by [], is the set {x A : R x}. For exmple, for the reltion bove, the equivlence clsses re [1] = [2] = {1,2} nd [3] = {3}. Reflexivity gurntees tht []. Symmetry gurntees tht if b [], then [b]. Finlly, trnsitivity gurntees tht if [b] nd b [c], then [c]. In prticulr, we hve the following proposition, whose proof is n exercise. Proposition If R is n equivlence reltion on set A, then every A is in exctly one equivlence clss. In prticulr, R b if nd only [] = [b]. Exmple : The set of rtionl numbers cn be defined s equivlence clsses of pir of n integer nd nturl number, tht is elements of Z N. The reltion is defined by (,b) (c,d) whenever d = bc. It is left s n exercise to prove tht is n equivlence reltion, Usully the equivlence clss [(,b)] is written s /b 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. For exmple {1,2,3} hs the sme crdinlity s {,b,c} by defining bijection f (1) :=, f (2) := b, f (3) := c. Clerly the bijection is not unique. A set A hs the sme crdinlity s the empty set if nd only if A itself is the empty set. We then write A := 0.

17 0.3. BASIC SET THEORY 17 Definition Suppose A hs the sme crdinlity s {1,2,3,...,n} for some n N. We then write A := n. If A is empty we write A := 0. In either cse we sy tht A is finite. We sy A is infinite or of infinite crdinlity if A is not finite. 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 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 Schröder 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 will distinguish two types of infinite crdinlity. 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: Let E N be the set of even nturl numbers. Given k E, write k = 2n for some n N. Then f (n) := 2n defines bijection f : N E. 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. Define bijection with N by letting 1 go to (1,1), 2 go to (1,2), nd so on. See Figure 4. 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. So the list would continue: 1/3, 3/1, 1/4, 2/3, etc.... For completeness, we mention the following sttements from the exercises. If A B nd B is countble, then A is countble. The contrpositive of the sttement is tht if A is uncountble, then B is uncountble. As consequence if A < N then A is finite. Similrly, if B is finite nd A B, then A is finite. For the fns of the TV show Futurm, there is movie theter in one episode clled n ℵ 0 -plex.

18 18 INTRODUCTION (1,1) (1,2) (1,3) (1,4) (2,1) (2,2) (2,3)... (3,1) (3,2)... (4,1)... Figure 4: Showing N N is countble. We give the first truly striking result. First, we need nottion for the set of ll subsets of set. Definition The power set of set A, denoted by P(A), is the set of ll subsets of A. For exmple, if A := {1,2}, then P(A) = { /0,{1},{2},{1,2} }. In prticulr, A = 2 nd P(A) = 4 = 2 2. In generl, for finite set A of crdinlity n, the crdinlity of P(A) is 2 n. This fct is left s n exercise. Hence, for finite set A, the crdinlity of P(A) is strictly lrger thn the crdinlity of A. 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 exists n injection f : A P(A). For ny x A, define f (x) := {x}. Therefore A P(A). To finish the proof, we must show tht no function g: A P(A) is surjection. Suppose g: A P(A) is function. So for x A, g(x) is subset of A. Define the set B := { x A : x / g(x) }. We clim tht B is not in the rnge of g nd hence g is not surjection. Suppose there exists n x 0 such tht g(x 0 ) = B. Either x 0 B or x 0 / B. If x 0 B, then x 0 / g(x 0 ) = B, which is contrdiction. If x 0 / B, then x 0 g(x 0 ) = B, which is gin contrdiction. Thus such n x 0 does not exist. Therefore, B is not in the rnge of g, nd g is not surjection. As g 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. A relted fct is 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.... Nmed fter the Germn mthemticin Georg Ferdinnd Ludwig Philipp Cntor ( ).

19 0.3. BASIC SET THEORY 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 nonempty nd finite, then there exists unique n 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}. 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 is bijective. Exercise : Prove by induction tht n < 2 n for ll n N. Exercise : Show tht for finite set A of crdinlity n, the crdinlity of P(A) is 2 n.

20 20 INTRODUCTION 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. 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 countbly infinite collection of finite sets A 1,A 2,..., whose union is not finite set. Exercise : Give n exmple of countbly infinite 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. Exercise : Suppose A B nd B is finite. Prove tht A is finite. Tht is, if A is nonempty, construct bijection of A to {1,2,...,n}. Exercise : Prove Proposition Tht is, prove tht if R is n equivlence reltion on set A, then every A is in exctly one equivlence clss. Then prove tht R b if nd only if [] = [b]. Exercise : Prove tht the reltion in Exmple is n equivlence reltion. Exercise : ) Suppose A B nd B is countbly infinite. By constructing bijection, show tht A is countble (tht is, A is empty, finite, or countbly infinite). b) Use prt ) to show tht if A < N, then A is finite. Exercise (Chllenging): Suppose N S, or in other words, S contins countbly infinite subset. Show tht there exists countbly infinite subset A S nd bijection between S \ A nd S.

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 the definitions of those properties. Definition An ordered set is set S, together with reltion < such tht (i) For ny x,y S, exctly one of x < y, x = y, or y < x holds. (ii) If x < y nd y < z, then x < z. We write x y if x < y or x = y. We define > nd in the obvious wy. The set of rtionl numbers Q is n ordered set by letting x < y if nd only if y x is positive rtionl number, tht is if y x = p/q where p,q N. Similrly, N nd Z re lso ordered sets. There re other ordered sets thn sets of numbers. For exmple, the set of countries cn be ordered by lndmss, so Indi > Lichtenstein. A typicl ordered set tht you hve used since primry school is the dictionry. It is the ordered set of words where the order is the so-clled lexicogrphic ordering. Such ordered sets often pper, for exmple, in computer science. In this book we will mostly be interested in ordered sets of numbers. Definition Let E S, where S is n ordered set. (i) If there exists b S 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 S 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. See Figure 1.1. We write sup E := b 0.

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. E upper bounds of E smller lest upper bound of E bigger Figure 1.1: A set E bounded bove nd the lest upper bound of E. A simple exmple: Let S := {,b,c,d,e} be ordered s < b < c < d < e, nd let E := {,c}. Then c, d, nd e re upper bounds of E, nd c is the lest upper bound or supremum of E. Supremum (or infimum) is utomticlly unique (if it exists): If b nd b re suprem of E, then b b nd b b, becuse both b nd b re the lest upper bounds, so b = b. A supremum or infimum for E (even if they exist) need not be in E. For exmple, the set E := {x Q : x < 1} hs lest upper bound of 1, but 1 is not in the set E itself. On the other hnd, if we tke G := {x Q : x 1}, then the lest upper bound of G is clerly lso 1, nd in this cse 1 G. On the other hnd, the set P := {x Q : x 0} hs no upper bound (why?) nd therefore it cnnot hve lest upper bound. On the other hnd 0 is the gretest lower bound of P. Definition An ordered set S hs the lest-upper-bound property if every nonempty subset E S tht is bounded bove hs lest upper bound, tht is sup E exists in S. The lest-upper-bound property is sometimes clled the completeness property or the Dedekind completeness property. As we will note in the next section, the rel numbers hve this property. Exmple 1.1.4: The set Q of rtionl numbers does not hve the lest-upper-bound property. The subset {x Q : x 2 < 2} does not hve supremum in Q. We will see lter tht the supremum is 2, which is not rtionl. Suppose x Q such tht x 2 = 2. 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. Write m = 2k nd so (2k) 2 = 2n 2. Divide by 2 nd note tht 2k 2 = n 2, nd hence n is divisible by 2. But tht is contrdiction s m/n is 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 us 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. Nmed fter the Germn mthemticin Julius Wilhelm Richrd Dedekind ( ). This is true for ll other roots of 2, nd interestingly, the fct tht k 2 is never rtionl for k > 1 implies no pino cn ever be perfectly tuned in ll keys. See for exmple:

23 1.1. BASIC PROPERTIES 23 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) x + y = y + x for ll x,y F. (A3) (ssocitivity of ddition) (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) xy = yx for ll x,y F. (M3) (ssocitivity of multipliction) (xy)z = x(yz) for ll x, y, z F. (M4) There exists n element 1 F (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. 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. For exmple, there is no x Z such tht 2x = 1, so (M5) is not stisfied. You cn check tht (M5) is the only property tht fils. We will ssume the bsic fcts bout fields tht re esily proved from the xioms. For exmple, 0x = 0 is esily proved by noting tht xx = (0 + x)x = 0x + xx, using (A4), (D), nd (M2). Then using (A5) on xx, long with (A2), (A3), nd (A4), 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. It cn be checked tht the rtionl numbers Q with the stndrd ordering is n ordered field. Proposition Let F be n ordered field nd x, y, z, w 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. (vi) If 0 < x < y, then x 2 < y 2. (vii) If x y nd z w, then x + z y + w. An lgebrist would sy tht Z is n ordered ring, or perhps more precisely commuttive ordered ring.

24 24 CHAPTER 1. REAL NUMBERS 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 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. Prts (vi) nd (vii) re left s exercises. The 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. Proposition Let x,y F where F is n ordered field. Suppose xy > 0. Then either both x nd y re positive, or both re negtive. Proof. Clerly both of the conclusions cn hppen. If either x nd y re zero, then xy is zero nd hence not positive. Hence we 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 x > 0 nd y < 0. Multiply y < 0 by x to get xy < 0x = 0. The result follows by contrpositive. Exmple : The reder my lso know bout the complex numbers, usully denoted by C. Tht is, C is the set of numbers of the form x + iy, where x nd y re rel numbers, nd i is the imginry number, number such tht i 2 = 1. The reder my remember from lgebr tht C is lso field, however, it is not n ordered field. While one cn mke C into n ordered set in some wy, it is not possible to put n order on C tht would mke it n ordered field: In ny ordered field 1 < 0 nd x 2 > 0 for ll nonzero x, but in C, i 2 = 1. Finlly, n ordered field tht hs the lest-upper-bound property hs the corresponding property for gretest lower bounds. Proposition Let F be n ordered field with the lest-upper-bound property. Let A F be nonempty set tht is bounded below. Then inf A exists. Proof. Let B := { x : x A}. Let b F be lower bound for A: if x A, then x b. In other words, x b. So b is n upper bound for B. Since F hs the lest-upper-bound property, c := sup B exists, nd c b. As y c for ll y B, then c x for ll x A. So c is lower bound for A. As c b, c is the gretest lower bound of A.

25 1.1. BASIC PROPERTIES Exercises Exercise 1.1.1: Prove prt (iii) of Proposition Tht is, let F be n ordered field nd x,y,z F. Prove If x < 0 nd y < z, then xy > xz. 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: Prove prt (vi) of Proposition Tht is, let x,y F, where F is n ordered field, such 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 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 b is n upper bound for A. Suppose 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 sup A exists nd sup A / A. Show tht A contins countbly infinite subset. 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. To keep things stright it might be good ide to use different nottion for the nonstndrd ordering such s n m. 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 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. Exercise : Let D be the ordered set of ll possible words (not just English words, ll strings of letters of rbitrry length) using the Ltin lphbet using only lower cse letters. The order is the lexicogrphic order s in dictionry (e.g. < < dog < door). Let A be the subset of D contining the words whose first letter is (e.g. A, bcd A). Show tht A hs supremum nd find wht it is. Exercise : Let F be n ordered field nd x,y,z,w F. ) Prove prt (vii) of Proposition Tht is, if x y nd z w, then x + z y + w. b) Prove tht if x < y nd z w, then x + z < y + w. Exercise : Prove tht ny ordered field must contin countbly infinite set. Exercise : Let N := N { }, where elements of N re ordered in the usul wy mongst themselves, nd k < for every k N. Show N is n ordered set nd tht every subset E N hs supremum in N (mke sure to lso hndle the cse of n empty set). Exercise : Let S := { k : k N} {b k : k N}, ordered such tht k < b j for ny k nd j, k < m whenever k < m, nd b k > b m whenever k < m. ) Show tht S is n ordered set. b) Show tht ny subset of S is bounded (both bove nd below). c) Find bounded subset of S which hs no lest upper bound.

26 26 CHAPTER 1. REAL NUMBERS 1.2 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. To simplify mtters, 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. We sw tht 1 > 0. By induction (exercise) we cn prove tht n > 0 for ll n N. Similrly, we verify simple sttements bout rtionl numbers. For exmple, we proved tht if n > 0, then 1/n > 0. Then m < k implies m/n < k/n. Let us prove one of the most bsic but useful results bout the rel numbers. The following proposition is essentilly how n nlyst proves n inequlity. Proposition If x R is such tht 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. Another useful version of this ide is the following equivlent sttement for nonnegtive numbers: If x 0 is such tht x ε for ll ε > 0, then x = 0. And to prove tht x 0 in the first plce, n nlyst might prove tht ll x ε for ll ε > 0. From now on, when we sy x 0 or ε > 0, we utomticlly men tht x R nd ε R. 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 tke Q nd throw in enough numbers to obtin R. We mentioned lredy tht R must contin elements tht re not in Q becuse of the lest-upperbound property. We sw there is no rtionl squre root of two. The set {x Q : x 2 < 2} implies the existence of the rel number 2, lthough this fct requires bit of work. See lso Exercise 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 (see Exercise 1.1.3), 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. 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. 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.