SUPPLEMENTARY NOTES ON INTRODUCTION TO ANALYSIS. Brooklyn College of the City University of New York. April 2002 Last Revised: January 2013

Transcription

1 SUPPLEMENTARY NOTES ON INTRODUCTION TO ANALYSIS Attil Máté Brooklyn College of the City University of New York April 2002 Lst Revised: Jnury 2013 c 2002, 2013 by Attil Máté Typeset by AMS-TEX

2 ii

3 Tble of Contents iii CONTENTS Prefce v 1. Interchnge of quntifiers 1 2. More on logic 1 3. The Axiom of Completeness 6 4. Supremum nd limits 8 5. Upper nd lower Limit Subspces nd compct sets Totlly bounded spces Compctness of closed intervls The method of intervl-hlving: The Heine Borel Theorem The Bolzno-Weierstrss Theorem The method of intervl-hlving: The Bolzno Weierstrss Theorem Continuous functions. An exmple Continuous functions The Intermedite-Vlue Theorem The method of intervl-hlving: the Intermedite-Vlue Theorem The Mximum-Vlue Theorem Uniform continuity, uniform convergence Differentibility nd continuity An Intermedite-Vlue Theorem for derivtives The chin rule for differentition The inverse of monotone function l Hospitl s rule The reminder term in Tylor s formul The binomil series The Riemnn integrl The Riemnn integrl: n exmple Integrbility of monotonic nd of continuous functions The Newton Leibniz formul Integrtion by prts Chnge of vrible Compctness product spces Interchnging integrtion nd differentition 73 Bibliogrphy 77 List of symbols 79 Subject index 81

4 iv

5 Prefce v Prefce These notes strted s set of hndouts to the students while teching course on introductory nlysis in the spring of 2002 t Brooklyn College of the City University of New York, using Rosenlicht s book [Ros]. 1 While I find this book relly excellent, there were some res where I wnted to give n emphsis to the mteril different from tht given in the book; for exmple, this being single-vrible nlysis, I wnted to give more emphsis to the rel line thn to bstrct metric spces. Hence the supplementry mteril enclosed here. The first drft of these notes were completed in April The dte of the present revision is Jnury New York, New York, Jnury 2013 Attil Máté 1 See the bibliogrphy t the end of these notes for the full title nd publiction dt.

6 vi

7 2. More on logic 1 1. Interchnge of quntifiers There re two kinds of quntifiers: universl:, mening for ll, nd existentil:, mening there is or there exists. Within quntifier, one my specify the kind of things the quntifier tlks bout, e.g., ( x : x is n integer), ( y : y is rel number), ( x Z), ( y R), etc.; here Z stnds for the set of integers (positive, negtive, or zero), nd R stnds for the set of rels. Two quntifiers of the sme kinds men either two universl quntifiers or two existentil quntifiers, while two quntifiers of different kinds refer to one universl nd one existentil quntifier. Two quntifiers of the sme kind re lwys interchngeble, but two quntifiers of different kinds re not. To see this, consider the following exmple: ( x : x is licensed driver)( y : y is cr) (x hs driven y). This sentence is entirely resonble, since it only sys tht every licensed driver hs driven cr, or, more precisely, every licensed driver hs driven t lest one cr. In rel life, there re exceptions even to such resonble sttements, but if you restrict your ttention to life in smll town, the sttement is most likely true. If one interchnges the quntifiers, the result is totlly bsurd: ( y : y is cr) ( x : x is licensed driver) (x hs driven y). Thissystht thereissinglecrthteverylicenseddriverhsdriven, ndthisisunlikely to be true even in smll town, unless everyone hs gone to the sme driving school within the lst few yers, nd got chnce of prctising on the sme cr. Another, more mthemticl, exmple is the following. Let x nd y run over integers. Then the formul ( x)( y)[x > y] is true, while the formul ( y)( x)[x > y] is flse. Indeed, the first formul is true. Given n rbitrry integer x, we cn pick y = x 1 to ensure tht x > y. On the other hnd, the second formul is not true. To see this, pick n rbitrry number y. Then ( x)[x > y] is certinly not true; for exmple, x > y is not true with x = y More on logic A sentence is sttement tht is either true or flse. 2 Sentences cn be connected by the logic opertions, lso clled sententil connectives nd, or, if... then, nd if nd only if, denoted in turn by &,,, nd. Insted of &, people often use the symbol. The logic opertions only connect sentences (or, more generlly, sttements see below). For 2 Tht is, true or flse in principle. If one tlks bout forml logic, then sentence would be declrtive sentence. Some declrtive sentences, however, cnnot be considered sentences in the sense of logic, since they cnnot be considered true or flse even in principle. For exmple, the sentence This sentence is flse cnnot be tken s true (since it then sys tht it is flse), nd it cnnot be tken s flse (since then it sys tht it is true). In mthemticl logic, wht is sentence or sttement is defined formlly by describing the rules how sttement cn be formed. The collection of these rules is clled syntx.

8 2 Notes on Anlysis exmple, the or in the sentence You cn hve coffee or te with your brekfst is not logic opertion. 3 The mening of the sententil connectives cn be illustrted by truth tbles. Writing T for true nd F for flse (T nd F re clled truth vlues) we hve for the opertion &, clled conjunction: A B A & B T T T T F F F T F F F F For the opertion, clled disjunction, we hve A B A B T T T T F T F T T F F F As seen from the truth tble, disjunction is lwys ment in the inclusive sense, tht is A B is true unless both A nd B re flse. This differs from colloquil usge, where one often uses the word or in the exclusive sense, where A or B in the exclusive sense is true if exctly one of A nd B is true. For the opertion, clled conditionl, we hve A B A B T T T T F F F T T F F T The mening of the conditionl is often differs from its colloquil use, where the mening of ifathenb isunclerincseaisflse. Inmthemtics, onestrictlyfollowsthetruthtble bove. For exmple, the sentence if 2 by 2 is 5 then the snow is blck is true sentence in mthemtics. 4 In colloquil speech one would consider this sentence meningless, or t best pointless. But it illustrtes n importnt point: in the conditionl, there does not need to be cusl connection between the constituents. 5 For the opertion A B, insted of sying if A then B, it is often more convenient to sy A only if B. In cse of this ltter sentence, the colloquil mening pproches more closely the mthemticl mening of the conditionl. Nmely, A only B mens tht A is llowed to be true only in cse B is lso true; indeed, when A is true nd B is flse, the truth tble entry for A B shows flse. One occsionlly reverses the rrow in the conditionl, using the symbol A B mening 3 However, on cn rewrite this s logic opertion by sying tht You cn hve coffee with your brekfst or you cn hve te with your sentence (lthough by doing this, one probbly chnges the mening, since in logic, or is ment in the inclusive sense (llowing to hve both coffee nd te), while on resturnt menu the mening is probbly exclusive (not llowing to hve both coffee nd te without pying extr). 4 This exmple is due to the Germn mthemticin Dvid Hilbert. Note tht the sentence if 2 by 2 is 5 then the snow is white is lso true. 5 Tht is, 2 by 2 being 5 does not cuse the snow to be blck. As for the constituents, or, with more technicl word, opernds, in the conditionl A B, A clled the ntecedent nd B is clled the consequence (the ltter is somewht misleding nme, since the nme seems to imply cusl connection).

9 2. More on logic 3 A if B, or if B then A : A B A B T T T T F T F T F F F T The conditionl A B (or B A) is usully clled the converse of A B. Finlly, the truth tble of the opertion, clled biconditionl, is A B A B T T T T F F F T F F F T A B is expressed s A if nd only if B, or, sometimes, s A iff B (but it is not cler how one should pronounce the word iff ). The word iff ws introduced by Pul Hlmos. A if nd only if B is short for sying A if B nd A only if B ; formlly, for (A B) & (A B). It is esy to check tht the truth tble for this formul is the sme s the one given for the biconditionl bove. These logic opertions re clled binry opertions, since they involve two constituents, clled opernds, A nd B in the bove truth tbles. The letters A nd B used in the bove formuls re often clled sententil vribles, i.e., vribles tht cn either be true or flse. Negtion. The opertion not is clled negtion. Not A mens tht it is not the cse tht A, or, more simply, A is not true. This is clled unry opertion, since it hs only one opernd. The truth tble for negtion is A A T F F T Tutologies. A tutology is logic expression (n expression involving the logicl opertions just defined) tht is lwys true, whether or not the sententil vribles in it re true or flse. Exmples for tutologies re ( (A & B)) (( A) ( B)). For better redbility, one cn drop severl pirs of prentheses here, to write (A & B) A B. To mke sense of this wy of writing the formul, one cn ssigns priority to the logic opertions in the order, &,,,,, mening tht one first try to perform the opertions with higher priority. 6 Most people consider the priority between & nd, nd between nd uncler, so it is best to use prentheses to void misunderstnding. To check tht the bove formul is indeed tutology, one cn use the truth tble to evlute 6 This is similr to the rule in lgebr tht (multipliction) hs higher priority thn + (ddition). Tht is, the formul mens 2 + (3 5), nd not (2 + 3) 5. + nd hve equl priority, so in the expression one performs the opertions from left to write, i.e., this expression mens (((2+3) 5) 6)+8 In computer science, one often uses the word precedence insted of priority; computer scientists unmbiguously ssign higher precedence to to logicl nd thn to logicl or; in mthemtics, this precedence is not lwys tken for grnted.

10 4 Notes on Anlysis it for ech choice of A nd B to find tht the formul is lwys true. One interprettion of the bove tutology is tht (A & B) nd A B men the sme thing. This is becuse the biconditionl is true exctly when the two sides hve the sme truth vlue; so the bove formul being lwys true mens tht the two sides on the biconditionl in it lwys hve the sme truth vlue. The bove formul is one of the two De Morgn identities. The other De Morgn identity is the tutology Another simple tutologies re (A B) A & B. (A B) A B; we used prentheses on the left here, since the sometimes nd is considered to hve equl priority. When one wnts to estblish the impliction 7 A B, one often tkes dvntge of the tutology (A B) ( B A), nd proves the impliction B A insted. The conditionl B A is clled the contrpositive of A B. A further tutology is (A B) ( A B). One often writes A B insted of (A B). The truth tble of A B is A B A B T T F T F T F T T F F F One might cll the opertion A B exclusive or, since it reflects the mening the word or often used colloquilly. However, it is best to void this term exclusive or, since in mthemtics the word or is lwys used in the inclusive sense, s defined by the logic opertion A B. The expression A B in mthemtics is often given s A if nd only if not B, reflecting the fct tht this this expression is equivlent to (i.e., true exctly the sme time s) the expression A B; tht is, the fct tht is tutology. (A B) (A B) Open sttements. An open sttement is sttement tht hs (zero or more) vribles in it; when one gives vlues to these vribles. For exmple, to sy tht x is greter thn 2 (where x denotes n unspecified rel number) is on open sttement. One cn tell its truth vlue only fter one specifies wht rel number x is. In mthemticl logic, one describes the rules how open sttements re formed; the collection of these rules re clled syntx. However, wnt to discuss mtters somewht informlly, so we will void detiled discussion of syntx. Occsionlly, open sttements will be denoted by script cpitl letters such s A or B. 7 One often clls conditionl n impliction, especilly in informl usge, since insted of if A then B one might sy A implies B. However, logicl impliction hs mening seprte from, though relted to, tht of the conditionl. Therefore, some consider clling conditionl n impliction objectionble.

11 2. More on logic 5 Negting quntified sttements. The formul ( x)a is the sme s ( x) A; here A is some open sttement. 8 Tht is, sying tht it is not true tht for ll x A holds mens tht there is n x for which A does not hold. One writes this by sying tht ( x) ( x) ; here mens tht wht re written on the two sides re equivlent, i.e., they men the sme thing, i.e. they cn replce ech other. 9 Similrly, the formul ( x)a is the sme s ( x) A. Tht is, to sy tht it is not true tht there exists n x for which A holds mens tht for ll x it is not true tht A holds. This cn be expressed by the rule ( x) ( x). One cn use these rules nd the tutologies mentioned bove to move negtion inside formul. For exmple, ( (1) ( ǫ > 0)( δ > 0)( p (0,1))( q (0,1)) p q < δ 1 p 1 q < ǫ)). Here p, q, ǫ, nd δ denote rel numbers, 10 nd (0,1) denotes the intervl {t : 0 < t < 1}. Wht this formul expresses is irrelevnt for our present purpose. 11 The quntifiers ( ǫ > 0) is clled restricted quntifier, since it sys tht ǫ rnges over positive rel numbers insted of ll rel numbers (where unrestricted vribles rnge in the present cse). Similrly, ( δ > 0), ( p (0,1) re restricted quntifiers. The importnt point t present is tht the bove rules of interchnging negtion nd quntifiers re lso true for restricted quntifiers. The bove formul cn be trnsformed s follows: ( ( ǫ > 0)( δ > 0)( p (0,1))( q (0,1)) p q < δ 1 p 1 ) q < ǫ ( ( ǫ > 0)( δ > 0)( p (0,1))( q (0,1)) p q < δ 1 p 1 ) q < ǫ ( ( ǫ > 0) ( δ > 0)( p (0,1))( q (0,1)) p q < δ 1 p 1 ) q < ǫ ( ( ǫ > 0)( δ > 0) ( p (0,1))( q (0,1)) p q < δ 1 p 1 ) q < ǫ ( ( ǫ > 0)( δ > 0)( p (0,1)) ( q (0,1)) p q < δ 1 p 1 ) q < ǫ ( ( ǫ > 0)( δ > 0)( p (0,1))( q (0,1)) p q < δ 1 p 1 ) q < ǫ ( ( ǫ > 0)( δ > 0)( p (0,1))( q (0,1)) p q < δ & 1 p 1 ) q < ǫ ( ( ǫ > 0)( δ > 0)( p (0,1))( q (0,1)) p q < δ & 1 p 1 ) q ǫ. 8 Presumbly, A depends on x, so one might be tempted to write A(x) insted. However, writing A(x) relly does not mke things clerer. 9 We do not regrd s logicl connective here. It cn be considered s rule ccording to which we cn chnge (or trnsform) formuls without chnging their menings. Tht is, it expresses trnsformtion rule; see below. 10 One lso sys tht these vribles rnge over rel numbers. 11 It expresses the sttement tht the function f(x) = 1/x is not uniformly continuous in the intervl (0, 1), discussed below in these notes.

12 6 Notes on Anlysis Restricted quntifiers. If E is set, nd A is n open sttement, quntifiers of the form ( x E) nd ( x E) re clled restricted quntifiers. More generlly, given n open sttement A, one cn consider the restricted quntifiers ( x : A nd x : A). If B is nother open sttement, then the formul ( x : A)B sys tht for ll x such tht A holds we (lso) hve B, nd for ( x : A)B mens tht for ll x for which A holds we lso hve B. It is esy to see tht ( x : A)B ( x)(a B), nd ( x : A)B ( x)(a & B). Therefore, restricted quntifiers re not strictly necessry, but they re often convenient to use. They frequently mke formuls simpler, nd, very importnt point, the rules discussed bove involving the interchnge of negtion nd quntifiers re lso true for restricted quntifiers. Finlly, for set E, ( x E) ( x : x E) nd ( x E) ( x : x E). 3. The Axiom of Completeness A cut is pir (A,B) such tht A nd B re nonempty subsets of the set R of rel numbers with A B = R, nd such tht for every x A nd every y B we hve x < y. If (A,B) is cut, it is cler tht the sets A nd B must be disjoint (since every element of A is less thn every element of B). For cut (A,B) nd rel number t we sy tht the cut determines the number t if for every x A we hve x t nd for every y B we hve t y. For exmple, cut tht determines the number 2 is the pir (A,B) with A = {t : t 2} nd B = {t : t > 2}. Another cut tht determines the number 2 is the pir (C,D) with C = {t : t < 2} nd D = {t : t 2}. Itisclerthtcutcnnotbedeterminemorethnonenumber. Assume, onthecontrry, tht the cut (A,B) determines the numbers t 1 nd t 2 nd t 1 t 2. Without loss of generlity, we my ssume tht t 1 < t 2. Then the number c = (t 1 +t 2 )/2 cnnot belong to A, since we cnnot hve t 1 < x for ny element of A, wheres t 1 < c. Similrly c = (t 1 +t 2 )/2 cnnot be n element of B, since we cnnot hve y < t 2 for ny element of B, wheres c < t 2. Furthermore, the bove exmple is of generl chrcter, in tht for every rel number u there re exctly two cuts, (A,B) nd (C,D) tht determine u, where A = {t : t u} nd B = {t : t > u}. C = {t : t < u} nd D = {t : t u}. The Axiom of Completeness is n importnt property of rel numbers: Axiom of Completeness. Every cut determines rel number. Ordinrily, one does not expect to prove this sttement, since xioms re bsic sttements tht one does not prove. However, one cn prove the Axiom of Completeness if one defines the rel numbers s infinite decimls However, defining the rels numbers s infinite decimls is inelegnt in tht the deciml number system cme bout only by historicl ccident. One could lso define the rels, however, s infinite binry frctions; i.e., deciml frctions in the binry number system. The binry number system is somewht more nturl thn the deciml system in tht it is the simplest of ll number systems.

13 3. The Axiom of Completeness 7 Proof of the Axiom of Completeness. Let (A,B) be cut. Assume first tht A contins positive number. Let n be positive integer, nd let x (n) = x (n) 0.x(n) 1 x(n) 2...x (n) n be the lrgest deciml frction with n digits fter the deciml point such tht x (n) is less thn or equl to every element of B; here x (n) 0 is nonnegtive integer, nd, for ech i with 1 i n, x (n) i is digit, i.e., one of the numbers 0, 1,..., There is such n x (n), since the set of deciml frctions with n digits fter the deciml point tht re nonnegtive nd less thn or equl to every element of B is finite. 14 It is esy to see tht if m nd n re positive integers nd m < n then for ech i with 0 i m we hve x (m) i = x (n) i. Indeed, we cnnot hve x (m) i < x (n) i. for ny i with 0 i m since then the number x = x (n) 0.x(n) 1 x(n) 2...x (n) m would be n m-digit deciml greter thn x (m) nd less thn or equl to every element of B. Furthermore, we cnnot hve x (m) i > x (n) i. for ny i with 0 i m, since then, with this i, the number x = x (m) 0.x (m) 1 x (n) 2...x (m) i (n i zeros t the end) would be n n-digit deciml greter thn x (n) nd less thn or equl to every element of B. Now, the infinite deciml r = x (0) 0.x(1) 1 x(2) 2...x (n) n... is the number determined by cut (A, B). Indeed, r x for every x A. Assume, on the contrry, tht r < x nd let n be such tht 10 n < x r. Then noting tht r x (n) < 10 n, the deciml number x (n) + 10 n r + 10 n < x would be less thn every element of B (since x is less thn every element of B), contrdiction the choice tht x (n) ws the lrgest such n-digit deciml frction. Furthermore, r y for every y B. Assume, on the contrry tht r > y for some y B, nd let n be such tht 10 n < r y. Then noting tht r x (n) < 10 n, we hve x (n) > r 10 n > y, contrdiction the ssumption tht x (n) y for every y B. The proof is complete in cse the cut (A,B) is such tht A contins positive number. If (A,B) is cut such tht B contins the negtive number, writing A = { x : x A} B = { y : y B}, the cut ( B, A) is such tht B contins positive number; hence, by the bove rgument, this cut determines number r. Then the cut (A,B) determines the number r. Finlly, if the cut (A,B) is such tht A does not contin positive number nd B does not contin negtive number, then it is cler tht the cut (A,B) determines the number 0. The proof is complete. TheAxiom ofcompleteness gurntees, forexmple, thtthenumber 2exists. Nmely, the cut (A,B) with A = {x : x < 0 or x 2 2} nd B = {x : x > 0 nd x 2 > 2} nd determines the number t such tht t 2 = 2. To show this, we will first show tht for every ǫ > 0 we hve t 2 2 < ǫ. In order to show this, we my ssume tht ǫ < 1. First note note tht there is n x A nd y B with x 0 nd y 0 (the ltter holds for every y B) such tht y x ǫ/6. To see this, consider the set S = {nǫ/6 : n 0 is n integer nd nǫ/6 A}. S is not empty, since 0 S. Furthermore, it is clerly finite set, nd so it hs lrgest element. Now, choose x to be the lrgest element of this set nd put y = x+ǫ/6 (clerly, y B, since if we hd y / B then we would hve y A, nd so y S, nd then x would not be the lrgest element of x. Observe tht x < 2 (becuse x A, nd so x 2 2 unless x < 0 by the definition of A) nd y = x+ǫ/6 2+1/6 < 3 (this is where we used the ssumption ǫ < 1). Furthermore, x t y, since t is the number determined by the cut (A,B). We hve t 2 2 t 2 x 2 = (t x)(t+x) (y x)(y +x) < ǫ 5 < ǫ; 6 13 i.e., if x (n) = then n = 2, x (n) 0 = 345, x (n) 1 = 8, nd x (n) 2 = The set of these deciml frctions is nonempty, since it contins the number 0. It is finite, becuse if b is n rbitrry element of B, there re t most 10 n b+1 n-digit nonnegtive deciml frctions less thn or equl to b.

14 8 Notes on Anlysis the third inequlity holds becuse we hve y x ǫ/6, x < 2, nd y < 3. Similrly, 2 t 2 y 2 t 2 = (y t)(y +t) (y x)(y +y) < ǫ 6 = ǫ. 6 These two inequlities together show tht t 2 2 < ǫ. s climed. Since ǫ > 0 ws rbitrry, this inequlity holds for every ǫ > 0. Now, ssume tht t 2 2. Then the number t 2 2 is positive. Choosing ǫ = t 2 2, the bove inequlity cnnot hold. This is contrdiction, showing tht we must hve t 2 = 2. Let S be nonempty set of rels. A number t such tht t x for every x S is clled n upper bound of S. The set A is clled bounded from bove if it hs n upper bound. The number c tht is n upper bound of S such tht c t for every upper bound t of S is clled the lest upper bound or supremum of S. The supremum of S is denoted s sups. Lemm. Let S be nonempty set S of rels tht is bounded from bove. Then S hs supremum. Proof. Let B be the set of upper bounds of S, nd let A be the set of those rels tht re not upper bounds of S (i.e., A = R\B). Then it is cler tht (A,B) is cut. Indeed, if we hve x y nd y B for the rels x nd y, then x is lso n upper bound of S (since it is t greter thn or equl to nother upper bound, nmely y). This shows tht for every x A nd y B we must hve x < y. A is not empty since S is not empty, so not every rel is n upper bound of S; B is not empty, since S is bounded from bove by ssumption. Let t be the rel determined by the cut (A,B). Then t is n upper bound of S. Assume, on the contrry, tht there is n s S is such tht s > t. Let u = (s+t)/2. Then u > t, so we must hve u B (since the cut (A,B) determines t). Yet u < s, so u is not n upper bound of S, contrdicting the reltion u B. Furthermore, t is the lest upper bound of S. Assume, on the contrry, tht y is n upper bound of S such tht y < t. Then y B (since B contins every upper bound of S). But then we must hve y t (since the cut (A,B) determines t). This contrdicts the reltion y < t. The proof is complete. For the empty set, one usully writes tht sup =, nd for set S tht is not bounded from bove, one writes tht sups = +. With this extension, the symbol sups will be meningful for ny subset of S rels. Let S be nonempty set of rels. A number t such tht t x for every x S is clled lower bound of S. The set A is clled bounded from below if it hs n lower bound. The number c tht is n lower bound of S such tht c t for every lower bound t of S is clled the gretest lower bound or infimum of S. The infimum of S is denoted s infs. Every nonempty set tht is bounded from below hs n infimum. The proof of this sttement is similr to the proof of the Lemm bove. Insted of crrying out this proof, one cn rgue more simply tht infs = sup( S), where S def = { s : s S}. One usully writes inf = +, nd if S is not bounded from below, then one writes infs =. 4. Supremum nd limits Lemm. Let S be nonempty set of rels tht is bounded from bove, nd let = sups. Let ǫ > 0. Then the intervl ( ǫ,] contins n element of S.

15 4. Supremum nd limits 9 Proof. As ǫ is not n upper bound of S ( being its lest upper bound), there must be n x S with x > ǫ. We hve x, since is n upper bound of S. Thus x S ( ǫ,]. Definition. A set S in the metric spce (E,d) is sid to be open if for every p S there is n ǫ > 0 such tht the open bll {q : d(p,q) < ǫ} is subset of S. Wht is ment in the bove definition is if nd only if (tht is, (E,d) is sid to be open if nd only if for every... ). In mthemticl definitions, it is customry to sy if in similr cses when one mens if nd only if. In other situtions in mthemtics, one mkes very creful distinction between if nd if nd only if. Definition. A set S in the metric spce (E,d) is sid to be closed if its complement S is open. Lemm. Let S be nonempty closed set of rels tht is bounded from bove. Then S hs mximum. In other words, there is u S such tht u s for every s S. Proof. As S is nonempty nd bounded from bove, it hs supremum; write u = sups. We will prove tht u S; then it will be cler tht u = mxs (i.e. tht u S nd x u for every x S). Assume, on the contrry, tht u / S. Then u S. As S is closed, its complement S is open; thus, there is n open bll (u ǫ,u+ǫ) with center u tht is included in S (ǫ > 0; in R, the open blls re open intervls). Hence S hs no elements in this intervl; since u is n upper bound of S (i.e., x u for every x S), this mens tht u ǫ is n upper bound of S (i.e., x u ǫ for every x S). This contrdicts the ssumption tht u is the lest upper bound of S. Therefore the ssumption u / S must be wrong, completing the proof. Lemm. Let S be set in metric spce (E,d) is not closed. Then there is point p / S of E nd sequence {p n } n=1 of elements of S tht converges to p. Proof. As S is not closed, its complement S is not open. Therefore, there is point p S such tht no open bll with center p is included in S. Tht is, for every ǫ > 0 the set {q : d(p,q) < ǫ} S is not empty (if it were, the open bll on the left would be included in S). Using this sttement with ǫ = 1/n, for ech positive integer n we cn find p n in this set; tht is p n S nd d(p,p n ) < 1/n. From the ltter inequlity we cn conclude tht limp n = p. Thus the sequence {p n } n=1 hs the desired properties. Lemm. Assume { n } n=1 is n incresing sequence of rels tht is bounded from bove. Then { n } n=1 is convergent. Proof. The set S = { n : 1 n < } is bounded from bove, so it hs supremum. Let be this supremum. We will show tht lim n =. To this end, let ǫ > 0 be rbitrry. Since ǫ is not n upper bound of S, the set S hs n element, sy N for some positive integer N, for which N > ǫ. As the sequence { n } n=1 is incresing, we hve n N for every n N. As we lso hve n (since n S nd is n upper bound of S), it follows tht n ( ǫ,+ǫ) whenever n N. Hence the sequence { n } n=1 converges to. Lemm. Let be such tht 0 < < 1. Then lim n n = 0. Proof. The ssumptions imply tht 0 < n+1 < n ; thus the sequence { n } n=1 is decresing nd bounded from below; therefore it hs limit. Write x = lim n. Clerly, we lso hve x = lim n+1. This cn esily be verified directly. Alterntively, one my observe

16 10 Notes on Anlysis tht { n+1 } n=1 is subsequence 15 of { n } n=1; hence, the former sequence must hve the sme limit s the ltter. Now, x = lim n = lim n = lim n+1 = x, i.e., x = x, or ( 1)x = 0. As 0 by our ssumptions, this is only possible if x = 0. Thus, lim n = 0, which is wht we wnted to prove. Second proof. The ssertion lim n = 0 cn lso be proved by using Bernoulli s Inequlity, sying tht (1+x) n 1+nx holds whenever x 1, for every positive integer n. For x 0, Bernoulli s Inequlity is direct consequence of the Binomil Theorem. The cse 1 x < 0 of Bernoulli s Inequlity is hrder to estblish, but this cse is not needed for proving lim n = 0. Writing x = 1 1, we hve x 0 nd = 1/(1+x), nd so (1/)n = (1+x) n 1+nx, i.e., n 1/(1 + nx). Given ǫ > 0, we will hve 1/(1 + nx) < 1/(nx) ǫ whenever n N = 1/(xǫ). For such n we will hve ǫ < n 0 < ǫ, showing tht lim n = 0. Lemm. Let A nd B be two nonempty sets of rels tht re bounded from bove. Then sup{x+y : x A nd y B} = supa+supb. Proof. Writing S = sup{x+y : x A nd y B}, = supa, nd b = supb, we will first show tht sups +b. To this end, let x+y be n element of S, where x A nd y B. Then x (since is n upper bound of A; in fct, it is its lest upper bound) nd y b (since b is n upper bound of B). Thus x+y +b. As x+y ws n rbitrry element of S, this shows tht +b is n upper bound of S. Therefore sups +b, since sups is the lest upper bound of S. Next we will show tht sups +b. We will do this by showing tht no number c < +b is n upper bound of S. So, let c < + b be rbitrry nd write ǫ = ( + b c)/2; then ǫ > 0. Now ǫ is not n upper bound of A (s is its lest upper bound), so there must be n x A with x > ǫ. Similrly, b ǫ is not n upper bound of B, so there must be y B with y > b ǫ. Then x+y > ( ǫ)+(b ǫ) = +b 2ǫ = c. As x+y S, this implies tht c is not n upper bound of S, s we wnted to show. As both sups +b nd sups +b hold, we must hve sups = +b. The proof is complete. Second proof. Using the nottion introduced in the first proof, we will give second proof of the inequlity sups + b (we will not give nother proof of the inequlity sups +b; the proof of this ltter inequlity will hve to be tken from the first solution). Let c be n upper bound of S; it will be enough to show tht c +b. Let x A nd y B be rbitrry. Then x+y S, nd so x+y c, since c is n upper bound of S. Tht is, x c y. Now, consider this inequlity for fixed y B. Then we cn see tht, for every x A, the inequlity x c y holds. I.e., c y is n upper bound of A, nd so c y, the lest upper bound of A. The lst inequlity cn lso be written s y c ; this inequlity holds for every element y B (since y B ws rbitrry). Thus c is n upper bound of B; i.e., c b, since b is the lest upper bound of this set. Thus c +b, s we wnted to show. 15 The sequence {b n} n=1 is clled subsequence of {n} n=1 if bn = f(n) for some strictly incresing function f tht is defined for ech positive integer nd hs positive integers s vlues. In the present cse, one cn put n = n nd b n = n+1 = n+1, tht is, f(n) = n+1.

17 5. Upper nd lower limit Upper nd lower limit Let { n } n=1 be sequence of rel numbers. The upper limit or limit superior of this sequence is defined s lim sup n = sup{x : x n n for infinitely mny n}. As the supremum of set ws lwys defined to be either rel number or + (when the set is not bounded from bove) or (when the set is empty), it follows every sequence hs n upper limit tht is either rel number or is + or. The lower limit or limit inferior of the bove sequence is defined s lim inf n n = inf{x : x n for infinitely mny n}. As the infimum of set ws lwys defined to be either number or (when the set is not bounded from below) or + (when the set is empty), it follows every sequence hs n lower limit tht is either rel number or is or +. We hve Lemm 1. Given sequence { n } n=1 of rel numbers, we hve lim inf n limsup n. n n Proof. Write s = liminf n n, S = limsup n n nd ssume s > S. Let c be rel number such tht S < c < s. 16 Then, noting tht c > S, it follows tht there re only finitely mny n s such tht c n ; tht is, the set {n : c n } is finite. 17 Similrly, noting tht c < s, it follows tht the set {n : c n } is finite. This is contrdiction, since the union of these two sets is the set of ll positive integers. Lemm 2. Given sequence { n } n=1 of rel numbers, if lim n n exists then we hve lim n n limsup n n. When sying tht lim n n exists we of course men tht lim n n is rel number. However, the reder my reflect tht the ssertion of the Lemm remins vlid even if we ssume lim n n =, nd it cn be proved similrly (with some modifictions, since infinite limits re defined differently from limits tht re equl to rel numbers). If we ssume lim n n = +, then the sttement is of course vlid, but in tht cse the sttement is uninteresting. Proof. Write L = lim n n, S = limsup n n, nd ssume tht L < S. Let c be rel number such tht L < c < S. 18 Write ǫ = c L; clerly, ǫ > 0. Then, noting tht L = lim n n, there is n N such tht n L < ǫ whenever n > N, i.e., such tht ǫ < n L < ǫ whenever n > N. Hence the set {n : x n } is finite for x c = L+ǫ; in fct, no n with n > N belongs to this set. Therefore, the supremum of the set {x : x n for infinitely mny n} is less thn or equl to c. On the other hnd, the supremum of this set is S = limsup n n, ccording to the definition of upper limit. This is contrdiction, since we hve S > c. The proof is complete. 16 If S nd s re rel numbers, one cn tke c = (s + S)/2, but this choice will not work if S or s is (positive or negtive) infinity. It is, however, esy to see tht n pproprite c cn be found even in this cse. 17 Note tht this rgument works even if S =. Similrly, the rgument next works even if s = +. We cnnot hve S = + or s = since S < s ccording to our ssumption. 18 If S is rel number then we cn tke c = (L+S)/2. If S = + then we cn tke c = L+1.

18 12 Notes on Anlysis Lemm 3. Given sequence { n } n=1 of rel numbers, if lim n n exists then we hve lim n n liminf n n. For the proof, one cn repet the rgument of Lemm 2, with pproprite chnges. However, it is esier to note tht lim n n = lim n ( n ) nd liminf n n = limsup n ( n ), According the Lemm 2, we hve lim ( n) limsup( n ), n n nd so the conclusion follows. Agin, the reder my reflect tht the ssertion of the Lemm remins vlid even if we ssume lim n n = +. If we ssume lim n n =, then the sttement is of course vlid, but in tht cse the sttement is uninteresting. Corollry. Given sequence { n } n=1 of rel numbers, if lim n n exists then we hve lim n = liminf n = limsup n. n n n This is n immedite consequence of Lemms 1, 2, nd 3, since ccording to these we hve lim n liminf n n n limsup n n lim n n. Agin, the sttement of the Corollry cn be extended to the cse when lim n is + or by using similr extensions of Lemms 2 nd In the converse direction we hve Lemm 4. Given sequence { n } n=1 of rel numbers, if limsup n n is rel number nd lim inf n = limsup n, n then lim n n exists. n Proof. Write L = liminf n n = limsup n n, nd let ǫ > 0 be rbitrry. Then noting tht L+ǫ > limsup n n, it follows tht there re only finitely mny n s such tht L+ǫ n ; tht is, the set {n : L+ǫ n } is finite. Then there is n integer N 1 such tht the set {n > N 1 : L+ǫ n } is empty. Similrly noting tht L ǫ < liminf n n, it follows tht there re only finitely mny n s such tht L ǫ n ; tht is, the set {n : L+ǫ n } is finite. Then there is n integer N 2 such tht the set {n > N 2 : L ǫ n } is empty. Writing N = mx{n 2,N 2 }, both the sets {n > N : L+ǫ n } nd {n > N : L ǫ n } re empty. Tht is, we hve L ǫ < n < L+ǫ for every n > N. Since ǫ > 0 ws rbitrry, it follows tht L = lim n n. The proof is complete. By modifiction of the rgument used here, it is esy to show tht if liminf n n = limsup n n = thenlim n n = ndifliminf n n = limsup n n = + then lim n n = +. The completeness of R. An esy consequence of the bove is the following 19 Lemm 1 need not be extended, since in tht Lemm we llowed liminf n n nd limsup n n to be + or.

19 Lemm 5. In R, ny Cuchy sequence is convergent. 6. Subspces nd compct sets 13 Proof. Let { n } n=1 be Cuchy sequence. Let ǫ > 0 be rbitrry, nd let N be n integer such tht m n < ǫ/3 whenever m,n > N. Fix k > N. Then the set {n 1 : k +ǫ/3 n } is finite; in fct, it does not contin ny integer n > N. Hence lim sup n = sup{x : x n n for infinitely mny n} k +ǫ/3. Similrly, the set {n 1 : k ǫ/3 n } is finite; in fct, it does not contin ny integer n > N. Hence lim inf n n = inf{x : x n for infinitely mny n} k ǫ/3. Thus Tht is, we hve lim sup n liminf n ( k +ǫ/3) ( k ǫ/3) = 2ǫ/3 < ǫ. n n lim sup n liminf n < ǫ. n n Sinceǫ > 0wsrbitrry, thisimpliesthtlimsup n n liminf n n 0. Indeed, ifwe hd limsup n n liminf n n > 0, then the choice ǫ = limsup n n liminf n n would men tht ǫ < ǫ ccording to the lst displyed inequlity, contrdiction. Thus, liminf n n limsup n n. Since, ccording to Lemm 1 we lso hve liminf n n limsup n n, the equlity liminf n n = limsup n n follows. Therefore, lim n n exists, ccording to Lemm 4. This completes the proof. 6. Subspces nd compct sets Let S be collection of sets ( collection is just synonym of set, so collection of sets is just set of sets). The union of S, lso clled the union of ll elements of S, is defined s S def = {x : y S [x y]}. Given metric spce (E,d), point p E, nd positive rel number r, we will write U E (p,r) for the open bll of rdius r centered t p. Tht is U E (p,r) def = {q E : d(p,q) < r}. If the spce E is understood from the context, we my omit it, nd write U(p,r) insted of U E (p,r). The following is simple chrcteriztion of open sets.

20 14 Notes on Anlysis Lemm. Let (E,d) be metric spce. The set S E is open if nd only if S = {U E (p,r) : p E & r > 0 & U E (p,r) S}. In plin lnguge, the Lemm sys tht the set S is open if nd only if it is the union of the open blls it includes. Sometimes one writes y S y insted of S, but the ltter nottion is clerly simpler. Using the former nottion, the set on the right-hnd side of the eqution in the Lemm could be described s S = p E & r>0 & U E(p,r) S U E (p,r). Proof of the Lemm. The set on the right-hnd side of the eqution of the Lemm, being union of open blls, is n open set. Hence, if the eqution does hold, then S is n open set. This proves the if prt of the Lemm. To prove the only if prt, first note tht S {U E (p,r) : p E & r > 0 & U E (p,r) S}, since union of subsets of S is tken on the right-hnd side. We will show tht S {U E (p,r) : p E & r > 0 & U E (p,r) S} lso holds in cse S is open. Indeed, ssume S is open, nd let p S be rbitrry. Then there is n r > 0 such tht U E (p,r) S; s p U E (p,r), this shows tht p is lso n element of the right-hnd side. Since p S ws rbitrry, it follows tht S is indeed subset of the union on the right-hnd side. Thus, both inclusions hold, nd so S must be equl to the union on the right-hnd side. This estblishes the only if prt. Thus, the proof of the Lemm is complete. Now, let (E,d) be metric spce nd let S be subset of E. It is esy to see tht then (S,d) is metric spce; this spce hs S s its set of points, nd its distnce function is the sme d s is the distnce function on E. 20 The spce (S,d) is clled subspce of the spce (E,d). The next lemm describes the open sets in the spce (S,d) in terms of those in (E,d). Lemm. Let (E,d) be metric spce, nd let S E be set. A set M S is open in (S,d) if nd only if there is set U open in (E,d) such tht M = U S. Similrly, set M S is closed in (S,d) if nd only if there is set F closed in (E,d) such tht M = F S. Note tht the fct tht the set M S is open (or closed) in (S,d) does not men tht M is open (or closed) in (E,d). For exmple tking E = R (R is the set of rel numbers, with the distnce function d(x,y) = x y ), nd tking S to be the semiclosed intervl [0,1), the set S is both open nd closed in (S,d) (since the whole spce is lwys both n open nd closed set), but S is neither open nor closed in (E,d). 20 Techniclly, this is not quite correct; the distnce function on S should be d defined s d (p,q) = { d(p,q) if p S nd q S, undefined otherwise (i.e., if p / S or q / S), since, clerly, the distnce function on S should not be defined outside S. d is clled the restriction of d to S (or, more precisely, to S S). It will, however, be simpler to sy, t the price of minor inccurcy, tht the distnce function on S is the sme d s the distnce function on E.

21 6. Subspces nd compct sets 15 Proof of the Lemm. Assume tht M is n open set in (S,d). Then, by the bove Lemm, M = {U S (p,r) : p S & r > 0 & U S (p,r) M}. As U S (p,r) = S U E (p,r) provided p S, 21 the set on the right-hnd side is equl to {S UE (p,r) : p S & r > 0 & U S (p,r) M} = S {U E (p,r) : p S & r > 0 & U S (p,r) M}; the second eqution here is true by distributive rule for the union of sets. 22 The set U def = {U E (p,r) : p S & r > 0 & U S (p,r) M} is n open set in the spce (E,d), since it is union of open blls in E. Since we hve M = S U, this completes the proof of the only if prt of the sttement on open sets in the lemm. As for the if prt, ssume M = S U, where U is open in (E,d), nd let p M be rbitrry. Then we lso hve p U. Thus, by the openness of U in (E,d), there is n r > 0 such tht U E (p,r) U. Then U S (p,r) = S U E (p,r) S U = M. Since p M ws rbitrry, this shows tht M is open in (S,d), completing proof of lso the if prt of the Lemm on open sets. The second sttement (i.e., the one bout closed sets) is esy to verify by going to complements. Tht is, if M is closed in (S,d), then its complement S \ M in S is open in (S,d). Thus by the sttement bout open sets, there is set U open in (E,d) such tht S\M = U S. Then, writing F for the complement in E of U (tht is, F = E\U), F is set closed in (E,d) such tht M = F S. Conversely, if M = F S where F is closed in (E,d), then U def = E \F is open in (E,d), so S \M = U S is open in (S,d) by the sttement bout open sets; thus M is closed in (S,d). The proof of the Lemm is complete. The lemm just proved hs importnt consequences for compct sets, defined next Definition. Let (E,d) be metric spce nd let S be subset of E. The set S is sid to be compct in (E,d) if the following holds: For ny collection U of open sets in (E,d), if S U, there is finite subcollection U U such tht S U. Further, the spce (E,d) is sid to be compct if E is compct set in (E,d). To sy tht U is collection of open sets in (E,d) mens tht U is collection every element of which is n open subset of (E,d). The inclusion S U is expressed by sying tht U is on open cover of S. One lso sys tht U covers S, or tht the elements of U cover S. If U U nd S U, then U is clled subcover of U (for S, if one wnts to void repeting the preposition of ). Thus, one cn sy shortly tht set is compct if nd only if ny of its open covers hs finite subcover. Exmple. The intervl (0, 1) is not compct. Indeed, (0,1) 21 The left-hnd side mkes no sense if p / S. 22 The distributive rule in question sys tht n=1 ( ) 1 n,1, X A = {X A : A} is vlid for ny set X nd ny collection A of sets. This eqution is esy to verify with the id of the definition of the union of sets.

22 16 Notes on Anlysis nd, for ny positive integer N, (0,1) N n=1 ( ) 1 n,1. Compctness is n intrinsic property; tht is, whether set is compct does not depend on which lrger spce it is considered. Tht is, the second Lemm bove hs the following Corollry. Let (E,d) be metric spce nd let S be subset of E. Then the set S is compct in (E,d) if nd only if (S,d) is compct spce. Proof. To estblish the if prt, ssume tht (S,d) is compct spce. Let U = {U ι : ι I} be collection of sets U ι, open in (E,d) such tht S U. 23 Then V def = {S U ι : ι I} is, by the second Lemm bove, collection of sets tht re open in (S,d). We hve S S U = V; tht is V is n open cover, in the sense of (S,d) of S. As the spce (S,d) is compct, V hs finite subcover, sy V ; i.e., S V. Since ech element V hs the form S U ι for some ι I, there is finite subset I of I such tht ( ) V = {S U ι : ι I } Then S V implies tht S {U ι : ι I }. In other words, ( ) U def = {U ι : ι I } is n finite open subcover, in the sense of (E,d), of U, showing tht S is compct in (E,d). To estblish the if prt, ssume tht S is compct set in (E,d), nd let V be n open cover, in the sense of (S,d), of S; then S V. Ech element V of V is open in the spce (S,d). Hence by the second Lemm bove, ech V in V hs form S U with some U open in (E,d). Thus, tking U def = {U : U is open in (E,d) & S U V}, U is n open cover, in the sense of (E,d), of S. As S is compct in (E,d), there is finite subset U of U tht is n open cover, in the sense of (E,d), of S. Then V def = {S U : U U } is finite subset of V tht is n open cover, in the sense of (S,d), of S, showing tht the spce (S,d) is compct. The proof is complete. Remrk. In the first prt of the proof, we indexed the elements of the set U = {U ι : ι I} with nother set, I. If one wnts to use function nottion insted of the customry, 23 The letter ι is the Greek lower cse iot, nd it is used insted of i, since often i is used to nme the elements of countble set, nd here it is not ssumed tht I is countble. We use the set I to index, or nme, the elements of the set U. The reson for this will be explined fter the proof.

23 6. Subspces nd compct sets 17 but perhps confusing, subscript nottion, one cn sy f(ι) insted of U ι. Then the set U equls the rnge of the function f (ssuming tht the domin of f is I, nd not lrger set). A set described this wy is sometimes clled n indexed set. Any set cn be described s n indexed set. For exmple, one cn tke I = U nd f s the identity function on U. It is not required tht the function f be one-to-one. In the bove proof, the purpose of the index set I ws to describe the set V given in ( ) in terms of finite index set I, since the sme index set I cn then be used to define the set U, s ws done in ( ). Without using the index set I, one might hve been tempted to sy tht let V be finite subcover of V, nd then write U def = {U : U U & V V [V = S U]}. This would not be correct though, since the set U cn be infinite. Tht is becuse, even though there re finitely mny choices for V V, for given V there my be infinitely mny choices for U U such tht V = S U. Tht is, for ech V V one needs to pick single U = g(v) in U such tht V = S U. Then one cn define U s U def = {g(v) : V V }. This definition of U ccomplishes the sme thing s the definition given in ( ), but the present one is somewht to messy to sy. The use of the index set I in the bove proof voids the need for this complicted circumlocution. In ny cse, difference between the two pproches (one using indexed sets, the other not) is only mtter of style, nd not of substnce. Next we show the following. Theorem. Let (E,d) be metric spce, nd let S E be compct. Then S is closed. Proof. Assume S is not closed. Then its complement, E \ S, is not open. Therefore, there is q E \S such tht for no r > 0 is the open bll U(q,r) subset of E \S. Tht is, we hve q / S for this q, nd for every r > 0 we hve S U(q,r). For ech p S, let r p = d(p,q)/2. 24 Then the open blls U(p,r p ) nd U(q,r p ) re disjoint, s is esily verified with the id of the tringle inequlity. Then S p SU(p,r p ), since ech point p of S is n element of the union on the right-hnd side (s p U(p,r p ). By compctness, the open cover on the right-hnd side hs finite subcover. Tht is, there is finite subset S of S such tht 25 Now, let S p S U(p,r p ). r = min{r p : p S }; the set on the right-hnd side is finite set of positive rels. Thus, the indicted minimum exists, nd is positive; i.e., r > 0. Since, s we remrked bove, the open blls U(p,r p ) nd 24 The dependence on q of r p is not indicted, since q is fixed throughout the rgument. 25 See the Remrk bove bout indexed sets. Here we chose S s the index set for the open cover {U(p,r p) : p S}.

24 18 Notes on Anlysis U(q,r p ) re disjoint for p S, the open blls U(p,r p ) nd U(q,r) re lso disjoint whenever p S, becuse r r p for such p. Thus, 26 U(q,r) p S U(p,r p ) =. This is, however, contrdiction, since the set on the right-hnd side of the intersection sign includes S, nd, s we indicted bove, the open bll on the left-hnd side of the intersection sign is not disjoint to S. This contrdiction completes the proof of the theorem. 7. Totlly bounded spces Our gol here is to give chrcteriztion of compct metric spces. To this end we introduce the following concept. Definition. The metric spce (E,d) is clled totlly bounded if for every ǫ > 0 it cn be covered by finitely mny closed blls of rdius ǫ. It will be convenient, but not essentil, tht we sid closed blls bove, since, clerly, if E cn be covered by finitely mny closed blls of rdius ǫ/2, then the open blls with the sme center nd rdius ǫ will lso cover E. One cn lso observe tht if (E,d) nd S E, then the subspce (S,d) is lso totlly bounded. This is lmost obvious, but there is the minor hitch: if we cover E by finitely mny closed blls of rdius ǫ, then we my not be ble use these sme closed blls to cover S, since the centers of these closed blls my not ll be in S, nd so they will not be closed blls in (S,d). Cover insted E by finitely mny closed blls of rdius ǫ/2, then tke only those blls B for which B S, tke p B S, nd replce B with closed bll B in S with center p nd rdius ǫ. As B S B, it is cler tht these new closed blls will cover S. A spce metric spce is clled bounded if it cn be included in single open bll. It is cler tht totlly bounded spce is lso bounded, since finitely mny closed blls cn lwys be covered by single open bll; the converse is, however, not true. For ny integer n 1 let R n be the set {(x 1,...,x n ) : x i R for 1 i n}; for x = (x 1,...,x n ) nd x = (x 1,...,x n ) write d(x,y) = ( n i=1 (x i y i ) 2) 1/2. Then one cn show tht (R n,d) is metric spce (see Rosenlicht [1, pp ]); this spce is clled the n-dimensionl Eucliden spce. It is esy to show tht ny bounded subspce of the n-dimensionl Eucliden spce is totlly bounded (see [1, p. 57]). Since this spce is lso complete (this is esy to conclude from the completeness of R; cf. [1, p. 53]), it follows from the Theorem below tht ny closed nd bounded subspce of the n-dimensionl Eucliden spce is compct. The spce l 2 is defined s the pir (E,d) where E = {(x 1,x 2,...) : x i R for i 1 nd i=1 x2 i < + } nd the distnce function d is defined s d(x,y) = ( i=1 (x i y i ) 2) 1/2 for x = (x1,x 2,...) nd y = (y 1,y 2,...) in E. It cn be shown tht l 2 is complete metric spce, nd tht no closed bll of positive rdius in l 2 is totlly bounded. We will need few more concepts. 26 For forml verifiction of the next equlity, one my use the distributive property mentioned bove: U(q,r) p S U(p,r p) = A common sense rgument my dmittedly be simpler. ( ) U(q,r) U(p,rp). p S

25 7. Totlly bounded spces 19 Definition. The metric spce is clled sequentilly compct if every sequence in it hs convergent subsequence. Definition. Let (E,d) be metric spce nd let p E nd S E. We sy tht p is cluster point of S if every open bll with center p in E contins infinitely mny elements of S. Observe tht we cn equivlently sy tht p is cluster point of S if every open bll with center p contins t lest one element of S different from p. Indeed, if the open bll with center p were to contin only finitely mny elements of S, ll these points, except p, wound be excluded from n open bll with center p nd n ppropritely smller rdius. The following result is simple. Lemm. Let (E,d) be metric spce, nd let S nd T be sets such tht T S E, S is compct, nd T is infinite. Then T hs cluster point in S. Proof. Assume, on the contrry, tht no point p S is cluster point of T. For ech p S we cn then tke n open bll B p with center p such tht B p T is finite. We hve S p SB p. Tht is, these open blls cover S. By the compctness of S, there re finitely mny mong these open blls tht lso cover S. In other words, there is finite set S S such tht We then hve ( T S T S p S B p p S B p. ) T p S (B p T). The sets B p T for p S re finite; the union of finitely mny of these is finite. This is, however, contrdiction, since T ws ssumed to be infinite, completing proof. Theorem. Let (E, d) be metric spce. The following re equivlent: (i) (E, d) is compct; (ii) (E, d) is sequentilly compct; (iii) (E, d) is complete nd totlly bounded. Proof. We will prove this result by showing the implictions (i) (ii), (ii) (iii), nd (iii) (i). Proof of (i) (ii). Assume (E,d) is compct. Let {p n } n=1 be sequence of points p n E. If the set {p n : n N} is finite, then there is p E such tht p = p n for infinitely mny n s. Then we cn tke subsequence {p nk } k=1 of {p n} n=1 such tht p nk = p for ll k N. This subsequence converges to p. If the set {p n : n N} is infinite, then it hs cluster point p by the bove Lemm. For ech k N the open bll U(p,1/k) contins infinitely elements of the set {p n : n N}. Let n k N be such tht p nk U(p,1/k) nd, in cse k > 1, we lso hve n k > n k 1. Then the subsequence {p nk } k=1 converges to p. Proof of (ii) (iii). Assume (E,d) is sequentilly compct. We will first show tht (E,d) is complete. Let {p n } n=1 be Cuchy sequence; then this sequence hs convergent subsequence. A Cuchy sequence tht hs convergent subsequence is itself convergent; so {p n } n=1 itself is convergent, showing tht (E,d) is complete. Next we show tht (E,d) is totlly bounded. Assume, on the contrry tht it is not totlly bounded; let ǫ > 0 be such tht (E,d) cnnot be covered by finitely mny closed