Further Topics in Analysis

Transcription

1 Further Topics in Anlysis Lecture Notes 2012/13 Lecturer: Prof. Jens Mrklof Notes by Dr. Vitly Moroz School of Mthemtics University of Bristol BS8 1TW Bristol, UK c University of Bristol 2013

2 Contents I Reltions nd Functions 1 1 Reltions 1 2 Functions 4 II Elements of Set Theory 6 3 Finite nd infinite sets. Countble sets. 6 4 Uncountble sets. Continuum Crdinlity 16 6 Power set. Hierrchy of crdinlities 18 III Convergence nd Continuity 20 7 Subsequences. Accumultion Points 20 8 The Bolzno-Weierstrss Theorem 24 9 Limit Superior nd Limit Inferior Cuchy sequences Uniformly Continuous Functions Pointwise nd Uniform convergence 35 IV The Riemnn Integrl Definition of the integrl Criterion of integrbility Clsses of integrble functions Inequlities nd the Men Vlue Property of the integrl Further properties of the integrl Integrtion s the inverse to differentition 49

3 Nottions N set of nturl numbers {0, 1, 2, 3,... } N + set of positive nturl numbers {1, 2, 3,... } Z set of integer numbers {..., 3, 2, 1, 0, 1, 2, 3,... } Q set of rtionl numbers {r = p q p Z, q N+, hcf(p, q) = 1} R set of rel numbers R + set of non-negtive rel numbers {x R x 0} In the following Theorem A etc. refers to Theorem in Lecture Notes in Anlysis 1 (MATH11006) by Vitli Liskevich. This mteril is copyright of the University of Bristol unless explicitly stted otherwise. It is provided exclusively for eductionl purposes t the University of Bristol nd is to be downloded or copied for your privte study only. c University of Bristol 2013

4 1 RELATIONS Prt I Reltions nd Functions Recommended Texts: 1. I. Stewrt, D. Tll The Foundtions of Mthemtics, Oxford University Press, S. Krntz Rel Anlysis nd Foundtions, Second Edition. Chpmn nd Hll/CRC Press, Reltions Definition 1.1 Let X, Y be sets. A set R X Y is clled reltion from X to Y. If (x, y) R, we sy tht x is in reltion R to y. We lso write xry. Exmple Let A = {1, 2, 3}, B = {3, 4, 5}. The set R = {(1, 3), (1, 5), (3, 3)} is reltion from A to B since R A B. 2. G = {(x, y) R R x > y} is reltion from R to R. Definition 1.3 Let R be reltion from X to Y. The domin of R is the set The rnge of R is the set D(R) = {x X y Y [(x, y) R]}. Rn(R) = {y Y x X [(x, y) R]}. The inverse of R is the reltion R 1 from Y to X defined s follows R 1 = {(y, x) Y X (x, y) R}. Definition 1.4 Let R be reltion from X to Y, S be reltion from Y to Z. The composition of S nd R is reltion from X to Z defined s follows S R = {(x, z) X Z y Y [(x, y) R] [(y, z) S]}. Theorem 1.5 Let R be reltion from X to Y, S be reltion from Y to Z, T be reltion from Z to V. Then 1. (R 1 ) 1 = R. 1

5 1 RELATIONS 2. D(R 1 ) = Rn(R). 3. Rn(R 1 ) = D(R). 4. T (S R) = (T S) R. 5. (S R) 1 = R 1 S 1. Proof. Exercise, or see [Stewrt nd Tll (1977)]. Next we tke look t some prticulr types of reltions. Let us consider reltions from X to X, i.e. subsets of X X. In this cse, we tlk bout reltion on X. A simple exmple of such reltion is the identity reltion on X which is defined by i X = {(x, y) X X x = y}. Definition A reltion R on X is sid to be reflexive if ( x X) (x, x) R. 2. R is sid to be symmetric if 3. R is sid to be trnsitive if Equivlence reltions ( x X)( y X) {[(x, y) R] [(y, x) R]}. ( x X)( y X)( z X){[((x, y) R) ((y, z) R)] [(x, z) R]}. A prticulrly importnt clss of reltions re equivlence reltions. Definition 1.7 A reltion R on X is sid to be n equivlence reltion if it is reflexive, symmetric nd trnsitive. Exmple Let X be the set of students. Define reltion on X by R = {(x, y) X X : x nd y re friends}. Then R is reflexive (ssuming tht ech student is friend to him or herself). It is lso symmetric. But it is not trnsitive. 2. Let X = R, be some positive number. Define R X X s R = {(x, y) x y }. R is reflexive, symmetric, but not trnsitive. 3. Let X = Z nd m Z. Define Then R is n equivlence reltion on Z. R := {(x, y) Z Z : ( k Z)[x y = km] }. 2

6 1 RELATIONS Definition 1.9 Let R be n equivlence reltion on X. Let x X. The equivlence clss of x with respect to R is the set [x] R = {y X (y, x) R}. Let us tke look t severl properties of equivlence clsses. Proposition 1.10 Let R be n equivlence reltion on X. Then 1. ( x X) x [x] R. 2. ( x X)( y X) [(y [x] R ) ([y] R = [x] R )]. Proof. 1. Since R is reflexive, (x, x) R, hence x [x] R. 2. First, let y [x] R, so tht (y, x) R. Suppose tht z [y] R. Then (z, y) R. By trnsitivity, (z, x) R; hence, z [x] R. This shows tht [y] R [x] R. The reverse inclusion [x] R [y] R cn be shown similrly. Therefore, [x] R = [y] R. The impliction ([y] R = [x] R ) (y [x] R ) is obvious. From the bove proposition it follows tht equivlence clsses [x] R nd [y] R either coincide or re disjoint, in other words, [x] R = [y] R or [x] R [y] R =. Also, every element of the set X belongs to n equivlence clss. This my be expressed [x] R = X. x X Exmple 1.11 Let R be the equivlence clss in Exmple 3 bove. Then, for x Z, [x] = { x + km : k Z}. This is clled the equivlence clss of integers congruent to x modulo m. Often we write y x to men y [x]. Remrk 1.12 For more on this topic, see [Stewrt nd Tll (1977)] 3

7 2 FUNCTIONS 2 Functions The notion of function is of fundmentl importnce in ll brnches of mthemtics. You met functions in your previous study of mthemtics, but without precise definition. Here we give definition nd mke connections to exmples you cme cross before. Definition 2.1 Let X nd Y be sets. Let F be reltion from X to Y. Then F is clled function from X to Y, if the following properties re stisfied: (i) ( x X)( y Y ) [(x, y) F ]. (ii) ( x X)( y Y )( z Y ){([(x, y) F ] [(x, z) F ]) (y = z)}. It is customry to write y = F (x), the imge of x under F. X is clled the domin of F nd Y is clled codomin. Let us consider severl exmples. Exmple 2.2 (i) Let X = {1, 2, 3}, Y = {4, 5, 6}. Define F X Y s F = {(1, 4), (2, 5), (3, 5)}. Then F is function. In contrst to this, define G X Y vi G = {(1, 4), (1, 5), (2, 6), (3, 6)}. Then G is not function. (ii) Let X = R nd Y = R. Define F X Y to be F = {(x, y) R R y = x 2 }. Then F is function from R to R. In contrst to this, define G X Y to be Then G is not function. G = {(x, y) R R x 2 + y 2 = 1}. We often use the nottion for function F from X to Y. F : X Y Note tht in order to define function F from X to Y we hve to define X, Y nd subset of X Y stisfying (i) nd (ii) of the definition. Note: one cn ssign vlue y to ech vlue x X by mens of rule or formul. But this notion is rther vgue. The present definition is more precise. Theorem 2.3 Let X, Y be sets nd F, G functions from X to Y. Then [( x X)(F (x) = G(x))] (F = G). 4

8 2 FUNCTIONS Proof. 1. ( ). Assume the LHS bove. Suppose tht F G. Let (x, y) F. Then y = F (x). So y = G(x); or, (x, y) G. The proof of the inclusion G F is similr. The impliction ( ) is trivil. Thus, two functions re equl if they shre the sme domin nd codomin nd elements in the domin hve equl imges. Exmple 2.4 Let f : R R, g : R R, h : R R +. non-negtive rel numbers. Suppose tht for ll x R, Here, R + denotes the set of f(x) = (x + 1) 2, g(x) = x 2 + 2x + 1, h(x) = (x + 1) 2. Then f nd g re equl, but f nd h re not since they hve different codomins. The definition of composition of reltions cn be lso pplied to functions. If f : X Y nd g : Y Z then g f = {(x, z) X Z ( y Y )[[(x, y) f] [(y, z) g]]}. Theorem 2.5 Let f : X Y nd g : Y Z. Then g f : X Z nd ( x X) [(g f)(x) = g(f(x))]. Proof. We know tht g f is reltion. So we must prove tht for every x X there exists unique element z Z such tht (x, z) g f. Existence: Let x X be rbitrry. As f is function, y Y such tht (x, y) f. As g is function, z Z such tht (y, z) g. So nd (x, z) g f. In short, ( y Y )[(x, y) F (y, z) g] ( x X)( z Z)[(x, z) g f]. Uniqueness: Let x X. Suppose tht (x, z 1 ) g f nd (x, z 2 ) g f. As (x, z 1 ) g f, ( y 1 Y )[(x, y 1 ) f (y 1, z 1 ) g]. As (x, z 2 ) g f, ( y 2 Y )[(x, y 2 ) f (y 2, z 2 ) g]. As f is function, y 1 = y 2. As g is function, z 1 = z 2. Exmple 2.6 Let f : R R, g : R R, Find (f g)(x) nd (g f)(x). f(x) = 1 x 2, g(x) = 2x Solution. (f g)(x) = f(g(x)) = 1 [g(x)] = 1 (2x 1) 2 + 2, (g f)(x) = g(f(x)) = 2f(x) 1 = 2 x Wrning: As is cler from the bove, f g g f in generl. 5

9 3 FINITE AND INFINITE SETS. COUNTABLE SETS. Prt II Elements of Set Theory Recommended Texts: 1. I. Stewrt, D. Tll The Foundtions of Mthemtics, Oxford University Press, S. Krntz Rel Anlysis nd Foundtions, Second Edition. Chpmn nd Hll/CRC Press, In this prt of the course we study the notion of crdinlity. The im is to formulte notion of size of n infinite set, or to count its elements. It is possible to sk questions such s how mny rtionl numbers re there? or is the set of rel numbers bigger then the set of nturl numbers? To do so, however, it is helpful to be ble to compre the number of elements in given set with those in nother. We sy tht sets X nd Y hve the sme crdinlity (intuitively, number of elements) if there is bijection f : X Y. This ide ws first conceived by the Germn mthemticin Georg Cntor ( ) t the end of the 19th century. His Nive Set Theory gives rise to vrious prdoxes (e.g. Russell s Prdox, see p. 19), which re resolved in the subsequent Axiomtic Set Theory. For our purposes, we tke the nive notion of set s collection of elements. 3 Finite nd infinite sets. Countble sets. Let X, Y be sets nd f : X Y function from X to Y. Recll tht the function f is n injection if ( x 1 X)( x 2 X)[(f(x 1 ) = f(x 2 )) (x 1 = x 2 )]. This mens tht f is one-to-one correspondence between X nd the rnge Rn(f). The function f is surjection if ( y Y )( x X)[f(x) = y]. This mens tht Rn(f) = Y. Finlly, f is bijection if f is n injection nd surjection. In this cse, f estblishes one-to-one correspondence between X nd Y. How do we decide how mny elements there re in set? One nswer is to count them. But wht does this men? This cn be described s follows. We tke n element of the set, nd think one. We then look t different element nd think two, nd so on. We need to be creful to mke sure tht ll elements hve been counted, nd none hs been counted twice. If this process termintes, we sy tht the set is finite. Definition 3.1 A set X is sid to be finite if either X = or for some n N + there exists bijection f : {1, 2,..., n} X. 6

10 3 FINITE AND INFINITE SETS. COUNTABLE SETS. In the former cse, we sy tht X hs zero elements; in the ltter, tht X hs n elements. X is sid to be infinite if it is not finite. A nturl wy to extend this process of counting to infinite sets is s follows. Definition 3.2 A set X is sid to be countble if there exists bijection f : N + X. Remrk 3.3 By Theorem A1.7.6, function f : N + X is bijection if nd only if the inverse function f 1 : X N + exists. In this cse, the inverse f 1 is bijection too. In prticulr, X is countble if nd only if there exists bijection f : X N +. Exmple 3.4 N = {0, 1, 2,... } is countble. Proof. Define f : N + N by f(n) = n 1. Then f is bijection n n 1... In this exmple, we observe the first remrkble property of infinite sets. N + is subset of N, so intuitively it should hve fewer elements. Yet N + nd N re both countble. The next exmple, constructed by Glileo in 1638, ws considered s prdox for more thn two centuries. Exmple 3.5 The set of perfect squres S := {1 2, 2 2, 3 2,..., n 2,... } is countble. Proof. Define f : N + S by f(n) = n 2. It is cler tht f is bijection n n 2... The reson Glileo s exmple might be considered prdox is tht the set S of perfect squres seems smller then N +. But S is still countble. Exmple 3.6 The set of integers Z is countble. Proof. Define f : N + Z by f(n) = The following digrm illustrtes f: { n 2 1 n 2, if n is even,, if n is odd n 2n n n... 7

11 3 FINITE AND INFINITE SETS. COUNTABLE SETS. The inverse of f is the function f 1 : Z N + given by { f 1 2m, if m 1, (m) = 1 2m, if m 0. By Theorem A1.7.6 we conclude tht f is bijection. Remrk 3.7 Notice tht lthough f in the lst exmple is bijection, it does not preserve order, in the sense tht m < n does not imply f(m) < f(n). Exmple 3.8 The set N + N + is countble. Proof. We write elements of N + N + in rectngulr rry (figure on the left). We then red them off long the cross digonls (figure on the right): first (1, 1), next (1, 2), (2, 1), then (1, 3), (2, 2), (3, 1) nd so on. (n, m) (1,1) (1,2) (1,3) (1,4)... 2 (2,1) (2,2) (2,3)... 3 (3,1) (3,2)... 4 (4,1) This correspondence cn lso be written f(n,m) (1, 1) (1, 2) (2, 1) (1, 3) (2, 2) (3, 1) In this wy we estblish bijection between N + N + nd N +. This bijection f : N + N + N + cn be represented nlyticlly by the formul f(n, m) = Therefore N + N + is countble. (n + m 2)(n + m 1) 2 + n. Remrk 3.9 Let X be countble set nd f : N + X bijection. Define x 1 := f(1), x 2 := f(2), x 3 := f(3),..., x n := f(n), n... x 1 x 2 x 3 x 4 x 5... x n... Thus ll elements of X cn be listed in sequence: X = {x 1, x 2, x 3, x 4, x 5,..., x n,... }. We now estblish severl importnt properties of countble sets. Lemm 3.10 If set X is countble then ny subset A X is finite or countble. 8

12 3 FINITE AND INFINITE SETS. COUNTABLE SETS. Proof. Assume A is infinite. We need to show tht A is countble. Since X is countble there exists bijection f : N + X. Define g : N + A inductively s follows. Set g(1) := f(min{k N + : f(k) A}); g(2) := f(min{k N + : f(k) A \ {g(1)}}); g(n + 1) := f(min{k N + : f(k) A \ {g(1),..., g(n)}}). It is cler tht g is bijection. Remrk 3.11 We sy tht set X is t most countble if X is finite or countble. Remrk 3.12 Lemm 3.10 is useful for proving tht given set is countble. For exmple, let P be the set of ll prime numbers. Then P N +. And it is known tht the set of primes is infinite. We my then conclude tht P is countble. Lemm 3.13 Any infinite set hs countble subset A X. Proof. First, X s X is infinite. So we my choose x 1 X. Define g 1 : {1} X; 1 x 1. Then g 1 is not onto, for otherwise X would be finite. So we my choose x 2 X \ {x 1 }. Define g 2 : {1, 2} X; j x j. Then g 2 is not onto, for otherwise X would be finite. So we my choose x 3 X \ {x 1, x 2 }. Continuing in this wy, we obtin sequence (x n ) n N + of distinct points in X. The mpping g : N + A := {x n X n N + }; n x n is then bijection from N + to A X. Thus A is countble subset of X. Definition 3.14 A set Y is sid to be proper subset of set X if Y X nd Y X. Proposition 3.15 If set X is infinite then there exists proper subset Y X nd bijection f : X Y. Proof. Let A := {x 1, x 2,..., x n,... } be countble subset of X, constructed s in Lemm Let B := {x 1 } A. Set Y := X \ B. Define f : X Y by f(x n ) = x n+1 for x n A nd Then f is bijection. f(x) = x for x X \ A. 9

13 3 FINITE AND INFINITE SETS. COUNTABLE SETS. Remrk 3.16 We proved tht if X is infinite nd B = {x 1 } X then Y := X \ B is infinite nd there is bijection f : X Y. Thus removing one element from n infinite set does not chnge its size. In similr wy, it is possible to select countble subset B X, set Y := X \ B nd yet construct bijection f : X Y. To do so, select countble subset A := {x 1, x 2,..., x n,... } of X, then let B := {x 2k 1 k N + } A, set Y := X \ B nd define f : X Y by f(x n ) = x 2n for x n A nd Then f is bijection. f(x) = x for x X \ A. Lemm 3.17 Let X nd Y be countble sets. Then the Crtesin product X Y is countble. Proof. Let f : X N + nd g : Y N + be bijections. Define the function h : X Y N + N + vi h(x, y) := (f(x), g(y)). To see tht h is n injection suppose tht h(x 1, y 1 ) = h(x 2, y 2 ). This mens tht (f(x 1 ), g(y 1 )) = (f(x 2 ), g(y 2 )). So f(x 1 ) = f(x 2 ) nd g(y 1 ) = g(y 2 ). Since f nd g re injections it follows tht x 1 = x 2 nd y 1 = y 2, tht is (x 1, y 1 ) = (x 2, y 2 ). To check tht h is surjection, suppose (n, m) N + N +. Then since f nd g re both surjections, we cn choose x X nd y Y such tht f(x) = n nd g(y) = m. Therefore, h(x, y) = (n, m), s required. Hence h : X Y N + N + is bijection. By Exmple 3.8, there is bijection ϕ : N + N + N +. It is esy to see tht the composition ϕ h : X Y N + is lso bijection. Therefore X Y is countble. Lemm 3.18 Let {X k } k N + be countble collection of pirwise disjoint countble sets. Then the union k=1 X k is countble. Proof. Since every set X k is countble we cn write the elements of X k in list: By definition of the union X 1 = {x 11, x 12, x 13,..., x 1n,... }, X 2 = {x 21, x 22, x 23,..., x 2n,... }, X 3 = {x 31, x 32, x 33,..., x 3n,... }, X k = {x k1, x k2, x k3,..., x kn,... }, k=1 X k = {x kn k N +, n N + }. 10

14 3 FINITE AND INFINITE SETS. COUNTABLE SETS. Define function h : k=1 X k N + N + by h(x kn ) := (k, n). Clerly h is bijection. Since N + N + is countble we conclude tht k=1 X k is countble too. Exmple 3.19 The set of rtionl numbers Q is countble. Proof. For n N +, let X k := { } i k i N+ be the set of ll positive frctions with denomintor k. It is cler tht for ech k N + the set X k is countble. Let X := k=1 X k. By Lemm 3.18 we conclude tht the set X is countble. Let Q + be the set of positive rtionls. Then Q + X nd Q + is infinite. By Lemm 3.10, Q + is countble. In the sme wy we cn prove tht the set Q of negtive rtionls is countble. Finlly, Q = Q {0} Q + is countble by Lemm

15 4 UNCOUNTABLE SETS. CONTINUUM. 4 Uncountble sets. Continuum. We hve shown tht the sets N + N +, Q, nd others, lthough seemingly lrger thn N +, re nevertheless countble. So mybe ll infinite sets re countble? The nswer is no. Definition 4.1 A set X is sid to be uncountble if X is infinite but not countble. Roughly speking, n uncountble set hs so mny elements tht they cnnot be listed, nor rrnged in sequence. Theorem 4.2 The open intervl (0, 1) is uncountble. Proof. We prove this by contrdiction. Assume tht the intervl (0, 1) is countble nd f : N + (0, 1) is bijection. Then we cn rrnge ll elements of (0, 1) in sequence {α 1, α 2, α 3, α 4, α 5,..., α k,... }. Represent ech α k (0, 1) s deciml expnsion α k = 0. k1 k2 k3 k4 k5..., where we gree tht if the deciml expnsion termintes 1 we will write it ending with sequence of zeros, not sequence of nines. Then elements of the intervl (0, 1) my be listed: α 1 = k... α 2 = k... α 3 = k... α 4 = k... α 5 = k = α k = 0. k1 k2 k3 k4 k5... kk = Let β (0, 1) be the rel number with the deciml expnsion where β := 0. b 1 b 2 b 3 b 4 b 5... b k... b k := { 1, if kk 1, 2, if kk = 1. Then β is different from α k for ny k N +. The reson is tht, for ech k N + the deciml expnsion of β differs from tht of α k in its kth plce. We thus rrive t contrdiction. Hence (0, 1) is uncountble. Remrk 4.3 The method used in the proof of Theorem 4.2 is often refered to s (Cntor s) digonliztion rgument. It is powerful technique tht plys role in mny other proofs. 1 See Proposition 4.11 below nd the discussion fter it. 12

16 4 UNCOUNTABLE SETS. CONTINUUM. Definition 4.4 A set X is sid to be continuum if there exists bijection f : (0, 1) X. Exmple 4.5 Let, b R nd < b. Then the open intervl (, b) is continuum. Proof. The function f : (0, 1) (, b) defined by the formul f(x) := (b )x + is bijection with inverse f 1 : (, b) (0, 1) given by f 1 (y) = y b. Exmple 4.6 The intervl [0, 1) is continuum (s is (0, 1] nd [0, 1]). Proof. (Compre this proof with the proof of Proposition 3.15!) The function f : [0, 1) (0, 1) defined by the formul { 1 1 f(x) := n+1, if x = 1 1 n, n N+, x, if x 1 1 n, n N+. is bijection with inverse { 1 f 1 1 (y) := n, if y = 1 1 n+1, n N+, y, if y 1 1 n+1, n N+. The intervls (0, 1] nd [0, 1] cn be considered in similr wy. Exmple 4.7 The rel line R is continuum. Proof. The function g : ( 1, 1) R defined by g(x) := x 1 x is bijection with inverse g 1 : R ( 1, 1) given by g 1 (y) = y 1 + y. By Exmple 4.5 there is bijection f : (0, 1) ( 1, 1). Then the composition g f : (0, 1) R is bijection. Exmple 4.8 [0, 1) [0, 1) is continuum. Ide of proof. Any x [0, 1) hs deciml expnsion x = We ssume tht the deciml expnsion does not contin string of consecutive 9 s with infinite length. The expnsion is then unique. The expnsion my be rewritten in blocks x = 0. α 1 α 2 α 3 13

17 4 UNCOUNTABLE SETS. CONTINUUM. where ech block α j hs the form α j = where stnds for one of the digits 0, 1, 2, 3, 4, 5, 6, 7, 8; or α j = ( finite block of 9 s followed by non-nine digit). Define f : [0, 1) [0, 1) [0, 1) vi f(x, y) := 0. α 1 β 1 α 2 β 2 α 3 β Here, we write x = = 0. α 1 α 2 α 3 ; y = 0. b 1 b 2 b 3 = 0. β 1 β 2 β 3 ; first in deciml form, then in block form, s described bove. Then f is bijection. To conclude, compose f with the bijection in Exmple 4.6. The intervl (0, 1) is uncountble nother proof. 2 Our proof of Theorem 4.2 relies on the deciml expnsion of the rel numbers. The originl Cntors s proof of Theorem 4.2, published in 1874, ws different. It ws bsed on the following nested intervls theorem, which is similr to Theorem A from Anlysis. Theorem 4.9 Let ([b k, c k ]) k N + be sequence of closed nonempty bounded intervls such tht ( k N + )([b k+1, c k+1 ] [b k, c k ]). Then k N +[b k, c k ]. Exercise 4.10 Prove Theorem 4.9. Hint. Modify proof of Theorem A from Anlysis. Originl Cntor s proof of Theorem 4.2. Let E (0, 1) be countble subset of (0, 1), so E cn be represented s E = {x 1, x 2, x 3,... }. Observe, tht given point x (0, 1) there exists nonempty intervl [b, c] (0, 1) such tht x [b, c]. Hence, we cn choose nonempty intervl [b 1, c 1 ] (0, 1) such tht x 1 [b 1, c 1 ]. Now we proceed inductively. Let n N +. Hving chosen nonempty intervl ([b k, c k ]) n k=1 such tht x k [b k, c k ] for k = 1,..., n, nd such tht [b k+1, c k+1 ] (b k, c k ) for k = 1,..., n 1, we cn choose nonempty intervl [b n+1, c n+1 ] such tht x n+1 [b n+1, c n+1 ] nd such tht [b n+1, c n+1 ] (b n, c n ). In such wy we inductively defined nested sequence of closed nonempty bounded intervls ([b k, c k ]) k N + so tht ( k N + )[([b k+1, c k+1 ] [b k, c k ]) (x k [b k, c k ])]. According to Theorem 4.9, there exists point x k N +[b k, c k ]. But since x n n N +[b k, c k ] for every n N +, we conclude tht x E. Therefore E (0, 1), which mens tht the intervl (0, 1) is uncountble. 2 The content of this prgrph will not be included in the exmintion pper. 14

18 4 UNCOUNTABLE SETS. CONTINUUM. Deciml expnsions. 3 Let ( k ) k N + be sequence of numbers chosen from the set {0, 1, 2,..., 9}. Then the series k 10 k converges by comprison with geometric progression to sum α [0, 1]. We write k=1 α = , nd sy tht the right hnd side is deciml expnsion of the rel number α. For exmple, 1 3 = , 1 7 = } {{} }{{} }{{}..., π 4 = Proposition 4.11 Every rel number α [0, 1] hs unique deciml expnsion unless it is rtionl number of the form α = m 10 for some m N + nd n N + in which cse α hs precisely two deciml n expnsions. Sketch of the proof. Let α [0, 1]. Suppose tht { 1, 2,..., k } hve been chosen from {0, 1, 2,..., 9} in such wy tht ( 1 ɛ k := α ) k 10 k stisfies 0 ɛ k < 1 10 k. Let k+1 be chosen from {0, 1, 2,..., 9} so tht k+1 is mximized subject to the constrint ɛ k+1 0. If k+1 < 9 then ɛ k+1 < 1 10 k+1, becuse otherwise we could replce k+1 by k If k+1 = 9 then ɛ k+1 = ɛ k 9 10 k+1 < 1 10 k 9 10 k+1 = 1 10 k+1. By the sndwich rule (A3.1.6) we conclude tht lim k ɛ k = 0. Therefore α = k=1 k 10 k, which gives the required deciml expnsion of α. We do not consider the issue of uniqueness of the deciml expnsion except for the observtion tht, where two distinct expnsions of α exist then α = m 10 for some m N + nd n N + nd one of n the expnsions consists of zeros from some point nd the other consists of nines from some point on. In this cse we sy tht deciml expnsion of α termintes. For exmple, 1 2 = 5 = = If we gree tht when deciml expnsion termintes we will write it ending with sequence of zeros, not sequence of nines, then every rel number α [0, 1] hs unique deciml expnsion. 3 The content of this prgrph will not be included in the exmintion pper. 15

19 5 CARDINALITY 5 Crdinlity In this section, we discuss method for compring the size of rbitrry sets. Definition 5.1 Let X, Y be sets. We sy tht X hs the sme crdinlity s Y if there exists bijection f : X Y. In this cse, we write X Y or crd(x) = crd(y ). Remrk 5.2 Intuitively, X Y or crd(x) = crd(y ) mens tht the sets X nd Y hve the sme size or the sme number of elements. Exmple 5.3 We lredy know tht N + Z Q N + N + nd (0, 1) [0, 1] R (0, 1) (0, 1). Proposition 5.4 For ny sets X, Y, Z: () X X (reflexivity); (b) if X Y then Y X (symmetry); (c) if X Y nd Y Z then X Z (trnsitivity). Remrk 5.5 Properties (), (b), (c) entil tht is n equivlence reltion on the collection of ll sets. We will see however tht the collection of ll sets cnnot be set. This mens tht is not reltion in the sense defined before, s its domin is not set. Even so, we cn still consider the equivlence clss [X] of given set X: the collection of sets which re equivlent to X. For exmple, the equivlence clss [N + ] of the set of nturl numbers N + is the collection of ll countble sets. Proof. () The identity function i X : X X is bijection. (b) Suppose X Y. Let f : X Y be bijection. Hence by Theorem A1.7.6, the inverse function f 1 : Y X exists. But now note tht (f 1 ) 1 = f : X Y is n inverse to f 1. So by Theorem A1.7.6 gin, f 1 : Y X is bijection. Therefore Y X. (c) Suppose X Y nd Y Z. Let f : X Y nd g : Y Z be bijections. Hence by Theorem A there exist inverse functions f 1 : Y X nd g 1 : Z Y. Consider the composition g f : X Z. It is esy to check tht (g f) 1 = f 1 g 1 : Z X is the inverse function of g f : X Z. Hence g f : X Z is bijection. Therefore X Z. Exercise 5.6 Suppose A B nd C D. Prove tht: i) (A C) (B D); ii) if A nd C re disjoint nd B nd D re disjoint, then (A C) (B D). Proof. See Exercises 3, Q 5. Definition 5.7 Let X, Y be sets. We sy tht the crdinlity of Y is greter thn or equl to the crdinlity of X (Y domintes X), if there exists n injection f : X Y. In this cse, we write X Y or crd(x) crd(y ). We sy tht the crdinlity of Y is strictly greter then the crdinlity of X, if there exists n injection f : X Y, 16

20 5 CARDINALITY but there is no injection (nd hence no bijection) from Y to X. In this cse, we write X Y or crd(x) < crd(y ). Remrk 5.8 () Intuitively, X Y (or crd(x) crd(y )) mens tht the set X hs fewer elements thn the set Y. (b) If X Y then X Y nd Y X. To see this, consider bijection f : X Y. Then f : X Y is n injection nd f 1 : Y X is n injection too. (c) If A X then the identity mp i A : A X is n injection. Hence if A X then A X. (d) If X Y then X Y nd X Y. Exmple 5.9 {1} {1, 2} {3, 4} {1, 2,..., n} N + Q R (0, 1). However {1, 2} {3, 4}, N + Q nd R (0, 1). Exercise 5.10 Prove tht for ny sets X, Y, Z: () X X (reflexivity); (b) if X Y nd Y Z then X Z (trnsitivity). Proof. Similr to the proof of Proposition 5.4. Exercise 5.11 Suppose A B nd C D. Prove tht: i) (A C) (B D); ii) If A nd C re disjoint nd B nd D re disjoint, then (A C) (B D). Proof. Similr to Exercises 3, Q5. Remrk 5.12 Properties () nd (b) of Exercise 5.10 men tht is reflexive nd trnsitive reltion on the collection of ll sets (see Remrk 5.5). However is not n order reltion, becuse it is not ntisymmetric. This mens tht X Y nd Y X does not imply tht X = Y. For exmple, N + Q, so N + Q nd Q N +, but N + Q. Theorem 5.13 (Cntor-Schröder-Bernstein) Let X nd Y be sets. If X Y nd Y X then X Y. Proof. See Stewrt nd Tll, Theorem 6 on p.238. The Cntor Schröder Bernstein is extremely useful when it is difficult to find bijection between two sets. Insted, it is enough to construct two injections. Exmple 5.14 [0, 1] (0, 1). A simple proof vi the Cntor Schröder Bernstein Theorem. Define f : (0, 1) [0, 1] nd g : [0, 1] (0, 1) by f(x) := x, g(x) := 1 2 x It is cler tht both f nd g re injections. 17

21 6 POWER SET. HIERARCHY OF CARDINALITIES 6 Power set. Hierrchy of crdinlities Definition 6.1 Let X be set. The power set of X, P(X) or 2 X, is the set of ll subsets of X. Exmple 6.2 Let X = {0, 1}. Then 2 X = {, {0}, {1}, {0, 1} }. Notice tht 2 = { }, so 2 hs one element. Exercise 6.3 Let X = {1, 2,..., N}. Then the set 2 X hs 2 N elements. Theorem 6.4 Let X be set. Then X 2 X. Proof. It is cler tht the function f : X 2 X given by f(x) := {x} is n injection, so X 2 X. We now show X 2 X. Assume tht f : X 2 X is bijection. Put Z := {x X x f(x)}. Then Z 2 X. As f is surjection, f(z) = Z for some z X. Now, z Z z f(z) z Z, z Z z f(z) z Z, contrdiction. So there is no bijection f : X 2 X. Applied to the set N +, the theorem sys tht 2 N+ is uncountble. Actully it my be shown tht 2 N+ is continuum. Proposition N+ R. Proof. See Stewrt nd Tll, pp Exercise 6.6 Let F = {f : [0, 1] R} be the set of ll functions from the intervl [0, 1] to R. Prove tht [0, 1] F. Hint. Consider the set F 0 = {f A : A [0, 1]} of ll chrcteristic functions f A : [0, 1] {0, 1} of the form { 1 if x A, f A (x) = 0 if x A. Here, A 2 [0,1] is subset of [0, 1]. Estblish bijection between the set F 0 nd the set 2 [0,1] of ll subsets of the intervl [0, 1]. Hierrchy of crdinlities. Theorem 6.4 leds us to hierrchy of crdinlities. We begin with crd(n + ). Then crd(2 N+ ) = crd(r) is strictly bigger. Then follows crd(2 R ), nd so on: {1} {1, 2} {1, 2,..., N} N + 2 N+ R 2 R 2 2R

22 6 POWER SET. HIERARCHY OF CARDINALITIES Continuum Hypothesis. In 1878, Cntor sked whether there set X such tht N + X R. He conjectured tht the nswer ws no, but ws unble to prove it. This conjecture is known s the Continuum Hypothesis. Let X be set. If N + X nd X R then X R. The Continuum Hypothesis ws resolved by Gödel nd Cohen in the middle of the 20th century. They proved tht in the frmework of Axiomtic Set Theory it is impossible to prove the continuum hypothesis nd it is lso impossible to disprove it: tht is, the continuum hypothesis is independent of the xioms. This mens tht it is possible to construct xiomtic models of set theory in which this hypothesis is true, nd models in which it is flse. Its sttus therefore is similr to tht of the prllel postulte in Eucliden geometry. Prdoxes of Nive Set Theory. In 1899 Cntor discovered prdox which rises if one considers the set of ll sets. Wht is the crdinl number of the set of ll sets? Clerly it must be the gretest possible crdinl, yet the crdinl of the set of ll subsets of set lwys hs greter crdinl thn the set itself. In 1902, Russell discovered nother prdox now known s Russell s prdox. Suppose tht we ssume the existence of the set y consisting of ll sets x which do not belong to themselves, i.e. y = {x : x x}. If y y then y must stisfy the defining property of y, i.e. y y. On the other hnd, if y y, then y stisfies the defining property of y nd then y y. In either cse, contrdiction is obtined. A populr version of this prdox concerns certin brber who shves everyone in his town who does not shve himself. The question is: who shves the brber? Ech nswer leds to contrdiction. The conclusion is tht there is no such brber. Similrly, it is meningless to spek of the set ll sets which do not belong to themselves. It is lso meningless to spek of the set ll sets. There is no such set. These prdoxes cn be resolved in the frmework of modern Axiomtic Set Theory. 19

23 7 SUBSEQUENCES. ACCUMULATION POINTS Prt III Convergence nd Continuity Recommended Texts: 1. J. Howie Rel Anlysis, Springer Verlg, S. Krntz Rel Anlysis nd Foundtions, Second Edition. Chpmn nd Hll/CRC Press, G. Wnner Anlysis by its History, Springer Verlg, Subsequences. Accumultion Points Definition 7.1 Let ( n ) n N + be sequence. A subsequence of ( n ) n N + is sequence ( m(k) ) k N + = ( m(1), m(2), m(3),..., m(k),... ), where m : N + N + is strictly incresing function, so tht 1 m(1) < m(2) < m(3) < < m(k) <.... Remrk 7.2 () The sequence ( n ) n N + is subsequence of itself (tke m(k) = k). (b) The monotonicity of the function m implies tht ( k N + )(m(k) k). So, m(k) s k. (c) Any sequence hs infinitely mny subsequences. Tke, for exmple, m 1 (k) = k, m 2 (k) = 2k, m 3 (k) = 3k,... for k N +. Exmple 7.3 Consider the sequence ( n ) n N + defined by the formul n = ( 1) n. Then ( 2k+1 ) k N + = ( 1, 1, 1, 1... ), ( 2k ) k N + = (1, 1, 1, 1... ), ( 3k ) k N + = ( 1, 1, 1, 1... ) re subsequences of ( n ) n N +. Note tht 2k+1 1 nd 2k 1 s k, while ( 3k ) k N + diverges. Exmple 7.4 Let X = {x 1, x 2,..., x s } R be finite set of distinct rel numbers. Consider the sequence ( n ) n N + = (x 1, x 2,..., x s }{{}, x 1, x 2,..., x s }{{}, x 1, x 2,..., x s }{{},... ). 20

24 7 SUBSEQUENCES. ACCUMULATION POINTS This sequence my be described by n = x 1 x 2. x s if n 1 (mod s), if n 2 (mod s),. if n 0 (mod s). Then sn+1 x 1, sn+2 x 2,..., nd sn x s s n. In short, the sequence ( n ) n N + possesses subsequences tht tend to ech element in X. Exmple 7.5 Consider the sequence ( n ) n N + = (}{{} 1, }{{} 1, 2, 1, 2, 3, 1, 2, 3, 4, 1, 2, 3, 4, 5 }{{}}{{}}{{},... ). Then for ny m N + there is subsequence of ( n ) n N + which tends to m s n. (m, m, m, m,... ) Lemm 7.6 Suppose tht the sequence ( n ) n N + converges to. Then ny subsequence ( m(k) ) k N + lso converges to. Proof. Let ε > 0. As n s n, we cn find N N + such tht Now, for k > N, m(k) > m(n) N. Thus n < ε for n > N. ( k N + ) [ (k > N) ( m(k) < ε) ]. Tht is, m(k) s k. Lemm 7.7 Let ( n ) n N + be bounded sequence. Then ny subsequence is bounded. Proof. Let ( m(k) ) k N + be subsequence of ( n ) n N +. As ( n ) n N + is bounded, there exists 0 M < such tht n M for n N +. But then ( k N + ) [ m(k) M ], nd so ( m(k) ) k N + is bounded. Definition 7.8 Let ( n ) n N + be sequence in R. We sy tht R is n ccumultion point of ( n ) n N + if there exists subsequence ( m(k) ) k N + which converges to. Exmple 7.9 () Consider the sequence ( 1, 1, 1, 1, 1, 1, 1, 1,... ). 21

25 7 SUBSEQUENCES. ACCUMULATION POINTS from Exmple 7.3. Then the set of ccumultion points of ( n ) n N + is { 1, 1}. (b) The set of ccumultion points of the sequence ( n ) n N + = (x 1, x 2,..., x s }{{}, x 1, x 2,..., x s }{{}, x 1, x 2,..., x s }{{},... ) from Exmple 7.4 is the set (c) Consider the sequence X = {x 1, x 2,..., x s }. ( n ) n N + = (}{{} 1, }{{} 1, 2, 1, 2, 3, 1, 2, 3, 4, 1, 2, 3, 4, 5 }{{}}{{}}{{},... ) from Exmple 7.5. The set of ccumultion points of ( n ) n N + is the set of nturl numbers N + = {1, 2, 3, 4, 5,... }. Exercise 7.10 Let ( n ) n N + be sequence tht converges to. Then the set of ccumultion points of ( n ) n N + comprises one element, nmely {}. Hint. Apply Lemm 7.6. Exercise 7.11 Let ( n ) n N + be bounded sequence. Then the set of ccumultion points of ( n ) n N + is bounded. Hint. Apply Lemm 7.7. Chrcteriztion of the set of ll ccumultion points of sequence. Proposition 7.12 A number R is n ccumultion point of the sequence ( n ) n N + if nd only if (1) for ny ε > 0, the set {n N + : ε < n < + ε} is infinite; equivlently, (2) ( ε > 0)( N N + )( n N + )[(n N) ( n < ε)]. We lso formulte the negtion of Proposition Negtion of 7.12 A number R is not n ccumultion point of the sequence ( n ) n N + if nd only if there is n ε > 0 such tht the set {n N + : ε < n < + ε} is finite. Proof. ( ) Assume tht R is n ccumultion point of ( n ) n N +. We prove tht (2) holds. Since R is n ccumultion point, there is subsequence ( m(k) ) k N + such tht lim k m(k) =. Let ε > 0 nd N N + be given. By definition of the limit ( K > 0)( k N + )[(k K) ( m(k) < ε)]. Now choose k N + such tht k mx{k, N} nd put n := m(k) ( k). Then [(n N) ( n < ε)]. 22

26 7 SUBSEQUENCES. ACCUMULATION POINTS ( ) Now ssume tht (2) holds. Set ε = 1 nd N = 1. By (2), Put m(1) = n. Set ε = 1/2 nd N = m(1) + 1. By (2), Put m(2) = n. Set ε = 1/3 nd N = m(2) + 1. By (2), ( n N + )[(n N) ( n < 1). ( n N + )[(n N) ( n < 1/2). ( n N + )[(n N) ( n < 1/3). Put m(3) = n. Continuing in this wy we construct subsequence ( m(k) ) k N + ( k N + )( m(k) < 1/k). such tht In prticulr, lim k m(k) =, nd is n ccumultion point of ( n ) n N +. Exmple 7.13 The set of ccumultion points of the sequence ( n ) n N + defined by n = ( 1) n + 1 n comprises { 1, 1}. Proof. As 2k 1 1 nd 2k 1 s k, 1 nd 1 re ccumultion points of ( n ) n N +. Clim: Ech b R \ { 1, 1} is not n ccumultion point of ( n ) n N +. Tht is, ( ε > 0)( N N + )( n N + ) [(n N) ( n b > ε ]. Put ε = (1/2) min{ 1 b, 1 + b }. Then n b = ( 1) n + 1/n b ( 1) n b 1/n min{ 1 b, 1 + b } 1/n = 2ε 1/n using the fct tht x y x y for ny x, y R. Choose N N + such tht N > 1/ɛ by the AP. Then for ny n N, n b 2ε 1/n 2ε 1/N > 2ε ε = ε. So b is not n ccumultion point. 23

27 8 THE BOLZANO-WEIERSTRASS THEOREM 8 The Bolzno-Weierstrss Theorem Theorem 8.1 (Bolzno-Weierstrss theorem) Any bounded sequence hs convergent subsequence. Proof. Let ( n ) n N + be bounded sequence; sy, n M for ll n N +, where 0 M <. Step 1 Set [b 1, c 1 ] := [ M, M] c 1 b 1 = 2M Step 2 Step 3 Step k Divide [b 1, c 1 ] into two equl intervls. At lest one intervl contins infinitely mny elements of ( n ) n N +. [b 2, c 2 ] [b 1, c 1 ] Denote this by [b 2, c 2 ]. c 2 b 2 = M Divide [b 2, c 2 ] into two equl intervls. At lest one intervl contins infinitely mny elements of ( n ) n N +. [b 3, c 3 ] [b 2, c 2 ] Denote this by [b 3, c 3 ]. c 3 b 3 = 2 1 M Divide [b k 1, c k 1 ] into two equl intervls. At lest one intervl contins infinitely mny elements of ( n ) n N +. [b k, c k ] [b k 1, c k 1 ] Denote this by [b k, c k ]. c k b k = 2 2 k M In summry, we hve constructed sequence of intervls ([b k, c k ]) k N + such tht: (i) ( k N + )([b k+1, c k+1 ] [b k, c k ]); (ii) lim k (c k b k ) = lim k 2 2 k M = 0. By the nested intervls theorem (Theorem A 3.2.3), there exists unique point α k N +[b k, c k ]; moreover, (3) lim k b k = lim k c k = α. We now construct subsequence ( m(k) ) k N + of the sequence ( n ) n N +. k = 1: Set m(1) = 1. k = 2: k = 3: Note tht Choose j [b 2, c 2 ] with j > m(1) (possible, s [b 2, c 2 ] contins -ly mny elements of ( n ) n N +). Set m(2) = j. Choose j [b 3, c 3 ] with j > m(2) (possible, s [b 3, c 3 ] contins -ly mny elements of ( n ) n N +). Set m(3) = j ( k N + )[b k m(k) c k ]. 24

28 8 THE BOLZANO-WEIERSTRASS THEOREM Since lim k b k = lim k c k = α, by the sndwich rule, we conclude tht lim m(k) = α. k Thus ( m(k) ) k N + is convergent subsequence of ( n ) n N +. Remrk 8.2 The following exmples show tht the Bolzno-Weierstrss theorem does not hold for unbounded sequences: () the sequence n = n diverges to + nd hs no convergent subsequences; (b) the sequence n = ( n) n diverges nd hs no convergent subsequences; (c) the sequence n = n + ( 1) n n diverges but hs convergent subsequence 2k+1 = 0. Exmple 8.3 Consider the sequence ( n ) n N + with n = cos(n). This sequence is bounded, s cos(n) 1 for ll n N +. The Bolzno-Weierstrss theorem gurntees tht ( n ) n N + contins convergent subsequence. It is by no mens strightforwrd, however, to produce such subsequence. Exercise 8.4 Prove tht ny unbounded sequence hs subsequence tht diverges to + or to. Hint. Imitte the proof of the Bolzno-Weierstrss theorem by selecting sequence of unbounded intervls of the form (, n] or [n, + ) tht contin infinitely mny members of the sequence. 25

29 9 LIMIT SUPERIOR AND LIMIT INFERIOR 9 Limit Superior nd Limit Inferior Definition 9.1 Let ( n ) n N + be bounded sequence. Let A denote the set of ccumultion points of ( n ) n N +. The limit superior lim sup n n of ( n ) n N + is defined by lim sup n := sup A. n The limit inferior lim inf n n of ( n ) n N + is defined by lim inf n n := inf A. Remrk 9.2 The limit superior nd the limit inferior of bounded sequence lwys exist. Indeed, let A be the set of ccumultion points of bounded sequence ( n ) n N +. By the Bolzno-Weierstrss theorem the set A is nonempty, nd by Exercise 7.11 the set A is bounded. Therefore from the Completeness Axiom nd Theorem A we conclude tht both sup A nd inf A exist. Exmple 9.3 ) Let n = 1 n. This sequence converges to zero. Hence the set of ll ccumultion points is A = {0}. Therefore lim sup n = lim inf n = 0. n n b) Let n = ( 1) n. The set of ll ccumultion points of this sequence is A = { 1, 1}. Hence lim sup n = 1, n lim inf n n = 1. c) Let n = ( 1) n + 1 n. The set of ll ccumultion points of this sequence is A = { 1, 1} (see Exmple 7.13). Hence lim sup n = 1, n lim inf n n = 1. d) Let n = n. This sequence is unbounded. The limit superior nd the limit inferior is undefined for n unbounded sequence. Exercise 9.4 Let ( n ) n N + be convergent sequence nd lim n n =. Prove tht lim sup n = lim inf n =. n n Hint. Use Exercise Theorem 9.5 Let ( n ) n N + be bounded sequence nd set α := lim sup n nd β := lim inf n. n n Then α nd β re ccumultion points of ( n ) n N +. 26

30 9 LIMIT SUPERIOR AND LIMIT INFERIOR Proof. Let A denote the set of ll ccumultion points of ( n ) n N +; then α = sup A. We prove tht α A. Fix ε > 0 nd N N +. Since α = sup A, As A, using Proposition 7.12, By the tringle inequlity, ( A)( > α ε 2 ). ( k N + )[(k N) ( k < ε 2 )]. α k = (α ) + ( k ) (α ) + }{{} k < ε. }{{} <ε/2 <ε/2 Theorem 7.12 now implies tht α is n ccumultion point of ( n ) n N +. Chrcteristion of the limit superior nd limit inferior Proposition 9.6 Let ( n ) n N + if for ny ε > 0: be bounded sequence. Then α = lim sup n n if nd only (i) the set {n N + : n > α + ε} is finite, (ii) the set {n N + : n > α ε} is infinite. Likewise, β = lim inf n n if nd only if for ny ε > 0: (iii) the set {n N + : n < β ε} is finite, (iv) the set {n N + : n < β + ε} is infinite. Proof. We show equivlence with (i) nd (ii). ( ) Let A be the set of ccumultion points of ( n ) n N + so tht α = lim sup n n = sup A. Fix ε > 0. Now α A by Theorem 9.5. So there exists subsequence ( m(k) ) k N + such tht lim k m(k) = α. By definition of the limit this mens tht In prticulr, ( K N + )( k N + )[(k > K) ( m(k) α < ε). ( k > K)( m(k) > α ε). So the set {n N + : n > α ε} is infinite nd condition (ii) holds. Assume, for contrdiction, tht (i) does not hold. This mens tht there exists ε > 0 nd subsequence ( m(k) ) k N + such tht ( k N + )( m(k) > α + ε). Since ( m(k) ) k N + is bounded, by the Bolzno-Weierstrss theorem nd Theorem 7.12 we conclude tht ( m(k) ) k N + hs n ccumultion point γ α+ε. This contrdicts the definition of α. 27

31 9 LIMIT SUPERIOR AND LIMIT INFERIOR ( ) Now let α R be such tht for ny ε > 0 properties (i) nd (ii) hold. We prove tht α = lim sup n n. First, we show tht α A. Indeed, from (i) nd (ii), it follows tht for ny ε > 0 the set {n N + : α ε < n < α + ε} is infinite. By Theorem 7.12 we conclude tht α A. We now prove tht α = sup A. Assume for contrdiction tht there exists γ A such tht γ > α. Set ε := (1/2)(γ α) > 0. As γ A, by Proposition 7.12, the set {n N + : γ ε < n < γ + ε} is infinite. But γ ε = γ (1/2)(γ α) = α + (1/2)(γ α) = α + ε. This contrdicts (i). We conclude tht α = sup A. Theorem 9.7 (formule for limit superior nd limit inferior) Let ( n ) n N + be bounded sequence. Then ( ) (4) lim sup n = lim sup k, n n k n ( ) (5) lim inf n = lim inf k. n n k n is decres- Remrk 9.8 For ech n N +, put c n := sup k n k. Then the sequence (c n ) n N + ing. In fct, for ny n N +, Now, A n+1 A n. Therefore, c n = sup A n where A n := { k : k n}. c n+1 = sup A n+1 sup A n = c n. The sequence (c n ) n N + is lso bounded below. For, s ( n ) n N + is bounded, for some 0 M <. So M n ( M) for ll n N + c n = sup{ k : k n} n M for ny n N +. Therefore, by Theorem A 3.2.2, the limit lim n c n = lim n ( supk n k ) exists in R. Likewise, the sequence (b n ) n N + with b n := inf k n k is incresing nd bounded bove. By Theorem A 3.2.1, the limit lim n b n = lim n (inf k n k ) exists in R. Proof. Put α := lim n ( supk n k ). This mens tht (6) ( ε > 0)( N N + )( n N + )[(n N) ( α sup k ) < ε)], k n or, equivlently, (7) ( ε > 0)( N N + )( n N + )[(n N) (α sup k < α + ε)]. k n 28

32 10 CAUCHY SEQUENCES Fix ε > 0. Choose N N + s bove. Then sup k N k < α + ε. Therefore, Consequently, the set {n N : n > α + ε} =. {n N + : n > α + ε} is finite, contining t most N 1 elements. On the other hnd, in view of (7), So the set ( n N)( k n)[ k α ε]. {n N + : n > α ε} is infinite. By Proposition 9.6 we conclude tht α = lim sup n n. The rgument in the cse of the limit inferior is similr. Exmple 9.9 Let n = ( 1) n + 1 n. Find lim sup n n nd lim inf n n by using (4) nd (5). Solution. Set A n := sup k n k nd B n := inf k n k. Then we obtin { A n = sup{ n, n+1, n+2, n+3,... } = n if n is even, n+1 if n is odd. So lim sup n n = lim n A n = 1. Similrly, So lim inf n n = lim n B n = 1. B n = inf{ n, n+1, n+2, n+3,... } = 1. Exercise 9.10 () Give n exmple of bounded sequences ( n ) n N + nd (b n ) n N + such tht lim sup n ( n + b n ) lim sup n n + lim sup b n. n (b) Prove tht for ny bounded sequences ( n ) n N + nd (b n ) n N + one hs lim sup n 10 Cuchy sequences ( n + b n ) lim sup n n + lim sup b n, n lim inf n ( n + b n ) lim inf n n + lim inf n b n. Definition 10.1 A sequence ( n ) n N + is clled Cuchy sequence if ( ε > 0)( N N + )( m, n N + )[(m N) (n N) ( m n < ε)]. Exmple 10.2 The sequence ( n ) n N + with n = 1 n is Cuchy sequence. 29

33 10 CAUCHY SEQUENCES Proof. Fix ε > 0 nd tke N N + such tht N > 1 ε. Let m, n N. Then m n = 1 m 1 ( 1 n mx m, 1 ) 1 n N < ε. So ( n ) n N + is Cuchy sequence. Exmple 10.3 The sequence ( n ) n N + with n = n2 +1 n 2 is Cuchy sequence. Proof. Fix ε > 0 nd tke N N + such tht N > 1 ε. Let m, n N. Then m n = m m 2 n2 + 1 n 2 = 1 m 2 1 { 1 n 2 mx m 2, 1 } n 2 1 N 2 < ε. This mens tht ( n ) n N + is Cuchy sequence. Theorem 10.4 Let ( n ) n N + be convergent sequence. Then ( n ) n N + is Cuchy sequence. Proof. Let ( n ) n N + be convergent sequence nd = lim n n. Fix ε > 0. Then ( N N + )( n N + )[(n N) ( n < ε 2 )]. Let m, n N. Then by the tringle inequlity, m n m + n < ε/2 + ε/2 = ε. Tht is, ( n ) n N + is Cuchy sequence. In fct, the converse to Theorem 10.4 is lso true. Theorem 10.5 (Cuchy s theorem - generl principle of convergence) Let ( n ) n N + be Cuchy sequence. Then ( n ) n N + converges. Proof. Let ( n ) n N + Clim. ( n ) n N + Fix ε = 1. Then be Cuchy sequence. is bounded. ( N N + )( m, n N + )[(m N) (n N) ( m n < 1)]. Let N + n N. By the tringle inequlity, Put M := mx{ 1, 2,..., N }. Then Therefore, ( n ) n N + Clim. ( n ) n N + is bounded. converges. n n N + N 1 + N. n M + 1 for ll n N +. 30

34 11 UNIFORMLY CONTINUOUS FUNCTIONS Since ( n ) n N + is bounded, by the Bolzno-Weierstrss theorem, ( n ) n N + hs convergent subsequence ( m(k) ) k N + with limit, sy. We prove tht ( n ) n N + lso converges to. Fix ε > 0. Then ( N N + )( m, n N + )[(m N) (n N) ( m n < ε 2 )], nd ( K N + )( k N + )[(k K) ( m(k) < ε 2 )]. Choose l N + such tht m(l) l mx{n, K}. Let N + n N. Then n n m(l) + m(l) < ε. }{{}}{{} <ε/2 <ε/2 In short, lim n n =. Remrk 10.6 Combining Theorems 10.4 nd 10.5 we see tht sequence ( n ) n N + converges if nd only if ( n ) n N + is Cuchy sequence. This convergence criterion is intrinsic in the sense tht it chrcterises convergence in terms of property of the elements of the sequence. Remrk 10.7 Theorem 10.5 my be useful when we need to show tht sequence diverges. Indeed, combining Theorems 10.4 nd 10.5 we see tht sequence ( n ) n N + diverges if nd only if ( n ) n N + is not Cuchy sequence. Exmple 10.8 Prove tht the sequence n = ( 1) n diverges. Solution. Let N N +. Then for N + m N we hve m m+1 = 2. So, if we fix ε := 1 (sy), nd choose n := m + 1, then m n > ε. So ( n ) n N + is not Cuchy sequence nd therefore ( n ) n N + diverges. 11 Uniformly Continuous Functions Definition 11.1 Let A R nd f : A R function. Then f is sid to be uniformly continuous on A if (8) ( ε > 0) ( δ > 0) ( x, y A)[( x y < δ) ( f(x) f(y) < ε)]. In the bove definition, δ = δ(ε) depends on ε but not on x, nor y. 31

35 11 UNIFORMLY CONTINUOUS FUNCTIONS Lemm 11.2 Let f : A R be uniformly continuous function on A R. continuous on A. Then f is Proof. We need to prove tht f is continuous t every point in A. Fix A. From (8), In prticulr, ( ε > 0)( δ > 0)( x, y A)[( x y < δ) ( f(x) f(y) < ε)]. ( ε > 0)( δ > 0)( y A)[( x < δ) ( f(x) f() < ε)]. This mens tht f is continuous t. As this rgument works for ny A, f is continuous on A. Exmple 11.3 Let < < b < +. Then the function f(x) = x is uniformly continuous on [, b]. Solution. Let ε > 0. Choose δ = ε. Then for ny x, y [, b], ( x y < δ) ( f(x) f(y) = x y < δ = ε). This mens tht f is uniformly continuous on [0, 1]. Exmple 11.4 Let < < b < +. Then the function f(x) = x 2 is uniformly continuous on [, b]. Solution. Set c := mx{, b }. Note tht for ny x, y [, b]. Hence, x + y x + y 2 c x 2 y 2 = x + y x y 2 c x y for ny x, y [, b]. Let ε > 0. Choose δ = ε/2 c. Then, for ny x, y [, b], ( x y < δ) ( f(x) f(y) = x 2 y 2 = x + y x y < 2 c δ = ε). }{{}}{{} 2c <δ This mens tht f is uniformly continuous on [, b]. From (11.1), the function f : A R is not uniformly continuous on A if (9) ( ε > 0)( δ > 0)( x, y A)[( x y < δ) ( f(x) f(y) ε)]. Alterntively, Negtion of Definition A function f : A R is not uniformly continuous on A if (10) ( ε > 0)( n N + )( x n, y n A)[( x n y n < 1/n) ( f(x n ) f(y n ) ε)]. Exercise 11.5 Prove tht (10) is equivlent to (9). 32

36 11 UNIFORMLY CONTINUOUS FUNCTIONS Exmple 11.6 The function f(x) = 1/x is continuous on (0, 1) but is not uniformly continuous on (0, 1). Solution. The fct tht f(x) = 1/x is continuous on (0, 1) follows from Theorem A We prove tht f is not uniformly continuous on (0, 1). Fix ε = 1. For ech n N +, choose x n = 1/2 n nd y n = 1/4 n. Then, for ny n N +, x n y n = 1/4 n < 1/n, nd f(x n ) f(y n ) = 1 1 x n = 2n 4n = 2n > ε = 1. By (10), f is not uniformly continuous on (0, 1). y n Exmple 11.7 The function f(x) = x 2 is not uniformly continuous on (0, ). Solution. Fix ε = 1. Then for ech n N +, choose x n = n nd y n = n + 1 x n y n = 1/2 n < 1/n nd ( f(x n ) f(y n ) = n2 n + 1 ) 2 2n = n 2 > 1 = ε. By (10), f is not uniformly continuous on (0, ). The following is fundmentl theorem in Anlysis. 2n. Then Theorem 11.8 (Cntor s theorem on uniform continuity) Let < < b < +. Let f : [, b] R be continuous function on [, b]. Then f is uniformly continuous on [, b]. Proof. Assume for contrdiction tht f is not uniformly continuous on [, b]. Then, ccording to (10), (11) ( ε > 0)( n N + )( x n, y n [, b])[( x n y n < 1/n) ( f(x n ) f(y n ) ε)]. Observe tht the sequence (x n ) n N + [, b] is bounded. By the Bolzno-Weierstrss theorem, we my extrct convergent subsequence (x m(k) ) k N + with limit x 0 (sy) in [, b]. Now, Also, by the tringle inequlity, In short, x m(k) y m(k) < 1/m(k) 0 s k. y m(k) x 0 y m(k) x m(k) + x m(k) x 0 0 s k. }{{}}{{} 0 0 y m(k) x 0 s k. Now, by hypothesis, f is continuous t x 0. Consequently, f(x m(k) ) f(x 0 ) nd f(y m(k) ) f(x 0 ) s k. 33