1 Introduction. 1.1 Notation and Terminology

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1 1 Itroductio You have already leared some cocepts of calculus such as limit of a sequece, limit, cotiuity, derivative, ad itegral of a fuctio etc. Real Aalysis studies them more rigorously usig a laguage based o axioms ad defiitios (remember ɛ δ defiitio of limit?). It may be challegig to may studets to get used to the laguage ad axiomatic approach. But beig comfortable with the machiery of Real Aalysis would defiitely make you a better mathematicia. Pure mathematics is, i its way, the poetry of logical ideas. Oe seeks the most geeral ideas of operatio which will brig together i simple, logical ad uified form the largest possible circle of formal relatioships. I this effort toward logical beauty spiritual formulas are discovered ecessary for the deeper peetratio ito the laws of ature. Albert Eistei, Obituary for Emmy Noether (May 5, 1935). 1.1 Notatio ad Termiology N or Z + = {1, 2, 3,...}, the set of atural umbers. Z = {0, ±1, ±2, ±3,...}, the set of itegers. { p Q = }, p, q Z ad q 0 the set of ratioal umbers. q R, the set of real umbers. a i = a 1 + a a. E.g., i=1 a i = a 1 a 2 a. E.g., i=1 i = i=1 i =:! i=1 ( + 1) 2 Frequetly used otatio i Real Aalysis: x S : read as x belogs to S or x is i S. x S : read as for all x i S. x S : read as there exists x i S. s. t. : read as such that. Set otatio: A B : read as A is a subset of B 1

2 A B : read as A is a proper subset of B A B = {x x A or x B} A B = {x x A ad x B} A \ B = {x x A ad x / B} 1.2 Subsets of R ad Cardiality Beside N, Z, ad Q, the followig are frequetly used subsets of R i Real Aalysis: Ope iterval : (a, b) = {x R a < x < b}. Closed iterval : [a, b] = {x R a x b}. Half-ope/ half-closed iterval : (a, b] = {x R a < x b} ad [a, b) = {x R a x < b}. The domai ad rage of the fuctio f defied by f(x) = x 1 are [1, ) ad (0, ) respectively. The precedig sets are all ifiite sets. Ifiite sets may be of differet kids i terms of coutig their elemets. Defiitio (Coutable set). A set S is coutable if it is fiite or there is a oe-to-oe fuctio from N oto S (equivaletly if there is a oe-to-oe fuctio from S oto a subset of N). 1. N is coutable because of the idetity map I : N N defied by I() =. 2. To show Z is coutable, defie f : N Z by { if is eve 2 f() = 1 if is odd 2 To show oe-to-oeess of f, let f(m) = f() for some m, N. Note that if both m ad are ot eve or both m ad are ot odd, the f(m) = f() = m + = 1, a cotradictio. So whe m ad are eve, f(m) = f() = m 2 = 2 = m =. Similarly whe m ad are odd, f(m) = f() = 1 m 2 = 1 2 = m =. To show otoess of f, let k Z. If k > 0, the f(2k) = 2k 2 k 0, the f(1 2k) = 1 (1 2k) = k where 1 2k N. 2 2 = k where 2k N. If

3 Theorem 1.1. Q is coutable. Proof. Sice Z is coutable, S q = { p q p Z} is coutable for all q N. Note that Q = S q. Sice coutable uio of coutable sets is coutable (log exercise), Q is coutable. q=1 Theorem 1.2. The iterval (0, 1) is ucoutable. Proof. (Cator s diagoal argumet, 1891) Suppose (0, 1) is ucoutable. Let (0, 1) = {x N} where each x has a decimal expasio (ot ecessarily fiite): x 1 = 0.a 11 a 12 a 13 a 14 x 2 = 0.a 21 a 22 a 23 a 24 x 3 = 0.a 31 a 32 a 33 a Now costruct a umber b = 0.b 1 b 2 b 3 b 4 (0, 1) by choosig b i a ii as follows: { 6 if a ii = 5 b i = 5 if a ii 5 The b x i for all i N but b (0, 1) cotradictig that (0, 1) = {x N}. By the precedig theorem, (0, 1) \ Q is ucoutable ad so are R ad R \ Q. Thus Q is a coutably ifiite set where R \ Q is a ucoutably ifiite set. To distiguish the sizes (cardiality) of ifiite sets such as N ad R, Georg Cator itroduced cardial umbers such as ℵ 0 (read aleph-aught or aleph-zero) < ℵ 1 < ℵ 2 < where ℵ 0 is the cardiality of N. The cardiality of R is deoted by c. By the precedig theorem, ℵ 0 < c. The cotiuum hypothesis states that c = ℵ 1, i.e., there is o set whose cardiality is strictly betwee that of N ad that of R. 3

4 2 Basic Properties of R 2.1 Order Structure We itroduce a order < o the field R ad impose the the followig order properties for all a, b, c R: (a) Either a < b, or a = b, or b < a. (Law of Trichotomy) (b) If a < b ad b < c, the a < c. (Trasitivity) (c) If a < b, the a + c < b + c c R. (Orderig after shiftig) (d) If a < b, the ca < cb c > 0 ad ca > cb c < 0. (Orderig after scalig) Axiom 2.1 (Well-orderig Property). Every oempty subset of N has a smallest elemet. The well-orderig property caot be proved just by the field ad order properties of R. Thus it is cosidered as a axiom or a priciple (uless we assume mathematical iductio). We itroduce a metric structure o the ordered field R by a distace fuctio (metric) give by the absolute value fuctio : x = x if x 0 ad x = x if x < 0. The distace betwee two real umbers x ad y is defied by x y. Amog kow properties of, the followig oe is frequetly used i real aalysis: Propositio (Triagle iequality). For all a, b R, a + b a + b. 2.2 Archimedea Property Theorem 2.2 (Archimedea Property). Let ε ad x be two positive real umbers. The ε > x for some N. ε ε x Although the Archimedea Property seems obvious ad = x ε, a formal proof is give i the ext sectio by the completeess property of R. The followig are some equivalet statemets of the well-orderig property: Corollary 2.3. (a) For positive real umbers ε (however small) ad x (however large), we have ε > x for some N. 4

5 (b) For a positive real umbers ε (however small), we have 1 < ε for some N. (c) For a positive real umber x (however large), we have > x for some N. (i.e., N is ubouded). Proof. (a) = (b) Set x = 1 i (a). (b) = (c) Set ε = 1/x i (b). (c) = (a) Set x = x/ε i (c). Theorem 2.4 (The desity of Q i R). Let a ad b be two real umbers such that a < b. The there exists a ratioal umber r such that a < r < b. Proof. By the Archimedea Property o ε = b a > 0 ad x = 1, we fid a atural umber such that (b a) > 1 which implies a + 1 < b. Similarly we fid a atural umber m such that m > a. Suppose m is the smallest iteger such that m > a. The m > a m 1 which implies a < m a + 1 < b. Dividig by we get a < m < b. Here r = m Q. 2.3 Ifimum ad Supremum I all the defiitios i this sectio, S R. Defiitio (Upper boud). A real umber M is a upper boud of S if M x for all x S. The S is bouded above by M. If a upper boud M of S belogs to S, the M is the maximum of S, deoted by M = max S. 1. S = [0, 1] is bouded above by the upper boud 1 ad max S = S = (0, 1) is bouded above by the upper boud 1 ad max S does ot exist as 1 / S. 3. S = {x Q x 2 2} is bouded above by 2 / S ad max S does ot exist. Suppose max S exists ad M = max S. The M Q ad sice 2 / Q, M < 2. Let M = 2+2M. Sice M Q, M Q. Note that M 2 2 = 2(M 2 2) < 0 which 2+M (2+M) 2 implies M < 2. Thus M S. Now M M = 2 M 2 > 0 which implies M > M. (2+M) 2 Thus M S ad M > M = max S, a cotradictio. (Alt. use the desity of Q i R) Defiitio (Lower boud). A real umber m is a lower boud of S if m x for all x S. The S is bouded below by m. If a lower boud m of S belogs to S, the m is the miimum of S, deoted by m = mi S. 1. S = [0, 1] is bouded below by the lower boud 0 ad mi S = 0. 5

6 2. S = (0, 1) is bouded below by the lower boud 0 ad mi S does ot exist as 0 / S. 3. S = {x Q x 2 2} is bouded below by 2 / S ad mi S does ot exist. Defiitio (Bouded Set). S is bouded if it is bouded above ad below. S is ubouded if it is ot bouded. Note that a fiite set S is always bouded because it is bouded above ad below by max S ad mi S respectively. But a ifiite set S may ot be bouded ad also max S ad mi S may ot exist. 1. S = (0, 1) is a ifiite bouded set. Note both max S ad mi S do ot exist. 2. S = [0, ) is a ubouded set. Note mi S = 0 but max S does ot exist. 3. S = { 1 N} is a ifiite bouded set. Note max S = 1 but mi S does ot exist. Defiitio (Supremum). If S is a oempty set bouded above by M ad M M for all upper bouds M of S, the M is the supremum or the least upper boud of S, deoted by M = sup S. Equivaletly M = sup S if ad oly if (a) M x for all x S ad (b) for every ε > 0, there exists x S such that x > M ε. S M ε x M Note that if S is ubouded above, we defie sup S =. Also we defie sup =. 1. For S = [0, 1], sup S = max S = For S = (0, 1), max S does ot exist but sup S = For S = { 1 N}, max S does ot exist but sup S = 0. Sice 1 < 0 N, x < 0 x S. Let ε > 0. By the Archimedea Property, we fid a atural umber N such that 1 < ε. The 1 > ε. So we foud x = 1 S N N N such that x > 0 ε. Thus sup S = For S = {x Q x 2 2}, max S does ot exist but sup S = 2. First ote x 2 x S. Let ε > 0. If ε 2 2, the 2 ε 2 < x = 0. Otherwise ε < 2 2. Sice there is a ratioal umber betwee two real umbers (the desity of Q i R), there exists a ratioal umber x such that 2 < 2 ε < x < 2. So we foud x S such that x > 2 ε. Thus sup S = 2. 6

7 Defiitio (Ifimum). If S is a oempty set bouded below by m ad m m for all lower bouds m of S, the m is the ifimum or the greatest lower boud of S, deoted by m = if S. Equivaletly m = if S if ad oly if (a) m x for all x S ad (b) for every ε > 0, there exists x S such that x < m + ε. m x m + ε S Note that if S is ubouded below, we defie if S =. Also we defie if =. 1. For S = [0, 1], if S = mi S = For S = (0, 1), mi S does ot exist but if S = For S = { 1 N}, mi S does ot exist but if S = For S = {x Q x 2 2}, mi S does ot exist but if S = 2. Axiom 2.5 (Completeess Property). Every oempty bouded above subset of R has a supremum (least upper boud) i R. The completeess property of R is cosidered as a axiom as it caot be proved just by the field ad order properties of R. The followig is equivalet to the Completeess Axiom: Theorem 2.6. Every oempty bouded below subset of R has a ifimum (greatest lower boud) i R. Proof. Suppose S R is oempty ad bouded below. Defie S = { x x R}. Sice S is bouded below, S is bouded above. The by the Completeess Axiom, sup( S) exists. The rest follows from the fact that if(s) = sup( S) (exercise). Note that the Archimedea Property follows from the completeess property of R. Proof of Theorem 2.2. Suppose x for all N. Cosider the set S = { N x}. ε ε Note that S = N ad it is bouded above by x. By the completeess property of R (see ε Axiom 2.5), N has a least upper boud, say M. The M 1 is ot a upper boud of N ad cosequetly M 1 < k for some k N. The M < k + 1 where k + 1 N cotradictig that M is a upper boud of N. A ordered field F is icomplete if it has a oempty bouded above subset with o supremum i F. For example, Q is icomplete because S = {x Q x 2 2} does ot have a supremum i Q. It ca be show that R is the uique complete ordered field meaig ay complete ordered field is isomorphic to R. 7

8 3 Topology of R We ofte deal with sets i R such as solutio sets of equatios which are ot simply ope or closed itervals. To aalyze the properties of the poits i a subset of R, we itroduce some geometric cocepts developed by Georg Cator i late ieteeth cetury. 3.1 Iterior, Boudary, Limit Poits A ope eighborhood of a poit x i R is a ope iterval (x ε, x + ε) for some ε > 0. Note (x ε, x + ε) is called a ε-eighborhood of x, deoted by N ε (x) or B ε (x). x ε x x + ε I all the defiitios i this sectio, S R. Defiitio (Iterior poit). A poit x of S is a iterior poit of S if there is a ope eighborhood of x cotaied i S, i.e., ε > 0 s. t. (x ε, x + ε) S. The set of all iterior poits of S is the iterior of S, deoted by it (S). 1. Each poit of (a, b] except b is a iterior poit. So it(a, b] = (a, b). Let x (a, b). Choose ε = mi{ x a, b x }. The (x ε, x + ε) (a, b]. So x it(a, b]. Thus (a, b) it(a, b]. To show (a, b) = it(a, b], let y it(a, b] \ (a, b). The y a or y b. Sice y it(a, b], ε > 0 s. t. (y ε, y + ε) (a, b] = a y ε < y < y + ε < b, which cotradicts that y a or y b. it(a, b] = (a, b). So it(a, b] \ (a, b) = ad cosequetly 2. Each poit of N is ot a iterior poit. So it (N) =. Defiitio (Boudary poit). A poit x (ot ecessarily i S) is a boudary poit of S if every ope eighborhood of x cotais oe poit of S ad oe poit ot i S, i.e., ε > 0 s. t. N ε (x) S ad N ε (x) (R \ S). The set of all boudary poits of S is the boudary of S, deoted by (S). 8

9 1. a ad b are boudary poits of (a, b] (choose ay ε > 0). So (a, b] = {a, b}. 2. Each poit of N is a boudary poit. So (N) = N. The followig are a special kid of boudary poits of a set. Defiitio (Isolated poit). A poit x of S is a isolated poit of S if there is a ope eighborhood of x which cotais oly oe poit of S, i.e., 1. No poit of (a, b] is a isolated poit. ε > 0 s. t. (x ε, x + ε) S = {x}. 2. Each poit of N is a isolated poit. For each N, ( 1, + 1) N = {}. 3. A isolated poit of a subset S of R is a boudary poit of S (exercise). There are poits of S that are ot isolated poits as the followig. Defiitio (Accumulatio or limit poit). A poit x (ot ecessarily i S) is a limit poit of S if every ope eighborhood of x cotais at least oe poit of S other tha x, i.e., ε > 0, (x ε, x + ε) (S \ {x}). The set of all limit poits of S is the derived set of S, deoted by S. 1. All poits of (a, b] ad a are limit poits of (a, b]. So (a, b] = [a, b]. 2. No poit of N is a limit poit. So N =. 3. The oly limit poit of S = { 1 N} is 0. So S = {0}. Note that x is a limit poit of a subset S of R if ad oly if there is a sequece {x } i S \ {x} covergig to the limit x (Hit. choose x N 1/ (x) (S \ {x})). Defiitio (Closure). The uio of S ad its derived set S is the closure of S, deoted by S or cl (S). So S = S S. 1. Sice (a, b] = [a, b], (a, b] = (a, b] (a, b] = [a, b]. 2. Sice N =, N = N N = N. 3. For S = { 1 N}, S = S S = { 1 N} {0}. 9

10 3.2 Ope, Closed, Compact Sets I all the defiitios i this sectio, S R. Defiitio (Ope set). S is a ope set if each poit of S is a iterior poit, i.e., it (S) = S. 1. Sice each poit of (a, b) is a iterior poit, i.e., it(a, b) = (a, b), (a, b) is a ope set. So are (a, ), (, a), ad (, ). 2. [a, b) is ot ope as a is ot a iterior poit of [a, b). Theorem 3.1. A arbitrary uio of ope sets is ope ad a fiite itersectio of ope sets is ope. Proof. Let {S i i I} be a arbitrary collectio ope sets. To show i I S i is ope, it suffices to show each of its poits is a iterior poit. Let x i I S i. The x S i for some i I. Sice S i is ope, x is a iterior poit of S i, i.e., there is a ope eighborhood N ε (x) of x cotaied i S i. The N ε (x) S i i I S i. Thus x is a iterior poit of i I S i. Let {S i i = 1, 2,..., } be a fiite collectio ope sets. To show show each of its poits is a iterior poit. Let x S i is ope, it suffices to i=1 S i. The x S i for all i = 1, 2,...,. Sice each S i is ope, x is a iterior poit of S i, i.e., there is a ope eighborhood N εi (x) of x cotaied i S i. Let ε = mi{ε 1, ε 2,..., ε }. The N ε (x) S i for all i = 1, 2,..., ad cosequetly N ε (x) S i. Thus x is a iterior poit of Note that S i. i=1 1. arbitrary itersectio of ope sets is ot ecessarily ope. For example, I = ( 1, 1 ) is ope for all N. But I = {0} is ot ope. =1 10 i=1 i=1

11 2. Noempty S is ope if ad oly if S = (why?). Defiitio (Closed set). S is a closed set if R \ S is a ope set. (a, b ) where itervals (a, b ) are disjoit =1 1. Sice R \ [a, b] = (, a) (b, ) is a ope set, [a, b] is a closed set. So are [a, ), (, a], ad {0}. 2. [a, b) is ot closed as R \ [a, b) = (, a) [b, ) is ot ope. Theorem 3.2. A arbitrary itersectio of closed sets is closed ad a fiite uio of closed sets is closed. Proof. Let {S i i I} be a arbitrary collectio closed sets. First ote that R \ S i is a ope set for all i I. To show S i is closed, it suffices to show R \ S i is ope. By De i I i I Morga s laws, R \ i I S i = i I(R \ S i ) is a ope set as a arbitrary uio of ope sets R \ S i. Let {S i i = 1, 2,..., } be a fiite collectio closed sets. First ote that R \ S i is a ope set for all i = 1, 2,...,. To show S i is closed, it suffices to show R \ S i is ope. By De Morga s laws, i=1 i=1 R \ S i = (R \ S i ) i=1 i=1 is a ope set as a fiite itersectio of ope sets R \ S i. Note that arbitrary uio of closed sets is ot ecessarily closed. For example, I = [ 1, 1] is closed for all N. But I = (0, 1] is ot closed. =1 Theorem 3.3. The followig are equivalet. (a) S is a closed set. (b) The closure of S is S itself, i.e., S = S. (c) S cotais all its limit poits, i.e., S S. 11

12 Proof. (b) (c) Sice S = S S, S = S if ad oly if S S. (a) = (c) Suppose S is a closed set. Let x S, i.e., x is a limit poit of S. It suffices to show x S. If ot, the x R \ S, a ope set. The ε > 0 s. t. (x ε, x + ε) R \ S which implies (x ε, x + ε) (S \ {x}) =. This cotradicts that x is a limit poit of S. Thus x S. (c) = (a) Suppose S S. We show that R \ S is ope. Let x R \ S. It suffices to show x it (R \ S). If ot, the ε > 0, (x ε, x + ε) is ot cotaied i R \ S which implies ε > 0, (x ε, x + ε) (S \ {x}). This implies x is a limit poit of S, i.e., x S S which cotradicts that x R \ S. Defiitio (Compact set). S is a compact set if it is closed ad bouded. Note that S is compact if ad oly if every sequece i S has a coverget subsequece with its limit i S. I a geeral topological space a set S is compact if each ope cover of S has a fiite subcover. 1. Sice [a, b] is is closed ad bouded, [a, b] is compact. 2. [a, b), [a, b), (, a], [a, ), ad (a, ) are ot compact sets. The Cator set C is a classical example of a compact set defied as follows: C = C, =1 where is C is a closed subset of [0, 1] obtaied by successively removig ope middle thirds of 2 1 closed subitervals of [0, 1]: C 1 = [0, 1/3] [2/3, 1] C 2 = [0, 1/9] [2/9, 1/3] [2/3, 7/9] [8/9, 1] C. = C 1 3 ( ) 2 + C N, C 0 = [0, 1]. Each C is closed as a fiite uio of closed sets. The C = C [0, 1], C is bouded. Thus C is compact. C is also closed. Sice We ca show that the Cator set has ucoutably may poits by usig terary expasio of each poit i it (how?). =1 12

13 4 Sequece ad Series A sequece {x } of real umbers is a fuctio f : N R such that f() = x. Sometimes it is deoted by {x } =1 or by a few terms with a certai patter {x 1, x 2, x 3,...}. A sequece ca also be defied recursively. 1. {( 1) }. 2. {1, 1 2, 1 3, 1 4,...}. 3. Fiboacci sequece {F } =0 is defied by F = F 1 + F 2, 2, F 0 = 0, F 1 = 1. A series is a sum of the terms of a sequece {x } which is deoted by the lecture otes we oly deal with real sequeces ad series. x. Throughout =1 4.1 Covergece Defiitio (Covergece). A sequece {x } coverges to a real umber L if for every ε > 0 there exists a positive iteger N such that x L < ε for all N. i.e., all terms of {x }, except x 1,..., x N 1, lie i (L ε, L + ε). It is deoted by lim x = L or x L as. The sequece {x } diverges if it is ot coverget. 1. Show that {x } = { } coverges to 1/2, i.e., lim = 1 2. Let ε > 0. We eed to fid a positive iteger N for which < ε for all N. Note that = 3 2(2 + 3) < 3 4 Note 3 < ε if ad oly if 3 <. By the Archimedea Property, we fid a atural 4 4ε umber N > 3. The 4ε < 3 < ε for all N. 4 13

14 2. Show that {x } = {( 1) } diverges. Suppose {x } = {( 1) } coverges to L. For ε = L + 1, x L = 1 L = 1 + L ε odd N. The there exists o N N such that x L < ε N, a cotradictio. If a sequece {x } diverges, its terms oscillate idefiitely like {( 1) } or it diverges to ±. To show lim x =, we prove that for ay positive umber L (however large), there exists N N such that x L for all N. Show that lim 2 =. Let L > 0. We eed to fid a positive iteger N for which 2 > L i.e., > L N. By the Archimedea Property, we fid a atural umber N > L. The 4.2 Algebra of Limits Algebra of Limits: 2 > L N. Theorem 4.1. The followig are true for ay coverget sequeces {x } ad {y }: (a) lim (x + y ) = lim x + lim y. (b) lim (cx ) = c lim x, c R. (c) lim x y = ( lim x )( lim y ). x (d) lim = y lim x lim y wheever lim y 0. Note that the covergece i the assumptio is crucial. For {x } = { 1 } ad {y } = {}, we have {x y } = {1}. The lim x y = 1 0 = ( lim x )( lim y ). Proof of Theorem 4.1. Suppose {x } ad {y } coverge to x ad y respectively. For (c) we show that {x y } coverges to xy. Let ε > 0. We eed to fid N N for which First ote by the triagle iequality that x y xy < ε for all N. x y xy = (x x)y + x(y y) y x x + x y y 14

15 Sice the coverget sequece {y } is bouded (see Theorem 4.4), M > 0 s. t. y M N. Thus x y xy M x x + x y y N. Sice {x } coverges to x, N 1 N such that x x < ε 2M N 1. Sice {y } coverges to y, N 2 N such that Let N = max{n 1, N 2 }. The y y < ε 2 x N 2. x y xy M x x + x y y < M ε 2M + x ε 2 x = ε N. (a), (b), ad (d) are exercises. Theorem 4.2. Suppose x y N. If both {x } ad {y } coverge, the lim x lim y. Proof. Let z = y x N. Sice lim z = lim (y x ) = lim y lim x, it suffices to show that z := lim z 0. Let ε > 0. The there exists N N such that Thus ε < z ε > 0 which implies z 0. 1 lim lim 0 z = z + (z z) < z + ε N. 2 1 because N (verify). Note that x < y N lim x < lim y. For example, { 1 } ad { 1 }. Theorem 4.3 (Sadwich Theorem). Suppose x z y N. If both {x } ad {y } coverge to L, the {z } also coverges to L. Proof. Suppose both {x } ad {y } coverge to L. Let ε > 0. The there exists N 1 N such that x L < ε N 1. Similarly there exists N 2 N such that Let N = max{n 1, N 2 }. The y L < ε N 2. ε < x L < z L < y L < ε N. Thus z L < ε N which implies {z } also coverges to L. 15

16 si(x) Show that lim = 0. First ote that 1 si(x) 1 for all, x R. The for all N, 1 Sice lim = lim 1 1 si(x) si(x) = 0, we have lim 1. = 0 by the Sadwich Theorem. 4.3 Covergece Theorems Defiitio (Bouded Sequece). A sequece {x } is bouded if there exists a real umber M such that x M N. 1. { 1 } is bouded sice 1 1 N. 2. We show {x } is ot bouded for x = Note that x 2 = ( ) ( = ) ( ( ) ( ) ( ( ) ( ) ( ) = ( ) 1 = ) ( + ) ( 1 2 ) Thus x /2 N. The for ay M > 0, x 2 2M 1 + 2M 2 > M. Thus {x } is ot bouded. Theorem 4.4. A coverget sequece is bouded. Proof. Cosider a sequece {x } covergig to L. For ε = 1, there exists N N such that The by the triagle iequality, Let Thus x M N. x L < 1 N. x = L + (x L) L + x L < L + 1 N. M = max{ x 1, x 2,..., x N 1, L + 1}. 16 )

17 1. Sice { 1 } is coverget, it is bouded. 2. Sice { } is ot bouded, it is ot coverget Defiitio (Icreasig/decreasig sequece). A sequece {x } is icreasig if x +1 > x N ad decreasig if x +1 < x N. A sequece {x } is o-decreasig if x +1 x N ad o-icreasig if x +1 x N. A sequece {x } is mootoe if it is either icreasig, decreasig, o-decreasig, or o-icreasig. Theorem 4.5 (Mootoe Covergece Theorem). (a) A icreasig sequece which is bouded above is coverget. (b) A decreasig sequece which is bouded below is coverget. Proof. (a) Suppose {x } is a icreasig sequece which is bouded above. Sice {x } is bouded above, the set {x N} is also bouded above. By the completeess property sup{x N} exists, which is say L. We prove that lim x = L. Let ε > 0. The there exists k N such that x k > L ε. Sice {x } is a icreasig, L ε < x k. Also sice L = sup{x N}, L x N. Combiig these facts we get, L ε < x L < L + ε k. Thus x L < ε k which implies {x } also coverges to L. (b) Exercise. Note that MCT is also true for o-decreasig ad o-icreasig sequeces respectively. 1. Sice { 1 } is a decreasig sequece (show) bouded below by 0, it is coverget by MCT. 2. {x } where x 1 = 2 ad x = 2 + x 1, 2. We show {x } is icreasig by iductio. First ote that x 1 = 2 < = 2 + x1 = x 2. Assume x 1 < x for some N. We show x < x +1. Note that x 2 +1 x 2 = (2 + x ) (2 + x 1 ) = x x 1 = x +1 x = (x x 1 )/(x +1 + x ) Sice x 1 < x ad x, x +1 > 0, we have x +1 x = (x x 1 )/(x +1 + x ) > 0, i.e., x < x +1. We show {x } is bouded above also by iductio. Note that x 1 = 2 < 3 ad x 2 = < 3. Assume x < 3 for some N. We show x +1 < 3. Note that x +1 = 2 + x < < 3. 17

18 By mathematical iductio x < 3 N. Sice {x } is a icreasig sequece bouded above by 3, it is coverget by MCT. Suppose lim x = L. To fid L, ote that x 2 = 2 + x 1, 2. Takig limit of both sides we get lim x2 = lim (2+x 1 ) = ( lim x ) 2 = 2+ lim x 1 = L 2 = 2+L = L = 1, 2. Sice x > 0 N, L 0. Thus L = 2, i.e., lim x = 2. Defiitio (Cauchy sequece). A sequece {x } is a Cauchy sequece if for every ε > 0 there exists a positive iteger N such that x m x < ε for all m, N. We show that { 1 } is a Cauchy sequece. Let ε > 0. We eed to fid a positive iteger N for which 1 m 1 < ε for all m, N. Note that 1 m 1 < 1 m + 1. Note 1 m, 1 < ε 2 if ad oly if 2 ε < m,. By the Archimedea Property, we fid a atural umber N > 2. The ε 1 m 1 < 1 m + 1 < ε 2 + ε 2 = ε m, N. Theorem 4.6 (Cauchy Covergece Criterio). A sequece is coverget if ad oly if it is a Cauchy sequece. Proof. Suppose {x } is a coverget sequece with limit L. Let ε > 0. The there exists a positive iteger N for which x L < ε 2 for all N. The by the triagle iequality, x m x = (x m L) (x L) x m L + x L < ε 2 + ε 2 = ε m, N. Thus {x } is a Cauchy sequece. We will prove the coverse after Bolzao-Weistrass Theorem (Theorem 4.8) usig the fact that a Cauchy sequece {x } is bouded: For ε = 1, there exists N N such that x x N < 1 N. 18

19 The by the triagle iequality, x = x N + (x x N ) x N + x x N < x N + 1 N. Let M = max{ x 1, x 2,..., x N 1, x N + 1}. Thus x M N. 4.4 Subsequeces Defiitio (Subsequece). A subsequece of a sequece {x } is {x 1, x 2, x 3,...} where { 1, 2, 3,...} is a icreasig sequece of atural umbers. { } ad { 1 2 } are two subsequeces of { 1 }. Note that it ca be show that if lim x = L, the lim k x k = L (exercise). Theorem 4.7. Every sequece has a mootoe subsequece. Proof. Cosider a sequece {x }. We defie a term x k of {x } to be a peak if x k x for all > k. So by defiitio, a icreasig sequece has o peaks. Now we have two cases: Case 1. {x } has ifiitely may peaks. Cosider the subsequece {x 1, x 2, x 3,...} of peaks. Sice 1 < 2 < 3 <, x 1 x 2 x 3. Thus {x 1, x 2, x 3,...} is a o-icreasig subsequece of {x }. Case 2. {x } has fiite umber of peaks. If there is o peak, {x } is a icreasig subsequece of itself. Otherwise cosider the last peak x k which is also the smallest peak of {x }. The x is ot a peak for all > k. The there exists 1 > k for which x 1 > x k. The x 1 is ot a peak ad cosequetly there exists 2 > 1 for which x 2 > x 1. Similarly there exists 3 > 2 for which x 3 > x 2. Cotiuig this process we costruct a icreasig subsequece {x 1, x 2, x 3,...} of {x }. {( 1) } has mootoe subsequeces {( 1) 2 } ad {( 1) 2 1 }. Theorem 4.8 (Bolzao-Weistrass Theorem). Every bouded sequece has a coverget subsequece. Proof. Suppose {x } is a bouded sequece. Sice every sequece has a mootoe subsequece, cosider a mootoe subsequece {x k } of {x }. Sice {x } is bouded, {x k } is also bouded. The {x k } is coverget by MCT. 19

20 Proof of Theorem 4.6. Coversely suppose {x } is a Cauchy sequece. We have proved before that {x } is bouded. The by Bolzao-Weistrass Theorem, {x } has a coverget subsequece {x k } with limit say L. We show that {x } also coverges to L. Let ε > 0. Sice {x } is a Cauchy sequece, there exists N N such that x m x < ε 2 m, N. (1) Sice {x k } coverges to L, there exists a positive iteger K { 1, 2, 3,...} such that x k L < ε 2 k K. Choose a iteger N { 1, 2, 3,...} such that N N, K. The x N L < ε ad 2 usig m = N i (1) we get x N x < ε 2 N. Combiig these facts we get x L = (x x N ) + (x N L) x x N + x N L < ε 2 + ε 2 = ε N. Theorem 4.9. A sequece {x } is coverget if ad oly if (a) {x } is bouded ad (b) every coverget subsequece of {x } coverges to the same limit. Proof. (= ) Exercise. ( =) Suppose {x } is bouded ad every coverget subsequece of {x } coverges to the limit L. Suppose that {x } does ot coverge to L. The there exists ε > 0 such that x L ε for ifiitley may N (thik a bit). So we ca fid a subsequece {x k } such that x k L ε k N. (2) Sice {x } is bouded, {x k } is also bouded. By Bolzao-Weistrass Theorem, {x k } has a coverget subsequece {x kj }. By (2), {x kj } does ot coverge to L. Note {x kj } is also a subsequece of {x } ad it does ot coverge to L which cotradicts our hypothesis. 4.5 Series* I might add the followig topics briefly i the ear (or far!) future: Covergece via sequece of partial sums Geometric series Oe covergece criterio Oe covergece test 20