MATH 337 Sequeces Dr. Neal, WKU Let X be a metric space with distace fuctio d. We shall defie the geeral cocept of sequece ad limit i a metric space, the apply the results i particular to some special real-valued sequeces. Defiitio 3.1. A sequece { x } =1 is a ifiite list of poits chose from the metric space X. Idividual poits x are called the terms of the sequece. For a fixed iteger N 1, the terms x will be called the tail ed of the sequece. { } = N (A sequece also ca be thought of as the list of fuctio values attaied for a fuctio f :ℵ X, where f () = x for 1.) We wat to make the otio of lim x = x as precise as possible. This cocept of limit meas that we ca make all the tail ed terms { x } = N as close as we like to the poit x X by choosig N large eough. Specifically, we ca make d (x, x) arbitrarily small for all of the tail ed terms provided N is sufficietly large eough. We formalize the cocept i the followig defiitio. Defiitio 3.2. We say lim x = x, which may be writte x x, provided the followig coditio holds: For every ε > 0 there exists a iteger N 1 such that d (x, x) < ε wheever N. I this case, we say that { x } coverges to x ad that x is the limit of { x }. If o such x X exists, the we say that { x } does ot coverge i X or that { x } diverges. O the real lie, lim x = x provided: For every ε > 0 there exists a iteger N 1 such that x x < ε wheever N. I ay metric space, we have x x i X if ad oly if d (x, x) 0 i R. x 1 x N +k x 4 x N +4 x 3 x 2 x 5 x N +3 x x N +2 x N x N +1 x 6 Arbitrarily small ε - eighborhood about x Evetually, all x for N (i.e., the terms x N, x N +1, x N +2,... ) are withi ε of x.
Example 3.1. O the real lie, lim + 1 = 1. Advaced Solutio. Let ε > 0 be give. Let N be the smallest iteger such that N 1 ad N > 1 1. The for N, we have ε thus + 1 1. ( + 1) 1 = +1 +1 1 = +1 1 N + 1 < 1 1 ε 1 +1 = ε ; The above proof is elegat ad precise, but it does ot idicate how N was foud. We shall ow give a alterate proof that icludes some more reasoig. 1 More Illustrated Solutio. We first ote that the sequece { +1} = =1 2, 2 3, 3 4,... icreasig ad the terms are always less tha 1. Thus the distace to 1 is +1 1 = 1 + 1 = 1 +1. { } is We wish to make this distace withi a arbitrary ε > 0 by choosig large eough. 1 So we eed + 1 < ε, which holds if 1 < + 1. Thus, ε +1 1 < ε provided > 1 ε 1. The followig result proves that 1 is the oly possible limit of { +1}. Theorem 3.1. I a metric space, limits are uique. That is, if x x ad x y, the x = y. Proof. To show that x = y, it suffices to show that d (x, y) < ε for all ε > 0. So let ε > 0 be give. Because x x ad x y, there exist itegers N 1 1 ad N 2 1 such that d (x, x) < ε / 2 for all N 1 ad d (x, y) < ε / 2 for all N 2. Now choose ay iteger max( N 1, N 2 ). The d (x, y) d( x, x ) + d(x, y) < ε / 2 + ε / 2 = ε. Example 3.2. O the real lie, { 6 + ( 1) } diverges. = 1
Solutio. The terms of the sequece { 6 + ( 1) } are 5, 7, 5, 7, 5,... which oscillate ad have o apparet limit. Also, successive terms are a distace of 2 apart. But suppose 6 + ( 1) L for some L R ad let ε = 1 / 2. The there would be a iteger N 1 such that 6 + ( 1) L < 1 / 2 wheever N. But the for all N, the triagle iequality gives the distace betwee successive terms to be (6 + ( 1) ) (6 + ( 1) +1 ) = (6 + ( 1) ) L + L (6 + ( 1) +1 ) 6 + ( 1) L + L (6 + ( 1) +1 ) < 1 / 2 +1 / 2 = 1, which cotradicts the fact that successive terms are a distace of 2 apart. Thus, o limit L exists ad 6 + ( 1) { } diverges. Defiitio 3.3. A sequece { x } is bouded if there exists a x X ad a real umber M > 0 such that d (x, x) M for all. The sequece { 6 + ( 1) } also gives a example of a bouded sequece. The terms are all withi 1 of 6: d (6 + ( 1), 6) = 6 + ( 1) 6 = ( 1) = 1 for all. So { 6 + ( 1) } is a bouded sequece but it does ot coverge. However all coverget sequeces are bouded as is prove ext. Theorem 3.2. If { x } coverges, the { x } is bouded. ( ) 1 for all ( ) for ( ) M for all. Proof. Assume x x. The there exists a iteger N 2 such that d x, x N. Now cosider the terms x 1,..., x N 1 ad the distaces d i = d x i, x i = 1,..., N 1. Let M = max{d 1,..., d N 1, 1). The d x, x Corollary 3.1. If { x } is ot bouded, the { x } diverges. { } be a sequece of real umbers. The { x } is bouded if ad Corollary 3.2. Let x oly if there exists a real umber c > 0 such that x c for all. { } is bouded. Now { } is bouded with x R ad M > 0 such that x x = d( x, x ) M for Proof. If such a c exists, the d (x, 0) = x c for all ; thus, x suppose x all. Now let c = M + x > 0. The for all, x = x x + x x x + x M + x = c.
Example 3.3. Cosider the sequece { 6 + ( 1) }, which is bouded because it has all terms withi 1 of 6. But i fact 6 + ( 1) = 6 + ( 1) 6 + 6 6 + ( 1) 6 + 6 = 1 + 6 = 7 ; thus, Thus, 6 + ( 1) 7 for all. Theorem 3.3. I a metric space X, let x X ad let E X. { } coverges to x if ad oly if every ε -eighborhood about x (a) The sequece x cotais all but a fiite umber of terms of { x }. (b) If x is a limit poit of E, the there exists a sequece { x } of distict poits i E such that lim x = x. Proof. (a) Suppose first that x x ad let N ε (x ) be give. The there exists a iteger N 1 such that d (x, x ) < ε for all N. That is, x N ε (x ) for all terms except possibly {x 1, x 2,..., x N 1 }. Next, suppose that N ε (x ) cotais all but a fiite umber of terms of { x } for every ε > 0. So for every ε > 0, there exists a fiite set of terms S ε = {x 1, x 2,..., x ε } such that x N ε (x ) for all { 1,..., ε }. Let N = max{ 1,..., ε } + 1. The d (x, x ) < ε for all N ; hece, x x. (b) Because x is a limit poit of E, every ε -eighborhood about x cotais a poit from E that is differet from x. Let ε 1 = 1. The there exists x 1 E, x 1 x, such that d (x 1, x) < ε 1. Let ε 2 = mi ( 1 / 2, d(x 1, x) ). The there exists x 2 E, x 2 x, such that d (x 2, x ) < ε 2. Because d (x 2, x ) < d( x 1, x ), we have x 2 x 1. Cotiuig for 3, we let ε = mi ( 1 /, d(x 1, x) ) ad fid a x E, x x, x {x 1, x 2,..., x 1 }, such that d (x, x ) < ε. The { x } is a sequece of distict poits i E ad we claim that x x. For if ε > 0 is give, we simply choose the smallest iteger N such that 1 / N < ε. The for N, we have d (x, x ) < ε 1 / 1 / N < ε. The ext results show how we ca always fid a sequece of ratioal umbers that coverge to a give irratioal umber. Corollary 3.3. Let x be a irratioal umber. There exists a sequece { x } of ratioal umbers such that x x. Proof. The ratioal umbers are dese o the real lie; thus, every irratioal umber is a limit poit of the Ratioals. The result follows from Theorem 3.3 (b).
Corollary 3.4. Let x = m. a 1 a 2 a 3... be the decimal form of a irratioal umber. Let x = m. a 1 a 2 a 3...a for 1. The x are ratioal ad x x. Proof. Each x is ratioal because it is writte as a termiated decimal. Now let ε > 0 be give. Choose the smallest iteger N such that 1 / 10 N < ε. The for N we have x x = 0. 0... a +1 a +2... 1 / 10 1 / 10 N < ε. Thus, x x. Theorem 3.4. lim y = y. The: The ext results are the commo limit theorems from calculus: Let { x } ad { y } be real-valued sequeces such that lim x = x ad (a) If x = c for all, the lim x = c. (The limit of a costat is that costat.) (b) lim c x = c x, for all c R; (Costats factor out of limits.) (c) lim (x + y ) = x + y ; (The limit of a sum is the sum of the limits.) (d) lim (x y ) = x y ; (The limit of a product is the product of the limits.) (e) lim 1 / x = 1/ x, provided x 0 for all ad x 0. Proof. (a) Let ε > 0 be give. For all, x c = c c = 0 < ε ; thus, x c. (b) If c = 0, the the result follows from (a). Otherwise, let ε > 0 be give. The there exists a iteger N 1 such that x x < ε / c for all N. The for N, we have c x c x = c x x < c ε / c = ε ; thus, c x c x. (c) Let ε > 0 be give. Because x x ad y y, there exist itegers N 1 1 ad N 2 1 such that x x < ε / 2 for all N 1 ad y y < ε / 2 for all N 2. Now let N = max (N 1, N 2 ). The for N, we have by the Triagle Iequality that ( x + y ) (x + y ) = (x x) + (y y ) x x + y y < ε / 2 + ε / 2 = ε. { } coverges, we kow that { x } is bouded. Thus, there exists M R, (d) Because x M > 0 such that x M for all. Because x x, we kow by (b) that y x y x. Now let ε > 0 be give. There exist itegers N 1 1 ad N 2 1 such that y y < ε / (2M ) for all N 1 ad y x y x < ε / 2 for all N 2. Now let N = max (N 1, N 2 ). The for N, we have by the Triagle Iequality that
x y x y = x y x y + x y x y x y x y + x y x y = x y y + y x y x M y y + y x y x < M ε 2M + ε 2 = ε. (e) Because { x } coverges to x 0, there exists a iteger N 1 1 such that x x < x 2 for N 1. The for N 1, we have x = x x + x x x + x < x 2 + x. By subtractig x / 2, we have x 2 < x ad the 1 < 2 x x for N 1. Let ε > 0 be give. There exists a iteger N 2 1 such that x x all N 2. Now let N = max (N 1, N 2 ). The for N, we have < ε x 2 / 2 for 1 1 x x = x x x x = 1 x x x x < 2 x ε x 2 / 2 x = ε. Thus, 1 / x 1/ x. Mootoe Real-Valued Sequeces We ow cosider the special case of real-valued sequeces { x } that are mootoe i the sese that either x 1 x 2 x 3... x... or x 1 x 2 x 3... x.... Defiitio 3.4. Let { x } be a sequece of real umbers. (a) We say that x the we say x (b) We say that x the we say x { } is mootoe icreasig if x x +1 for all. If x < x +1 for all, { } is strictly icreasig. { } is mootoe decreasig if x x +1 for all. If x > x +1 for all, { } is strictly decreasig. Theorem 3.5. Let { x } be a sequece of real umbers such that { x } = K is mootoe icreasig ad bouded above for some K, or mootoe decreasig ad bouded below for some K. The there exists a real umber x such that x x.
Proof. Suppose { x } = K is mootoe icreasig ad is bouded above. The there exists a real umber M such that x M for all K. That is, x K x K +1 x K +2... M. By the Least Upper Boud Property of the Reals, x = sup{ x } = K exists ad x R. We assert that x x. Let ε > 0 be give. The x ε < x, so x ε caot be a upper boud of { x } = K because x is the least upper boud of { x } = K. So there exists a idex N K such that x ε < x N x. The for all N, we have x ε < x N x x ; thus, x x = x x < x ( x ε ) = ε. Hece, x x. We leave the proof for decreasig sequeces that are bouded below as a exercise. Provig that a Sequece is Mootoe from Some Poit We ow describe a formal method for provig that a sequece of real umbers is mootoe. If we wat to show that x K x K +1 x K +2... for some K, the we wat to prove that 1 x +1 for all K. So we simply solve for the smallest idex that x satisfies the iequality. Alterately, we ca solve the iequality x +1 x 0. Likewise, if we wat to show that x K x K +1 x K +2..., the we solve for the smallest idex that satisfies the iequality 1 x +1 x, or also writte as +1 1. x x Alterately, we ca solve the iequality x +1 x 0. Example 3.4. Cosider the sequeces below. Prove that each is strictly icreasig or strictly decreasig from some poit. The determie whether or ot the sequece is bouded. If the sequece is mootoe ad bouded, the fid its limit. (a) +1 { } =1 (b) 4 1 2 =1 (c)! 6 = 0 (d) 3 2 = 0 Solutio. (a) The terms of +1 { } are 2, 3/2, 4/3, 5/4,... which appear to be strictly =1 decreasig from the oset. Claim: x +1 < x for 1. Now let x = + 1, ad we shall show that x +1 x < 1. This ratio becomes x +1 x = + 2 + 1 + 1 = + 2 +1 +1 = 2 + 2 2 + 2 + 1 < 1 for all 1 because the umerator is less tha the deomiator ad both are positive. Thus, x +1 < x for all 1, ad the etire sequece is strictly decreasig.
+1 All the terms of { } are positive; thus, this sequece is clearly bouded =1 below by 0. But i fact, 0 < x x 1 = 2 ; thus, x 2 for all 1. So this sequece is bouded. We ow assert that + 1 1 (which also may be writte as + 1 1 ). Let ε > 0 be give. Let N be the smallest iteger such that N > 1. The for N we have ε +1 1 = 1 1 N < ε. - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - (b) The terms of 4 1 2 are (4 1), (4 1/4), (4 1/9), (4 1/16),... which =1 appear to be strictly icreasig from the oset ad earig 4. Claim: x < x +1 for all 1. It is clear that 3 x < 4 for all 1; thus, the sequece is bouded. Now let x = 4 1 2, ad we shall try to show that x +1 we shall istead show that x +1 > x or that x +1 x > 0. We the have x +1 x = 4 1 ( +1) 2 x > 1 for 1. I this case, 4 1 2 = 1 2 1 ( + 1) 2 > 0 for 1 because 0 < 2 < ( + 1) 2 which makes 1 / 2 > 1 / ( + 1) 2. We ow assert that 4 1 2 4. To prove this, we shall show istead that 1 2 0 ad the apply limit theorems. Let ε > 0 be give. Let N be the smallest iteger such that N > 1 ε. The N2 > 1. The for N we have ε Thus, 1 2 0 ; hece, lim 4 1 2 1 2 0 = 1 2 1 N 2 < ε. = lim 4 + ( 1) lim 1 2 = 4 + ( 1)0 = 4. - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -! (c) We claim that the sequece becomes strictly icreasig after a certai 6 = 0 poit. Let x =!. The cosider the iequality 6 1 < x +1 x = ( + 1)! 6 +1 ( +1)!! = 6 6 +1 6! = + 1 6.
This iequality holds if ad oly if 6 < + 1 if ad oly if 5 <. Thus for 6, we have x < x +1. The first few terms of this sequece are x 0 = 1, x 1 = 1 / 6, x 2 = 2 / 36, x 3 = 6 / 216, x 4 = 24 / 1296, x 5 = 120 / 7776, x 6 = 720 / 46656, x 7 = 5040 / 279936,... Note that x 0 > x 1 > x 2 > x 3 > x 4 > x 5 = x 6, but the x 6 < x 7 < x 8 <....! Next, we assert that 6 is ot bouded. Suppose this sequece were = 0 bouded by M. The ote that 1 < x 14 x for 14. Now adjust the iequality above: M < x +1 = +1, which is satisfied for 6M 1 <. The for max(6m, 14), x 6 because x > 1 we have x +1 > x +1 > M, which cotradicts the fact that M bouds the x sequece. Hece the sequece is ot bouded ad therefore must diverge. - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 3 (d) We claim that the sequece 2 becomes strictly decreasig after a certai = 0 poit. Let x = 3. The cosider the iequality 2 1 > x +1 x = ( + 1) 3 2 +1 3 2 = ( +1)3 2 3 = 3 + 3 2 + 3 + 1 2 3. This iequality holds if ad oly if 2 3 > 3 + 3 2 + 3 + 1 if ad oly if 3 3 2 3 1 > 0. Because the cubic polyomial 3 3 2 3 1 has a positive leadig coefficiet, it grows without boud; thus, there is a iteger N such that 3 3 2 3 1 > 0 for all N. We simply eed to fid the smallest such iteger N. By trial ad error, we see that for 4, we have 3 3 2 3 1 > 0 ad thus x > x +1. 3 We ow kow that 2 is strictly decreasig. All the terms are positive; thus = 4 we have 0 < x x 4 = 4 for all 4. I fact, 0 < x 4 for = 0, 1, 2, 3 also; thus, the 3 origial sequece is bouded. Because 2 is strictly decreasig ad bouded = 4 below by 0, it must have a limit. I fact 3 0. This result follows from L'Hopital s 2 Rule which we shall prove later.
Diverget Real-Valued Sequeces If { x } does ot coverge, the it diverges. Oe type of divergece is a oscillatio for which the sequece remais bouded but varies back ad forth betwee several terms ad ever attais a limit. For example, cosider {( 1) } = 0. The the terms are 1, 1, 1, 1, 1, 1,..., which ever have a limit. Aother example is { si( π / 4) } = 0, which has terms 0, 2 / 2, 1, 2 / 2, 0, 2 / 2, 1, 2 / 2, 0, 2 / 2, 1,... Aother type of divergece occurs whe a sequece is ubouded. (Recall: A coverget sequece must be bouded.) Whe terms ted to get larger ad larger, or ted to get smaller ad smaller, they may be said to have a ifiite limit. Defiitio 3.5. Let { x } be a sequece of real umbers. (a) We say that lim x = + provided for every positive real umber M, there exists a iteger N 1 such that x > M wheever N. (b) We say that lim x = provided for every egative real umber M, there exists a iteger N 1 such that x < M wheever N. Example 3.6. (a) Prove that the followig limits diverge to either + or. (i) {2 } = 0 (ii) { 3 } = 0 (b) Explai why {( 1) 2 } = 0 diverges but does ot have a ifiite limit. Solutio. (a) (i) Let M > 0. We wat to make 2 > M, so we eed > l M / l 2. Let N be the smallest iteger such that N > l M / l 2. The 2 > M for all N. We ote that {2 } is i fact a strictly icreasig sequece; thus, 2. (ii) Let M < 0. We wat to make 3 < M, so we eed > M. Let N be the smallest 3 iteger such that N > M. The 3 < M for all N. We ote that { 3 } is i fact a strictly decreasig sequece; thus, 3. { } =0 = {(2k ) 2 } k =0 (b) The terms of ( 1) 2 are 0, 1, 4, 9, 16, 25,... 3 There is a subsequece {( 1) 2k (2k ) 2 } = {0, 4, 16,... } that is ubouded; thus, this k= 0 subsequece diverges. The origial sequece must also diverge, because if it coverged, the every subsequece would also coverge (Theorem 3.9 (a) to come). Because the terms of the sequece oscillate from positive to egative, there is o way to make x > 1000 for all N o matter the N. Thus, the sequece caot have a limit of +. Likewise, the limit caot be.
Theorem 3.6. If lim x = + or if lim x =, the lim 1 / x = 0. Proof. Assume lim x = +. Let ε > 0 be give. There exist a iteger N 1 such that x > 1 / ε > 0 wheever N. But the ε > 1 / x > 0 or 1 / x 0 < ε for all N ; thus, 1 / x 0. The proof is similar if lim x =. Some Special Real-Valued Sequeces Here are the limits of a few commo real-valued sequeces: (i) For p > 0, lim p = + ad lim 1 / p = 0. (ii) lim! = + ad lim 1 /! = 0 (iii) (a) If 1 < x < 1, the x 0 (b) If x = 1, the x 1 (c) If x > 1, the x + (d) If x 1, the x diverges (oscillates) (iv) For x > 0, lim x = 1. (v) lim = 1 Ratioale (i) For ay M > 0, we kow p > M for > M 1/ p ad ( +1) p > p for all 1; thus p { } is ubouded ad strictly icreasig for 1. Hece, p ad 1 / p 0. - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - (ii) We kow! = ( 1)... 1 > 0 ad ( +1)! = ( + 1)! >! for all 1; thus! { } is ubouded ad strictly icreasig for 1. Hece,! ad 1 /! 0. - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - (iii) (a) Let 1 < x < 1. If x = 0, the x = 0 for all 1; thus, x 0. Now suppose x 0, so that 0 < x < 1 ad l x < 0. The let ε > 0 be give. We wat x 0 = x 0. x = x < ε for large eough. We simply eed > l ε / l x. Thus,
(b) If x = 1, the x = 1 for all 0; thus, x 1. (c) Suppose x > 1. The for ay M > 0, we kow x > M for > l M / l x ad because x > 1, we have x +1 = x x > x for all 1; thus { x } is ubouded ad strictly icreasig for 1. Hece, x (ad 1 / x 0 ). (d) For x 1, the terms of { x } oscillate betwee positive ad egative for eve ad odd expoets of. Moreover, the terms x = x are also ubouded for x < 1 (they { } is oscillatigly diverget. are bouded though whe x = 1). So it is clear that x - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - (iv) Assume x > 0, ad cosider x = x 1/. We take the atural log of this expressio to obtai 1 l x. Because 1 0 by sequece (i) above ad l x is a costat, we have 1 l x 0. Thus, x e 0 = 1. (Note: This argumet uses a result about cotiuous fuctios to be prove later: If l( x ) x, the x e x.) - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - (v) To show lim = 1, we also eed the fact that l 0 (which ca be obtaied from L Hopital s Rule to come later). The l = 1 l 0 ; hece, e 0 = 1. - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - Cauchy Sequeces A special type of sequece is oe i which the terms become closer ad closer together. Our ext mai goal is to show that such sequeces i R will always have a limit. Defiitio 3.5. Let { x } be a sequece i a metric space X with metric d. The { x } is called a Cauchy sequece provided for every ε > 0, there exists a iteger N 1 such that d( x, x m ) < ε wheever, m N. x 1 A Cauchy Sequece x 3 x 2 evetually all close together x N x N +1 x N +2 x N +k
Theorem 3.7. If { x } coverges, the { x } is Cauchy. Proof. Let x x. The there exists a iteger N 1 such that d( x, x) < ε / 2 wheever N. The for all, m N, we have d( x, x m ) d ( x, x) + d( x, x m ) < ε / 2 + ε / 2 = ε. Theorem 3.8. A Cauchy sequece is bouded. { } be Cauchy, ad choose the smallest iteger N 2 such that ( ) < 1 wheever, m N. Let M = max{d( x 1, x N ),...,d ( x N 1, x N ), 1}. The ( ) M, so all poits of { x } are withi M of x N. Thus, { x } is bouded. Proof. Let x d x, x m d x, x N Corollary 3.5. (Completeess of the Reals) Let umbers. There exists x R such that x x. { x } be a Cauchy sequece of real { x } be the set of distict terms i the sequece. Suppose first that E is Proof. Let E = fiite with S = { x 1, x 2,..., x k } beig the fiite set of distict terms labeled with icreasig idices. The there are oly a fiite umber of pairs of distict elemets that ca be chose from S. So let ε = mi{d (x i, x j ) : x i, x j S, x i x j }. Because the ( ) < ε / 2 for all sequece is Cauchy, we ca choose a iteger N > k such that d x, x m, m N. But x N must equal oe of the elemets i S as must x for N. Because d( x, x N ) < ε for N, we must have x = x N for all N. Hece, x x N R. Next, suppose that E = { x } is ifiite. Because { x } is Cauchy, E is bouded. So by the Bolzao-Weierstrass Theorem, there exists x R such that x is a limit poit of E. We claim that x x. Let ε > 0 be give. There exists a iteger N 1 such that x x m < ε / 2 for all, m N. We ow wat to choose a smaller eighborhood aroud x. Let ε 1 > 0 such that ε 1 < ε / 2 ad ε 1 < mi{ x x i : 1 i < N, x x i }. Because x is a limit poit of E, there exists a poit x K from E (differet from x ) such that x x K < ε 1. By the choice of ε 1, we must have K N. The for all N, x x = x x K + x K x x x K + x K x < ε 1 + ε / 2 < ε Hece, x x. Thus, o the real lie R, every Cauchy sequece coverges. Ay metric space that has this property is said to be complete. Thus, R is complete.
Subsequeces Let { x } be a sequece i a metric space. The terms are idexed as x 1, x 2, x 3,.., x... Suppose we wat to choose a partial ifiite list of these, such as x 3, x 6, x 9,..., x 3k... Such a partial list is called a subsequece ad is deoted by { x k }. The terms of the subsequece are x 1, x 2, x 3,..., x k... So 1 is the smallest idex chose from the origial sequece that puts x 1 i the subsequece. So 1 1. The 2 is the ext smallest idex chose from the origial sequece that puts x 2 i the subsequece. Thus 2 2 (ad 2 > 1 ). I geeral, we always must have k k ad ( k +1 > k ). Example 3.7. Let x = 2 for 1. Let x k be the square of the k th prime umber. The x 1 = 1 2 x 2 = 2 2 x 3 = 3 2 x 4 = 4 2 x 5 = 5 2 x 6 = 6 2 x 7 = 7 2 x 8 = 8 2... x 1 = 2 2 x 2 = 3 2 x 3 = 5 2 x 4 = 7 2 So, 1 = 2 1 2 = 3 2 3 = 5 3... k = kth prime k Theorem 3.9. Let { x } be a sequece i a metric space. (a) If lim x every subsequece { x k }, we have lim x k k = x. = x, the for (b) If { x } is Cauchy ad lim x k k = x for some subsequece { x k }, the lim x = x. Proof. (a) Let ε > 0 be give. There exists a iteger N 1 such that d (x, x) < ε for all N. The N th idex N of a subsequece must be at least N ; that is, N N. So for k N, we have k N N ad therefore d (x k, x ) < ε. Hece, lim k x k = x. (b) Let ε > 0 be give. There exists a iteger K 1 1 such that d (x k, x ) < ε / 2 for all k K 1. There also exists a iteger K 2 such that d (x, x m ) < ε / 2 for all, m K 2. Now let N = max (K 1, K 2 ) ad let N. The N N K 2 ad N K 1, thus d (x, x) d (x, x N ) + d(x N, x ) < ε.
Note: Just because a subsequece coverges, it does ot mea that the origial o- Cauchy sequece coverges. For example, let x = 6 + ( 1) for 1: 5, 7, 5, 7, 5, 7,... The let x k = x 2k = 6 + ( 1) 2k for k 1: 7, 7, 7, 7,... The x k 7 but x diverges. { } Theorem 3.10. Let x be a limit poit of a sequece { x } i a metric space. The there exists a subsequece x k { } such that lim k x k = x. Proof. Because x is a limit poit of a { x }, every ε -eighborhood about x cotais ifiitely may poits from { x }. Let ε 1 = 1 / 2 ad let 1 be the first idex such that x 1 N ε1 (x ). Next let ε 2 = 1 / 4 ad choose the first idex 2 > 1 such that x 2 N ε 2 (x ). Cotiuig, let ε k +1 = 1 / 2 k+1 ad choose the first idex k +1 > k such that x k+1 N εk+1 ( x). The d (x, x k ) < ε k = 1 / 2 k for k 1, which implies lim k x k = x. Corollary 3.6. I R, every bouded sequece has a coverget subsequece. Proof. Suppose first that there are oly a fiite umber of distict terms i { x }. The at least oe term must recur a ifiite umber of times i the list x 1, x 2, x 3,... We let x deote the value of this recurrig term, the choose 1 < 2 < 3 <... such that x k = x for all k 1. The lim x k k = x. Next suppose that there are a ifiite umber of terms i { x }. The the sequece creates a ifiite, bouded set of real umbers. By the Bolzao-Weierstrass Theorem, this set has a limit poit x. The result ow follows from Theorem 3.10. Lim sup ad Lim if { } be a sequece of real umbers. The { x } may or may ot coverge, but some Let x subsequeces may coverge. If { x } has some coverget subsequeces, the let E be the set of all limits of all coverget subsequeces of { x } icludig those limits that are ±. The E may or may ot be bouded. If E has o lower boud, the let if E =. If E has o upper boud, the let sup E = +. If bouds exist, the if E ad sup E exist as real umbers by the glb ad lub properties of R. { } be a sequece of real umbers ad let E be the set of all limits { } by Defiitio 3.7. Let x of all coverget subsequeces of { x }. We defie the lim if ad lim sup of x x = lim if x = if E ad x = lim sup x = sup E.
Example 3.8. (a) Let { x } be the set of all ratioal umbers. Every irratioal umber x is a limit poit of the Ratioals, thus there exists a subsequece of the Ratioals that coverges to x. So the set E of all limits of all coverget subsequeces of { x } is ot bouded; thus, x = if E = ad x = sup E = +. (b) Let x = si( π / 4) for 0. The E = 1, 2 2, 0, 2 2, 1. Thus, x x = sup E = 1 = if E = 1 We ext give a result that relates a limit with the lim if ad lim sup: Theorem 3.11. Let { x } be a sequece of real umbers ad let x R. The lim if ad oly if lim if x = x = lim sup x. x = x Proof. Assume lim x = x, ad let E be the set of all limits of all coverget subsequeces of { x }, icludig possibly ±. By Theorem 3.9 (a), E = { x }; thus, lim if x = if E = x = sup E = lim sup x. Next, assume that lim if assert that x a subsequece x = x = lim sup x. We must show that x x. We first { } is bouded. For if { x } were ot bouded above, the there would be { x k } that icreases to + ad the lim sup x = sup E = + R. So { x } must be bouded above, ad likewise must be bouded below. Next, let E be the set of all limits of all coverget subsequeces of { x }. We assert that E = { x }. For if there were two poits x 1, x 2 E with x 1 < x 2, the we would have lim if x = if E x 1 < x 2 sup E = lim sup x. Thus, E = { x }. Now suppose x / x. The there exists a ε > 0 such that for every iteger N 1 there exists a N such that x x ε. For this ε, choose the smallest iteger 1 1 such that x 1 x ε. Havig chose k, let k +1 be the smallest iteger with { } with all k +1 k + 1 such that x k+1 x ε. We the obtai a subsequece x k terms i (, x ε ] [x + ε, ). At least oe of these rays must have a ifiite umber of terms of { x k }. If [ x + ε, ) has a ifiite umber of the terms, the these terms i [ x + ε, ) are a ifiite, bouded set. By Bolzao-Weierstrass, there must be a real-valued limit poit y to this set, ad by Theorem 3.10, there must be a further subsequece of the terms that coverges to y. Thus, y E = {x}. But because all the terms are i [ x + ε, ), we must have x + ε y, which is a cotradictio. Likewise, we obtai a cotradictio if (, x ε ] has a ifiite umber of terms. Hece, we must have x x.
Exercises 1. Let { x } ad { y } be real-valued sequeces such that lim x where x 0 for all ad x 0. Prove that lim y / x = y / x. = x ad lim y = y 2. Let { x } be a sequece of real umbers such that { x } = K is mootoe decreasig ad bouded below for some K. Prove that there exists a x R such that x x. 3. Let { x } be a sequece of real umbers. { } coverges, the { x } coverges. { } coverges, does it ecessarily follow that { x } coverges? Prove or (a) Prove that if x (b) If x disprove. (c) Prove that if { x } coverges to 0, the { x } coverges to 0. 4. Let { x } be a sequece of real umbers that is bouded above by M ad such that x x. Prove that x M. 5. Let { x } ad { y } be real-valued sequeces. (a) Suppose x 0 ad y (b) Suppose x + ad y { } is bouded. Prove that x y 0. { } is bouded. Prove that y / x 0. 6. I a metric space, prove that every subsequece of a Cauchy sequece is Cauchy. 7. O the real lie, suppose x x ad (x y ) 0. Prove that y x. 8. Prove that the followig sequeces are mootoe from some poit. The determie if the sequeces are bouded. If a sequece is bouded, compute its limit. (a) 2 { + 6 } =1 (b)! 6 2 =1 (c) ( 3) 2 2 =1 (d) 7 e 2 =1 9. (a) Suppose x. Prove that {x } is ot bouded. (b) Suppose {x } is ot bouded above. Prove that there exists a subsequece {x k } such that lim k x k = +.
10. Let {a } ad {b } be real-valued sequeces with 0 a b for all. Prove the followig results: (a) If b 0, the a 0. (b) If a, the b. 11. Let {a }, {b }, ad {c } be real-valued sequeces with a b c for all. Suppose a x ad c x. Prove that b x. 12. Use the formal defiitios of limit to prove the followig: (a) lim 9 + 1 = 3 (b) lim 8 = + (c) lim 3/2 = 0 13. Give the reaso for covergece or divergece of each sequece. (a) { ( 5) } = 1 (b) 2 3 =1 (c) si cos {! } = 1 (d) 1 3 2 3 +1 =1 14. Let u = ( x 1,, x 2,,..., x k, ) deote the th term of a sequece i R k. Prove that u u = ( x 1, x 2,..., x k ) i R k if ad oly if x i, x i i R for all i = 1, 2,..., k. 15. Apply Exercise 13 to prove that R k is complete.