Part A, for both Section 200 and Section 501

Transcription

1 Istructios Please write your solutios o your ow paper. These problems should be treated as essay questios. A problem that says give a example or determie requires a supportig explaatio. I all problems, you should explai your reasoig i complete seteces. Studets i Sectio 501 should aswer questios 1 6 i Parts A ad B. Studets i Sectio 200 (the hoors sectio) should aswer questios 1 3 i Part A ad questios 7 9 i Part C. Part A, for both Sectio 200 ad Sectio Prove by iductio that.1 C x/ 1 C x whe 1 < x < 0 ad 2 N. Solutio. Whe D 1, equality holds i the idicated statemet: amely,.1 C x/ 1 D 1 C 1x. Thus the basis step of the iductio argumet is valid. Next suppose it has bee established that.1 C x/ 1 C x for a certai atural umber. Multiplyig both sides by the positive quatity 1 C x (positive by the hypothesis that 1 < x) leads to aother true iequality: amely,.1 C x/ C1.1 C x/.1 C x/ D 1 C. C 1/x C x 2 : Now x 2 is a positive quatity, beig the product of the two positive umbers ad x 2. Therefore 1C.C1/x Cx 2 > 1C.C1/x. By trasitivity of iequality, it follows that.1 C x/ C1 > 1 C. C 1/x. Thus validity of the idicated statemet for a certai atural umber implies validity of the statemet for the ext atural umber. Sice the statemet holds too whe D 1, the priciple of mathematical iductio implies that the statemet holds for every atural umber. Remarks The argumet proves more tha what is stated. Namely, strict iequality holds for every atural umber larger tha 1. Moreover, the argumet did ot use the hypothesis that x < 0, so the iequality actually holds whe 1 < x (or eve whe 1 x). The iequality is kow as Beroulli s iequality. February 22, 2011 Page 1 of 5 Dr. Boas

2 2. (a) State the defiitio of what lim!1 x D L meas. Solutio. For every positive real umber ", there exists a atural umber N such that jx Lj < " whe N. (b) Use the defiitio to prove that lim!1 2 C 1 D 0. Solutio. Suppose " is a arbitrary positive real umber. Take N to be d1="e (a value that I determied by a side calculatio). Here is the verificatio that this value of N satisfies the defiitio. (Notice that L D 0 i this problem.) If N, the 2 C 1 < 2 D 1 1 N 1 1=" D ": (a smaller deomiator meas a larger fractio) (a smaller deomiator meas a larger fractio) Thus ˇ 0 2 C 1 ˇ D < " whe N. Accordigly, the 2 C 1 value of the limit has bee verified by the defiitio of limit. 3. Give a cocrete example of a o-empty set of real umbers that has o accumulatio poit. Solutio. Every eighborhood of a accumulatio poit has to cotai ifiitely may elemets of the set. Therefore a fiite set has o accumulatio poit. The simplest cocrete example of a fiite set is a sigleto set, such as f5g. A example of a ifiite set with o accumulatio poit is N, the set of atural umbers. This set has o accumulatio poit because a eighborhood of width 1=3 cotais at most oe atural umber. O the other had, the Bolzao Weierstrass theorem implies that every bouded ifiite set does have a accumulatio poit. February 22, 2011 Page 2 of 5 Dr. Boas

3 Part B, for Sectio 501 oly 4. Determie the supremum of the set f x 2 R W jx 2 3j < 6 g. Solutio. The idicated iequality meas that 6 < x 2 3 < 6, or, equivaletly, 3 < x 2 < 9. The left-had part of the iequality is vacuous, sice x 2 is ever egative. Therefore the iequality reduces to the statemet that x 2 < 9, or 3 < x < 3. The supremum (least upper boud) of the set of such umbers is equal to 3; the ifimum equals Suppose that a sequece fx g is defied recursively as follows: x 1 D 1; ad x C1 D 1 1 C x whe 1. Does this sequece have a coverget subsequece? Explai why or why ot. Solutio. Evidetly all the terms of the sequece are positive umbers. (Namely, the first term is positive, ad if x is positive, the so is 1Cx ad so is the reciprocal 1=.1 C x /; by iductio, all the terms of the sequece are positive.) Therefore 1 C x > 1, ad the reciprocal, which equals x C1, is less tha 1. I other words, this sequece is bouded: all the terms lie betwee 0 ad 1. By the Bolzao Weierstrass theorem, there is a coverget subsequece. Remark Actually, this sequece coverges, so every subsequece is coverget (ad to the same limit). But to prove covergece of the whole sequece is a bit tricky. The subsequece fx 2 g is icreasig, while the subsequece fx 2C1 g is decreasig, ad it turs out that these two mootoic sequeces coverge to the same limit. The problem is related to the sequece of Fiboacci umbers 1, 1, 2, 3, 5, 8, 13,..., each term beig the sum of the precedig two terms. The sequece i the problem starts 1, 1=2, 2=3, 3=5, 5=8, 8=13, 13=21,.... These fractios are ratios of cosecutive Fiboacci umbers. The limit of these ratios turs out to be. 1 C p 5 /=2. February 22, 2011 Page 3 of 5 Dr. Boas

4 6. Give a cocrete example of a bouded, o-empty set of real umbers that has o iterior poits ad is ot compact. Solutio. A compact set is both closed ad bouded. To produce a bouded ocompact set, the, you eed to look at sets that fail to be closed, that is, sets that fail to cotai some accumulatio poit. The simplest coceivable example is a coverget sequece that does ot cotai its limit. A cocrete example is the sequece f1=g cosistig of reciprocals of atural umbers. The set is bouded sice each umber 1= is betwee 0 ad 1; each poit of the set is isolated; ad the limit 0 is ot i the set. Part C, for Sectio 200 oly 7. (a) Whe E is a coutable set of real umbers, is the complemetary set R E always ucoutable? (b) Whe E is a ucoutable set of real umbers, is the complemetary set R E always coutable? Explai why or why ot. Solutio. (a) The uio of two coutable sets is coutable. Cosequetly, if E is coutable, the R E must be ucoutable, for otherwise the uio of these two sets, which is all of R, would be coutable. But R is ucoutable. (b) O the other had, if E D. 1; 0/, the both E ad R E are ucoutable. 8. Does the limit superior respect absolute value? I other words, is ˇˇlim sup x ˇˇ!1 always equal to lim sup jx j whe fx g is a sequece of real umbers? Prove!1 or give a couterexample, whichever is appropriate. February 22, 2011 Page 4 of 5 Dr. Boas

5 Solutio. Couterexample: If ( x D 1; whe is eve; 2; whe is odd; the lim sup!1 x D 1, so j lim sup!1 x j D 1. O the other had, lim sup!1 jx j D Cosider the followig property that a set E of real umbers might or might ot have: Wheever a family of closed sets covers the set E, some fiite subcollectio of those closed sets covers E. (The Heie Borel property is the correspodig statemet for coverigs by ope sets.) Is this ew coverig property equivalet to compactess? Explai. Solutio. The property is ot equivalet to compactess. Ideed, the closed iterval Œ0; 1 is compact but lacks the idicated property. Cosider the collectio of closed itervals f Œ1=; 1 W 2 N g together with Œ 1; 0. This family of closed sets covers Œ0; 1, but o fiite subfamily covers Œ0; 1. Remark sets! The oly sets that have the idicated coverig property are fiite Ideed, a ubouded ifiite set is covered by the family of closed sets f Œ ; W 2 N g, ad o fiite subfamily covers a ubouded set. The Bolzao Weierstrass theorem implies that a bouded ifiite set has a accumulatio poit, call it c. The family of closed sets cosistig of f x 2 R W jx cj 1= g for each atural umber together with the sigleto set fcg forms a cover of R, yet o fiite subfamily covers a set that has c as a accumulatio poit. February 22, 2011 Page 5 of 5 Dr. Boas