Week 3 Solutions Page 1 n2−1 Exercise (2.4.1). Prove that n2 is Cauchy using directly the definition of Cauchy sequences. r2 Proof. Given > 0, let M 2 be such that < M. N Then, for any m; n ≥ M, 2 2 m − 1 n − 1 jxm − xnj = − m2 n2 1 1 = − n2 m2 1 1 ≤ + n2 m2 1 1 ≤ + M 2 M 2 2 = M 2 < . n2−1 Therefore, n2 is a Cauchy sequence. Exercise (2.4.2). Let fxng be a sequence such that there exists a 0 < C < 1 such that jxn+1 − xnj ≤ Cjxn − xn−1j: Prove that fxng is Cauchy. Hint: You can freely use the formula (for C 6= 1) 1 − Cn+1 1 + C + C2 + ··· + Cn = : 1 − C Proof. Let > 0 be given. Note that jx3 − x2j ≤ Cjx2 − x1j 2 jx4 − x3j ≤ Cjx3 − x2j ≤ C · Cjx2 − x1j = C jx2 − x1j and in general, one could prove that 2 n−1 jxn+1 − xnj ≤ Cjxn − xn−1j ≤ C jxn−1 − xn−2j ≤ · · · ≤ C jx2 − x1j: Week 3 Solutions Page 2 Now, for m > n, we can evaluate the quantity jxm − xnj ≤ jxm − xm−1j + jxm−1 − xm−2j + ··· + jxn+1 − xnj m−2 m−3 n−1 ≤ C jx2 − x1j + C jx2 − x1j + ··· + C jx2 − x1j m−2 n−1 = (C + ··· + C )jx2 − x1j n−1 m−n−1 = C (1 + C + ··· + C )jx2 − x1j 1 − Cm−n = Cn−1 jx − x j 1 − C 2 1 1 ≤ Cn−1 jx − x j 1 − C 2 1 Now, since 0 < C < 1, Cn−1 ! 0 as n ! 1. Therefore, there exists N 2 N such that whenever n ≥ N, jCn−1 − 0j < : 1 1−C jx2 − x1j For this same N, whenever m > n ≥ N 1 jx − x j ≤ Cn−1 jx − x j m n 1 − C 2 1 1 < · jx2 − x1j 1 1 − C 1−C jx2 − x1j = . Therefore, fxng is a Cauchy sequence. Exercise. Prove the following statement using Bolzano-Weierstrass theorem. Assume that (xn)n2N is a bounded sequence in R and that there exists x 2 R such that any convergent subsequence (xni )i2N converges to x. Then limn!1 xn = x. Proof. Assume for contradiction that xn 6! x. Then 9 > 0 such that jxn −xj ≥ for infinitely many n. From this, we can create a subsequence fxnj g such that jxnj − xj ≥ for all j 2 N. Since our original sequence is bounded, this subsequence is bounded, and so, by Bolzano-Weierstrass, there is a convergent subsequence of this subsequence, fx g. By assumption, fx g converges to x. However, this is a contradiction njk njk since jx − xj ≥ for all k 2 . njk N Hence, we must have xn ! x as n ! 1. Week 3 Solutions Page 3 Exercise. Show that a) the set Z = f:::; −1; 0; 1;:::g has no cluster points. b) every point in R is a cluster point of Q. Proof. a) If x 2 Z, then (x − 1=2; x + 1=2) \ Znfxg = ; and so x is not a cluster point of Z. If x 62 Z, then 9k 2 Z such that x 2 (k; k + 1). Choose = minfjx − kj; jx − (k + 1)jg. Then (x − , x + ) \ Znfxg = ;, and so x is not a cluster point of Z. Therefore Z has no cluster points. b) Let x 2 R. Let > 0. By the density of Q, 9r 2 Q such that x < r < x + . Then (x − , x + ) \ Qnfxg 6= ;. Since was arbitrary, this shows that x is a cluster point of Q. Since x was arbitrary, every point in R is a cluster point of Q. Exercise. In the lecture we have shown any Cauchy sequence (xn)n2N ⊂ R has a limit in R, i.e. there exists x 2 with lim xn = x. (1) R n!1 The same statement is false in Q, the following is false: any Cauchy sequence (xn)n2N ⊂ Q has a limit in Q, i.e. there exists x 2 with lim xn = x. (2) Q n!1 a) Give a counterexample to (2). b) Which part of the proof of (1) (from the lecture) fails when we attempt to prove (2)? Proof. a)p Define a sequencep fxng as follows. For all n 2 N, choose xn 2 Q such that 2 − 1=n < xn < 2. Then fxng is a Cauchyp sequence in Q that does not have a limit in Q (since the limit \should be" 2). b) Depending on which proof you are looking at, you can either point out that the Bolzano-Weierstrass Theorem does not guarantee a subsequence that converges to some x 2 Q . or . lim sup xn and lim inf xn may not be members of Q. Week 3 Solutions Page 4.