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 ∈ be such that < M. N  Then, for any m, n ≥ M,

2 2 m − 1 n − 1 |xm − xn| = − 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 {xn} be a sequence such that there exists a 0 < C < 1 such that

|xn+1 − xn| ≤ C|xn − xn−1|.

Prove that {xn} 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

|x3 − x2| ≤ C|x2 − x1| 2 |x4 − x3| ≤ C|x3 − x2| ≤ C · C|x2 − x1| = C |x2 − x1| and in general, one could prove that

2 n−1 |xn+1 − xn| ≤ C|xn − xn−1| ≤ C |xn−1 − xn−2| ≤ · · · ≤ C |x2 − x1|. Week 3 Solutions Page 2

Now, for m > n, we can evaluate the quantity

|xm − xn| ≤ |xm − xm−1| + |xm−1 − xm−2| + ··· + |xn+1 − xn| m−2 m−3 n−1 ≤ C |x2 − x1| + C |x2 − x1| + ··· + C |x2 − x1| m−2 n−1 = (C + ··· + C )|x2 − x1| n−1 m−n−1 = C (1 + C + ··· + C )|x2 − x1| 1 − Cm−n  = Cn−1 |x − x | 1 − C 2 1  1  ≤ Cn−1 |x − x | 1 − C 2 1

Now, since 0 < C < 1, Cn−1 → 0 as n → ∞. Therefore, there exists N ∈ N such that whenever n ≥ N,  |Cn−1 − 0| < .  1  1−C |x2 − x1|

For this same N, whenever m > n ≥ N

 1  |x − x | ≤ Cn−1 |x − x | m n 1 − C 2 1   1  < · |x2 − x1|  1  1 − C 1−C |x2 − x1| = .

Therefore, {xn} is a Cauchy sequence.

Exercise. Prove the following statement using Bolzano-Weierstrass theorem.

Assume that (xn)n∈N is a bounded sequence in R and that there exists x ∈ R such that any convergent subsequence (xni )i∈N converges to x. Then limn→∞ xn = x.

Proof. Assume for contradiction that xn 6→ x. Then ∃ > 0 such that

|xn −x| ≥  for infinitely many n. From this, we can create a subsequence {xnj } such that |xnj − x| ≥  for all j ∈ N. Since our original sequence is bounded, this subsequence is bounded, and so, by Bolzano-Weierstrass, there is a convergent subsequence of this subsequence, {x }. By assumption, {x } converges to x. However, this is a contradiction njk njk since |x − x| ≥  for all k ∈ . njk N

Hence, we must have xn → x as n → ∞. Week 3 Solutions Page 3

Exercise. Show that a) the set Z = {..., −1, 0, 1,...} has no cluster points. b) every point in R is a cluster point of Q.

Proof. a) If x ∈ Z, then (x − 1/2, x + 1/2) ∩ Z\{x} = ∅ and so x is not a cluster point of Z. If x 6∈ Z, then ∃k ∈ Z such that x ∈ (k, k + 1). Choose  = min{|x − k|, |x − (k + 1)|}. Then (x − , x + ) ∩ Z\{x} = ∅, and so x is not a cluster point of Z. Therefore Z has no cluster points. b) Let x ∈ R. Let  > 0. By the density of Q, ∃r ∈ Q such that x < r < x + . Then (x − , x + ) ∩ Q\{x}= 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)n∈N ⊂ R has a limit in R, i.e. there exists x ∈ with lim xn = x. (1) R n→∞

The same statement is false in Q, the following is false:

any Cauchy sequence (xn)n∈N ⊂ Q has a limit in Q, i.e. there exists x ∈ with lim xn = x. (2) Q n→∞ 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)√ Define a sequence√ {xn} as follows. For all n ∈ N, choose xn ∈ Q such that 2 − 1/n < xn < 2. Then {xn} is a Cauchy√ 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 ∈ Q . . . or . . .

lim sup xn and lim inf xn may not be members of Q. Week 3 Solutions Page 4