Homework Solutions for MTH 471

2.15 Let X be a metric space and let 〈xn : n ∈ ω〉 be a Cauchy sequence in X. Prove that if 〈xn : n ∈ ω〉 has aconvergent subsequence then 〈xn : n ∈ ω〉 converges.

Solution. Let 〈xnk: k ∈ ω〉 be a subsequence of 〈xn : n ∈〉 which converges to some x ∈ X. Then 〈nk : k ∈ ω〉

is a strictly increasing sequence in ω so k ≤ nk for all k ∈ ω. Let ≥> 0. There is N ∈ ω such thatd(xm, xn) <≥ /2 for all m,n ≥ N . There is K ∈ ω such that d(xnk

, x) <≥ /2 for all k ≥ K. Letk ≥ max{N, K}. If n ≥ N then d(xn, x) ≤ d(xn, xnk

) + d(xnk, x) <≥ /2+ ≥ /2 =≥. Thus xn → x.

2.16 Let X be a metric space and D ⊆ X. Prove that D is dense in X if and only if U ∩D 6= ∅ for every nonemptyopen subset U of X.

Solution. Assume that D is dense in X and let U be a nonempty open subset of X. Let a ∈ D. Then a ∈ Dbecause D is closed in X. Therefore U ∩D 6= ∅ by Theorem 2.25.

Now assume that if U is a nonempty open subset of X then U ∩ D 6= ∅. Let a ∈ X. If U is an opensubset of X and a ∈ U then U ∩ D 6= ∅. Therefore a ∈ D by Theorem 2.25. Thus D = X and D is densein X.

2.17 Let X be a metric space and U an open subset of X. Prove that if U is dense in X then X \ U is nowheredense in X.

Solution. Let V be a nonempty open subset of X. Then V ∩ U 6= ∅ because U is dense in X. ThereforeV * X \ U so Int(X \ U = ∅ and X \ U is nowhere dense in X.

2.18 Prove that Q is not the intersection of a countable number of open dense subsets of R.

Solution. Let Un be an open dense subset of R for every n ∈ ω. Assume that Q =⋂

n∈ω Un. For every r ∈ Qlet Vr = R \ {r}. Then W = {Un : n ∈ ω} ∪ {Vr : r ∈ Q} is a countable collection of open dense subsets of R.We can write W as {Wn : n ∈ ω}. Then

⋂n∈ω Wn = ∅ because for every irrational number x there is n ∈ ω

such that x /∈ Wn and for every rational number x there is n ∈ ω such that x /∈ Wn. This contradicts theBaire Category Theorem, so Q cannot equal

⋂n∈ω Un.

2.19 Prove that Z under the usual metric is a complete metric space, but it is the union of a countable number ofsingletons. Why doesn’t this contradict the Baire Category Theorem?

Solution. The usual metric on Z is given by d(m,n) = |m − n|. Note that if m,n ∈ Z with m 6= n thend(m, n) ≥ 1. Let 〈xn : n ∈ ω〉 be a Cauchy sequence in Z. We will show that there is N ∈ ω such thatxm = xn for all m,n ≥ N . Let N ∈ ω such that xm, xn < 1/2 for all m,n ≥ N . Then xm = xn for allm,n ≥ N . Therefore 〈xn : n ∈ ω〉 converges to xN . Since every Cauchy sequence in X converges, Z is acomplete metric space. But Z =

⋃n∈Z{n}. This does not contradict the Baire Category Theorem because

{n} = B(n, 1/2), so {n} is open and not nowhere dense for every n ∈ Z.

3.1 Give Z its usual metric.

(a) Describe the topology of Z.Solution. As we have seen in problem 1.19 every singleton of Z is open. Therefore every subset of Z isopen.

(b) Let X be an arbitrary metric space. Which functions f : Z→ X are continuous?Solution. Every function is continuous because if U is an open subset of X then f−1[U ] is automaticallyan open subset of Z.

3.2 Let D be a set with the discrete metric and let X be a metric space. Which functions f : D → X arecontinuous?

Solution. Since D has the discrete metric, every element of D is open. If x ∈ D then {x} = BD(x, 1/2).Therefore every subset of D is open and every function f : D → X is continuous.

3.3 Let X, Y , and Z be metric spaces. Prove that if f : X → Y and g : Y → Z are both continuous then g ◦ f iscontinuous.

Solution. You can do this using sequences or using the inverse image of open sets. We will do it first usingsequences. Let a ∈ X and let 〈an〉 be a sequence in X which converges to a. Then 〈f(an)〉 is a sequence in Yand f(an) → f(a) because f is continuous at a. Also, 〈g(f(an))〉 is a sequence in Z and g(f(an)) → g(f(a))because g is continuous at f(a). Therefore g ◦ f is continuous.

Now we will prove it using the inverse images of open sets. Let V be an open subset of Z. Since g iscontinuous, g−1[V ] is an open subset of Y . Since f is continuous, f−1[g−1[V ]] is an open subset of X. But(g ◦ f)−1[V ] = f−1[g−1[V ]], so (g ◦ f)−1[V ] is an open subset of X. Therefore g ◦ f is continuous.

3.4 Let X and Y be metric spaces and let f be a continuous function from X onto Y . Prove that if D is a densesubset of X then f [D] is a dense subset of Y . What happens if we do not require that f is onto?

Solution. Let b ∈ Y . Since f is onto there is a ∈ X such that f(a) = b. Let V be an open subset of Ywith b ∈ V . Then f−1[V ] is a nonempty open subset of X and f−1[V ] ∩ D 6= ∅. If c ∈ f−1[V ] ∩ D thenf(c) ∈ V ∩ f [D]. Therefore V ∩ f [D] 6= ∅. If follows that b ∈ f [D] and that f [D] is dense in Y . You can alsoget this result using sequences.

If f is not onto, then f [D] need not be continuous. For example, f(x) = x2 is a continuous function from Rto R and R is certainly dense in itself, but f [R] = [0,∞) so f [R] is not dense in R.

3.10 Let X and Y be metric spaces and let f be a continuous function from X onto Y . Prove that if every sequencein X has a convergent subsequence then every sequence in Y has a convergent subsequence.

Solution. Let 〈bn〉 be a sequence in Y . For every n ∈ ω let an ∈ X such that f(an) = bn. Then 〈an〉 musthave a convergent subsequence 〈ank

〉. Let a ∈ X such that a = limk→∞ ank= a. Now 〈bnk

〉 = 〈f(ank)〉 is

a subsequence of 〈bn〉 and f(a) = f(limk→∞ ank) = limk→∞ f(ank

) = limk→∞ bnk, so 〈bnk

〉 is a convergingsubsequence of 〈bn〉.