10.1. METRIC SPACES 247

Proof: =⇒ Let H be closed and let p be a limit point. We need to verify that p ∈ H. Ifit is not, then since H is closed, its complement is open and so there exists δ > 0 such thatB(p,δ )∩H = /0. However, this prevents p from being a limit point.⇐= Next suppose H has all of its limit points. Why is HC open? If p ∈ HC then it is

not a limit point and so there exists δ > 0 such that B(p,δ ) has no points of H. In otherwords, HC is open. Hence H is closed. ■

Corollary 10.1.10 A set H is closed if and only if whenever {hn} is a sequence of pointsof H which converges to a point x, it follows that x ∈ H.

Proof: =⇒ Suppose H is closed and hn→ x. If x ∈ H there is nothing left to show. Ifx /∈ H, then from the definition of limit, it is a limit point of H. Hence x ∈ H after all.⇐= Suppose the limit condition holds, why is H closed? Let x ∈ H ′ the set of limit

points of H. By Theorem 10.1.7 there exists a sequence of points of H, {hn} such thathn → x. Then by assumption, x ∈ H. Thus H contains all of its limit points and so it isclosed by Theorem 10.1.9. ■

Next is the important concept of a subsequence.

Definition 10.1.11 Let {xn}∞

n=1 be a sequence. Then if n1 < n2 < · · · is a strictly increasingsequence of indices, we say

{xnk

}∞

k=1 is a subsequence of {xn}∞

n=1.

The really important thing about subsequences is that they preserve convergence.

Theorem 10.1.12 Let{

xnk

}be a subsequence of a convergent sequence {xn} where xn→

x. Then limk→∞ xnk = x also.

Proof: Let ε > 0 be given. Then there exists N such that d (xn,x) < ε if n ≥ N. Itfollows that if k ≥ N, then nk ≥ N and so d

(xnk ,x

)< ε if k ≥ N. This is what it means to

say limk→∞ xnk = x. ■Another useful idea is the distance to a set.

Definition 10.1.13 Let (X ,d) be a metric space and let S be a nonempty set in X. Then

dist(x,S)≡ inf{d (x,y) : y ∈ S} .

The following lemma is the fundamental result.

Lemma 10.1.14 The function, x→ dist(x,S) is continuous and in fact satisfies

|dist(x,S)−dist(y,S)| ≤ d (x,y) .

Proof: Suppose dist(x,S) is as least as large as dist(y,S). Then pick z ∈ S such thatd (y,z)≤ dist(y,S)+ ε. Then

|dist(x,S)−dist(y,S)|= dist(x,S)−dist(y,S)≤ d (x,z)− (d (y,z)− ε)

= d (x,z)−d (y,z)+ ε ≤ d (x,y)+d (y,z)−d (y,z)+ ε = d (x,y)+ ε.

Since ε > 0 is arbitrary, this proves the lemma. It is similar if dist(x,S) ≤ dist(y,S). Justswitch the roles of x and y. ■