Answer to Question #201323 in Functional Analysis for smi

Question

Q. Prove that a compact subset of a metric space is closed and bounded.

Answer 1

Let K ⊂ X be compact. If (x_n) is a convergent sequence in K with limit x ∈ X, then every subsequence of (x_n) converges to x. Since K is compact, some subsequence of (x_n) converges to a limit in K, so x ∈ K and K is closed.

Suppose that K is not bounded, and let x₁ ∈ K. Then for every r > 0 there exists x ∈ K such that d(x₁, x) ≥ r. Choose a sequence (x_n) in K as follows. Pick x₂ ∈ K such that d(x₁, x₂) ≥ 1. Given {x₁, x₂, . . . , x_n}, pick x_n+1 ∈ K such that

$d(x_1, x_{n+1}) ≥ 1 + \displaystyle{\max_ {1≤k≤n}} d(x_1, x_k)$

By the triangle inequality:

d(x_k, x_n+1) ≥ d(x₁, x_n+1) − d(x₁, x_k) ≥ 1 for every 1 ≤ k ≤ n.

It follows that d(x_m, x_n) ≥ 1 for every $m \ne n$ , so (x_n) has no Cauchy subsequences, and therefore no convergent subsequences, so K is not compact.