Show that a totally bounded complete metric space X is compact.

I can use the fact that sequentially compact ⇔ compact.

Attempt: Complete ⟹ every Cauchy sequence converges. Totally bounded ⟹ ∀ϵ>0, X can be covered by a finite number of balls of radius ϵ. I’m trying to show that all sequences in X have a subsequence that converges to an element in X. I don’t see how to go from convergent Cauchy sequences and totally bounded to subsequence convergent in X.

**Answer**

You need to show that if X is totally bounded, every sequence in X has a Cauchy subsequence. Let σ=⟨xn:n∈N⟩ be a sequence in X. For each n∈N let Dn be a finite subset of X such that the open balls of radius 2−n centred at the points of Dn cover X. D0 is finite, so there is a point y0∈D0 such that infinitely many terms of σ are in B(y0,1). Let A0={n∈N:xn∈B(y0,1)}, so that A0 is infinite. Now D1 is finite, so there is a y1∈D1 such that A1={n∈A0:xn∈B(y1,2−1)} is infinite. Repeat: if Ak is an infinite subset of N, there must be a yk+1∈Dk+1 such that Ak+1={n∈Ak:xn∈B(yk+1,2−(k+1))} is infinite, and the process can continue.

Now choose a strictly increasing sequence \langle n_k:k\in\mathbb{N}\rangle of natural numbers in such a way that n_k\in A_k for every k\in\mathbb{N}. Can you show that \langle x_{n_k}:k\in\mathbb{N}\rangle is Cauchy?

**Attribution***Source : Link , Question Author : Emir , Answer Author : Brian M. Scott*