Totally bounded, complete $\implies$ compact

Question

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

I can use the fact that sequentially compact $\Leftrightarrow$ compact.

Attempt: Complete $\implies$ every Cauchy sequence converges. Totally bounded $\implies$ $\forall\epsilon>0$, $X$ can be covered by a finite number of balls of radius $\epsilon$. 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 1

You need to show that if $X$ is totally bounded, every sequence in $X$ has a Cauchy subsequence. Let $\sigma=\langle x_n:n\in\mathbb{N}\rangle$ be a sequence in $X$. For each $n\in\mathbb{N}$ let $D_n$ be a finite subset of $X$ such that the open balls of radius $2^{-n}$ centred at the points of $D_n$ cover $X$. $D_0$ is finite, so there is a point $y_0\in D_0$ such that infinitely many terms of $\sigma$ are in $B(y_0,1)$. Let $$A_0=\{n\in\mathbb{N}:x_n\in B(y_0,1)\}\;,$$ so that $A_0$ is infinite. Now $D_1$ is finite, so there is a $y_1\in D_1$ such that $$A_1=\{n\in A_0:x_n\in B(y_1,2^{-1})\}$$ is infinite. Repeat: if $A_k$ is an infinite subset of $\mathbb{N}$, there must be a $y_{k+1}\in D_{k+1}$ such that $$A_{k+1}=\{n\in A_k:x_n\in B(y_{k+1},2^{-(k+1)})\}$$ 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?