If $X$ is complete and totally bounded, then $X$ is compact

Question

Let $X$ be a metric space. Whar is your favorite way to show:

If $X$ is complete and totally bounded, then $X$ is compact?

Thanks for your help.

Answer 1

planetmath.org : Let $X$ is complete and totally bounded. For metric spaces, compact and sequentially compact are equivalent; we prove that $X$ is sequentially compact. Choose a sequence $p_n\in X$; we will find a Cauchy subsequence (and hence a convergent subsequence, since $X$ is complete).

Cover $X$ by finitely many balls of radius $1$ (since $X$ is totally bounded). At least one of those balls must contain an infinite number of the $p_i$. Call that ball $B_1$, and let $S_1$ be the set of integers $i$ for which $p_i\in B_1$.

Proceeding inductively, it is clear that we can define, for each positive integer $k>1$, a ball $B_k$ of radius $1/k$ containing an infinite number of the $p_i$ for which $i\in S_{k-1}$; define $S_k$ to be the set of such $i$.

Each of the $S_k$ is infinite, so we can choose a sequence $n_k\in S_k$ with $n_k < n_{k+1}$ for all $k$. Since the $S_k$ are nested, we have that whenever $i,j\geq k$, then $n_i,n_j\in S_k$. Thus for all $i,j\geq k$, $p_{n_i}$ and $p_{n_j}$ are both contained in a ball of radius $1/k$. Hence the sequence $p_{n_k}$ is Cauchy.