Metric Space Compact iff Complete and Totally Bounded

Lemma
A metric space is compact if and only if it is complete and totally bounded.

Proof
First suppose that $(S,d)$ is a sequentially compact metric space.

We first show that any Cauchy sequence $(a_k)$ in $(S,d)$ has a limit.

Indeed, by (sequential) compactness the sequence has a subsequence converging to some point $a\in S$; but since the sequence is Cauchy, this implies that the entire sequence has limit $a$. So $(S,d)$ is complete.

To show that a compact space is totally bounded, let $\varepsilon>0$.

Then the family of open $\varepsilon$-balls $\{N_{\varepsilon}(x):x\in S\}$ forms an open cover of $S$.

By compactness, there exists a finite subcover.

That is, there are points $x_0,\dots,x_n$ such that:
 * $\displaystyle S = \bigcup_{0\leq i\leq n} N_{\varepsilon}(x),$

as required.

Now assume that $(S,d)$ is complete and totally bounded; we will show that $(S,d)$ is sequentially compact.

This will prove that $(S,d)$ is in fact compact, as a sequentially compact metric space is also compact.

We first claim that total boundedness implies the following:


 * If $(a_k)$ is any sequence in $S$ and $\varepsilon>0$, then there is some $x\in S$ such that $d(a_k,x)\leq \varepsilon$ for infinitely many $k$.

Indeed, let $x_0,\dots,x_n$ be as in the definition of total boundedness.

Then for every $k$, there is some $j_k\in\{0,\dots,n\}$ such that $d(a_k,x_{j_k})\leq\varepsilon$.

For some $j$, we must have $j_k=j$ for infinitely many $k$, and the claim follows by setting $x:=x_{j_k}$.

Now let $(a_k)$ be an arbitrary sequence in $S$.

By the claim, there is some $x_1\in S$ such that $d(a_k,x_1)\leq 1/2$ for infinitely many $k$.

Now we can apply the claim to the subsequence of $(a_k)$ consisting of those elements for which $d(a_k,x_1)\leq 1/2$, to find $x_2\in S$ such that infinitely many $k$ satisfy both $d(a_k,x_2)\leq 1/4$ and $d(a_k,x_2)\leq 1/2$.

Now we proceed inductively, to obtain a sequence $(x_m)$ with the property that there exist infinitely many $k$ such that, for $1\leq j\leq m$:
 * $(1) \qquad d(a_k,x_j)\leq 2^{-j}$

Now define a subsequence $(a_{k_m})$ inductively by letting $k_0$ be arbitrary, and choosing $k_{m+1}$ minimal such that $k_{m+1}>k_m$ and such that $(1)$ holds for $k=k_m$ and all $1\leq j\leq m$.

We claim that this subsequence is a Cauchy sequence.

Indeed, let $\varepsilon>0$, and choose $n$ sufficiently large that $1/2^{n-1}<\varepsilon$.

Then:
 * $d(a_{k_j},a_{k_{j'}}) \leq d(a_{k_j},x_n) + d(a_{k_{j'}},x_n)) \leq 2\cdot 2^{-n} < \varepsilon$

whenever $j,j'\geq n$.

Because $(S,d)$ is complete by assumption, we see that $(a_k)$ has a convergent subsequence, as required.