Compact iff complete and totally bounded

Let $(X,d)$ be a metric space and let $K\subseteq X$ with the subspace metric.

Theorem: The following are equivalent:

• $K$ is compact .
• $K$ is complete and totally bounded .

This theorem is the fundamental metric characterization of compactness: “no missing limits” (completeness) plus “finite $\varepsilon$-approximation at every scale” (total boundedness).

Proof sketch: ($\Rightarrow$) If $K$ is compact, then $K$ is complete (every Cauchy sequence has a convergent subsequence , and hence the full sequence converges) and totally bounded (the balls $\{$B$ (x,\varepsilon):x\in K\}$ form an open cover ; take a finite subcover).

($\Leftarrow$) If $K$ is totally bounded, then every sequence in $K$ has a Cauchy subsequence . If $K$ is complete, that Cauchy subsequence converges to a point in $K$. Hence $K$ is sequentially compact . In metric spaces, sequential compactness is equivalent to compactness.