Compact iff complete and totally bounded

Compact iff complete and totally bounded: Let $(X,d)$ be a metric space . Then the following are equivalent:

1. $X$ is a compact set .
2. $X$ is a complete metric space and totally bounded .

Equivalently, a subset $K\subseteq X$ is compact in the subspace topology if and only if $(K,d|_{K\times K})$ is complete and totally bounded.

This characterization packages compactness implies completeness and compactness implies total boundedness into a single criterion that is often easier to verify than the open cover definition.