Compactness implies completeness

Compactness implies completeness: Let $(X,d)$ be a metric space and let $K\subseteq X$ be compact (with the induced metric). Then $K$ is a complete metric space : every Cauchy sequence in $K$ is a convergent sequence with limit in $K$.

Together with compactness implies total boundedness , this yields the metric equivalence compact iff complete and totally bounded .