Let $(X,d)$ be a [metric space](/page/Metric%20Space), let $(\widehat X,\widehat d)$ be a metric completion of $X$, and let $i:X\to\widehat X$ be the completion embedding, so that $i$ is an isometry, $\widehat X$ is complete, and $i(X)$ is dense in $\widehat X$. For $A\subset X$, define
to be the closure of $i(A)$ in $\widehat X$. Then $\overline{i(A)}^{\,\widehat X}$ is compact in $\widehat X$ if and only if $A$ is totally bounded in $X$, meaning that for every $\varepsilon>0$ there exist finitely many points $x_1,\dots,x_n\in X$ such that
Consequently, if a subset of a metric space is called precompact precisely when its image has compact closure in the completion, then $A$ is precompact if and only if $A$ is totally bounded.