Let \((X,d)\) be a metric space and let \(E\subseteq X\), with the metric induced by \(d\). The following are equivalent [@Fol13]:
\(E\) is complete and totally bounded: for every \(\varepsilon>0\), \(E\) is covered by finitely many balls of radius \(\varepsilon\).
\(E\) is sequentially compact: every sequence in \(E\) has a subsequence that converges to a point of \(E\).
\(E\) is compact: every cover of \(E\) by open subsets of \(X\) has a finite subcover.
If \(X\) is complete, then \(E\) is complete if and only if \(E\) is closed in \(X\). In \(\RR^n\) with the Euclidean metric, \(E\) is totally bounded if and only if \(E\) is bounded; combined with the theorem, this gives the Heine–Borel theorem. In a general metric space, a closed bounded subset need not be compact: in an infinite set with the discrete metric \(d(x,y)=1\) for \(x\neq y\), the whole space is closed and bounded but not totally bounded.