Heine-Borel theorem

One implication is general: every compact set in a metric space is closed and bounded. It is closed because a metric space is Hausdorff, and bounded because the open balls \(B(x_0,n)\) cover it and finitely many suffice. The hard implication is the converse, which uses the completeness of \(\mathbb{R}\).

In \(\mathbb{R}\), the outline is: a closed bounded set fits in some \([a,b]\); this is proved compact by bisection (if it had no finite subcover, one half would not either, and the nested intervals converge to a point covered by an open set of the cover, a contradiction); and a closed subset of a compact set is compact. For \(\mathbb{R}^n\) one uses that a finite product of compact sets is compact.

The theorem is what makes compactness manageable in analysis: \(\{x:g(x)\le c\}\) with \(g\) continuous is closed, and if it is also bounded it is compact; this guarantees maxima and minima. Examples: the sphere \(S^{n-1}\) and closed balls are compact, while \(\mathbb{R}^n\) and open balls are not.

Outside \(\mathbb{R}^n\) the equivalence fails. The interval \((0,1)\) with the usual distance is closed and bounded in itself but not compact, and an infinite set with the discrete distance (\(d=1\) between distinct points) is bounded and closed without being compact. In metric spaces the right characterization is: complete and totally bounded.