A topological space X is said to be Hausdorff if, for any two distinct x,y in X, there are disjoint open sets U,V containing x,y respectively. (It follows automatically that single-point sets {x} are closed: their complements are the unions of the open sets V obtained by varying y over all of X except x.)

Note that this differs from the definition I gave in class, which is incorrect. It is still true that a metric space is automatically Hausdorff (why?), but there are non-metrizable Hausdorff spaces.

It is now easy to show that a compact subset of a Hausdorff space is closed. Let K be a compact subset of the Hausdorff space X. We must show that for each p not in X there exists an open set containing p and disjoint from K. For each x in K, let Ux and Vx be disjoint neighborhoods of x,p. The Ux constitute an open cover of K. Extract a finite subcover, and consider the intersection V of the corresponding Vx. As a finite intersection of open sets, V is open. Since the union of the Ux in the subcover contains K, and is disjoint from V, we conclude that V is disjoint from K. Since V clearly contains p we are done.