Hausdorff implies KC

Statement
There are three equivalent formulations:


 * 1) Any compact subset of a Hausdorff space is closed.
 * 2) The property of being a Hausdorff space is stronger than the property of being a KC-space
 * 3) The property of being a compact space is stronger than the property of being a H-closed space

Related facts

 * Hausdorff implies US
 * KC implies US
 * KC not implies Hausdorff
 * US not implies KC

Proof
Suppose $$X$$ is a Hausdorff space and $$A$$ is a closed subset of $$X$$ that is compact in the subspace topology. We need to show that $$A$$ is closed in $$X$$. In order to do this, it suffices to show that the complement of $$A$$ in $$X$$ is an open subset. In fact, it suffices to show that for any $$x \in A \setminus X$$, there is an open subset $$U$$ of $$X$$ containing $$x$$, and disjoint from $$A$$.

Let's do this:


 * For every $$y \in A$$, use the Hausdorffness of $$X$$ to find disjoint open subsets $$U_y \ni x$$ and $$V_y \ni y$$
 * Now the $$V_y$$s, as $$y \in A$$, cover $$A$$, hence $$V_y \cap A$$ form an open cover of $$A$$ in the subspace topology. By compactness of $$A$$, we can find a finite collection $$y_1, y_2, \ldots, y_n \in A$$ such that the union of $$V_{y_1}, V_{y_2}, \ldots, V_{y_n}$$ contains $$A$$
 * Now consider the intersection:

$$U = \bigcap_{i=1}^n U_{y_i}$$

This is an open set containing $$x$$, because it is a finite intersection of open sets, and it is disjoint from each of the $$V_{y_i}$$s, hence it is disjoint from $$A$$. This completes the proof.

Note that the crucial thing is to pass to a finite subcover of $$A$$, so that we need to intersect only finitely many open subsets containing $$x$$, because the axioms for open subsets only guarantee that finite intersections of open subsets are open.

Textbook references

 * , Page 165 (Theorem 26.3) Also see Lemma 26.4, that provides a slightly more general formulation useful in other proofs, for instance, compact Hausdorff implies normal
 * , Page 27