Compact Hausdorff Topology is Minimal Hausdorff

Theorem
Let $T = \struct {S, \tau}$ be a Hausdorff space which is compact.

Then $\tau$ is the minimal subset of the power set $\powerset S$ such that $T$ is a Hausdorff space.

Proof
there exists a topology $\tau'$ on $S$ such that:
 * $\tau' \subseteq \tau$ but $\tau' \ne \tau$
 * $\tau'$ is a Hausdorff space.

From Equivalence of Definitions of Finer Topology:
 * the identity mapping $I_S: \struct {S, \tau} \to \struct {S, \tau'}$ is continuous.

Let $A \in \tau$.

Then $S \setminus A \subseteq S$ is closed in $\struct {S, \tau}$.

By Closed Subspace of Compact Space is Compact, $S \setminus A$ is compact in $\struct {S, \tau}$.

From Continuous Image of Compact Space is Compact:
 * $I_S \sqbrk {S \setminus A}$ is also compact.

By hypothesis, $\struct {S, \tau'}$ is a Hausdorff space.

From Compact Subspace of Hausdorff Space is Closed:
 * $I_S \sqbrk {S \setminus A}$ is closed in $\struct {S, \tau'}$.

Hence:
 * $A = S \setminus I_S \sqbrk {S \setminus A} \in \tau'$

By definition of subset:
 * $\tau \subseteq \tau'$

Thus we have that $\tau \subseteq \tau'$ and $\tau' \subseteq \tau$.

Hence by definition of set equality:
 * $\tau' = \tau$

But this contradicts our hypothesis that $\tau' \ne \tau$.

By Proof by Contradiction, it follows that no topology which is strictly coarser than $\tau$ can be Hausdorff.