# 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

Aiming for a contradiction, suppose there exists a topology $\tau'$ on $S$ such that:

$\tau' \subseteq \tau$ but $\tau' \ne \tau$
$\tau'$ is a Hausdorff space.
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 $\struct {S, \tau}$.

$I_S \sqbrk {S \setminus A}$ is also compact.

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

$I_S \sqbrk {S \setminus A}$ 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.

$\blacksquare$