Compact Hausdorff Topology is Minimal Hausdorff

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Theorem

Let $T = \left({S, \tau}\right)$ be a Hausdorff space which is compact.


Then $\tau$ is the minimal subset of the power set $\mathcal P \left({S}\right)$ 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.

From Identity Mapping is Continuous:

the identity mapping $I_S: \left({S, \tau}\right) \to \left({S, \tau'}\right)$ is continuous.


Let $A \subseteq S$ be closed in $\left({S, \tau}\right)$.

By Closed Subspace of Compact Space is Compact, if $A$ is compact in $\left({S, \tau}\right)$.

From Continuous Image of Compact Space is Compact:

$I_S \left({A}\right)$ is also compact.

By hypothesis, $\left({S, \tau'}\right)$ is a Hausdorff space.

From Compact Subspace of Hausdorff Space is Closed:

$I_S \left({A}\right)$ is closed in $\left({S, \tau'}\right)$.

Thus $I_S$ is a closed mapping and so:

$\tau \subseteq \tau'$



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.

$\blacksquare$


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