Compact Hausdorff Topology is Maximally Compact

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Theorem

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


Then $\tau$ is maximally compact.


Proof

Let $\tau'$ be a topology on $S$ such that $\tau \subseteq \tau'$ but that $\tau \ne \tau'$.

Consider the identity mapping $I_S: \struct {S, \tau'} \to \struct {S, \tau}$.

From Separation Properties Preserved in Subspace, $I_S$ is a continuous bijection from a Hausdorff space to a compact Hausdorff space.

Aiming for a contradiction, suppose $\struct {S, \tau'}$ is compact.

Then $I_S$ is also open, and so $\tau' \subseteq \tau$.


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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 finer than $\tau$ can be compact.

$\blacksquare$


Sources

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