Compact Hausdorff Topology is Maximally Compact

From ProofWiki
Jump to navigation Jump to search

Theorem

Let $T = \left({S, \tau}\right)$ 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 $I_S: \left({S, \tau'}\right) \to \left({S, \tau}\right)$.

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 $\left({S, \tau'}\right)$ is compact.

Then $I_S$ is also open, 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 finer than $\tau$ can be compact.

$\blacksquare$


Sources