Compact Hausdorff Topology is Minimal Hausdorff

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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.


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 $I_S: \struct {S, \tau} \to \struct {S, \tau'}$ is continuous.

Let $A \subseteq S$ be closed in $\struct {S, \tau}$.

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

From Continuous Image of Compact Space is Compact:

$I_S \sqbrk A$ is also compact.

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

From Compact Subspace of Hausdorff Space is Closed:

$I_S \sqbrk A$ is closed in $\struct {S, \tau'}$.

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.