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.
From Equivalence of Definitions of Finer Topology:
- 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 $\struct {S, \tau}$.
From Continuous Image of Compact Space is Compact:
- $I_S \sqbrk {S \setminus A}$ is also compact.
By hypothesis, $\struct {S, \tau'}$ is a Hausdorff space.
From Compact Subspace of Hausdorff Space is Closed:
- $I_S \sqbrk {S \setminus A}$ 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.
$\blacksquare$
Sources
- 1978: Lynn Arthur Steen and J. Arthur Seebach, Jr.: Counterexamples in Topology (2nd ed.) ... (previous) ... (next): Part $\text I$: Basic Definitions: Section $3$: Compactness: Compactness Properties and the $T_i$ Axioms