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If $X$ is a compact Hausdorff space with respect to the topologies $T$ and $T'$, then either $T=T'$ or they are incomparable.

Here is my proof. Suppose they are comparable, and without loss of generality take $T \subset T'$. However, by way of contradiction, suppose that $T$ is strictly contained in $T'$. This means there is a closed set in $T'$ that isn't closed in $T$. Now, since $C$ is closed in $T'$, it must also be compact; but since $C$ is not closed in $T$, and $X$ is Hausdorff, $C$ cannot be compact $T$. This means there exists an open cover $\mathcal{A}$ in $T$ for $C$ but which no finite subcollection of $\cal{A}$ that covers $C$. But by hypothesis $\cal{A}$ is also be an open cover in $T'$. This is a contradiction.

Does this seem right? Is there a way of making this a direct proof?

EDIT: I believe the following a more direct proof, but in the same spirit as the above, and also different than the one mentioned by Nate Eldredge. WLOG, suppose that $T \subseteq T'$. Let $C$ be a closed set in $T'$. Since $X$ is a compact Hausdorff space with respect to $T'$, $C$ must also be compact. But compactness is preserved when moving to coarser topologies, which is to say that $C$ is also compact with respect to $T$. Again, since $X$ is compact Hausdorff with respect to $T$, this compact set $C$ must also be closed. Hence, we have $T' \subseteq T$, completing the proof.

user193319
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1 Answers1

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The same logic is expressed more briefly in the following fact:

A continuous bijection from a compact space to a $T_2$ space is always a homeomorphism

So if $T \subset T'$, consider $(X,T)$ and $(X,T')$ as two topological spaces. If $i : X \to X$ is the identity map, then $i : (X,T') \to (X,T)$ is a continuous bijection...

Nate Eldredge
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