topology on a compact hausdorff not comparable? The original problem:
Show that if $X$ is compact hausdorff under both $\tau$ and $\tau'$, then either $\tau = \tau'$ or they are not comparable?
Most answer on the internet base on this logic:
Assuming they are comparable, we can prove they must be equal. Then end their proof. (which I think is not complete)
My question is how to show there are two topology that are not comparable on $X$ and both make $X$ compact hausdorff?
Can anyone give some examples ?

A very simple example is to let $X=\Bbb N$ and define the following topologies. (Note that for me $0\in\Bbb N$.) Let
$$\tau_1=\wp(\Bbb Z^+)\cup\{U\subseteq\Bbb N:0\in U\text{ and }\Bbb N\setminus U\text{ is finite}\}\;;$$
the only non-isolated point is $0$, and every nbhd of it contains all but finitely many points of $\Bbb N$, so $\langle\Bbb N,\tau_1\rangle$ is compact and Hausdorff. Now let $E=\{2n:n\in\Bbb N\}$, the set of even non-negative integers, and let $O=\Bbb N\setminus E$, the set of odd positive integers. Let $\tau_2$ be the topology generated by the following base:
$$\begin{align*}
\big\{\{n\}:n\in\Bbb N\setminus\{0,1\}\big\}&\cup\{U\subseteq E:0\in E\text{ and }E\setminus U\text{ is finite}\}\\
&\cup\{U\subseteq O:1\in E\text{ and }O\setminus U\text{ is finite}\}\;.
\end{align*}$$
Then $\{0\}\cup E$ and $\{1\}\cup O$ are clopen subspaces of $\langle\Bbb N,\tau_2\rangle$ that are homeomorphic to $\langle\Bbb N,\tau_1\rangle$, so $\langle\Bbb N,\tau_2\rangle$ is also compact Hausdorff. However, $\langle\Bbb N,\tau_2\rangle$ has two non-isolated points, while $\langle\Bbb N,\tau_1\rangle$ has only one, so the two spaces are not homeomorphic.

As for the original theorem, assuming that two compact Hausdorff topologies on a set are comparable and proving from that assumption that they must be equal is a complete proof that any two compact Hausdorff topologies on a set are either incomparable or equal; that’s elementary logic.
