Is it possible to construct Hausdorff compact topology on every set?

Question

I'd like to know if it's possible to construct Hausdorff compact topology on every set. Assume the axiom of choice if needed. Thanks for ideas.

Answer 1

Yes. Let $X$ be any non-empty set. Perhaps the easiest way is to fix a point $p\in X$, let $Y=X\setminus\{p\}$, and set

$$\tau=\wp(Y)\cup\{X\setminus F:F\subseteq Y\text{ and }F\text{ is finite}\}\;.$$

Then $\tau$ is a compact Hausdorff topology on $X$. It’s easy to check that it’s Hausdorff. If $\mathscr{U}$ is any open cover of $X$, there is a $U_0\in\mathscr{U}$ containing $p$, and $X\setminus U_0$ is finite and therefore covered by just finitely many more members of $\mathscr{U}$.