Compact sets closed in Hausdorff spaces without choice?

Question

An elementary proof that compact sets are closed in Hausdorff spaces involves making arbitrary choices based on the Hausdorff property. Is there a way to avoid invoking choice?

Answer 1

Yes, there is.

Let $\langle X,\tau\rangle$ be a Hausdorff space, and let $K$ be a compact subset of $X$. Suppose that $K$ is not closed. Then we can pick $p\in(\operatorname{cl}K)\setminus K$. (Note that this does not require $\mathsf{AC}$: it’s a single choice.) Let

$$\mathscr{U}=\{U\in\tau:p\notin\operatorname{cl}U\}\;.$$

• $K\subseteq\bigcup\mathscr{U}$.

If not, pick $x\in K\setminus\bigcup\mathscr{U}$. $X$ is Hausdorff, so there are disjoint $U,V\in\tau$ such that $x\in U$ and $p\in V$. But then $U\in\mathscr{U}$, contradicting the choice of $x$. Note that I picked only three things here, $x$, $U$, and $V$; this does not require $\mathsf{AC}$.

$K$ is compact, so there is a finite $\{U_1,\ldots,U_n\}\subseteq\mathscr{U}$ such that $K\subseteq\bigcup_{k=1}^nU_k$. Then

$$p\notin\bigcup_{k=1}^n\operatorname{cl}U_k=\operatorname{cl}\bigcup_{k=1}^nU_k\supseteq\operatorname{cl}K\;,$$

contradicting the choice of $p$. Therefore $K$ is closed.