AB does not imply KC

Question

We say that a space $X$ is:

1)AB provided that $X$ is $T_1$ and for each pair $A,B$ of compact disjoint subsets of $X$ there is $U$ an open subset of $X$ such that either $A\subseteq U$ and $U\cap B=\emptyset$ or $B\subseteq U$ and $U\cap A=\emptyset$.

2)KC provided that every compact subset of X is closed.

I am looking for a space that is AB but not KC.

Thanks

Answer 1

Let $Y=\{0\}\cup\{2^{-n}:n\in\Bbb N\}$, topologized as a subset of $\Bbb R$ with the usual topology. Let $p$ be a non-principal ultrafilter on $\Bbb N$, and let $X=\{p\}\cup Y$; $Y$ is an open subset of $X$, and if $p\in A\subseteq X$, then $A$ is a nbhd of $p$ (not necessarily open) iff $\{n\in\Bbb N:2^{-n}\in U\}\in p$.

• $X$ is not $KC$, because $Y$ is a compact subset of $X$ that isn’t closed.
• $X$ is $AB$, because $K\subseteq X$ is compact iff $K$ is finite or $0\in K$. Thus, if $H$ and $K$ are disjoint compact subsets of $X$, at least one of them is finite, and its complement is the desired open set.