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This is from a discussion me and my friend was having regarding compact implies closed.
If $X$ is hausdorff then every compact subset of $X$ is closed. But Hausdorffness is more than sufficient since there are examples which are not hausdorff but still the conclusion holds (e.g. $\mathbb{R}$ with cocountable topology.
Now we want to know does the conclusion hold in weakly Hausdorff spaces. That is if a space is weakly Hausdorff then is this true that every compact subset is closed?
Note: A space $X$ is weakly Hausdorff if for any compact Hausdorff space $Y$ and a continuous map $f : Y \to X$ we have $f(Y)$ is closed in $X$

Amrit Awasthi
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    A search on $\pi$-base yields one counterexample: https://topology.pi-base.org/spaces/S000031 – David Gao Apr 17 '24 at 16:28
  • See also: https://math.stackexchange.com/questions/632320/weak-hausdorff-space-not-kc and https://math.stackexchange.com/questions/465521/if-a-is-compact-and-b-is-lindel%C3%B6f-space-will-be-a-cup-b-lindel%C3%B6f/466564#466564 – David Gao Apr 17 '24 at 16:39

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As noted by David Gao, https://topology.pi-base.org/properties/P000100 is tracked by pi-Base and you can find several necessary and sufficient conditions there.

I have two recent preprints exploring these $T_1$-not-$T_2$ properties, one more expository, and one more technical.

The only property weaker than $T_2$ I'm aware of from the literature that properly implies KC is $k_1$-Hausdorff: all compact subspaces are Hausdorff. https://topology.pi-base.org/properties/P000170

While not implied by $T_2$, here's an observation: $T_1$ with anticompact implies $k_1$-Hausdorff. https://topology.pi-base.org/spaces?q=Anticompact%2B%24T_1%24%2B%7E%24k_1%24-Hausdorff But the only $k_1$H space known to pi-Base today is anticompact.

Steven Clontz
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