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It's a very standard topological proof that in Hausdorff spaces compact sets are closed. But is Hausdorff condition necessary, is there an example of topological space $X$ so that all compact sets are closed but space is not Hausdorff? If yes, is there a weaker condition which is necessary for all compact sets to be closed?

PatrickR
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Amey Joshi
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    I don't understand the last sentence: do you want a necessary condition for the assertion "$X$ is a topological space such that every compact subspace is closed", or do you want a more general sufficient condition than being Hausdorff? –  Sep 14 '19 at 06:09
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    Compact does not imply closed in the line with two origins! Take $[-1,1]$ with only one of the origins, and it is compact but not closed. – Eric Wofsey Sep 14 '19 at 06:11
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    By "stronger" do you mean "weaker"? Any condition stronger than Hausdorff will work. – bof Sep 14 '19 at 06:18
  • Spaces for which all compact sets are closed are called kc (compact closed) spaces. It is hardly a useful distinction. – William Elliot Sep 14 '19 at 07:45
  • @WilliamElliot can you give example of kc space which is not hausdorff? – Amey Joshi Sep 14 '19 at 08:23

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In the classical paper between $T_1$ and $T_2$ Wilansky introduces the notion of a KC space:

$X$ is called a KC space if every compact subset of $X$ is closed in $X$.

And as you note in the question, A Hausdorff ($T_2$) space is KC. And a KC space is clearly $T_1$ as all finite subsets are always compact and thus closed in a KC space.

The notion of US (unique sequential limits) is also introduced:

$X$ is US iff for all sequences $(x_n)$ from $X$, if $x_n \to x$ and $x_n \to x'$ then $x=x'$.

And again $T_2 \implies US \implies T_1$ (for the latter consider constant sequences).

A KC space that is not Hausdorff is $\alpha\Bbb Q$, the Alexandroff extension (one-point compactification) of the rationals.

When such spaces are compact they turn out to be the maximally compact spaces, see this answer, e.g. They're of some interest (look in Google scholar etc. for papers on them).

Henno Brandsma
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