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Every KC, i.e. "Kompacts are Closed", (and thus every $T_2$) space has the property I'll call IKK: the Intersection of any family of Kompact subsets is itself Kompact.

Not all IKK spaces are $T_1$: consider the right-ray topology on $\omega$, where the non-empty open sets are exactly $\{n,n+1,\dots\}$. In such a $T_0$-not-$T_1$ space, every subset is compact (covering the least element covers the subset), so the space is IKK.

But are all $T_1$ IKK spaces $KC$, or some weaker strengthening of $T_1$? See e.g. https://math.stackexchange.com/a/4761212/ for a few examples of properties between $T_1$ and $T_2$.

Steven Clontz
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  • The finite complement topology is $T_1$ and IKK (since every subset is compact) but not KC, right? – M W Nov 27 '23 at 02:40
  • Nice observation: every anticompact (compacts are finite) space is IKK. Interestingly, all $T_1$ anticompact spaces are $k_1$-Hausdroff. – Steven Clontz Nov 27 '23 at 02:43
  • Oh I misread (was thinking countable complement, which is anticompact). The finite complement topology on an infinite set is not even US: every non-trivial sequence converges to every point. So that's the answer, if you want to write it up. – Steven Clontz Nov 27 '23 at 02:48
  • Is there a good example of a $T_1$ space that fails IKK? I know its easy without $T_1$, just take $\mathbb R$ with Alexandrov topology, subsets are compact if and only if they have a least element, so consider ${0}\cup (1,\infty)$ and $[1,\infty)$. But I couldn't immediately think of a $T_1$ counterexample. – M W Nov 27 '23 at 03:06
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    I made one here (the inspiration for this question): https://math.stackexchange.com/a/4815048/86887 – Steven Clontz Nov 27 '23 at 03:30
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    Oh I guess the doubled line per this answer also works https://math.stackexchange.com/a/1276727/1210477 – M W Nov 27 '23 at 03:31

1 Answers1

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Just summarizing from the comments. The finite complement topology on an infinite set is $T_1$, and $IKK$ since indeed, every subset is compact. However, for the same reason, it is not $KC$, since the only nontrivial closed sets are finite.

Moreover, as also remarked in the comments, the finite complement topology on an infinite set fails to satisfy even the weakest property at the linked post, namely the property $US$ (every convergent sequence has exactly one limit), since every sequence that has no duplicate points must converge to every point in the space.

Remark

Conversely, the property $US$ does not imply $IKK$. For a counter-example, consider $[0,\omega_1]$ with a doubled endpoint $\omega_1'$, then $[0,\omega_1]$ and $[0,\omega_1)\cup \{\omega_1'\}$ are compact, but their intersection $[0,\omega_1)$ is not.

M W
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