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We know that all closed subsets of a compact topological space $(X,\tau)$ are compact. If we add the Hausdorff condition on the topology $\tau$ we can see the equivalence of these conditions on a subset (i.e., in a compact Hausdorff space, closed subsets are the same as compact subsets).

For asking the converse of the above fact we could attempt to omit the compactness of the space $(X,\tau)$ as follows:

  • (STATEMENT) If all compact subsets of a topological space $(X,\tau)$ are closed then $(X,\tau)$ is Hausdorff. Is this true?

  • If the above statement is not valid, is there a separation axiom weaker than Hausdorffness on the space $X$, related to compact subsets being closed?

For the first statement, if we add the condition of compactness of $(X,\tau)$, it changes as follows:

  • Is the space $(X,\tau)$ Hausdorff, if closed subsets and compact subsets are equivalent in $X$?
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  • $\begingroup$ It may be valuable to look up the notion of weak hausdorff which appears frequently in homotopy theory. $\endgroup$
    – David White
    Commented Jun 1, 2012 at 20:06
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    $\begingroup$ And, here's an example that STATEMENT is false: mathoverflow.net/questions/88420/… $\endgroup$
    – David White
    Commented Jun 1, 2012 at 20:08
  • $\begingroup$ Thank you very much dear White. But You only show to me that The first statement is false. But as you have seen, the next question asked that for which category of spaces, compact subsets are closed? I didn't see anything about Asking for it.(best wishes) $\endgroup$
    – Ali Reza
    Commented Jun 1, 2012 at 20:20
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    $\begingroup$ Some people consider Weak Hausdorff to be a separation axiom (weaker than Hausdorff, of course). See for instance: mathoverflow.net/questions/86812/separation-axioms. That post claims that Weak Hausdorff implies T1, so if you are only interested in T0-T6 then I guess T1 should do the job. Also, nLab discusses in some detail how Weak Hausdorff implies T0: ncatlab.org/nlab/show/compactly+generated+topological+space $\endgroup$
    – David White
    Commented Jun 1, 2012 at 20:40
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    $\begingroup$ Spaces where compact sets are closed are called KC-spaces. Here is a monthly article which discusses the topologies between $T_1$ and $T_2$. jstor.org/stable/2316017 $\endgroup$
    – Gjergji Zaimi
    Commented Jun 1, 2012 at 21:00

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The Statement in the question is false. A counter-example can be found in this MO answer.

As Gjergji points out, spaces where compact subsets are closed are called KC-spaces. Hausdorff implies KC, but not conversely (this answers the OP's third question).

As for a separation axiom weaker than Hausdorff that is implied by the KC property, one such notion is that of a weak Hausdorff space. That link also defines the notion of a $k$-Hausdorff space and mentions that Hausdorff implies $k$-Hausdorff implies KC.

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  • $\begingroup$ I just wanted to turn my comments into an answer, so this wouldn't come back to the front-page. I should have done that in the first place, way back when this was first asked. $\endgroup$
    – David White
    Commented Jan 12, 2013 at 18:57
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    $\begingroup$ @MW I edited to clarify. $\endgroup$
    – David White
    Commented Feb 21 at 14:04
  • $\begingroup$ Note there are at least two notions of k-Hausdorff in the literature, one stronger than weak Hausdorff, and one weaker. topology.pi-base.org/properties/P000170 topology.pi-base.org/properties/P000171 $\endgroup$
    – Steven Clontz
    Commented Feb 24 at 13:41

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