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It is known that a $T_2$ space $X$ is $KC$, i.e. every compact subset of $X$ is closed. The space $[0, \omega_1]$ is $T_5$ but not $T_6$ and the subset $[0, \omega_1)$ is sequentially compact (and thus countably compact) but not closed. Is there an example of a $T_6$ space with a sequentially compact (or countably compact) subspace which is not closed?

Remark: Spaces $X$ in which countably compact subsets are closed were investigated by Ismail and Nyikos. They called such space C-closed. Examples for C-closed spaces are $T_2$ sequential spaces or even $T_2$ spaces in which every countably compact subspace is sequential (e.g. $C([0,1])$ with pointwise convergence). From above it follows that $[0, \omega_1]$ is not a C-space (it is moreover not sequential).

The Arens-Fort space is $T_6$ and not sequential and I hope that one can show that it is also not C-closed.

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    $\begingroup$ The Arens-Fort space, $X = \omega \times \omega$, is C-closed. If $A \subseteq X$ is not closed, then $A$ is infinite and the special point $(0,0)$ cannot belong to $A$. But then $A$ is a discrete subspace of $X$, and since it is infinite it cannot be countably compact. $\endgroup$ – the thesaurus saurus Feb 7 '16 at 9:59
  • $\begingroup$ In the paper by Ismail we have the proposition that if points of $X$ are $G_\delta$ and $X$ is regular, then $X$ is $C$-closed. Proof: if $A$ is countably compact and not closed, pick $x \in \overline{A}\setminus A$. Then $B = A \cup \{x\}$ is also countably compact and $B$ is first countable from points being $G_\delta$. But then a sequence $x_n$ from $A$ coverging to $x$ has no accumulation point in $A$. Contradiction. So we need a space where not all points are $G_\delta$. So countable spaces are out. $\endgroup$ – Henno Brandsma Feb 7 '16 at 11:56
  • $\begingroup$ @Henno: In a $T_6$ space all closed sets are $G_\delta$-sets, so that proposition shows that every $T_6$ space is $C$-closed. $\endgroup$ – Brian M. Scott Feb 7 '16 at 13:05
  • $\begingroup$ @BrianM.Scott True. I'll make the remark an answer. $\endgroup$ – Henno Brandsma Feb 7 '16 at 13:07
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In the original paper by Ismail we have the proposition that if $X$ is regular and every point is a $G_\delta$, then $X$ is $C$-closed.

Proof: if $A$ is countably compact and $A$ is not closed (striving for a contradiction) then pick $x \in \overline{A}\setminus A$, and consider $B = A \cup \{x\}$. Then $B$ is countably compact and also has the property that all points are $G_\delta$, and by standard facts on countably compact spaces, $B$ is firts countable. So there exists a sequence $(x_n)$ from $A$ that converges to $x$, but then $\{x_n: n \in \omega\}$ is a subset of $A$ without an accumulation point, contradicting countable compactness. So $A$ must be closed and $X$ is $C$-closed.

In particular (as Brian M. Scott rightly remarked) this implies that all $T_6$ spaces (which are regular, $T_1$ and have all closed subsets $G_\delta$, so singleton sets in particular) are always $C$-closed.

So no space as asked for can exist.

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  • $\begingroup$ Ah right, I completely forgot the characterization of $T_6$ by $T_4$ + all closed sets are $G_\delta$ (sometimes called "perfect"). $\endgroup$ – yadaddy Feb 7 '16 at 14:03
  • $\begingroup$ @yadaddy what else can the definition be? $\endgroup$ – Henno Brandsma Feb 7 '16 at 14:11
  • $\begingroup$ Separation of closed sets by continuous functions in an exact way in contrast to the $T_5$ separation. But then the preimage of $0$ is of course $G_\delta$ by perfectness of $\mathbb{R}$. $\endgroup$ – yadaddy Feb 7 '16 at 14:25
  • $\begingroup$ I'm used to that being a theorem (using Tietze and series). The intrinsic view is nicer I think (no need for the reals). But it's indeed equivalent. $\endgroup$ – Henno Brandsma Feb 7 '16 at 15:34

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