Let $X$ be a Tychonoff space with no isolated points such that the boundary of any subset of $X$ is compact. Does it mean that $X$ is compact ? (If $X$ is a resolvable space then it is clearly compact.)
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Your property implies in particular:
"Every nowhere dense closed subset of $X$ is compact."
(A nowhere dense closed subset is the boundary of its complement.)
This in turn is equivalent to compactness for $T_1$ spaces with no isolated points, as shown by Katetov in 1947. Interestingly, the result holds as well for Lindelöf spaces, or more generally for $[\lambda,\kappa]$-compact spaces (a space is $[\lambda,\kappa]$-compact if every cover of it by $\le\kappa$ open sets has a subcover of cardinality $<\lambda$). This is due to Mills and Wattel, a very nice short proof can be found in Blair's "Some nowhere densely generated topological properties" (easily found online).
Completely unaware of these results, I actually recently published a paper whose main result is weaker than Mills and Wattel's, and with a more complicated proof than Blair's.
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$\begingroup$ Can you give the full citations of the papers you mention here? $\endgroup$ Commented Sep 28, 2016 at 6:42
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$\begingroup$ You can find all the references in Blair's paper, which can be found here, for instance: eudml.org/doc/17268?lang=cs&limit=10 $\endgroup$ Commented Sep 28, 2016 at 6:56
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$\begingroup$ +1 Thank you. Any informatin for general Tychonoff spaces? I mean not necessary spaces without isolated points. $\endgroup$– Lisa_KCommented Sep 28, 2016 at 8:43
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1$\begingroup$ If $X$ is a $T_1$ space each of whose nowhere dense subsets is compact, then either $X$ is compact, or there is a clopen infinite discrete subset in $X$. Let me sketch the proof (following Blair). Take a cover $C$ of $X$. Take a maximal disjoint family $F$ of open sets such that each one is contained in a member of $C$. Take one point $x_U$ in each $U$ in $F$. Then $X-\cup F$ is closed and nowhere dense, cover it with finitely members of $C$. The $x_U$ not covered form a closed discrete set. (cont.) $\endgroup$ Commented Sep 28, 2016 at 9:03
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1$\begingroup$ If there are only finitely many of them, take the members of the cover that contain the corresponding $U$s, this gives a finite cover of $X$. If there are infinitely many, either there is a infinite subset such that $\{x_U\}$ is open, so there is a clopen discrete subset, or infinitely many points are non-isolated, so they form an infinite discrete closed nowhere dense set, which is thus not compact. $\endgroup$ Commented Sep 28, 2016 at 9:07