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I am struggling with this problem. I couldn't obtain anything from an arbitrary open covering of the space. I couldn't think of any counterexample. I tried googling it, but there were no results. The problem is:

Is a topological space $X$ compact if every nontrivial closed subset of $X$ is compact?


marked as duplicate by Jason DeVito, Moishe Kohan, MJD, Eric Wofsey general-topology Jan 28 at 18:24

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  • $\begingroup$ Seems like the countable complement topology or something like it should be a counterexample. Haven't really thought about it, though. $\endgroup$ – Randall Jan 28 at 17:02
  • $\begingroup$ Do you mean finite complement topology rather? Countable subsets do not seem likely to be compacts in the countable complement topology (a countable intersection of closed subsets of them can be empty with all finite intersections nonempty). $\endgroup$ – Mindlack Jan 28 at 17:11
  • $\begingroup$ Interesting that this is actually true. $\endgroup$ – Randall Jan 28 at 17:14
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    $\begingroup$ Actually, I take that back, it is crystal clear (after reading the open-set comment in the accepted answer in the dupe). $\endgroup$ – Randall Jan 28 at 17:15
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    $\begingroup$ Direct proof from covers: suppose we have an open cover of $X$. Let $O$ be a non-empty element of that cover. If $X=O$ we have a finite subcover. If not $X\setminus O$ is a proper closed subset of $X$ and the remaining elements of the open cover cover it, so by assumption these have a finite subcover. Add $O$ and we have a finite subcover of the original cover. Hence $X$ is compact. $\endgroup$ – Henno Brandsma Jan 28 at 20:01

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