Is there a connected topological space that is maximal compact, but not $T_2$? (A space $(X,\tau)$ is said to be maximal compact if for any topology $\tau'$ on $X$ with $\tau'\supseteq \tau$ and $\tau'\neq \tau$ we have that $(X,\tau')$ is not compact.)
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asked Apr 17, 2015 at 6:43
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2 Answers
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Let $X$ be any compactly generated connected Hausdorff space that is not locally compact and let $Y=X\cup\{\infty\}$ be its one-point compactification. Then $Y$ is compact, connected, and not Hausdorff. To show $Y$ is maximal compact, we must show that every compact subset $K\subseteq Y$ is closed in $Y$. If $K\subseteq X$, then $K$ is closed in $Y$ by definition. If $\infty\in K$, it suffices to show that $K\cap X$ is closed in $X$. Since $X$ is compactly generated, it suffices to show that $K\cap A$ is closed for every compact subset $A\subset X$. But any such $A$ is closed in $Y$, so $K\cap A$ is closed in $K$ and hence compact. Since $X$ is Hausdorff, this implies $K\cap A$ is closed in $X$.
answered Apr 17, 2015 at 8:30
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Yes. Let Y be any compactly generated connected Hausdorff space such that Y fails to be locally compact. Let X be the one point compactification. X is maximal compact (since compact subsets of X are closed), but X is not Hausdorff.
See for example, Example 99 from Counterexamples in Topology by Steen and Seebach.
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$\begingroup$ The one-point compactification will not be maximal compact unless $Y$ is compactly generated. Indeed, if $A\subset Y$ has closed intersection with every compact set but is not closed, then $A\cap \{\infty\}$ is compact but not closed in the one-point compactification. $\endgroup$ Commented Apr 17, 2015 at 9:14