Assume that we are in the following situation: a compact Lie group $G$ acts on a compact space $X$ which is not necessarily Hausdorff. $X$ is assumed to be compactly generated and weakly Hausdorff, though. The quotient space $X/G$ is Hausdorff.

Does this imply that $X$ is Hausdorff? I do not think this is true but I'm also unable to find a counterexample since I cannot come up with a non-Hausdorff space that admits a non-trivial $G$-action in the first place.

The reason I'm thinking about this is the following: If the $G$-action on $X$ was additionally free (and $X$ was Hausdorff), then $X$, being compact and Hausdorff, would be completely regular, and hence the quotient map $X \to X/G$ would be a fiber bundle by a result of Bredon. I'm also interested in whether $X \to X/G$ is a fiber bundle if $X,G,X/G$ are as above and the $G$-action is free.

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    $\begingroup$ $X$ weakly Hausdorff: all images of compact Hausdorff spaces in $X$ are closed (implies $T_1$: points are closed). Compactly generated: en.wikipedia.org/wiki/Compactly_generated_space $\endgroup$ – YCor Dec 26 '16 at 3:16
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    $\begingroup$ One reason it may be hard to think of an example by hand is because you assume the quotient is Hausdorff, and then compactness of $G$ implies that each orbit subspace must be Hausdorff. With a Hausdorff "base" and Hausdorff "fibers," non-Hausdorffness in the "total space" has to come from something weird. $\endgroup$ – Gabriel C. Drummond-Cole Dec 26 '16 at 16:38
  • $\begingroup$ @GabrielC.Drummond-Cole: Thanks, I had not realized that the weak Hausdorff property on $X$ is sufficient to have Hausdorff orbits until now. I agree that this makes a potential example even less likely to be found. $\endgroup$ – Alexander Körschgen Jan 5 '17 at 0:08
  • $\begingroup$ It seems that a weakly Haudorff first-countable space is Hausdorff, so the answer is affirmative for first-countable spaces $X$. By the way, what is the answer for finite acting group $G$? $\endgroup$ – Taras Banakh Aug 23 '17 at 10:21
  • $\begingroup$ @TarasBanakh why does WH + first-countable imply Hausdorff? I wasn't able to find this statement, maybe you're confusing this with the fact that every first-countable space is compactly generated (see Strickland's notes, for example). $\endgroup$ – Alexander Körschgen Aug 24 '17 at 16:36

How about $X=\{ a,b,x,y\}$ with nontrivial open sets $\{a,b\}$ and $\{ x,y\}$ and $G= Z/2$, discrete. The action exchanges $a$ with $ b$ and $x$ with $y$.

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    $\begingroup$ Thank you for your comment. Unfortunately, the space $X$ is not weak Hausdorff. If we endow $K := \{c,d\}$ with the discrete topology, $K$ is compact Hausdorff. However, the map $K \to X$ given by $c \mapsto a, d \mapsto x$ is continuous while its image $\{a,x\}$ is not closed in $X$. $\endgroup$ – Alexander Körschgen Nov 28 '16 at 18:57

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