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Let $G$ be a second countable locally compact Hausdorff group and $X$ be a second countable locally compact almost-Hausdorff $G$-space. If $X$ is totally disconnected and the orbit space X/G is Hausdorff, can we conclude that X/G is also totally disconnected?

We may assume that $G$ is totally disconnected.

If $X$ is Hausforff, then $X$ has a compact-open basis. It is not difficult to see that X/G is totally disconnected.

Recall that a topological space X is called almost-Hausdorff if every non-empty, closed subset $A$ of X contains a non-empty, relatively open Hausdorff subset.

See Definition 6.1, Theorem 6.2 and Lemma 6.3 of the following book about almost-Hausdorff. https://books.google.dk/books?id=0aBS-gYTWboC&pg=PA172&lpg=PA172&dq=almost-Hausdorff&source=bl&ots=e2VfXO0NYf&sig=pzVDwS5kpVMLO8gZ8QysuNi8n18&hl=en&sa=X&ei=heCfVIzkDMSeywOM-oCQCw&ved=0CDMQ6AEwAg#v=onepage&q=almost-Hausdorff&f=false

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  • $\begingroup$ Should you have ... different from that closed set after the last phrase ...Hausdorff subset? Let $\ A\subseteq X.\ $ Is, by definition, $\ B\subseteq A\ $ relatively open $\ \Leftarrow:\Rightarrow\ $ there exists open $\ H\ $ such that $\ B=A\cap H\ $ ? $\endgroup$ – Włodzimierz Holsztyński Dec 27 '14 at 23:53
  • $\begingroup$ I once asked a similar question, it might be interesting.mathoverflow.net/questions/41027/… $\endgroup$ – HenrikRüping Dec 28 '14 at 0:36
  • $\begingroup$ @Włodzimierz: normally $A$ is not Hausdorff, but by definition $A$ contains a relatively open Hausdorff subset. So in general they are different. $\endgroup$ – m07kl Dec 28 '14 at 10:47

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