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I can't find a counterexample to this sentence:

Let $X$ be a $T2$ topological space, and $G \subset Hom(X)$ be a group of homeomorphisms which acts properly discontinuously on X (i.e. $\forall x \in X \ \ \exists U$ neighborhood of $x$ such that $\forall g \in G\setminus \{id\} \ \ g(U) \cap U=\varnothing).$ Then the quotient $X/G$ is $T2.$

The following theorem gives me some hints about what to avoid while making the counterexample :

Let $A \subset X$ (with $X$ Hausdorff) be an open set such that: $A$ intersects every orbit of $G,$ and $A \cap g(A) \neq \varnothing$ for a finite quantity of $g \in G.$ Then $X/G$ is $T2.$

Clearly $G$ cannot be finite, otherwise the quotient is automatically $T2,$ taking $A=X.$ Moreover, this condition is somehow similar to the hypothesis of properly discontinuous action, and I can't find an example where they don't "coincide". Do you have some hints? Thanks to everybody.

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    $\begingroup$ See here. Not the same question, but the answer there contains what you want, I think. (I'm not quite sure what definition of "properly discontinuous action" you use.) $\endgroup$ – Daniel Fischer May 29 '17 at 22:02
  • $\begingroup$ I've edited the question with my definition of properly discontinuous action. Thank you very much! $\endgroup$ – Riccardo Ceccon May 29 '17 at 22:13
  • $\begingroup$ Sorry, I've thought about it for a while but I don't really understand why the quotient is not Hausdorff. :( $\endgroup$ – Riccardo Ceccon May 30 '17 at 9:09
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    $\begingroup$ Consider the orbits of $(1,0)$ and of $(0,1)$. If we could separate them with disjoint neighbourhoods, there'd be neighbourhoods $U$ of $(1,0)$ and $V$ of $(0,1)$ so that $n\cdot U \cap V = \varnothing$ for all $n\in\mathbb{Z}$. By shrinking, we can assume that $U$ and $V$ are balls with radius $\varepsilon$. Pick $n\in\mathbb{N}$ such that $2^{-n} < \varepsilon$. Then $a = (2^{-n},1) \in V$ and $n\cdot a = (1,2^{-n}) \in U$. Hence every neighbourhood of the orbit of $(1,0)$ intersects every neighbourhood of the orbit of $(0,1)$. $\endgroup$ – Daniel Fischer May 30 '17 at 10:19
  • $\begingroup$ @DanielFischer thank you! $\endgroup$ – Riccardo Ceccon May 30 '17 at 13:37
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This contradicts the theorem you posted, so I'm not sure if our definitions of properly discontinuous actions are the same. Anyhow,

$\mathbb{Z}/2$ acts on the disjoint union of $\mathbb{R} \times \{a\}$ and $\mathbb{R} \times \{b\}$ by $g \cdot (0, a) = (0, a), g \cdot (0, b) = (0, b)$, and otherwise $g \cdot (t, a) = (t, b)$ and vice-versa. This is a properly discontinuous action (the group is finite). Furthermore, the original space is certainly Hausdorff in the disjoint union topology; the resulting space is the line with two origins which is not Hausdorff at the origins.

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  • $\begingroup$ I was thinking now about that, but there's a problem: the action must be a homeomorphism, and the counter example gives an action which is not a homeomorphism. :( $\endgroup$ – Riccardo Ceccon May 29 '17 at 20:47
  • $\begingroup$ However, what I mean is that G is a subgroup of the group of homeomorphisms of X, and $\forall x \in X \ \exists \ U$ neighborhood of $x$ such that $g(U) \cap U=\varnothing \forall g \in G.$ Maybe we had different definitions of properly discontinuous actions :). $\endgroup$ – Riccardo Ceccon May 29 '17 at 20:49

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