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Let $G$ be a group acting on a topological space $X.$

I wanted to check the following statement.

"Recall that the action is called even (or classically properly discontinuous) if for every point $x \in X$ (topological space) there exists an open set $U \in X$ such that $x \in U$ and $gU \cap U = \emptyset;$ unless $g = e:$ Under this condition, later, we shall see that the quotient map is a very special kind of map called a covering projection. Right now, you can directly verify that if $X$ is Hausdorff then $X/G$ is Hausdorff."

But, the problem is, I can't prove this. But, I can prove this with extra conditions like

$1)$ $G$ is finite or

$2)$ for any $x,y$ in different orbit there exists corresponding $W,V$ open and $gW \cap V = \emptyset$ for all (most ?) $g \in G.$

Is the statement wrong or I am not considering something?

Thank You.

Air Mike
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Sagnik Biswas
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  • Typically $X$ is assumed to be locally compact and "properly discontinuous" is defined for compact subsets, not just points (although with local compactness this may be equivalent?). Anyway see this: https://www.math.brown.edu/~res/M141/hausdorff.pdf . I don't think the statement is true otherwise. – freakish Sep 28 '20 at 09:05
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    See here https://math.stackexchange.com/questions/1082834/properly-discontinuous-action-equivalent-definitions and https://mathoverflow.net/questions/55726/properly-discontinuous-action – Sumanta Sep 28 '20 at 09:14

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