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Let $G$ be a finite group acting on some Hausdorff space $X$. I have some feeling that the action of $G$ is always properly discontinuous, i.e. for all $x\in X$, there is an open neighbourhood $U$ of $x$ such that $g\cdot U\cap U=\emptyset$ for all $g\neq e_G\in G$. Is this true, or under what additional conditions might this be true?

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    $\begingroup$ What you are trying to prove us actually false. Think of an action which is not free. The mistake us to use a wrong definition of proper discontinuity $\endgroup$ – Moishe Kohan Apr 2 at 23:38
  • $\begingroup$ What would be a correct statement to proof? $\endgroup$ – Lucas Smits Apr 3 at 8:45
  • $\begingroup$ The point is that it may in general happen that $g ⋅ x = x$ for some $x$ and $g ≠ e_G$. Consider the trivial action as an example. In that case you have no chance to find the desired $U$ for that $x$. Actions where this does not happen (i.e. for every $x$ and $g ≠ e_G$ we have $g ⋅ x ≠ x$) are called free. If you additionally suppose that your action is free, you may prove your statement. Think about my answer as a hint. $\endgroup$ – user87690 Apr 3 at 9:55
  • $\begingroup$ Start with a correct definition of properness. $\endgroup$ – YCor Apr 3 at 12:46
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Try to prove the following observation: If $g⋅x ≠ x$, then there is $U_g$ a neighborhood of $x$ such that $g ⋅ U_g ∩ U_g = ∅$. Moreover, for every smaller neighborhood $V$ of $x$, the same holds.

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