# Finite group action on Hausdorff space

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?

• 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 – Moishe Kohan Apr 2 at 23:38
• What would be a correct statement to proof? – Lucas Smits Apr 3 at 8:45
• 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. – user87690 Apr 3 at 9:55
• Start with a correct definition of properness. – YCor Apr 3 at 12:46

## 1 Answer

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.