Finite group action on Hausdorff space Let $G$ be a finite group acting on some Hausdorff space $X$ such that the action is free. 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?
 A: 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.
A: We first show that given finitely many points $a_1,a_2,\cdots,a_n$ in a Hausdorff space $Y$, there exist open sets $G_1,G_2,\cdots,G_n$ such that $a_i\in G_i$ for each $i$ and $G_i\cap G_j=\emptyset$ for $i\neq j$. We show this by induction on $n$. For $n=2$, this follows from the definition of Hausdorff space. Let us assume that the statement is true for some $n=m$. Thus, given distinct points $a_1,a_2,\cdots,a_m$ in $Y$, there are mutually disjoint open sets $G_1,G_2,\cdots,G_m$, such that $a_i\in G_i$ for each $i$. Now we choose a point $a_{m+1}$ which is distinct from each of $a_1,a_2,\cdots,a_m$. Then by Hausdorffness, we get open sets $A_1,A_2,\cdots,A_m$ and $B_1,B_2,\cdots,B_m$ such that $a_{m+1}\in A_i$ for each $i$, $a_i\in B_i$ for each $i$, and $A_i\cap B_i=\emptyset$. If we set $G_{m+1}^{'}=\cap_{i=1}^m A_i$ and $G_i^{'}=G_i\cap B_i$ for $i=1(1)m$, then we see that the $G_i^{'}$'s are mutually disjoint. Thus the statement is true for $n=m+1$, and thus by induction, for any natural number $n\geq 2$.
Now we start with a $y\in Y$. Then, since by the given problem, the kernel of the action is $\{1\}$, so, for any $g,h\in G$ with $g\neq h$, we must have $g\cdot y\neq h\cdot y$. Thus if we write the finite group $G$ as $G=\{g_1,g_2,\cdots,g_n\}$, taking $g_1=1$, we have distinct points $y_1=g_1\cdot y,y_2=g_2\cdot y,\cdots,y_n=g_n\cdot y$. By whatever we have proved above, we get mutually disjoint open sets $E_1,E_2,\cdots,E_n$, such that $y_i\in E_i$ for each $i=1(1)n$. We consider the set $E=\cap_{i=1}^n g_i^{-1}\cdot E_i$. Action by each $g_i$ being a homeomorphism, $E$ is a finite intersection of open sets each containing $y$, and is thus an open set containing $y$. Also for each $i$, $g_i\cdot E\subseteq g_i\dot(g_i^{-1}\cdot E_i)=E_i$. Thus given $y\in Y$ we obtain an open set $E$ such that $y\in E$ and given $i\neq j$, $(g_i\cdot E)\cap (g_j\cdot E)\subseteq E_i\cap E_j=\emptyset$ (again by homeomorphism of group action by $g_i$), which proves that the action is even, i.e, properly discontinuous.
