# $G$ finite group acts freely on top. sp. $X$, can we find for every $x\in X$ an open neighborhood such that:

Let $G$ be finite topological group, and acts freely over the hausdorff topological space $X$, i want to prove that every element $x$ in $X$ has an open neighborhood $U_x$ such that:

$g\star U_{x}\cap U_{x} \neq \phi$ and $g\in G$ implies that $g=e_{G}$

my best regards

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For each $g \ne e$ in $G$ and $x \in X$, the fact that $G$ acts freely means $gx \ne x$. Since $X$ is Hausdorff, there are disjoint open neighbourhoods $V_{g,x}$ and $W_{g,x}$ of $gx$ and $x$ respectively.
Take $U_x$ to be the intersection of $g^{-1}V_{g,x} \cap W_{g,x}$ for all $g \ne e$. I'll let you verify that this satisfies the requirements.
It has to be assumed that the maps $x \to gx$ are continuous. This should be part of the definition of a group acting on a topological space. –  Robert Israel Oct 19 '12 at 16:01