Let $X$ be a Hausdorff topological space and $G$ a group acting continuously on $X$. We denote the group action as $(g,x) \rightarrow gx$. Let $x\in X$ and $U$ a neighborhood of $x$ such that the set $P(x,U)=\{g \in G: gx \in U\}$ is relatively compact. Then, I claim, $G$ is locally compact.

Proof. It suffices to prove that $P(x,U)$ is a neighborhood of the identity $e \in G$. The action map $f:G \times X \rightarrow X$ is continuous so the map $F:G \rightarrow X: g \rightarrow gx$ is continuous too, mapping $e$ to $x$. Thus, since $U$ is a neighborhood of $x$ there exists a neighborhood $V \subset G$ of $e$ such that $F(V) \subset U$, that is $Vx \subset U$. So, if $g \in V$, then $gx \in U$, which gives $g \in P$. Thus, $V \subset P$, meaning $P$ is a neighborhood of $e$. Since it is relatively compact, we are done.

Is this correct?

  • 1
    $\begingroup$ It seems right to me. $\endgroup$
    – Javi
    Jul 20, 2018 at 19:29


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