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Suppose that $X$ is locally compact and $G$ acting on $X$ is proper. Show that the quotient $X/G$ is Hausdorff.

I am working through some notes on Geometric Group Theory and I am having a hard time with this problem. Some help would be great.

A topological group action $μ : G$ acts on $X$ is called proper if for every compact subsets $K_1,K_2 ⊂ X,$ the set is $\{g \in G | g(K_1)\cap K_2 \neq \emptyset \}$ contained in $G$ is compact.

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  • $\begingroup$ What are $X$ and $G$ here and what do you mean by "proper"? $\endgroup$ – Cameron Williams Aug 4 '14 at 16:30
  • $\begingroup$ You may find the following fact useful: for any topological space $X$ (Hausdorff or not!), the quotient of $X$ by an equivalence relation $R \subseteq X \times X$ is Hausdorff if and only if $R$ is closed as a subset of $X \times X$. $\endgroup$ – Zhen Lin Aug 4 '14 at 16:34
  • $\begingroup$ Related. The proof works in the given setting. $\endgroup$ – Daniel Fischer Aug 4 '14 at 17:09
  • $\begingroup$ One usually uses the following fact to prove this: suppose the action of $G$ on $X$ is proper, then for every $x\in X$ there exists an open neighbourhood (connected even) $U_x$ with compact closure such that $gU_x \cap U_x$ is nonempty iff $g$ fixes $x$ (where $g\in G$). $\endgroup$ – sdf Aug 4 '14 at 18:00

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