I am trying to prove the following:

Let $G$ be a topological group acting properly on a Hausdorff locally compact space $X$, i.e. preimages of compacts sets by the map $$G\times X\to X\times X$$ $$(g,x)\mapsto (gx,x)$$ are compacts. Then the quotient $X/G$ is Hausdorff.

Take $x,x'\in X$ such that $Gx\not=Gx'$.

  1. By using that X is Hausdorff locally compact and property of the action, I found $V_1,V_2$ disjoint open set with compact closure and $H\subset G$ compact, such that $x\in V_1,y'\in V_2$ and $$(G-H)\overline{V_1}\cap \overline{V_2}=\varnothing.$$
  2. So now we need to work on $H$. Since $x'\in (Hx)^c\cap V_2$ open, there exists $x'A_2\subset V_2$ such that $$A_2\cap Hx=\varnothing.$$ It implies that $x\in (H^{-1}A_2)^c\cap V_1$.
  3. Now, I'd like to find as before $x\in A_1\subset V_1$ open such that $$A_1\cap H^{-1}A_2=\varnothing,$$ which would be enough to conclude. But $H^{-1}A_2$ is not closed like $Hx$, so I'm stuck...

Are my first ideas correct ? Is there a way to exit this dead end ?

  • 1
    $\begingroup$ Perhaps the proof of Lemma 10 of my answer to the following question helps: math.stackexchange.com/questions/241711/… $\endgroup$ – Makoto Kato Jan 7 '13 at 23:50
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    $\begingroup$ What do you know about proper maps? Do you know that under adequate topological conditions, proper maps are closed? (usually proper maps are defined with this in mind: closed maps s.t. the preimage of any point is compact) If you do it follows from very basic topology. $\endgroup$ – Olivier Bégassat Jan 7 '13 at 23:54
  • $\begingroup$ @OlivierBégassat Thank you, I think I've been able to prove that (see the answer below). However, I'm still looking for a way to conclude with my original approach. $\endgroup$ – Klaus Jan 11 '13 at 13:19

So using that locally compact spaces are compactly generated, it is easy to show that a proper map $f: A\to B$ with $B$ locally compact is closed.

Therefore, $f: G\times X\to X\times X$ as defined above is closed. But the quotient map $\pi: X\to X/G$ is closed, so $$\tilde f=\pi\circ f: G\times X\to X/G\times X/G$$ is also closed. Note that $\tilde f(g,x)=([gx],[x])=([x],[x])$, so the diagonal $$\Delta(X/G)=\tilde f(G\times X)$$ is closed, i.e. $X/G$ is Hausdorff.

Is that correct ?

  • 2
    $\begingroup$ Why is the quotient map $\pi$ closed? (this is not true, if $G = \mathbb{R}$ acts on $X = \mathbb{R}^2$ by translation on the first coordinate, the action is proper, but the corresponding quotient map $\pi$ is not closed). However, you can check that $\pi$ is always open (since you have a group acting). Compare Makoto Kato's Lemma 9. $\endgroup$ – Martin Jan 11 '13 at 13:27
  • $\begingroup$ Whoups, thanks. That solved everything! $\endgroup$ – Klaus Jan 11 '13 at 15:02

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