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Let $G$ be a compact Hausdorff group. Let $X$ be a Hausdorff space. Suppose $G$ acts continuously on $X$. Is the orbit space $X/G$ Hausdorff? If not, I would like to know an counter-example.

Remark As my answer to this question shows, if $X$ is a locally compact Hausdorff space, $X/G$ is Hausdorff.

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    $\begingroup$ This should be useful: math.stackexchange.com/questions/50044/… $\endgroup$ – Rankeya Nov 22 '12 at 6:19
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    $\begingroup$ Yes. This follows from lemma 9 of your answer: the projection $X \to X/G$ is open and the orbit equivalence relation $\Gamma$ is closed in $X \times X$ because $G$ is compact and $X$ is Hausdorff. $\endgroup$ – commenter Nov 22 '12 at 8:24
  • $\begingroup$ @commenter Could you explain why $\Gamma$ is closed? $\endgroup$ – Makoto Kato Nov 22 '12 at 8:35
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In view of lemma 9 of your answer it suffices to prove that the orbit equivalence relation $\Gamma$ is closed:

Let $G$ be a compact group acting continuously by homeomorphisms on a Hausdorff space $X$. Then the orbit equivalence relation $\Gamma \subset X \times X$ is closed.

Suppose $(x_i, y_i) \to (x,y)$ is a convergent net in $X \times X$ with $(x_i,y_i) \in \Gamma$. Then $x_i = g_i y_i$ for some net $g_i \in G$. Since $G$ is compact, there is a subnet $g_j$ which converges, say $g_j \to g$. Since $y_j \to y$ and $g_j \to g$ we have $x_j = g_j y_j \to gy$. But by assumption $x_j \to x$, so $x = gy$ because $X$ is Hausdorff and hence $(x,y) \in \Gamma$.

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  • $\begingroup$ Am I wrong or in order to deduce that $x_j=g_jy_j\rightarrow gy$ you also need that $X$ and $G$ are second-countable? $\endgroup$ – W4cc0 Dec 23 '18 at 14:50

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