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Let $G$ be a finite group acting on a Hausdorff topological space $X$. Prove that $X/G$ is Hausdorff. Deduce that the projective space $P^n$ is Hausdorff for all $n$.

My Try:

Consider the quotient projection $p:X\rightarrow X/G$. Let $Gx\neq Gy$. Then, $x\neq y$. There exists neighborhoods $U$ of $x$ and $V$ of $y$, such that $U\cap V=\emptyset$. $p$ is an open map. So, I was going to prove that $p(U)\cap p(V)=\emptyset$. But, in order to prove it, I need to have $p^{-1}(p(U))=U$ and $p^{-1}(p(V))=V$. But, I could not prove it. My question is, is it true? Then how may I prove it? Moreover, I am confused with $P^n$ here. What is it? Any help is appreciated.

Rob Arthan
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Extremal
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    You need to use that the fibres of $p$ are finite. – Daniel Fischer Nov 01 '15 at 20:26
  • The result is false unless the group acts freely. Look at the first counterexample. – Najib Idrissi Nov 01 '15 at 20:26
  • @Daniel Fischer: What do you mean by fibres of $p$? – Extremal Nov 01 '15 at 20:26
  • @EpsilonDelta If $f \colon A \to B$, the fibres of $f$ are the sets $f^{-1}(b)$ for $b\in B$. – Daniel Fischer Nov 01 '15 at 20:30
  • Ok, but still I am confused. – Extremal Nov 01 '15 at 20:33
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    @NajibIdrissi I have a problem with the first counterexample there. When one talks of the action of a group $G$ on a topological space $X$, isn't it understood that the maps $x \mapsto gx$ are continuous (and hence homeomorphisms) for all $g\in G$? Otherwise, the group just acts on the underlying set, doesn't it? – Daniel Fischer Nov 01 '15 at 20:38
  • @DanielFischer That's a good point... I actually mixed up my definitions (it's properly discontinuous that implies Hausdorff quotient, and a finite, continuous group action is always properly discontinuous). – Najib Idrissi Nov 01 '15 at 20:56
  • Still I am confused on how to prove $X/G$ is Hausdorff. Can anyone please help me? – Extremal Nov 01 '15 at 21:03
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    You can use the argument of the answer to your previous question. – Daniel Fischer Nov 01 '15 at 21:12
  • Got it. Thanks. By the way what is the meaning of $P^n$ in this question? – Extremal Nov 01 '15 at 21:32
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    $P^n$ is presumably real $n$-dimensional projective space $S^n/{\pm 1}$. – Rob Arthan Nov 01 '15 at 21:35
  • @EpsilonDelta In fact a more general result is true. If $G$ is compact and $X$ is Hausdorff then $X/G$ is Hausdorff. – R_D Nov 30 '15 at 12:29
  • @DanielFischer Just to confirm, the result in the question statement does not require local-compactness of $X$, right? – MathematicsStudent1122 Dec 14 '20 at 13:19
  • @TheDayBeforeDawn Right. For finite $G$ and Hausdorff $X$ it is straightforward to separate two orbits by disjoint invariant neighbourhoods. First separate them by disjoint neighbourhoods $U$ and $V$ (uses only that $X$ is Hausdorff, and orbits are finite [and disjoint]). Then replace $U$ with $U' = \bigcap_{g \in G} gU$ and analogusly for $V$. Then $U' \cap V' = \varnothing$ since $U' \subset U$, $V' \subset V$, and $U', V'$ are neighbourhoods of the respective orbits because each $gU$ ($gV$) is, and there are only finitely many to intersect. – Daniel Fischer Dec 14 '20 at 16:09

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Hint: given $x \neq y$, try to find neighbourhoods $U$ of $x$ and $V$ of $y$ such that $GU \cap GV = \emptyset$.

Rob Arthan
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