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Let $R$ be an equivalence relation on a Hausdorff space $X$ such that $\forall x \in X$, $x/R$ is closed. Show that $X/R$ is Hausdorff.


Firstly, I am not familiar with the notation $x/R$. Does it stand for the equivalence class $[x]_R:=\{y\in X: yRx\}$?

If so, here are my thoughts.

Take $[x]_R\neq[y]_R\in X/R$, where $x,y\in X$. We want to construct $U,V\subseteq X/R$ open such that $[x]_R\in U, [y]_R\in V$ and $U \cap V=\varnothing$. As $x\neq y \in X$ and $X$ is Hausdorff, we obtain $A,B\subseteq X$ open such that $x\in A,y\in B$ and $A \cap B=\varnothing$.

My idea was now to set $A':=A-[y]_R=A\cap(X-[y]_R)$, which is open in $X$ as an intersection of two open sets. Similarly, define $B':=B-[x]_R$. Then one still has $x\in A'$, $y\in B'$ and $A'\cap B'=\varnothing$.

Afterwards, for the canonical surjection $\pi: X\rightarrow X/R,\, \pi(z):=[z]_R$, let $U:=\pi(A')$ and $V:=\pi(B')$.

But this is where I don't know how to proceed. I cannot prove the wanted properties of $U$ and $V$, leading me to think that this is not the right approach. To be honest, the way I defined $A'$ and $B'$ was rather arbitrary, as I wanted to include the fact that the sets $[z]_R, z\in X,$ are closed in $X$.

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    $\begingroup$ I think that indeed $x{/}R$ and $[x]_R$ are the same thing. $\endgroup$ Dec 8, 2019 at 9:03
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    $\begingroup$ Does this answer your question? $X/{\sim}$ is Hausdorff if and only if $\sim$ is closed in $X \times X$ $\endgroup$ Dec 8, 2019 at 9:08
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    $\begingroup$ @ArcticChar Is it equivalent to say that (i) $R$ is closed in $X^2$, respectively that (ii) $\forall x \in X: x/R$ is closed? Otherwise, I don't see the connection. $\endgroup$
    – Zuy
    Dec 8, 2019 at 9:31
  • $\begingroup$ No these are not equivalent. In my example $R$ is not closed in $X^2$ while $\forall x \in R x/R$ is closed. $\endgroup$ Dec 8, 2019 at 9:41

1 Answer 1

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Let $X$ be a Hausdorff space that is not regular. Let $p \in X$ and $F \subseteq X$ (closed) be witnesses to non-regularity of $X$. (e.g. take $\Bbb R_K$ from Munkres' text, with $F=K$ and $p=0$ for definiteness.)

Then identify $F$ to a point, i.e. take the equivalence relation $R$ on $X$ with classes

$$F, \{x\}, x \notin F$$

And it's easy to see that all equivalence classes are closed, $X$ is Hausdorff but $X{/}R$ is not Hausdorff (as this would imply we could separate $p$ and $F$ in $X$, which we cannot). So the statement you want to show is false as interpreted.

But we do get that $X/R$ is also Hausdorff if $R \subseteq X \times X$ is closed, see this question, e.g. and this is stronger than just the classes being closed, and we also need some extra open map condition in that case.

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  • $\begingroup$ In the linked question, one adds the condition that the surjection $\pi$ is open. Does this follow from the assumption that the classes $[x]_R$ are closed? $\endgroup$
    – Zuy
    Dec 8, 2019 at 10:09
  • $\begingroup$ @Zuy I don’t think so. $\endgroup$ Dec 8, 2019 at 11:03
  • $\begingroup$ But then how does the linked question relate to the situation where $X$ is Hausdorff, $R\subseteq X^2$ is closed and $\forall x \in X: [x]_R$ is closed? $\endgroup$
    – Zuy
    Dec 8, 2019 at 11:07
  • $\begingroup$ @Zuy it does not directly, you need that $p$ is open too. In my example $R$ is actually closed in $X^2$ too. So an extra condition is needed. $\endgroup$ Dec 8, 2019 at 14:33

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