# Let $R$ be a closed equivalence relation on a Hausdorff space $X$. Show that $X/R$ is Hausdorff. [duplicate]

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 $$X$$.

• I think that indeed $x{/}R$ and $[x]_R$ are the same thing. Dec 8, 2019 at 9:03
• Does this answer your question? $X/{\sim}$ is Hausdorff if and only if $\sim$ is closed in $X \times X$ Dec 8, 2019 at 9:08
• @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.
– Zuy
Dec 8, 2019 at 9:31
• No these are not equivalent. In my example $R$ is not closed in $X^2$ while $\forall x \in R x/R$ is closed. Dec 8, 2019 at 9:41

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.

• 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?
– Zuy
Dec 8, 2019 at 10:09
• @Zuy I don’t think so. Dec 8, 2019 at 11:03
• 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?
– Zuy
Dec 8, 2019 at 11:07
• @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. Dec 8, 2019 at 14:33