# $X/\sim$ is Hausdorff if and only if $\sim$ is closed in $X \times X$

$X$ is a Hausdorff space and $\sim$ is an equivalence relation.

If the quotient map is open, then $X/ \sim$ is a Hausdorff space if and only if $\sim$ is a closed subset of the product space $X \times X$.

Necessity is obvious, but I don't know how to prove the other side. That is, $\sim$ is a closed subset of the product space $X \times X$ $\Rightarrow$ $X/ \sim$ is a Hausdorff space. Any advices and comments will be appreciated.

-
I wonder if we can remove the condition that the quotient map is open. In that case, necessity is also obvious, is the sufficiency also true? Or is there any counterexamples? –  user22111 Jan 2 '12 at 5:51
@Jingren: You should post this as a separate question. Anyway: Yes, the condition that the quotient map be open is necessary. Consider $X/A$ where $X$ is a non-regular Hausdorff space and $x$ is a point that cannot be separated from the closed set $A$. In the quotient $X/A$ the image of the point $x$ can't be separated from the point corresponding to $A$ while the equivalence relation is obviously closed. –  t.b. Jan 2 '12 at 6:07
@yaoxiao Very nice post. –  fpqc Apr 3 '12 at 13:54
Let $\pi:X\to X/\!\!\!\sim\;$ denote the projection map associated with $\sim$. (That is, for any $x\in X$, $\pi(x)$ is the $\sim$-equivalence class that $x$ belongs to.) Let $\nsim\; \subseteq X \times X$ be shorthand for the complement of $\;\;\sim\;\;$ in $X \times X\;$, i.e. $\nsim\;\;=\;(X \times X\;) \;\;-\; \sim\;$.
Suppose that $\pi(x) \neq \pi(y)\;$. (Here I'm relying on the fact that, since $\pi$ is surjective, any element $\widetilde{z}\in X/\!\!\!\sim\;$ may be written in the form $\pi(z)$, for some $z \in X$.) We must show that there exist open sets $U_{\pi(x)}, U_{\pi(y)} \subseteq X/\!\!\!\sim\;$ such that ${\pi(x)} \in U_{\pi(x)}$, ${\pi(y)} \in U_{\pi(y)}$, and $U_{\pi(x)} \cap U_{\pi(y)} = \varnothing\;$.
By assumption, $\;\sim\; \subseteq X \times X$ is closed, so $\nsim\; \subseteq X \times X$ is open. Therefore there exist open neighborhoods $N_x$ and $N_y$ of $x$ and $y$, respectively, such that $(x,\;y)\in N_x \times N_y \subseteq$$\;\;\nsim\;. (This is because the family of all pairwise products of open subsets of X is a basis for the product topology on X \times X.) For any v, w \in X,$$ (v,\;w) \;\in \;\nsim \;\;\;\;\;\Leftrightarrow\;\;\;\;\; \pi(v) \neq \pi(w)\;\;. $$Therefore,$$ N_x \times N_y \subseteq \;\;\nsim\;\;\;\;\Leftrightarrow\;\;\;\; \forall (v, w) \in N_x \times N_y \;[\pi(v) \neq \pi(w)] \;\;\;\;\Leftrightarrow\;\;\;\; \pi[N_x] \cap \pi[N_y] = \varnothing$$Furthermore, since$\pi$is open (by assumption), the image sets$\pi[N_x], \pi[N_y] \subseteq X/\!\!\!\sim\;$are open neighborhoods of${\pi(x)}$and${\pi(y)}$, respectively. Therefore,$\pi[N_x]$and$\pi[N_y]$are the desired open neighborhoods$U_{\pi(x)} \ni {\pi(x)}, U_{\pi(y)} \ni {\pi(y)}$. - add comment Let$R$be the subset of$X \times X$which gives the equivalence relation$\sim$, and let$f\colon X \to X/{\sim}$be the quotient map. Let$x, y \in X$be points not equivalent under the relation, i.e.$(x, y) \notin R$. Since$R$is closed and$X \times X$has the product topology, there exist open sets$U, V$in$X$such that$(x, y) \in U \times V$and$U \times V$does not meet$R$. Can you separate$f(x)$and$f(y)$using$U$and$V$? Remember that$f$is assumed to be an open map. [This is a lot like the proof of the fact that Alex is using: that a space$X$is Hausdorff if and only if the diagonal is closed in$X \times X$.] - add comment Since the map$\pi:X\to X/\sim$is open, it's clear that the map$g:X^2\to (X/\sim)^2$given by$g(x,y)=(\pi(x),\pi(y))$is open. What we claim is that$g(X^2-\sim)=(X/\sim)^2-\Delta_{X/\sim}$. Indeed, if$x\nsim y$then$\pi(x)\ne\pi(y)$which tells us that$g\left(X^2-\sim\right)\subseteq (X/\sim)^2-\Delta_{X/\sim}$. That said, if$(\pi(x),\pi(y))\notin\Delta_{X/\sim}$then$\pi(x)\ne \pi(y)$so that$x\nsim y$so that$(x,y)\in X^2-\sim$and clearly$g(x,y)=(\pi(x),\pi(y))$. Thus,$g(X^2-\sim)=(X/\sim)^2-\Delta_{X/\sim}$as claimed. But, since$X^2-\sim$is open by assumption, and$g$is an open map we have that$(X/\sim)^2-\Delta_{X/\sim}$is open, and so$\Delta_{X/\sim}$is closed. This gives us$T_2$ness. - add comment Start with a point$(x,y)$with$x$and$y$not related. Then, as the relation is reflexive, it contains the diagonal. Now, as the relation is closed, its complement is open and there is a neighbourood of$(x,y)\$ which does not intersect it. Next think about what a base for the product topology might look like...