# $X/{\sim}$ 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. – user38268 Apr 3 '12 at 13:54
• @t.b. You could post this as an answer to the new question math.stackexchange.com/questions/1903343/… – Stefan Hamcke Aug 25 '16 at 19:20
• Is the result still true if the quotient map is not open? – Sachchidanand Prasad Jan 13 '18 at 23:47

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.
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 $X \times X$.]
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$U_{\pi(x)} \ni {\pi(x)}, U_{\pi(y)} \ni {\pi(y)}$. 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... Let$Y$be the quotient. This is a question about the diagonal$D$in$Y\times Y$. Note that the product map$X\times X \to Y \times Y$is also a quotient map (this is where open-ness is needed), and that your set$\sim$is precisely$\pi\times\pi^{-1}(D)$. Thus$\sim$is closed if and only if$D$is closed. Now,$D$is closed if and only if$Y$is Hausdorff, for any topological space$Y$. This follows by considering disjoint neighborhoods$U$and$V$of any two distinct points$p$and$q$:$(p,q)\in U\times V$which is open and disjoint from$D$, so$D^c$is closed. Conversely, the closedness of$D$implies the existence of some product open set$U\times V$disjoint from$D$and containing$(p,q)$, and thus$U$and$V$are open sets separating$p$and$q\$.