# If $X$ is Hausdorff and the quotient map $q\colon X\to X/\mathord{\sim}$ is closed, must $\sim$ be closed in $X\times X$?

Let $$\sim$$ be an equivalence relation on a Hausdorff space $$X$$. Let $$X/\mathord{\sim}$$ be the corresponding quotient space. Assume that the quotient mapping $$q\colon X\to X/\mathord{\sim}$$ is closed.

Question: Is $$\sim$$ necessarily a closed subset of $$X\times X$$ ?

Let us note that the elements of $$X/\mathord{\sim}$$ are the equivalence classes, and the topology of $$X/\mathord{\sim}$$ consists of all sets $$\mathcal{U}\subseteq X/{\mathord{\sim}}$$ such that $$\bigcup\mathcal{U}$$ is an open subset of $$X$$. The quotient mapping $$q\colon X\to X/\mathord{\sim}$$ assigns to each $$x$$ its corresponding equivalence class $$[x]$$.

We show that if the answer is 'no' then the example must be a non-regular space $$X$$ and an equivalence $$\sim$$ must be such that $$X/\mathord{\sim}$$ is not Hausdorff.

We denote by $$[A]$$ the saturation of set $$A\subseteq X$$ with respect to the equivalence relation $$\sim$$, that is, $$[A]=\{x\in X\colon (\exists y\in A)\,x\sim y\}$$.

Proposition: Let $$X$$ be a Hausdorff space and let $$\sim$$ be an equivalence relation on $$X$$ such that the quotient mapping $$q\colon X\to X/\mathord{\sim}$$ is closed and $$\sim$$ is not closed in $$X\times X$$. Then $$X$$ is not regular and $$X/\mathord{\sim}$$ is not Hausdorff.

Proof: The second part follows easily since if $$X/\mathord{\sim}$$ is Hausdorff then its diagonal $$D=\{([x],[x])\colon\,x\in X\}$$ is closed in $$(X/\mathord{\sim})\times(X/\mathord{\sim})$$ and $$\sim$$ is the preimage of $$D$$ under continuous mapping $$q\times q\colon (x,y)\mapsto([x],[y])$$, hence $$\sim$$ is closed in $$X\times X$$.

To show the first part, assume that $$X$$ is regular, $$q$$ is a closed map and $$\sim$$ is not closed in $$X\times X$$. Then $$[x]$$ is closed for every $$x$$. There exist $$u,v\in X$$ such that $$u\nsim v$$ but for every neighbourhood $$W\subseteq X\times X$$ of $$(u,v)$$ there exists $$(u_W,v_W)\in W$$ such that $$u_W\sim v_W$$. We have $$u\notin [v]$$ and $$v\notin [u]$$ hence by regularity of $$X$$ there exist open sets $$U_0,U_1,V_0,V_1$$ such that $$u\in U_0$$, $$v\in V_0$$, $$[u]\subseteq U_1$$, $$[v]\subseteq V_1$$, and $$U_0\cap V_1=U_1\cap V_0=\emptyset$$. Denote by $$\mathcal{W}$$ the family of all open sets $$W\subseteq U_0\times V_0$$ containing $$(u,v)$$, and let $$A=\{u_W\colon W\in\mathcal{W}\}$$, $$B=\{v_W\colon W\in\mathcal{W}\}$$. Then $$u\in\mathrm{cl}{A}$$, $$v\in\mathrm{cl}{B}$$, and $$[A]=[B]$$. Since $$A\subseteq U_0$$ and $$B\subseteq V_0$$, we have $$A\cap [v]=B\cap [u]=\emptyset$$, hence $$u,v\notin [A]=[B]$$. Since $$q$$ is a closed map, sets $$[\mathrm{cl}{A}]$$, $$[\mathrm{cl}{B}]$$ are closed, hence $$[\mathrm{cl}{A}]\supseteq\mathrm{cl}[A]=\mathrm{cl}[B]\supseteq \mathrm{cl}{B}$$, similarly also $$[\mathrm{cl}{B}]\supseteq\mathrm{cl}{A}$$, hence $$[\mathrm{cl}{A}]=[\mathrm{cl}{B}]$$. It follows that there exist $$u'\in\mathrm{cl}{A}$$ and $$v'\in\mathrm{cl}{B}$$ such that $$u'\sim v$$ and $$u\sim v'$$, hence $$u'\in [v]\subseteq V_1$$. But we have $$\mathrm{cl}{A}\cap V_1=\emptyset$$, a contradiction. q.e.d.

Note 1: This question was originally entitled A closed quotient mapping such that the corresponding equivalence is not closed. The question was completely rewritten but remains equivalent.

Note 2: This question was mentioned in the comments to another question: When is a quotient by closed equivalence relation Hausdorff.

Note 3: Exercise 2.4.c (a) in Engelking's General Topology (1989) asks to find a topological space $$X$$ (without the requirement of Hausdorffness) and an equivalence relation $$\sim$$ on $$X$$ such that the quotient mapping $$q\colon X\to X/\mathord{\sim}$$ is closed but $$\sim$$ is not a closed subset of $$X\times X$$. There is no hint to the exercise.

Note 4: A closely related question is the following: Is the image of a Hausdorff space under a closed continuous mapping necessarily Hausdorff ? If this is true then we have also a positive answer to the original question. If the answer is 'no' and there exists a closed continuous surjection $$f$$ from a Hausdorff space $$X$$ onto a non-Hausdorff space $$Y$$ then this surjection is necessarily a quotient mapping, but it is not clear whether the equivalence relation $$\sim$$ defined by $$x\sim y$$ iff $$f(x)=f(y)$$ must be a closed subset of $$X\times X$$.

• note 3' s problem has a trivial solution: $\sim$ the identity, $X$ an infinite set in the cofinite topology. $q$ is a homeomorphism, hence closed, but $\sim$, which is just $\Delta(X)$, is not closed in $X^2$. To exclude it, we must restrict ourselves to Hausdorff $X$. – Henno Brandsma Jan 2 at 19:04

It seems that I have finally found an answer. By this post, there exists an open closed continuous mapping from a Hausdorff space onto a space that is not Hausdorff. It is well known that a closed continuous surjection is a quotient mapping. By this post, if $$X$$ is Hausdorff and the quotient map $$q\colon X\to X/\mathord{\sim}$$ is open, then $$X/\mathord{\sim}$$ is Hausdorff iff $$\sim$$ is closed in $$X\times X$$. Together we obtain an example of a Hausdorff space $$X$$ and a closed quotient mapping $$q\colon X\to X/\mathord{\sim}$$ such that the equivalence relation $$\sim$$ is not closed in $$X\times X$$. So the answer to the original question is 'NO'.