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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 $X\times X$.

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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'.

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