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Let $\sim$ be an equivalence relation on a topological space $X$ such that $\{(x,y)\!:x\sim y\}$ is a closed subset of the product space $X\times X$. It is known that if $X$ is a compact Hausdorff space then the quotient space $X/\mathord{\sim}$ is normal, and hence Hausdorff. It is also known that $X/\mathord{\sim}$ need not to be Hausdorff if $X$ is not normal. But what about $X=\mathbb{R}$? Is non-normality a necessary condition in the last example?

Question: Is the quotient space $\mathbb{R}/\mathord{\sim}$ Hausdorff if $\sim$ is an equivalence relation on $\mathbb{R}$ such that $\{(x,y)\!:x\sim y\}$ is a closed subset of $\mathbb{R}\times\mathbb{R}$?

If an equivalence relation $\sim$ is closed in $X\times X$ then each equivalence class is closed, and hence $X/\mathord{\sim}$ is $T_1$. The Hausdorffness of $X/\mathord{\sim}$ is equivalent to the property that any two points of $X$ can be separated by open saturated neighborhoods (a set $A\subseteq X$ is saturated iff $(x\in A\wedge x\sim y)\Rightarrow y\in A$ holds true for all $x,y\in X$).

First attempt to find a counterexample failed. Assuming that $\sim$ is closed in $X\times X$, one can show that $X/\mathord{\sim}$ is Hausdorff if $X$ is Hausdorff and the quotient mapping $q\colon X\to X/\mathord{\sim}$ is open (here), or if $X$ is normal and $q$ is closed (this is essentially proved here). It is not difficult to find an equivalence relation $\sim$ that is closed in $\mathbb{R}\times\mathbb{R}$ and the quotient mapping $q\colon\mathbb{R}\to\mathbb{R}/\mathord{\sim}$ is neither open nor closed. An example is $x\sim y$ iff $x=y\vee(\sin x=\sin y\le 0)\vee(\sin x=\sin y=1)$, but $\mathbb{R}/\mathord{\sim}$ is Hausdorff in this case.

I will appreciate any advices, suggestions, or references.

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The answer is "yes", and the result can be found in Bourbaki's General Topology, Part I, Exercise 19 to $\S$ 10 of Chapter 1, on page 153 of the English translation published by Springer. The exercise has a hint, but it took me quite a time to fully elaborate the proof. So I try to present the argument with some more details included.

Let us first make clear some terminology. A neighbourhood of a set $A$ is any set containing an open set containing $A$. A space $X$ is compact if any open cover of $X$ has a finite subcover. A space is locally compact if it is Hausdorff and any point has a compact neighbourhood; it is $\sigma$-compact if it is a countable union of compact sets. A $\sigma$-locally compact space is a space which is both $\sigma$-compact and locally compact. A normal space is a space in which any two disjoint closed sets have disjoint neighbourhoods.

It is well known that every compact Hausdorff space is normal, and in a normal space, any two disjoint closed sets have disjoint closed neighbourhoods. A normal space in which every singleton is closed, is necessarily Hausdorff. If $X$ is compact and $Y$ is arbitrary then the projection to the second coordinate $\textit{pr}_2\colon X\times Y\to Y$ is a closed mapping. It is also known that every $\sigma$-locally compact Hausdorff space $X$ can be expressed as $\bigcup_{n\in\omega}X_n$, where $X_n$ are open, $\overline{X_n}$ are compact, and $\overline{X_n}\subseteq X_{n+1}$, for every $n$.

Let $E$ be an equivalence relation on $X$. Denote by $[A]$ the saturation of a set $A\subseteq X$, that is, $[A]=\{y\in X\!:(\exists x\in A)\,(x,y)\in E\}$. If $Y\subseteq X$ then $E_Y=E\cap(Y\times Y)$ is an equivalence relation on $Y$. The saturation of a set $A\subseteq Y$ with respect to $E_Y$ is $[A]\cap Y$. If $A,B\subseteq Y$ are disjoint sets saturated with respect to $E_Y$ then their saturations $[A]$, $[B]$ with respect to $E$ are disjoint as well.

Lemma. Let $X$ be a compact Hausdorff space and let $E$ be an equivalence relation on $X$ such that $E$ is a closed subset of $X\times X$. Then the quotient mapping $q\colon x\mapsto[x]$ is closed.

Proof. Let $A$ be a closed subset of $X$. Then $A$ is compact and $[A]=\textit{pr}_2(E\cap(A\times X))$. Since $E\cap(A\times X)$ is closed in $A\times X$ and $\textit{pr}_2$ is a closed mapping, $[A]$ is a closed subset of $X$. Hence $q$ is closed. q.e.d.

Lemma. Let $X$ be a normal space and let $E$ be an equivalence relation on $X$ such that the quotient mapping $q\colon x\mapsto [x]$ is closed. Then the quotient space $X/E$ is normal.

Proof. Let $A,B$ are disjoint closed saturated subsets of $X$. Since $X$ is normal, there exist disjoint open sets $U,V$ such that $A\subseteq U$ and $B\subseteq V$. Then $[X\setminus U]$ and $[X\setminus V]$ are closed saturated sets disjoint from $A$ and $B$, respectively, hence $U'=X\setminus [X\setminus U]$ and $V'=X\setminus [X\setminus V]$ are open saturated neighbourhoods of $A$ and $B$, respectively. If $z\in U'\cap V'$ then $[z]$ is disjoint from both $X\setminus U$ and $X\setminus V$, hence $z\in U\cap V$, a contradiction. It follows that $U',V'$ are disjoint, so any two disjoint closed subsets of $X/R$ have disjoint open neighbourhoods. q.e.d.

Theorem. Let $X$ be a $\sigma$-locally compact Hausdorff space and let $E$ be an equivalence relation on $X$ such that $E$ is a closed subset of $X\times X$. Then the quotient space $X/E$ is normal and Hausdorff.

Proof. Let $X=\bigcup_{n\in\omega}X_n$, where $X_n$ are open sets such that $\overline{X_n}$ are compact and $\overline{X_n}\subseteq X_{n+1}$ for every $n$. For every $n$ let us consider the quotient space $Y_n=\overline{X_n}/E_n$, where $E_n=E\cap(\overline{X_n}\times\overline{X_n})$. Since $\overline{X_n}$ is a compact Hausdorff space, the corresponding quotient mapping $q_n\colon\overline{X_n}\to Y_n$ is closed and hence the space $Y_n$ is normal.

Let $A,B$ be disjoint closed saturated subsets of $X$. We prove that there exist disjoint open saturated sets $U,V$ such that $A\subseteq U$ and $B\subseteq V$. For $n\in\omega$, define sets $A_n$, $B_n$, $U_n$, $V_n$ by induction as follows. Start with $A_0=A\cap\overline{X_0}$, $B_0=B\cap\overline{X_0}$. In $n$-th step, assume that $A_n$, $B_n$ are disjoint closed subsets of $\overline{X_n}$ saturated with respect to $E_n$. Assume also that $A\cap\overline{X_n}\subseteq A_n$ and $B\cap\overline{X_n}\subseteq B_n$. Then, by normality of $Y_n$, there exist disjoint relatively open subsets $U_n$, $V_n$ of $\overline{X_n}$ such that $A_n\subseteq U_n$, $B_n\subseteq V_n$, $\overline{U_n}\cap\overline{V_n}=\emptyset$, and $U_n$, $V_n$, $\overline{U_n}$, $\overline{V_n}$ are saturated with respect to $E_n$. We have $\overline{V_n}\subseteq\overline{X_n}$, hence $$A\cap\overline{V_n}\subseteq A\cap\overline{X_n}\cap\overline{V_n}\subseteq A_n\cap\overline{V_n}\subseteq U_n\cap\overline{V_n}=\emptyset,$$ similarly $B\cap\overline{U_n}=\emptyset$. Since $A,B$ are saturated with respect to $E$, we also have $A\cap\big[\,\overline{V_n}\,\big]=B\cap\big[\,\overline{U_n}\,\big]=\emptyset$. Let us take $A_{n+1}=\big(A\cup\big[\,\overline{U_n}\,\big]\big)\cap\overline{X_{n+1}}$, $B_{n+1}=\big(B\cup\big[\,\overline{V_n}\,\big]\big)\cap\overline{X_{n+1}}$. It is easy to check that $(A\cap\overline{X_{n+1}})\cup U_n\subseteq A_{n+1}$, $(B\cap\overline{X_{n+1}})\cup V_n\subseteq B_{n+1}$, and that $A_{n+1}$, $B_{n+1}$ are closed and saturated with respect to $E_n$. Since $\overline{U_n}$, $\overline{V_n}$ are disjoint subsets of $\overline{X_n}$ saturated with respect to $E_n$, their saturations $\big[\,\overline{U_n}\,\big]$, $\big[\,\overline{V_n}\,\big]$ with respect to $E$ are disjoint as well, and we obtain that $A_{n+1}\cap B_{n+1}=\emptyset$. Hence we can proceed with the induction.

This way we can define increasing sequences $\{U_n\}_{n\in\omega}$, $\{V_n\}_{n\in\omega}$ such that $U_n$, $V_n$ are relatively open subsets of $\overline{X_n}$, $A\cap\overline{X_n}\subseteq U_n$, $B\cap\overline{X_n}\subseteq V_n$, $\overline{U_n}\cap\overline{V_n}=\emptyset$, and $U_n$, $V_n$, $\overline{U_n}$, $\overline{V_n}$ are saturated with respect to $E_n$. Set $U=\bigcup_{n\in\omega}U_n$, $V=\bigcup_{n\in\omega}V_n$. If $x\in U$ then there exists $n$ such that $x\in U_n\cap X_n\subseteq U$, where $U_n\cap X_n$ is open, hence $U$ is open. Also, if $x\in[U]$ then there exists $n$ such that $x\in\big[\,\overline{U_n}\,\big]\cap\overline{X_{n+1}}$, hence $x\in U$. So $U$ is an open set saturated with respect to $E$, similarly for $V$. Finally, $U$ and $V$ are disjoint since if $x\in U\cap V$ then there exists $n$ such that $x\in U_n\cap V_n$, which is impossible.

We have proved that the quotient space $X/E$ is normal. To see that it is also Hausdorff it suffices to show that every singleton in $X/E$ is a closed set. This is clear since if $E$ is a closed subset of $X\times X$ then every equivalence class is closed. q.e.d.

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  • $\begingroup$ How do you know $\overline{U_n}$ is saturated? Closure of saturated set need not be saturated. $\endgroup$ – YuiTo Cheng Aug 7 at 1:46
  • $\begingroup$ @YuiToCheng To any closed saturated set $S\subseteq\overline{X_n}$ corresponds a closed set of equivalence classes in the space $Y_n=\overline{X_n}/E_n$: $S^*=\{\{y\in\overline{X_n}\!:(x,y)\in E_n\}\!:x\in S\}$. Moreover, if $T\subseteq Y_n$ is closed then $T=S^*$ where $S=\bigcup T$. Now, if $A_n$, $B_n$ are disjoint closed saturated subsets of $\overline{X_n}$ then $A^*_n$, $B^*_n$ are disjoint closed subsets of $Y_n$, hence by normality of $Y_n$ they have disjoint closed neighbourhoods $C_n\supseteq A^*_n$, $D_n\supseteq B^*_n$. We take $U_n=\bigcup C_n$, $V_n=\bigcup D_n$. $\endgroup$ – Peter Elias Aug 7 at 10:03
  • $\begingroup$ Then why is $\overline{U_n}=\overline{\bigcup{C_n}}$ saturated? $\endgroup$ – YuiTo Cheng Aug 7 at 11:03
  • $\begingroup$ @YuiToCheng Each element of $C_n$ is a point of $Y_n$, and hence a saturated subset of $\overline{X_n}$. $U_n$ is the union of these sets, hence it is saturated. Ah, and now I see the point in your question. $U_n$ is also a closed subset of $\overline{X_n}$ since $C_n$ is closed in $Y_n$. So $\overline{U_n}=U_n$. $\endgroup$ – Peter Elias Aug 7 at 14:09
  • $\begingroup$ But infinite union of closed sets need not be closed... $\endgroup$ – YuiTo Cheng Aug 7 at 14:13
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This is not a solution, but here are some ideas to consider.

let $S=\{(x,y): x \sim y\})$. For sake of argument, we can define $\sim$ is given by $x \sim y$ if $y=2x$ for simplicity, and then attempt to generalize this for an arbitrary equivalence relation. Suppose we had some continuous map $\phi: \mathbb{R}\setminus \sim\rightarrow \mathbb{R^2}$. If we let $S$ be closed, then $\phi^{-1}(S)\subset \mathbb{R}\setminus \sim$ is closed by continuity. There are a few possible approaches we can consider.

$(1)$ To show that $\mathbb{R}\setminus \sim$ is Hausdorff (or not), we need to argue that every point can be separated by open sets (or not). We can consider singleton sets $\phi^{-1}(\{s\})\subset \mathbb{R}\setminus \sim$, where $s\in S$, and try to find some topological argument as to why these sets can be separated by open sets or not.

$(2)$ We can try to use the fact that limits of sequences in a Hausdorff space are unique (somehow)?

$(3)$ Somehow use the theorem that a space $X$ is Hausdorff iff the diagonal $D=X\times X$ is closed, taking the equivalence relation into account. Note that the condition given is not sufficient to argue if $S$ is closed, then $D$ is closed, as we don't know if $D=\mathbb{R}\setminus \sim \times \mathbb{R}\setminus \sim$ is necessarily closed. However, if you can show it is closed, then $\mathbb{R}\setminus \sim$ is necessarily Hausdorff.

Hope this helps!

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