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Let $\sim$ be an equivalence relation on a topological space $X$, and let $Y = X/\sim$ be equipped with the quotient topology. How to show that if $X$ is Hausdorff and the set $\big\{ (x, y) : x \sim y \big\} \subseteq X \times X$ is closed, then $Y$ is Hausdorff.

My main question would be that is this true without the assumption that the quotient map is open.

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The result isn’t true in general.

Let $X$ be any Hausdorff space that it not normal, and let $H$ and $K$ be disjoint closed subsets of $X$ that cannot be separated by disjoint open sets. For $x,y\in X$ let $x\sim y$ if and only if $x=y$, or $x,y\in H$, or $x,y\in K$. Then

$$\{\langle x,y\rangle\in X\times X:x\sim y\}=\Delta_X\cup(H\times H)\cup(K\times K)\;,$$

which is closed in $X\times X$.

Let $p_H$ and $p_K$ be the points of $Y=X/{\sim}$ corresponding to the sets $H$ and $K$, respectively, and let $q:X\to Y$ be the quotient map. Suppose that $U$ and $V$ are open nbhds of $p_H$ and $p_K$, respectively. Then $q^{-1}[U]$ and $q^{-1}[V]$ are open nbhds of $H$ and $K$, respectively, so $q^{-1}[U]\cap q^{-1}[K]\ne\varnothing$, and therefore $U\cap V\ne\varnothing$. Thus, $Y$ is not Hausdorff.

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  • $\begingroup$ @BrianMScott thanks for that, I have been stuck on this question for ages, I was told it was true. $\endgroup$ – Topology guy May 16 '16 at 19:11
  • $\begingroup$ @Topologyguy: You’re welcome. $\endgroup$ – Brian M. Scott May 16 '16 at 19:13

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