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$X$ 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.

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    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?2012-01-02
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    @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.2012-01-02
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    @yaoxiao Very nice post.2012-04-03
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    @t.b. You could post this as an answer to the new question http://math.stackexchange.com/questions/1903343/give-an-example-of-a-non-compact-hausdorff-space-such-that-delta-is-closed-bu2016-08-25
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    Is the result still true if the quotient map is not open?2018-01-13

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