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Suppose $X$ is a topological space, $\sim$ is an equivalence relation and the quotient space $X/\sim$ is Hausdorff. How does this imply that $\{(x,y)\in X\times X|x\sim y\}$ is closed in $X\times X$?


That would mean that $(X\times X)\backslash R$ is open. But how does this follow from the fact that the quotient space is Hausdorff?

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  • $\begingroup$ @AymanHourieh This is not a duplicate, as the OP of the other post dismisses the direction Nicholas is interested in as “obvious.” $\endgroup$ – triple_sec Jun 11 '15 at 19:38
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A space $Y$ is Hausdorff iff the diagonal in $Y\times Y$ is closed. Note that $R$ actually is the diagonal in $(X/{\sim})\times (X/{\sim})$, just lifted up to $X\times X$.

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  • $\begingroup$ What exactly do you mean by "lifted up"? $\endgroup$ – RBS Jun 11 '15 at 19:32
  • $\begingroup$ For $R$ to be the diagonal line in (X/∼)×(X/∼), it seems that we need to have an open quotient map? thought it may not be a necessary condition. $\endgroup$ – kousaka May 15 '16 at 12:05
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Yes. To see this, suppose that $(x_{\alpha},y_{\alpha})_{\alpha\in A}$ is any net in $R$ converging (with respect to the product topology) to some $(x,y)\in X\times X$, where $(A,\succsim)$ is a non-empty directed index set. If we show that $(x,y)\in R$, that is, $x\sim y$, then the proof will be complete. (Basically, what we're trying to show here is that the closure of $R$, which is fully characterized by nets in general topological spaces, is included in $R$.)

For the sake of contradiction, suppose that $x\nsim y$. This implies that $\pi(x)\neq \pi(y)$, where $\pi:X\to X/\sim$ is the map sending each element to its equivalence class. Since $X/\sim$ is Hausdorff, there are disjoint open subsets $U$ and $V$ of $X/\sim$ such that $\pi(x)\in U$ and $\pi(y)\in V$. By the definition of the quotient topology, $\pi^{-1}(U)$ and $\pi^{-1}(V)$ are open in $X$, and $x\in\pi^{-1}(U)$ and $y\in\pi^{-1}(V)$. Since $x_{\alpha}\to x$ and $y_{\alpha}\to y$ in $X$ (a net converges in the product topology if and only if the component nets separately converge), one can find some $\alpha_0\in A$ such that $x_{\alpha_0}\in\pi^{-1}(U)$ and $y_{\alpha_0}\in\pi^{-1}(V)$. That is, $\pi(x_{\alpha_0})\in U$ and $\pi(y_{\alpha_0})\in V$. But $U$ and $V$ are disjoint, so $\pi(x_{\alpha_0})\neq\pi(y_{\alpha_0})$, and this contradicts $(x_{\alpha_0},y_{\alpha_0})\in R$.

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To elaborate on Hagen's answer. If the quotient space, $Y$, is hausdorff, its diagonal is closed.

Now, a quotient space is associated with a continuous quotient map,

$\pi : X \rightarrow Y$.

From continuity, an open set $U \in Y$, $\pi^{-1}(U)$ is open in $X$.

Now we an create a map from $f: X \times X \rightarrow Y \times Y$ with

$f(x_1,x_2) = \pi(x_1),\pi(x_2)$ which is also continous.

Now the off-diagonal elements in $Y\times Y$ form an open set as it is the complement of the closed diagonal.

So now $f^{-1}(Y\times Y \setminus D)$ is open, where $D$ is the diagonal.

This is also the complement of the set $X \times X : x_1 \sim x_2$. Therefore $X \times X : x_1 \sim x_2$ is closed.

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