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Let $X$ be a topological space. The diagonal of $X \times X$ is the subset $$D = \{(x,x)\in X\times X\mid x \in X\}.$$
Show that $X$ is Hausdorff if and only if $D$ is closed in $X \times X$.

First, I tried to show that $X \times X \setminus D$ is open using the fact that $X \times X$ is Hausdorff (because $X$ is Hausdorff), but I couldn't find an open set that contains a point outside $D$ and is disjoint to it...

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    $\begingroup$ One part is here and a generalization of this question can be found here. $\endgroup$
    – Asaf Karagila
    Apr 25, 2012 at 19:37
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    $\begingroup$ you should add that topology on $X \times X$ is product topology $\endgroup$ Jun 4, 2016 at 19:30

4 Answers 4

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You have two things to show: that if $D$ is closed, then $X$ is Hausdorff, and that if $X$ is Hausdorff, then $D$ is closed.

Suppose first that $D$ is closed in $X\times X$. To show that $X$ is Hausdorff, you must show that if $x$ and $y$ are any two points of $X$, then there are open sets $U$ and $V$ in $X$ such that $x\in U$, $y\in V$, and $U\cap V=\varnothing$. The trick is to look at the point $p=\langle x,y\rangle\in X\times X$. Because $x\ne y$, $p\notin D$. This means that $p$ is in the open set $(X\times X)\setminus D$. Thus, there must be a basic open set $B$ in the product topology such that $p\in B\subseteq(X\times X)\setminus D$. Basic open sets in the product topology are open boxes, i.e., sets of the form $U\times V$, where $U$ and $V$ are open in $X$, so let $B=U\times V$ for such $U,V\subseteq X$. Can you see how to use this to get the desired open neighborhoods of $x$ and $y$?

Now suppose that $X$ is Hausdorff. To show that $D$ is closed in $X\times X$, you need only show that $(X\times X)\setminus D$ is open. To do this, just take any point $p\in(X\times X)\setminus D$ and show that it has an open neighborhood disjoint from $D$. I suggest that you try to reverse what I did above. First, $p=\langle x,y\rangle$ for some $x,y\in X$, and since $p\notin D$, $x\ne y$. Now use disjoint open neighborhoods of $x$ and $y$ to build a basic open box around $p$ that is disjoint from $D$.

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    $\begingroup$ Suppose that $U$ is a neighborhood of $x$, $V$ is a neighborhood of $y$, then $U\times V$ is open and contains in $X\times X\setminus D$. How can I show that $X\times X\setminus D$ is open from that? (because I think that a closed set can contains open set) $\endgroup$
    – Arsenaler
    Jun 24, 2013 at 2:39
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    $\begingroup$ @Arsenaler: A set is open if it contains an open nbhd of each of its points. The argument of the last paragraph shows that $(X\times X)\setminus D$ contains an open nbhd of each of its points. $\endgroup$ Jun 24, 2013 at 4:45
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    $\begingroup$ @Tim: Since $\langle x,y\rangle\in U\times V$, we have $x\in U$ and $y\in V$. If there were a point $z\in U\cap V$, then $\langle z,z\rangle$ would be in $U\times V$. But $B\cap D=\varnothing$, so this is impossible. $\endgroup$ Jan 24, 2016 at 21:58
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    $\begingroup$ @Tim: Because $B$ was chosen to miss the diagonal: $B\subseteq(X\times X)\setminus D$. We can do this because the diagonal is closed. $\endgroup$ Jan 24, 2016 at 22:03
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    $\begingroup$ @kmecpp: If $|X|\le 1$, the diagonal is either empty or $X\times X$, so it’s automatically closed (and $X$ is vacuously Hausdorff). $\endgroup$ Oct 2, 2020 at 23:16
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There is a related problem:

Let $f,g:A\to B$ be two continuous maps, $B$ is a Hausdorff space and $A$ is an arbitrary space. Then the set $C(f,g)=\{x\in A| f(x)=g(x)\}$ is the coincidence set of $f$ and $g$. Prove $C(f,g)$ is closed.

to proof this, for every point $x_0$ not in $C(f,g)$, $y_f=f(x_0)$ and $y_g=g(x_0)$ are two distinguished point in $B$, because $B$ is Hausdorff, you get two disjoint open neighborhood $U(y_f)$ and $U(y_g)$, because $f,g$ is continuous, $O=f^{-1}(U(y_f))\cap g^{-1}(U(y_g))$ is open in $A$. You can verify that $O$ is a neighborhood of $x_0$ disjoint of $C(f,g)$, for if $c\in C(f,g)$ and $c\in O$, then $c\in f^{-1}(U(y_f))$ and $c\in g^{-1}(U(y_g))$, this is equal to say $f(c)=g(c)$ is in both $U(y_f)$ and $U(y_g)$, which is impossible because the two neighborhood is disjoint.

Set $A=X\times X$, $B=X$, $f=\pi_1$, $g=\pi_2$ be the projection map, you get the desired result.

Hope this can illustrate the problem you ask.

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    $\begingroup$ Do you mean this one? $\endgroup$
    – user636532
    Apr 25, 2019 at 10:51
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Let $(x,y)\in X\times X$, with $x\neq y$. Since $x,y\in X$, $x\neq y$, and $X$ is Hausdorff, there exist open set $\mathcal{U}$ and $\mathcal{V}$ such that $x\in\mathcal{U}$, $y\in\mathcal{V}$, and $\mathcal{U}\cap\mathcal{V}=\varnothing$. Now, $\mathcal{U}\times\mathcal{V}$ is an open subset of $X\times X$. It contains $(x,y)$. Does it intersect $D$?

Conversely, suppose $D$ is closed, and let $x,y\in X$, $x\neq y$. Then $(x,y)\notin D$, so there is an open subset $\mathscr{O}$ such that $(x,y)\in\mathscr{O}\subseteq X\times X - D$. Do you know something about open sets of a special type in $X\times X$ that might let you obtain open sets of $X$ $\mathcal{U}$ and $\mathcal{V}$ as in the previous paragraph?

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  • $\begingroup$ Suppose it didn't intersect $D$, why would I care? $\endgroup$ Aug 15, 2020 at 20:24
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    $\begingroup$ @JosvanNieuwman If you are asking about the first paragraph: you care because, presumably, you are trying to prove that $D$ is closed. So you are trying to find an open set that contains $(x,y)$ and does not intersect $D$, in order to show the complement of $D$ is open. What wasn’t clear about that? $\endgroup$ Aug 15, 2020 at 20:33
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    $\begingroup$ @JosvanNieuwman: No, that’s not what is happening. A set $U$ is open if and only if for every $x\in U$ there exists an open neighborhood $V$ of $x$ (a set $V$ that is open and contains $x$) such that $V\subseteq U$. A set is closed if and only if its complement is open. So you can prove that $D$ is closed by showing that for every $x\notin D$, there is an open set $V$ that contains $x$ and does not intersect $V$. It’s not a single open set $U$, it’s a set $V_x$ that depends on $x$: one for every point in the complement. $\endgroup$ Aug 15, 2020 at 20:58
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    $\begingroup$ @JosvanNieuwman: Your point of view is incomplete. What was proven is that for every point $(x,y)$ in the complement, there is an open set that contains the point and is contained in the complement. Once you put all those open sets together, then you conclude the complement is open, and hence the set is closed. $\endgroup$ Aug 15, 2020 at 20:59
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    $\begingroup$ @JosvanNieuwman: The result being used is: “If for every $\mathbf{x}\notin D$ there exists an open set $U\subseteq X$ such that $\mathbf{x}\in U$ and $U\cap D=\varnothing$, then $D$ is closed.” $\endgroup$ Aug 15, 2020 at 21:02
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Let X be Hausdorff, then if $x\ne y$ there are neighborhoods $V_x$ and $V_y$ such that $V_x \cap V_y = \emptyset$. Therefore $V_x\times V_y \cap D=\emptyset$ and the complement of $D$ is open. Now, assume that the latter is true. Then, for any point $(x,y)$, $x\ne y$, there is an open set around it that does not intersect $D$. Therefore, there are two sets $x\in V_x$ and $y\in V_y$ such that $V_x\times V_y$ doesn't intersect $D$, therefore $V_x \cap V_y = \emptyset$.

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