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 '12 at 19:37
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    $\begingroup$ you should add that topology on $X \times X$ is product topology $\endgroup$ – Filip Parker Jun 4 '16 at 19:30

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 $D$.

  • $\begingroup$ 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 '13 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. $\endgroup$ – Brian M. Scott Jun 24 '13 at 4:45
  • $\begingroup$ If $\left(U\times V\right)\cap\Delta=\phi$ can we immediately conclude that $U\cap V=\phi$? $\endgroup$ – JEM Oct 2 '14 at 20:19
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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. $\endgroup$ – Brian M. Scott Jan 24 '16 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. $\endgroup$ – Brian M. Scott Jan 24 '16 at 22:03

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.


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?


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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