a space $X$ is Hausdorff if and only if $\mathrm{\Delta }(X)$ is closed

Theorem.

A space $X$ is Hausdorff if and only if

$$\{(x,x)\in X\times X\mid x\in X\}$$

is closed in $X\times X$ under the product topology.

Proof.

First, some preliminaries: Recall that the diagonal map $\mathrm{\Delta }:X\to X\times X$ is defined as $x\stackrel{\mathrm{\Delta }}{⟼}(x,x)$. Also recall that in a topology generated by a basis (like the product topology), a set $Y$ is open if and only if, for every point $y\in Y$, there’s a basis element $B$ with $y\in B\subset Y$. Basis elements for $X\times X$ have the form $U\times V$ where $U,V$ are open sets in $X$.

Now, suppose that $X$ is Hausdorff. We’d like to show its image under $\mathrm{\Delta }$ is closed. We can do that by showing that its complement $\mathrm{\Delta }{(X)}^c$ is open. $\mathrm{\Delta }(X)$ consists of points with equal coordinates, so $\mathrm{\Delta }{(X)}^c$ consists of points $(x,y)$ with $x$ and $y$ distinct.

For any $(x,y)\in \mathrm{\Delta }{(X)}^c$, the Hausdorff condition gives us disjoint open $U,V\subset X$ with $x\in U,y\in V$. Then $U\times V$ is a basis element containing $(x,y)$. $U$ and $V$ have no points in common, so $U\times V$ contains nothing in the image of the diagonal map: $U\times V$ is contained in $\mathrm{\Delta }{(X)}^c$. So $\mathrm{\Delta }{(X)}^c$ is open, making $\mathrm{\Delta }(X)$ closed.

Now let’s suppose $\mathrm{\Delta }(X)$ is closed. Then $\mathrm{\Delta }{(X)}^c$ is open. Given any $(x,y)\in \mathrm{\Delta }{(X)}^c$, there’s a basis element $U\times V$ with $(x,y)\in U\times V\subset \mathrm{\Delta }{(X)}^c$. $U\times V$ lying in $\mathrm{\Delta }{(X)}^c$ implies that $U$ and $V$ are disjoint.

If we have $x\ne y$ in $X$, then $(x,y)$ is in $\mathrm{\Delta }{(X)}^c$. The basis element containing $(x,y)$ gives us open, disjoint $U,V$ with $x\in U,y\in V$. $X$ is Hausdorff, just like we wanted.∎