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

Proof. First, some preliminaries: Recall that the diagonal map $\Delta\colon\thinspace X\to X\times X$ is defined as $x\stackrel{\Delta}{\longmapsto}(x,x)$ . Also recall that in a topology generated by a basis (like the product topology), a set

is open if and only if, for every point $y\in Y$ , there's a basis element

with $y\in B\subset Y$ . Basis elements for $X\times X$ have the form $U\times V$ where

are open sets in

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

For any $(x,y)\in \Delta(X)^c$ , 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 . and have no points in common, so $U\times V$ contains nothing in the image of the diagonal map: $U\times V$ is contained in $\Delta(X)^c$ . So $\Delta(X)^c$ is open, making $\Delta(X)$ closed.

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

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