Show that X is Hausdorff if and only if the diagonal $\Delta$= {x $\times$ x : x$\in$ X} is closed in X $\times$ X.

Let X be Hausdorff.
We have to show that $\Delta$ is closed in X$\times$X .
Let x $\times$ y$ \in$ X$ \times $Y such that $x \neq y$ .

Aim:  x$\times$y is not a limit point of of A .

Since X is Hausdorff and x $\neq$ y , so $\exists$ neighbourhoods U and V of x and y respectively such that U $\cap$ V= $\phi$.

Clearly U $\times$ V is a neighbourhood of x$\times$y and U $\times$ V doesn't intersect A (If U $\times$ V intersects A , then some p $\times$ p $\in$U $\times$ V, ie p$\in$U and p$\in$V, which contradicts U $\cap$ V = $\phi$).

$\therefore$  x$\times$y is not a limit point of of A.
$\therefore$ A is closed in X$\times$X .

Conversely , let A be closed on X$\times$X .

Aim : X is Hausdorff.

Let x,y $\in$ X such that x $\neq$ y .

x$\times$y $\not\in$ A. Since A is closed , x$\times$y is not a limit point of of A.

So $\exists$ neighbourhoods U $\times$ V of x $\times$ y respectively such that U $\times$ V doesn't intersect A , which means U $\cap$ V= $\phi$ .

(Since if p $\in$ U $\cap$ V $\Rightarrow$ p$\in$ U and p$\in$ V $\Rightarrow$ p$\times$p $\in$ U$\times$V $\Rightarrow$ U $\times$ V intersects A, a contradiction).

That means $\exists$ neighbourhoods U and V of x and y respectively such that U $\cap$ V= $\phi$.

$\therefore$ X is Hausdorff.