The diagonal is not closed in $T_1$ space.

Question

Let $X$ be a topological space and let $\Delta = \{(x,x) : x\in X \}$ be the diagonal of $X\times X$ (with the product topology).

we know that $X$ is ${\rm T}_1$ if and only if $\Delta$ can be written as intersection of open subsets of $X\times X$.

I need a Counterexample for : $X$ is ${\rm T}_1$ ، but $\Delta = \{(x,x) : x\in X \}$ is not closed in $X\times X$ (with the product topology).

we know the cofinite topology on a set X is the coarsest topology on X that satisfies the ${\rm T}_1$ separation axiom.

how we can show $\Delta = \{(x,x) : x\in X \}$ is not close in $X\times X$ (with the product topology) and cofinite topology on a set X?

Answer 1

Let $X=\mathbb N$ equiped with the cofinite topology.

Then $X$ is $T_1$, but $\Delta$ is not closed in $X\times X$ as this property is equivalent to being a $T_2$ space, which $X$ isn’t.

Answer 2

It is well-known that the diagonal is closed iff $X$ is Hausdorff. Thus each non-Hausdorff $T_1$-space is a counterexample. The cofinite topology on an infinite set will do.

Answer 3

In particular, take two points $x,y$ in a non-$T_2$ space which cannot be separated by open sets. Then every neighborhood of $(x,y)$ intersects the diagonal.