Theorem. A space is Hausdorff iff its diagonal map is closed [math-000B]

The proposition says a space is Hausdorff if and only if its diagonal map $\Delta : X \to X \times X$ to product space is closed, which given an alternative definition. Notice that close means its complement $\Delta^\complement$ is open. The definition of $\Delta$ is

Not hard to see below argument also applies to all finite $X$ -product spaces.

Proof

(Hausdorff => diagonal map is closed) Hausdorff means every two different points has a pair of disjoint neighborhoods $(U \in \mathcal{N}_x, V \in \mathcal{N}_y)$ , where $x \in U$ and $y \in V$ . Therefore, every pair $(x, y)$ not line on the diagonal has $U \times V$ cover them. The union of all these open sets $U \times V$ covers $\Delta^\complement$ , so $\Delta$ the complement of the union is closed.

(diagonal map is closed => Hausdorff) Since

is open, which implies for all $x, y$ the pair $(x, y) \in \Delta^\complement$ . Since $\mathcal{N}_{(x, y)} = \mathcal{N}_x \times \mathcal{N}_y$ , this implies the fact that $\Delta^\complement \in \mathcal{N}_x \times \mathcal{N}_y$ .

Also, for all $U \in \mathcal{N}_x$ and $V \in \mathcal{N}_y$ , it's natural that $U \times V \subseteq \Delta^\complement$ , since the open set $U \times V$ at most cover $\Delta^\complement$ .

Consider that reversely again, that means for all $(x, x) \in \Delta$ , pair $(x, x) \notin U \times V$ (i.e. $U \times V$ will not cover any part of diagonal), that implies $U \cap V = \emptyset$ as desired.