Hausdorff Space iff Diagonal Set on Product is Closed

Theorem

Let $T = \struct{S, \tau}$ be a topological space.

Let $\Delta_S$ be the diagonal set on $S$:

$\Delta_S = \set{\tuple{x, x} \in S \times S: x \in S}$

where $S \times S$ is the Cartesian product of $S$ with itself.

Let $T^2 = \struct{S \times S, \TT}$ be the product space with Tychonoff topology $\TT$.

Then $T$ is a $T_2$ (Hausdorff) space iff $\Delta_S$ is closed in $T^2$.

Proof

Suppose $T$ is a $T_2$ (Hausdorff) space.

Let $\Delta: S \to S \times S$ be the diagonal mapping on $S$.

We have that $\map \Delta S$ consists of all the elements of $S \times S$ whose coordinates are equal.

$\map \Delta S = \set{\tuple{x, x} \in S \times S: x \in S} = \set{\tuple{x, y} \in S \times S: x = y}$

By definition, $\Delta_S$ is the same set as $\map \Delta S$.

Now consider $H := \relcomp {S \times S} {\map \Delta S}$, its complement in $S \times S$.

We have that $H$ consists of all the elements of $S \times S$ whose coordinates are different:

$H = \set{\tuple{x, y} \in S \times S: x \ne y}$

Let $\tuple{x, y} \in H$.

Because $T$ is a $T_2$ space:

$\exists U, V \in \tau: x \in U, y \in V: U \cap V = \O$

Because $U \cap V = \O$ it follows that $U \times V \cap \map \Delta S = \O$.

That is:

$U \times V \subseteq H$

Now from Natural Basis of Tychonoff Topology of Finite Product, we have that $U \times V$ is an element of the natural basis of $T^2 = \struct{S \times S, \TT}$.

Hence $H$ is open in $T^2$.

Therefore, by definition, $\Delta_S$ is closed in $T^2 = \struct{S \times S, \TT}$.

$\Box$

Now suppose $\Delta_S$ is closed in $T^2 = \struct{S \times S, \TT}$.

Then $H = \set{\tuple{x, y} \in S \times S: x \ne y}$ is open in $T^2$.

Let $\tuple{x, y} \in H$.

We have that $x \ne y$ by the nature of $H$.

Then there exists an element $U \times V$ in the natural basis of $T^2$, and therefore open in $T^2$.

As $U \times V \subseteq H$ it follows that $U \cap V = \O$.

From Natural Basis of Tychonoff Topology of Finite Product, as $U \times V$ is open in $T^2$, then $U$ and $V$ are both open in $T$.

So, for any $\tuple{x, y}$ we have $U, V \subseteq S$ such that $x \in U, y \in V, U \cap V = \O$.

So $T$ is a $T_2$ (Hausdorff) space.

$\blacksquare$