Discrete Space iff Diagonal Set on Product is Open

Theorem
Let $\struct {X, \tau}$ be a topological space.

Endow $X \times X$ with the product topology.

Let:


 * $\Delta = \set {\tuple {x, x} : x \in X}$

Then:


 * $\tau = X$ $\Delta$ is open in $X$.

That is:


 * the topology on $X$ is discrete $\Delta$ is open in $X$.

Sufficient Condition
Suppose that $\tau = X$.

Then for each $x \in X$, the set:


 * $\set {\tuple {x, x} }$

is open.

Note that we can write:


 * $\ds \Delta = \bigcup_{x \in X} \set {\tuple {x, x} }$

Then from the definition of a topology, we have that:


 * $\Delta$ is open.

Necessary Condition
Suppose that $\Delta$ is open.

Let $x \in X$ so that $\tuple {x, x} \in \Delta$.

From the definition of the product topology:


 * there exists open sets $S, R \subseteq X$ such that $\tuple {x, x} \in S \times R \subseteq \Delta$.

We want to conclude that:


 * $S = R = \set x$

Certainly:


 * $\set x \subseteq S$

and:


 * $\set x \subseteq R$

Suppose that:


 * $S \ne \set x$

then:


 * there exists some $y \in S$ with $y \ne x$.

But then:


 * $\tuple {y, x} \in S \times R$

while, since $y \ne x$, we have:


 * $\tuple {y, x} \not\in \Delta$

This contradicts:


 * $S \times R \subseteq \Delta$

So we must have:


 * $S = \set x$

So, since $x$ was arbitrary:


 * for each $x \in X$, the set $\set x$ is open.

So, from Topology Discrete iff All Singletons Open, we have that:


 * the topology on $X$ is discrete.