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:

$\tau = \powerset X$

if and only if:

$\Delta$ is open in $X$.

That is:

the topology on $X$ is discrete if and only if $\Delta$ is open in $X$.

Proof

Sufficient Condition

Suppose that $\tau = \powerset X$.

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

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

is open.

Note that we can write:

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

This is the union of open sets, so from the definition of a topology we have that:

$\Delta$ is open.

$\Box$

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 is Discrete iff All Singletons are Open, we have that:

the topology on $X$ is discrete.

$\blacksquare$