Point in Topological Space is Open iff Isolated

Theorem
Let $T = \left({S, \tau}\right)$ be a topological space.

Let $x \in S$.

Then $\left\{{x}\right\}$ is open in $T$ $x$ is an isolated point of $T$.

Proof
Let $\left\{{x}\right\}$ be open in $T$.

Then we have that:
 * $\exists \left\{{x}\right\} \in \tau: x \in \left\{{x}\right\}\subseteq S$

This is precisely the condition which ensures that $x$ is an isolated point of $T$.

Now suppose that $x$ is an isolated point of $T$.

Then by definition there exists a open set of $T$ containing no points other than $x$:
 * $\exists U \in \tau: U = \left\{{x}\right\}$

That is, if $x$ is an isolated point of $T$ then $\left\{{x}\right\}$ is open in $T$.