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 $T$ iff $x$ is an isolated point of $T$.

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

The definition of a neighborhood $N_x$ of $\left\{{x}\right\}$ is:
 * $\exists U \in \tau: x \in U \subseteq N_x \subseteq S$

As $\left\{{x}\right\}$ is open we have that:
 * $\exists \left\{{x}\right\} \in \vartheta: x \in \left\{{x}\right\}\subseteq S$

demonstrating that $\left\{{x}\right\}$ itself is a neighborhood of $x$.

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 neighborhood of $x$ in $T$ containing no points other than $x$:
 * $\exists U \in \tau: x \in U \subseteq \left\{{x}\right\} \subseteq S$

from which it follows that $U = \left\{{x}\right\}$ and so $\left\{{x}\right\} \in \tau$.

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