Space with Open Point is Non-Meager

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

Let $x \in S$ be an open point.

Then $T$ is a non-meager space.

Proof
Suppose $x \in S$ is an open point of $T$.

That is, $\left\{{x}\right\} \in \tau$.

A topological space is non-meager if it is not meager.

A topological space is meager iff it is a countable union of subsets of $S$ which are nowhere dense in $S$.

Suppose $T$ were meager.

Let:
 * $\displaystyle T = \bigcup \mathcal S$

where $\mathcal S$ is a countable set of subsets of $S$ which are nowhere dense in $S$.

Then $\exists H \in \mathcal S: x \in H$, and so $\left\{{x}\right\} \subseteq H$.

As $H$ is nowhere dense in $T$ its closure $H^-$ contains no open set of $T$ which is not empty.

But from Set is Subset of its Topological Closure we have that $H \subseteq H^-$ and so by Subset Relation is Transitive $\left\{{x}\right\} \subseteq H^-$.

So $H$ is not nowhere dense and so $T$ can not be the union of a countable set of subsets of $S$ which are nowhere dense in $S$.

That is, $T$ is not meager.

Hence the result by definition of non-meager.