Discrete Space is Non-Meager/Proof 1

Theorem
Let $T = \left({S, \tau}\right)$ be a topological space where $\tau$ is the discrete topology on $S$.

Then $T$ is second category.

Proof
Let $U \subseteq S$ such that $U \ne \varnothing$.

From Interior Equals Closure of Subset of Discrete Space, we have:
 * $U^\circ = U = U^-$

where $U^\circ$ is the interior and $H^-$ the closure of $U$.

So $\left({U^-}\right)^\circ = U \ne \varnothing$.

Thus, by definition, no non-null subset of $S$ is nowhere dense.

So $S$ can not be the union (countable or otherwise) of nowhere dense subsets.

So by definition $S$ can not be first category.

Hence the result.