Discrete Space is Non-Meager/Proof 1

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Let $T = \struct {S, \tau}$ be a discrete topological space.

Then $T$ is non-meager.


Let $U \subseteq S$ such that $U \ne \O$.

From Interior Equals Closure of Subset of Discrete Space, we have:

$U^\circ = U = U^-$

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


$\paren {U^-}^\circ = U \ne \O$

Thus, by definition, no non-empty 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 meager.

Hence the result.