Discrete Space is Non-Meager/Proof 1

Theorem

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

Then $T$ is non-meager.

Proof

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 $U$.

So:

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

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.

$\blacksquare$

Sources

• 1978: Lynn Arthur Steen and J. Arthur Seebach, Jr.: Counterexamples in Topology (2nd ed.) ... (previous) ... (next): Part $\text {II}$: Counterexamples: $1 \text { - } 3$. Discrete Topology: $9$