# Indiscrete Space is Non-Meager

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## Theorem

Let $T = \struct {S, \set {\O, S} }$ be an indiscrete topological space.

Then $T$ is non-meager.

## Proof

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

From Limit Points of Indiscrete Space, every point of $S$ is a limit point of $H$.

So, by definition:

- $H^- = S$

where $H^-$ is the closure of $H$ in $T$.

Now the interior of $S$ is $S$ itself (trivially, by definition).

So:

- $\paren {H^-}^\circ = S \ne \O$

where $\paren {H^-}^\circ$ is the interior of the closure of $H$.

So, by definition, $H$ is not nowhere dense.

(Note that if $H = \O$, then Empty Set is Nowhere Dense applies, which is why the stipulation above that $H \ne \O$.)

It follows that $H$ can not be the countable union of nowhere dense subsets of $S$.

So $H$ is, by definition, not meager, so must be non-meager.

$\blacksquare$

## Sources

- 1978: Lynn Arthur Steen and J. Arthur Seebach, Jr.:
*Counterexamples in Topology*(2nd ed.) ... (previous) ... (next): Part $\text {II}$: Counterexamples: $4$. Indiscrete Topology: $5$