Discrete Space is Non-Meager/Proof 1
Jump to navigation
Jump to search
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: $9$