Space with Open Point is Non-Meager
Theorem
Let $T = \struct {S, \tau}$ be a topological space.
Let $x \in S$ be an open point.
Then $T$ is a non-meager space.
Proof
Let $x \in S$ be an open point of $T$.
That is:
- $\set x \in \tau$
Recall that:
- a topological space is non-meager if it is not meager
and:
- a topological space is meager if and only if it is a countable union of subsets of $S$ which are nowhere dense in $S$.
Aiming for a contradiction, suppose that $T$ is meager.
Let:
- $\ds T = \bigcup \SS$
where $\SS$ is a countable set of subsets of $S$ which are nowhere dense in $S$.
Then:
- $\exists H \in \S S: x \in H$
and so:
- $\set x \subseteq H$
We have that $H$ is nowhere dense in $T$.
By definition, its closure $H^-$ contains no open set of $T$ which is non-empty.
But from Set is Subset of its Topological Closure we have that:
- $H \subseteq H^-$
So by Subset Relation is Transitive:
- $\set x \subseteq H^-$
So $H$ is not nowhere dense.
Therefore $T$ cannot be a countable union of subsets of $S$ which are nowhere dense in $S$.
That is, $T$ is not meager.
Hence the result by definition of non-meager.
$\blacksquare$
Sources
- 1978: Lynn Arthur Steen and J. Arthur Seebach, Jr.: Counterexamples in Topology (2nd ed.) ... (previous) ... (next): Problems: Section $1: \ 4$