Space with Open Point is Non-Meager

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Theorem

Let $T = \left({S, \tau}\right)$ 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:

$\left\{{x}\right\} \in \tau$


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:

$\displaystyle T = \bigcup \mathcal S$

where $\mathcal S$ is a countable set of subsets of $S$ which are nowhere dense in $S$.


Then:

$\exists H \in \mathcal S: x \in H$

and so:

$\left\{{x}\right\} \subseteq H$


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:

$\left\{{x}\right\} \subseteq H^-$


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