Meager Sets in Arens-Fort Space

Theorem
Let $T = \left({S, \tau}\right)$ be the Arens-Fort space.

Let $A \subseteq S$.

Then $A$ is meager $A = \left\{{\left({0, 0}\right)}\right\}$.

Proof
First let $A = \left\{{\left({0, 0}\right)}\right\}$.

From the definition of Arens-Fort space, $\left\{{\left({0, 0}\right)}\right\}$ is closed because $S \setminus \left\{{\left({0, 0}\right)}\right\}$ is open.

From Closed Set Equals its Closure:
 * $\left\{{\left({0, 0}\right)}\right\}^- = \left\{{\left({0, 0}\right)}\right\}$

where $\left\{{\left({0, 0}\right)}\right\}^-$ denotes the closure of $\left\{{\left({0, 0}\right)}\right\}$.

From the definition of Arens-Fort space $\left\{{\left({0, 0}\right)}\right\}$ is not open.

Therefore the smallest open set contained in $\left\{{\left({0, 0}\right)}\right\}$ is $\varnothing$.

Hence:
 * $\left\{{\left({0, 0}\right)}\right\}^\circ = \varnothing$

where $\left\{{\left({0, 0}\right)}\right\}^\circ$ denotes the interior of $\left\{{\left({0, 0}\right)}\right\}$.

Thus we have that:
 * $\left({\left\{{\left({0, 0}\right)}\right\}^-}\right)^\circ = \varnothing$

and so by definition $\left\{{\left({0, 0}\right)}\right\}$ is nowhere dense in $T$.

From Union of Singleton it trivially follows that $\left\{{\left({0, 0}\right)}\right\}$ is the union of a countable set of subsets of $S$ (that is: just the one) which are nowhere dense in $S$.

Hence by definition $\left\{{\left({0, 0}\right)}\right\}$ is meager.

Now assume $A \ne \left\{{\left({0, 0}\right)}\right\}$.

Then $\exists x \in A: x \ne \left({0, 0}\right)$.

By definition of the Arens-Fort space, $\left\{{x}\right\}$ is open in $T$.

Thus, from Interior of Open Set $\left\{{x}\right\}^\circ = \left\{{x}\right\}$.

Hence $\left\{{x}\right\} \subseteq \left({\left\{{x}\right\}^-}\right)^\circ$ and by definition is not nowhere dense in $T$.

So $A$ is not the union of a countable set of subsets of $S$ which are nowhere dense in $S$.

So, by definition, $A$ is not meager.