Arens-Fort Space is Expansion of Countable Fort Space

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

Let $T_p = \left({S, \tau_p}\right)$ be the Fort space on $S$ where $p = \left({0, 0}\right)$.

Then $\tau$ is an expansion of $\tau_p$.

Furthermore, $S$ is countably infinite, so $T_p$ is a countable Fort space.

Proof
Let $H \in \tau_p$ where $p = \left({0, 0}\right)$.

Then either:
 * $(1): \quad \left({0, 0}\right) \in \complement_S \left({H}\right)$

or:
 * $(2): \quad H$ is cofinite in $S$, i.e. $\complement_S \left({H}\right)$ is finite.

Case $(1)$ means that $\left({0, 0}\right) \notin H$ and so $H \in \tau$ by definition of the Arens-Fort space.

Suppose case $(2)$ applies.

Then for all $m \in \Z_+$, the sets $S_m$ defined as $S_m = \left\{{n: \left({m, n}\right) \notin H}\right\}$ are finite.

Thus in a finite number of these (i.e. none of them) the set $S_m$ is infinite.

So $H \in \tau$.

So, if $H \in \tau_p$ it follows that $H \in \tau$ and so $\tau_p \subseteq \tau$.

Hence the result by definition of expansion.

Finally, note that $\Z_+$ is straightforwardly countable.

From Cartesian Product of Countable Sets it follows that $S = \Z_+ \times \Z_+$ is likewise countable.