Countable Fort Space is Second-Countable

Theorem

Let $T = \struct {S, \tau_p}$ be a Fort space on a countably infinite set $S$.

Let $f: \N \to S$ be an enumeration of $S \setminus \set p$.

For brevity, let us write $s_n$ for $\map f n$.

Now define, for $n \in \N$, $S_n = S \setminus \set {s_0, \ldots, s_{n - 1} }$.

Note that $s_n \ne p$ for all $n \in \N$, so that $p \in S_n$ for all $n$.

By Relative Complement of Relative Complement, $S \setminus S_n = \set {s_0, \ldots, s_{n - 1} }$ and so $S_n \in \tau_p$.

Also, define $S'_n = \set {s_n}$.

Since $s_n \ne p$, $S'_n$ is open in $\tau_p$ as well.

From Finite Union of Countable Sets is Countable, $\BB := \set {S_n : n \in \N} \cup \set {S'_n : n \in \N}$ is countable.

Let us verify that $\BB$ forms a basis for $\tau_p$.

So let $U \subseteq S$ be open in $\tau_p$.

If $p \notin U$ then $\ds U = \bigcup \set {S'_n: s_n \in U}$ because the $s_n$ form an enumeration of $S \setminus \set p$.

Now if $p \in U$, then by definition of the Fort topology, $S \setminus U$ must be finite.

It follows that $A = \set {n \in \N: s_n \in S \setminus U}$ is also finite.

From Subset of Naturals is Finite iff Bounded, there exists an $N \in \N$ such that $N - 1 > n$ for all $n \in A$.

Therefore, $A \subseteq \set {0, \ldots, N - 1}$ and hence:

$S \setminus U \subseteq \set {s_0, \ldots, s_{N - 1} }$

$S_N = S \setminus \set {s_0, \ldots, s_{N - 1} } \subseteq U$

It follows that $\BB$ is a basis for $\tau_p$.

Since $\BB$ is countable, we conclude $T$ is second-countable.

$\blacksquare$