Countable Fort Space is Second-Countable

Theorem
Let $T = \left({S, \tau_p}\right)$ be a Fort space on a countably infinite set $S$.

Then $T$ is a second-countable space.

Proof
From Subset of Countable Set, $S \setminus \left\{{p}\right\}$ is countable.

Let $f: \N \to S$ be an enumeration of $S \setminus \left\{{p}\right\}$.

For brevity, let us write $s_n$ for $f \left({n}\right)$.

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

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, $S \setminus S_n = \left\{{s_0, \ldots, s_{n-1}}\right\}$ and so $S_n \in \tau_p$.

Also, define $S'_n = \left\{{s_n}\right\}$. Since $s_n \ne p$, $S'_n$ is open in $\tau_p$ as well.

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

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

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

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

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

It follows that $A = \left\{{n \in \N: s_n \in S \setminus U}\right\}$ 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, $A \subseteq \left\{{0, \ldots, N - 1}\right\}$ and hence:


 * $S \setminus U \subseteq \left\{{s_0, \ldots, s_{N-1}}\right\}$

From Complements Invert Subsets, also:


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

It follows that $\mathcal B$ is a basis for $\tau_p$.

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