Fortissimo Space is Excluded Point Space with Countable Complement Space

Theorem
Let $T = \left({S, \tau_p}\right)$ be a Fortissimo space.

Then $\tau_p$ is the minimal topology that is generated by the excluded point topology and the countable complement topology.

Proof
Let $T_1 = \left({S, \tau_1}\right)$ be the excluded point space on $S$ from $p$.

Let $T_2 = \left({S, \tau_2}\right)$ be the countable complement space on $S$.

By definition:
 * $\tau_1 = \left\{{H \subseteq S: p \in \complement_S \left({H}\right)}\right\} \cup \left\{{S}\right\}$


 * $\tau_2 = \left\{{H \subseteq S: \complement_S \left({H}\right)}\right.$ is countable $\left.{}\right\} \cup \left\{{\varnothing}\right\}$

By definition of Fortissimo space, we have:


 * $U \in \tau_1 \implies U \in \tau_p$


 * $U \in \tau_2 \implies U \in \tau_p$

So $\tau_1 \cup \tau_2 \subseteq \tau_p$.

Similarly:
 * $U \in \tau_p \implies U \in \tau_1 \lor U \in \tau_2$

and so $\tau_p \subseteq \tau_1 \cup \tau_2$.

So $\tau_p = \tau_1 \cup \tau_2$ and the result follows from Union Smallest.