# Fortissimo Space is not First-Countable

## Theorem

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

Then $T$ is not a first-countable space.

## Proof

This proof follows the proof from Countable Complement Space is not First-Countable.

Aiming for a contradiction, suppose that $p \in S$ has a countable local basis.

That means:

there exists a countable set of sets $\mathcal B_p \subseteq \tau$

such that:

$\forall B \in \mathcal B_p: p \in B$

and such that:

every open neighborhood of $p$ contains some $B \in \mathcal B_p$.

So:

 $\displaystyle \bigcap \mathcal B_p$ $=$ $\displaystyle \left\{ {p}\right\}$ $\displaystyle \implies \ \$ $\displaystyle S \setminus \left\{ {p}\right\}$ $=$ $\displaystyle S \setminus \bigcap \mathcal B_p$ $\displaystyle$ $=$ $\displaystyle \bigcup_{B \mathop \in \mathcal B_p} \left({S \setminus B}\right)$ De Morgan's Laws: Difference with Intersection

By definition of the fortissimo space, each of $S \setminus B$ is countable.

From Countable Union of Countable Sets is Countable it follows that $\displaystyle \bigcup_{B \mathop \in \mathcal B_p} \left({S \setminus B}\right)$ is also countable.

So $S \setminus \left\{ {p}\right\}$ and therefore $S$ is also countable.

From this contradiction (as we have specified that $S$ is uncountable) it follows that our assumption that $p \in S$ has a countable local basis must be false.

Hence by definition $T$ can not be first-countable.

$\blacksquare$