# Fortissimo Space is not First-Countable

## Theorem

Let $T = \struct {S, \tau_p}$ 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 $\BB_p \subseteq \tau$

such that:

- $\forall B \in \BB_p: p \in B$

and such that:

- every open neighborhood of $p$ contains some $B \in \BB_p$.

So:

\(\displaystyle \bigcap \BB_p\) | \(=\) | \(\displaystyle \set p\) | |||||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle S \setminus \set p\) | \(=\) | \(\displaystyle S \setminus \bigcap \BB_p\) | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \bigcup_{B \mathop \in \BB_p} \paren {S \setminus B}\) | 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 $\displaystyle \bigcup_{B \mathop \in \BB_p} \paren {S \setminus B}$ is also countable.

So $S \setminus \set p$ 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.

$\blacksquare$

## Sources

- 1978: Lynn Arthur Steen and J. Arthur Seebach, Jr.:
*Counterexamples in Topology*(2nd ed.) ... (previous) ... (next): Part $\text {II}$: Counterexamples: $25$. Fortissimo Space: $1$