# Fortissimo Topology is Topology

## Theorem

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

Then $\tau_p$ is a topology on $T$.

## Proof

We have that $p \in \complement_S \left({\varnothing}\right) = S$ so $\varnothing \in \tau_p$.

We have that $\complement_S \left({S}\right) = \varnothing$ so $\varnothing \in \tau_p$.

Now consider $A, B \in \tau_p$, and let $H = A \cap B$.

If $p \notin A$ or $p \notin B$ then $p \notin A \cap B$ and so $H \in \tau_p$.

Now suppose $p \in A$ and $p \in B$.

Then:

 $\displaystyle H$ $=$ $\displaystyle A \cap B$ $\displaystyle \implies \ \$ $\displaystyle \complement_S \left({H}\right)$ $=$ $\displaystyle \complement_S \left({A \cap B}\right)$ $\displaystyle$ $=$ $\displaystyle \complement_S \left({A}\right) \cup \complement_S \left({B}\right)$ De Morgan's Laws: Complement of Intersection

In order for $A$ and $B$ to be open sets it follows that $\complement_S \left({A}\right)$ and $\complement_S \left({B}\right)$ are both countable.

From Countable Union of Countable Sets is Countable, their union is also countable and so $\complement_S \left({H}\right)$ is countable.

So $H = A \cap B \in \tau_p$ as its complement is countable.

Now let $\mathcal U \subseteq \tau_p$.

Then:

$\displaystyle \complement_S \left({\bigcup \mathcal U}\right) = \bigcap_{U \mathop \in \mathcal U} \complement_S \left({U}\right)$

We have either of two options:

$(1): \quad \forall U \in \mathcal U: p \in \complement_S \left({U}\right)$

in which case:

$\displaystyle p \in \bigcap_{U \mathop \in \mathcal U} \complement_S \left({U}\right)$

Or:

$(2): \quad \exists U \in \mathcal U: \complement_S \left({U}\right)$ is countable

in which case:

$\displaystyle \bigcap_{U \mathop \in \mathcal U} \complement_S \left({U}\right)$ is countable, from Intersection is Subset.

So in either case $\displaystyle \bigcup \mathcal U \in \tau_p$.

$\blacksquare$