Fortissimo Topology is Topology

Theorem
Let $T = \struct {S, \tau_p}$ be a Fortissimo space.

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

Proof
We have that $p \in \relcomp S \O = S$ so $\O \in \tau_p$.

We have that $\relcomp S S = \O$ so $\O \in \tau_p$.

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:

In order for $A$ and $B$ to be open sets it follows that $\relcomp S A$ and $\relcomp S B$ are both countable.

From Countable Union of Countable Sets is Countable, their union is also countable and so $\relcomp S H$ is countable.

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

Now let $\UU \subseteq \tau_p$.

Then:
 * $\displaystyle \relcomp S {\bigcup \UU} = \bigcap_{U \mathop \in \UU} \relcomp S U$

by De Morgan's Laws: Complement of Union.

We have either of two options:
 * $(1): \quad \forall U \in \UU: p \in \relcomp S U$

in which case:
 * $\displaystyle p \in \bigcap_{U \mathop \in \UU} \relcomp S U$

Or:
 * $(2): \quad \exists U \in \UU: \relcomp S U$ is countable

in which case:
 * $\displaystyle \bigcap_{U \mathop \in \UU} \relcomp S U$ is countable, from Intersection is Subset.

So in either case $\displaystyle \bigcup \UU \in \tau_p$.