Fort Topology is Topology

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

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

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

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

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

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

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

Then:

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 finite.

Hence their union is also finite and so $\complement_S \left({H}\right)$ is finite.

So $H = A \cap B \in \tau$ as its complement is finite.

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

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

by De Morgan's laws.

We have either of two options:
 * $(1): \quad \forall U \in \mathcal U: p \in \complement_S \left({U}\right)$

in which case:
 * $\displaystyle p \in \bigcap_{U \in \mathcal U} \complement_S \left({U}\right)$

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

in which case:
 * $\displaystyle \bigcap_{U \in \mathcal U} \complement_S \left({U}\right)$ is finite, from Intersection Subset.

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