Modified Fort Topology is Topology

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

Then $\tau_{a, b}$ is a topology on $S$.

Proof
Let $S = N \cup \left\{{a, b}\right\}$ where $N$ is infinite, $a \ne b$ and $a, b \notin N$.

We have that $\varnothing \subseteq N$ so $\varnothing \in \tau_{a, b}$.

We have that $a, b \in S, S \setminus S = \varnothing$ and $\varnothing$ is trivially finite, so $S \in \tau_{a, b}$.

Now consider $A, B \in \tau_{a, b}$, and let $H = A \cap B$.

If $A \subseteq N$ and $B \subseteq N$ then $A \cap B \subseteq N$ from Intersection is Largest Subset.

So by definition $A \cap B \in \tau_{a, b}$

Now suppose $A \cap \left\{{a, b}\right\} \ne \varnothing$ and $B \cap \left\{{a, b}\right\} \ne \varnothing$.

Then:

In order for $A$ and $B$ to be open sets we have that $N \setminus A$ and $N \setminus B$ are both finite.

Hence their union is also finite and so $N \setminus \left ({A \cap B}\right)$ is finite.

So $H = A \cap B \in \tau_{a, b}$ as its complement is finite.

Now let $\mathcal U \subseteq \tau_{a, b}$.

Then from De Morgan's Laws: Difference with Union:
 * $\displaystyle N \setminus \left({\bigcup \mathcal U}\right) = \bigcap_{U \in \mathcal U} \left({N \setminus U}\right)$

We have either of two options:
 * $(1): \quad \forall U \in \mathcal U: U \subseteq N$

in which case:
 * $\displaystyle \bigcup \mathcal U \subseteq N$

Or:
 * $(2): \quad \exists U \in \mathcal U: N \setminus U$ is finite

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

So in either case $\displaystyle \bigcup \mathcal U \in \tau_{a, b}$.