Fort Topology is Topology

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Theorem

Let $T = \left({S, \tau_p}\right)$ be a Fort 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 Union


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


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

Then from De Morgan's Laws: Complement of Union:

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


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 finite

in which case:

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


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

$\blacksquare$


Sources