Modified Fort Topology is Topology

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Theorem

Let $T = \struct {S, \tau_{a, b} }$ be a modified Fort space.


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


Proof

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


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


We have that $a, b \in S, S \setminus S = \O$ and $\O$ 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 $A \cap \set {a, b} \ne \O$ and $B \cap \set {a, b} \ne \O$.

Then:

\(\displaystyle H\) \(=\) \(\displaystyle A \cap B\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle N \setminus H\) \(=\) \(\displaystyle N \setminus \paren {A \cap B}\)
\(\displaystyle \) \(=\) \(\displaystyle \paren {N \setminus A} \cup \paren {N \setminus B}\) De Morgan's Laws: Difference with Intersection


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 $N \setminus \paren {A \cap B}$ is finite.

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


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

Then from De Morgan's Laws: Difference with Union:

$\displaystyle N \setminus \paren {\bigcup \UU} = \bigcap_{U \mathop \in \UU} \paren {N \setminus U}$


We have either of two options:

$(1): \quad \forall U \in \UU: U \subseteq N$

in which case:

$\displaystyle \bigcup \UU \subseteq N$


Or:

$(2): \quad \exists U \in \UU: N \setminus U$ is finite

in which case:

$\displaystyle \bigcap_{U \mathop \in \UU} \paren {N \setminus U}$ is finite, from Intersection is Subset.


So in either case:

$\displaystyle \bigcup \UU \in \tau_{a, b}$

$\blacksquare$


Sources