# Fort Topology is Topology

## Theorem

Let $T = \struct {S, \tau_p}$ be a Fort 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 $\relcomp S S = \O$ which is finite, so $S \in \tau_p$.

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:

\(\displaystyle H\) | \(=\) | \(\displaystyle A \cap B\) | |||||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle \relcomp S H\) | \(=\) | \(\displaystyle \relcomp S {A \cap B}\) | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \relcomp S A \cup \relcomp S B\) | De Morgan's Laws: Complement of Union |

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

Hence their union is also finite and so $\relcomp S H$ is finite.

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

Now let $\UU \subseteq \tau_p$.

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

- $\displaystyle \relcomp S {\bigcup \UU} = \bigcap_{U \mathop \in \UU} \relcomp S U$

We have either of two options:

- $(1): \quad \forall U \in \U U: p \in \relcomp S U$

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 finite

in which case:

- $\displaystyle \bigcap_{U \mathop \in \UU} \relcomp S U$ is finite, from Intersection is Subset.

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

$\blacksquare$

## Sources

- 1970: Lynn Arthur Steen and J. Arthur Seebach, Jr.:
*Counterexamples in Topology*... (previous) ... (next): $\text{II}: \ 23 - 24$