# Arens-Fort Topology is Topology

## Theorem

Let $T = \left({S, \tau_p}\right)$ be the Arens-Fort space.

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

## Proof

We have that $\left({0, 0}\right) \notin \varnothing$ so $\varnothing \in \tau$.

We have that $\forall m: \left\{{n: \left({m, n}\right) \notin S}\right\} = \varnothing$ which is finite, so $S \in \tau$.

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

If $\left({0, 0}\right) \notin A$ or $\left({0, 0}\right) \notin B$ then $\left({0, 0}\right) \notin A \cap B$ and so $H \in \tau_p$.

Now suppose $\left({0, 0}\right) \in A$ and $\left({0, 0}\right) \in B$.

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 Intersection

In order for $A$ and $B$ to be open sets it follows that:

At most a finite number $m_A$ of sets $S_{m_A} = \left\{{n: \left({m_A, n}\right) \notin A}\right\}$ are infinite
At most a finite number $m_B$ of sets $S_{m_B} = \left\{{n: \left({m_B, n}\right) \notin B}\right\}$ are infinite.

Let $r \in \Z_{\ge 0}$ and let:

$S_{r_A} = \left\{{n: \left({r, n}\right) \notin A}\right\}$
$S_{r_B} = \left\{{n: \left({r, n}\right) \notin B}\right\}$

We have that:

 $\displaystyle$  $\displaystyle S_{r_{A \mathop \cap B} }$ $\displaystyle$ $=$ $\displaystyle \left\{ {n: \left({r, n}\right) \notin A \cap B}\right\}$ $\displaystyle$ $=$ $\displaystyle \left\{ {n: \left({r, n}\right) \notin A}\right\} \cup \left\{ {n: \left({r, n}\right) \notin B}\right\}$ De Morgan's Laws: Difference with Intersection $\displaystyle$ $=$ $\displaystyle S_{r_A} \cup S_{r_B}$

If $S_{r_A}$ and $S_{r_B}$ are finite then so is $S_{r_A} \cup S_{r_B}$.

If only a finite number of $S_{m_A}$ and $S_{m_B}$ are infinite then it follows that only a finite number of $S_{m_{A \cap B}}$ are infinite.

So $A \cap B \in \tau$.

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

Then:

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

We have either of two options:

$(1): \quad \forall U \in \mathcal U: \left({0, 0}\right) \notin U$

in which case:

$\displaystyle \left({0, 0}\right) \notin \bigcap \mathcal U$

Or:

$(2): \quad$ At most a finite number $m$ of sets $S_m = \left\{{n: \left({m, n}\right) \notin U}\right\}$ are infinite

in which case:

At most a finite number $m$ of sets $S_m = \left\{{n: \left({m, n}\right) \notin \bigcap \mathcal U}\right\}$ are infinite.

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

$\blacksquare$