# Arens-Fort Topology is Topology

## Theorem

Let $T = \struct {S, \tau}$ be the Arens-Fort space.

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

## Proof

We have that $\tuple {0, 0} \notin \O$ so $\O \in \tau$.

We have that $\forall m: \set {n: \tuple {m, n} \notin S} = \O$ which is finite, so $S \in \tau$.

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

If $\tuple {0, 0} \notin A$ or $\tuple {0, 0} \notin B$ then $\tuple {0, 0} \notin A \cap B$ and so $H \in \tau_p$.

Now suppose $\tuple {0, 0} \in A$ and $\tuple {0, 0} \in B$.

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 Intersection

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

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

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

$S_{r_A} = \set {n: \tuple {r, n} \notin A}$
$S_{r_B} = \set {n: \tuple {r, n} \notin B}$

We have that:

 $\displaystyle$  $\displaystyle S_{r_{A \mathop \cap B} }$ $\displaystyle$ $=$ $\displaystyle \set {n: \tuple {r, n} \notin A \cap B}$ $\displaystyle$ $=$ $\displaystyle \set {n: \tuple {r, n} \notin A} \cup \set {n: \tuple {r, n} \notin B}$ 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 $\UU \subseteq \tau$.

Then:

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

We have either of two options:

$(1): \quad \forall U \in \UU: \tuple {0, 0} \notin U$

in which case:

$\displaystyle \tuple {0, 0} \notin \bigcap \UU$

Or:

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

in which case:

At most a finite number $m$ of sets $S_m = \set {n: \tuple {m, n} \notin \bigcap \UU}$ are infinite.

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

$\blacksquare$