# Finite Complement Topology is Topology

## Theorem

Let $T = \struct {S, \tau}$ be a finite complement space.

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

## Proof

By definition, we have that $\O \in \tau$.

We also have that $S \in \tau$ as $\relcomp S S = \O$ which is trivially finite.

$\Box$

Let $A, B \in \tau$.

Let $H = A \cap B$.

Then:

 $\ds H$ $=$ $\ds A \cap B$ $\ds \leadsto \ \$ $\ds \relcomp S H$ $=$ $\ds \relcomp S {A \cap B}$ $\ds$ $=$ $\ds \relcomp S A \cup \relcomp S B$ De Morgan's laws: Complement of Intersection

But as $A, B \in \tau$ it follows that $\relcomp S A$ and $\relcomp S B$ are both finite.

Hence their union is also finite.

Thus $\relcomp S H$ is finite.

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

$\Box$

Let $\UU \subseteq \tau$.

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

But as:

$\forall U \in \UU: \relcomp S U \in \tau$

each of the $\relcomp S U$ is finite.

Hence so is their intersection.

So $\ds \relcomp S {\bigcup \UU}$ is finite.

So by definition:

$\ds \bigcup \UU \in \tau$

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

$\blacksquare$