Countable Complement Topology is Topology

Theorem
Let $T = \struct {S, \tau}$ be a countable 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 $\complement_S \left({S}\right) = \O$ which is trivially countable.

Now suppose $A, B \in \tau$.

Let $H = A \cap B$.

Then:

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

Hence from Countable Union of Countable Sets is Countable their union is also countable and so $\relcomp S H$ is countable.

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

Now let $\UU \subseteq \tau$.

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

by De Morgan's Laws: Complement of Union.

But as:
 * $\forall U \in \UU: \relcomp S U \in \tau$

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

Hence so is their intersection.

So $\displaystyle \relcomp S {\bigcup \UU}$ is countable which means:
 * $\displaystyle \bigcup \UU \in \tau$

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