Countable Complement Topology is Topology

Theorem
Let $T = \left({S, \tau}\right)$ be a countable complement space.

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

Proof
By definition, we have that $\varnothing \in \tau$.

We also have that $S \in \tau$ as $\complement_S \left({S}\right) = \varnothing$ 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 $\complement_S \left({A}\right)$ and $\complement_S \left({B}\right)$ are both countable.

Hence from Countable Union of Countable Sets is Countable their union is also countable and so $\complement_S \left({H}\right)$ is countable.

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

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

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

by De Morgan's laws.

But as:
 * $\forall U \in \mathcal U: \complement_S \left({U}\right) \in \tau$

each of the $\complement_S \left({U}\right)$ is countable.

Hence so is their intersection.

So $\displaystyle \complement_S \left({\bigcup \mathcal U}\right)$ is countable which means $\displaystyle \bigcup \mathcal U \in \tau$.

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