# Countable Complement Space Satisfies Countable Chain Condition

## Theorem

Let $T = \left({S, \tau}\right)$ be a countable complement topology on an uncountable set $S$.

Then $T$ satisfies the countable chain condition.

## Proof

We have that a Countable Complement Space is Irreducible.

Therefore there is no disjoint set of open sets of $T$.

So the result holds vacuously.

$\blacksquare$