Countable Complement Space Satisfies Countable Chain Condition

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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.

$\blacksquare$


Sources