Countable Complement Space Satisfies Countable Chain Condition

From ProofWiki
Jump to navigation Jump to search

Theorem

Let $T = \struct {S, \tau}$ 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