# Separable Space satisfies Countable Chain Condition

## Theorem

Let $T = \struct {S, \tau}$ be a separable space.

Then $T$ satisfies the countable chain condition.

## Proof

Let $T = \struct {S, \tau}$ be a separable space.

Then by definition there exists a countable subset of $X$ which is everywhere dense.

We need to show that every disjoint set of open sets of $T$ is countable.