# Separable Space satisfies Countable Chain Condition

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## 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.

## Sources

- 1978: Lynn Arthur Steen and J. Arthur Seebach, Jr.:
*Counterexamples in Topology*(2nd ed.) ... (previous) ... (next): Part $\text I$: Basic Definitions: Section $3$: Compactness: Countability Axioms and Separability