# Existence of Topological Space satisfying Countable Chain Condition which is not Separable

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## Theorem

There exists at least one example of a topological space which satisfies the countable chain condition which is not also a separable space.

## Proof

Let $T$ be a countable complement topology on an uncountable set.

From Countable Complement Space Satisfies Countable Chain Condition, $T$ satisfies the countable chain condition.

From Countable Complement Space is not Separable, $T$ is not a separable space.

Hence the result.

$\blacksquare$