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.
$\blacksquare$
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