I'm trying to find a proof behind a small proposition. Recall that a topological space satisfies the countable chain condition if each disjoint collection of open sets is countable.
Why is it the case that every separable spaces satisfies the CCC, but the converse is not true?
Thanks.
Best Answer
The Suslin line is a topological space which has the CCC property but is not separable.
However, proving that the Suslin line exists cannot be done within $ZFC$. Why?
Assuming $V=L$ (The axiom of constructibility) implies certain combinatorial properties from which we can construct a Suslin line, however assuming a different axiom $MA$ (Martin's axiom) we can prove that no Suslin line exists.
The result of this is that we cannot prove from ZFC alone that every CCC space is separable.
Added: (To make this answer complete, I'll add the right answer given by Henno Brandsma in the comments)
We cannot prove in ZFC that CCC spaces are separable because $\{0,1\}^X$ has CCC for any $X$, but is only separable for $|X|\le\frak c$. In particular, taking $X=P(\mathbb R)$ gives us a CCC space which is not separable.