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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?


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If $\{U_\alpha\}$ is a disjoint collection of open sets, choose an $x_\alpha$ in each $U_\alpha$ where $x_\alpha$ is taken from a countable dense subset $\{x_\alpha\}$ of the space. – David Mitra Dec 11 '11 at 11:53
Here,, it is claimed that $\{0,1\}^{2^{2^{\aleph_0}}}$ with the product toplology satisfies CCC, but is not separable. – David Mitra Dec 11 '11 at 12:14
Thanks for the links. – General Mustard Dec 11 '11 at 12:36
You can also see this MathOverflow question. – arjafi Dec 11 '11 at 12:37
I think that JDH answer is better suited for this question. You should accept it instead of mine. – Asaf Karagila Dec 11 '11 at 12:49
up vote 4 down vote accepted

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.

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we cannot prove it because $\{0,1\}^I$ is ccc for all $I$ but not separable for $|I| > |\mathbb{R}|$. So there is a ccc counterexample. – Henno Brandsma Dec 11 '11 at 17:37
a Suslin line is an ordered space (hence line) that is ccc but not separable. Not just any space. The problem came from trying to generalize a characterization of the real line. – Henno Brandsma Dec 11 '11 at 17:40
@Henno: Yes, this is true. The Suslin line was the first example which came to my mind. By the time I returned to the thread JDH's answer was better and mine was accepted. – Asaf Karagila Dec 11 '11 at 17:41

It is explained in this MathOverflow question that there are many spaces that are ccc but not separable, including any products of ccc separable spaces with sufficiently many nontrivial factors. In particular, there is no ZFC independence on this question: we can prove that there are ccc spaces that are not separable.

Other examples include the numerous instances of ccc forcing notions that have no countable dense set, which are quite commonly considered in forcing arguments.

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My understanding is that in the class of all topological spaces, there exist ccc spaces which are not separable (and there are examples (*) of these). But, if we restrict our attention to spaces with the order topology, the question is undecidable in ZFC (See the Suslin Hypothesis).

Assuming Martin's Axiom (+$\neg$CH), SH is true (for spaces with the order topology, ccc implies separable). It can be shown that a product of non-separable ccc spaces is not ccc, yet MA implies products of ccc spaces are ccc.

So in particular, the spaces in the examples * must have topologies which are not equivalent with any order topology on the underlying set.

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I've been trying to come up with a simpler example to answer your question, although the answes of course complete and great, I still fancied a simpler example. So here it is:

Every separable space is CCC. The converse is not true.

For example:

Let $X$ be a space with $|X|= \aleph_1$, let $\tau_X= \lbrace U: X\setminus U \text{ is countable } \rbrace$. This space is CCC, but not separable.

Proof: $X$ is not separable: for any countable set $A \subset X$, clearly, $U=X\setminus A$ is open and $U \cap A=\emptyset$.

$X$ is CCC: if $X$ has uncountable disjoint open sets $\lbrace \cal U_\xi: \xi \in \aleph_1\rbrace$. Pick one open set, for example, $U_0$. Because $X \setminus U_0$ is uncountable, it is a contradiction with $U_0$ is open.

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