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Why does "separable" imply the "countable chain condition"?

Thanks for any help.

t.b.
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Paul
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  • @Dylan: while the question might count a duplicate, the answers given don't address this easy implication but rather focus on the more subtle converse, so I don't think this question should be closed. – t.b. Jul 14 '12 at 13:20
  • @tb I had the same reservation, but the other direction is given by David Mitra in the comments. [I'm not sure why the answers ignore it.] I can add a CW answer addressing it there, if that helps, but I don't see this question as adding much new. – Dylan Moreland Jul 14 '12 at 13:23
  • @Dylan: I see. Thanks, I missed that comment. A comment upvote should suffice, I think (in order to make it more visible). – t.b. Jul 14 '12 at 13:25
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    @Dylan, t.b. I wish to withdraw my close vote. I think it is better to have the answer (even CW) here and not there. – Asaf Karagila Jul 14 '12 at 13:31
  • @Asaf Okay, sure. I'll vote to reopen if this does get closed, in that case. Do you think that the other question has become too high-level because of the answers? – Dylan Moreland Jul 14 '12 at 13:41
  • @Dylan: I think that the other question was looking for one thing, and by luck there was a comment addressing this thing. Adding this answer there would be answering a different question. Had it been mentioned before, it would be fine to close. However if we intend to add that... I think it's better to add here. – Asaf Karagila Jul 14 '12 at 13:46
  • @Asaf I got the impression from the linked question (the part highlighted in gray, even) that the OP wanted both a proof of this direction and a counterexample for the other. – Dylan Moreland Jul 14 '12 at 13:48

2 Answers2

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The reason is that separability implies that there is a countable subset $D$ that for every non-empty open set $U$, $D\cap U\neq\varnothing$.

If $X$ does not have CCC then there is an uncountable family $\{U_i\mid i\in I\}$ of pairwise disjoint open sets. Any countable set can intersect only countably many of those, but never all of them, therefore $X$ is not separable.

Alternatively, you can argue directly: If $\mathcal{U}$ is a family of pairwise disjoint open sets, each $U \in \mathcal{U}$ must contain a point $d \in D$ by density, so there's a surjection from some subset of $D$ onto $\mathcal{U}$, hence $\mathcal U$ is countable.

The other direction is not true since there are CCC spaces which are not separable.

An example of a non-separable but CCC space would be a sufficiently high power $\{0,1\}^\kappa$ ($\kappa \gt \mathfrak{c}$ is enough): indeed, an arbitrary product $\prod_{i \in I} X_i$ of topological spaces is CCC if all finite products $\prod_{j \in J} X_j$ with $J \subset I$, $\# J \lt \infty$ are CCC.

t.b.
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Asaf Karagila
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Let $D$ be a dense subset of $X$ and let $\{ U_i : i \in I \}$ be a pairwise disjoint family of non-empty open sets indexed by $I$.

Define a map $f: I \rightarrow D$ by picking $f(i) \in U_i \cap D$, which can be done as each $U_i$ is non-empty and open and $D$ is dense. For a countable $D$ we can choose the one with minimal index in some fixed enumeration of $D$, for definiteness.

The function $f$ is 1-1, because if $i \neq j$ then $f(i) \in U_i$ and $f(j) \in U_j$ but as $U_i \cap U_j = \emptyset$, $f(i) \neq f(j)$.

Hence we have an injection from $I$ into $D$ and so $|I| \le |D|$ as cardinal numbers.

If $X$ is separable we can fix some countable dense subset $D$ and this then shows that all pairwise disjoint families of non-empty open sets are at most countable, or $X$ is ccc.

Henno Brandsma
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