7
$\begingroup$

I am interested in the relation between the property of countable chain condition (ccc) and the property of separable. Could someone recommend some papers or books about this to me? thanks in advance.

$\endgroup$

5 Answers 5

13
$\begingroup$

Nathan's answer seems to indicate that there are some strange spaces that are ccc but not separable. But in fact, such spaces are rather common:

All products of separable Hausdorff spaces are ccc, but if the spaces have at least two different points, then products with more than $2^{\aleph_0}$ factors are not separable anymore. You can find this is basic books on topology, such as Engelking's book.

Also, in set-theoretic forcing, partial orders with the ccc are very common. If you take the completion of a ccc po, you get a ccc Boolean algebra whose Stone space is ccc. Separability of the Stone space translates into $\sigma$-centeredness of p.o.s. And there are a lot of partial orders used in forcing that are ccc but not $\sigma$-centered.

$\endgroup$
6
$\begingroup$

The following is a special case of the last two sentences of Stefan Geschke's answer, but it may be more accessible than the general case to non-set-theorists. The quotient of the Boolean algebra of Borel sets of reals modulo the ideal of sets of measure 0 is an example of a ccc but not $\sigma$-centered Boolean algebra. So its Stone space is an answer to the original question.

$\endgroup$
4
$\begingroup$

Lest there be any question on this point, ZFC alone implies the existence of some non-separable ccc spaces. The tables in the back of Counterexamples in Topology list four of them. Probably the most interesting is #63, the Countable Complement Extension Topology. This is the minimal extension of the Euclidean topology on the real line given by letting all countable sets be closed.

$\endgroup$
3
$\begingroup$

You will want to look at Suslin lines, which are examples of ccc non-separable orders. The existence of Suslin lines is independent of the axioms of ZFC, and this material is covered in any of the usual graduate set theory texts, such as Jech's book Set Theory.

The real line $\langle\mathbb{R},\lt\rangle$ was known classically to be uniquely determined by the following properties:

  • The real line is a dense linear order with no endpoints
  • the real line has the LUB property
  • the real line has a countable dense subset.

Suslin inquired whether the final condition can be weakened to the ccc property, asserting that every family of pairwise disjoint intervals is countable, and still retain its characterization of $\mathbb{R}$. A counterexample to this latter property is known as a Suslin line. The existence of a Suslin line is equivalent to the existence of a Suslin tree, a well-founded tree of height $\omega_1$ with no uncountable branches or antichains. Much of the set-theoretic development is centered on the Suslin tree concept, which seems to clarify certain issues better than the Suslin lines.

Suslin himself struggled with the question, and died before coming to learn the answer, which is:

  • It is consistent with ZFC that Suslin lines exist. This is true in the constructable universe, and indeed, the existence of Suslin trees and hence Suslin lines is a consequence of the combinatorial principle known as $\Diamond$. Furthermore, every model of set theory has a forcing extension in which $\Diamond$ holds and hence in which Suslin trees exist.

  • It is also consistent with ZFC that there are no Suslin trees. The solution of this problem was the main motivating example for the development of iterated forcing by Solovay and Tennenbaum, proved in the early 1960s. It is a conseqeunce of Martin's axiom at $\omega_1$ that there are no Suslin trees.

$\endgroup$
4
  • $\begingroup$ Was that an answer in record time? $\endgroup$ Oct 20, 2011 at 4:10
  • $\begingroup$ David, I don't think it is a record---some MO people are really quick! For my part in this instance, it was just happy coincidence, since I return to my hotel from a nice dinner, happen to log on to MO, and find a set theory question just posted... $\endgroup$ Oct 20, 2011 at 4:23
  • 1
    $\begingroup$ You can hover over the timestamps (or ask Anton Geraschenko to grep the database) to check server times. I don't think it is a record time. I think after dinner Joel relaxes and takes a minute at least to think before typing. You should see him doing wind sprints before breakfast. Gerhard "Think Before Typing? Who, Me?" Paseman, 2011.10.19 $\endgroup$ Oct 20, 2011 at 6:35
  • 5
    $\begingroup$ Justin Moore once pointed out to me that the Suslin problem, i.e., the question whether there are Suslin lines or not, has actually motivated several very important development in current set theory, perhaps even more than the question whether CH holds: Combinatorial principles in L: $\lozenge$ and $\square$ (Jensen). Iterated forcing and forcing axioms: MA (Solovay and Tennenbaum, Martin) Countable support iteration and the use of master conditions as in proper forcing: Jensen's model of CH without Suslin trees. That is a remarkable achievement of a single problem. $\endgroup$ Oct 20, 2011 at 7:26
3
$\begingroup$

If you're interested in the relationship between the ccc and separability, you should read Stevo Todorcevic's survey "Chain condition methods in topology".

http://www.sciencedirect.com/science/article/pii/S0166864198001126

It's a very convincing pamphlet on the power of chain conditions, clarifying that relationship with such beautiful theorems as:

1) (Todorcevic) Let X be compact Hausdorff. If every subspace of $X^2$ has the ccc then $X$ is separable.

2) (Rosenthal) A compact Hausdorff space $X$ is ccc if and only if every weakly compact subspace of $C(X)$ is separable.

Also, there's a whole book dedicated to chain conditions. It's called "Chain conditions in topology", by Wistar Comfort and Stelios Negropontis (Cambridge Tracts in Mathematics #79, Cambridge University Press).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.