# A topological space for which having the ccc is independent of ZFC?

It is well known that a generalized Cantor space $2^A$ is separable if and only if $|A| \leq 2^{\aleph_0}$. This means that one cannot decide in $ZFC$ whether the space $2^{\omega_2}$ is separable or not. On the other hand it is also well known that every $2^A$ has the ccc.

Does anyone know a topological space for which its ccc status cannot be decided in $ZFC$?

(A topological space has the ccc if any pair-wise disjoint collection of open sets is at most countable. A space is separable if it has a countable dense subset)

Note: I´m not looking for ad hoc examples like: "let $X$ be a dense subset of $2^{\omega_2}$ of minimum size and give $X$ the discrete topology".

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Hi Ramiro. Are you looking for a "nicely definable" example? There are certainly examples where the ccc property of a space can be destroyed by ccc forcing, but I'd hesitate to claim that this sort of thing answers your question... – Todd Eisworth Oct 5 '11 at 18:51
@Todd: Yes, I guess I´m looking for a "some-way definable" example, so that the following question makes sense. In the examples you are refering to: is the space "the same space" in the extension? if so, how could you prove that it had ccc initially? – Ramiro de la Vega Oct 5 '11 at 19:33
Well, it's the "same space" in the extension, in the sense that it's got the same underlying set of points, and the topology in the extension is generated by the topology from the ground model. What I had in mind is that CH implies the existence of a compact Hausdorff ccc space X whose square isn't ccc. If we then force MA in the usual way, the ccc-ness of X must be destroyed, as MA implies products of ccc spaces are ccc (since the forcing doesn't collapse cardinals, we know that $X^2$ doesn't satisfy ccc). – Todd Eisworth Oct 5 '11 at 19:37
@Todd: Oh, I see. This is not quite what I´m looking for. By "the same space" I meant something like in the $2^{\omega_2}$ example. Under $CH$ it is not separable and then in some ccc extension it ("THE space $2^{\omega_2}$") becomes separable. Along these lines we could ask: is it possible to construct in ZFC a space $X$ such that $X^2$ is not ccc (still in ZFC) and such that some additional axiom (e.g. $CH$) implies that $X$ is ccc? (and of course MA would imply that it is not). – Ramiro de la Vega Oct 5 '11 at 20:43
If you relax the CCC condition to $\omega_2$-cc, then the stone space of $\mathcal{P}(\omega_1)/NS$ might be an example. – Michael Blackmon Oct 5 '11 at 21:06

Let $T$ be the $L$-least Suslin tree of $L$, with the usual cone topology. Thus, in $L$, this is a Suslin tree, a tree of height $\omega_1$ wtih all countable levels, and satisfying the countable chain condition. As a topological space, it is c.c.c. in $L$.

This tree is absolutely definable, in the sense that the definition, "the $L$-least Suslin tree in $L$," picks out exactly the same object in the universe as it does in all inner and outer models of the universe.

But meanwhile, it is independent of ZFC whether this space is ccc or not, since if $V=L$, it definitely is c.c.c., but in the forcing extension where we have forced over this tree, then this tree is no longer c.c.c.

There are many other examples in the same vein. For example,

• the partial order $\text{Coll}(\omega,\omega_1^L)$, consisting of finite partial functions from $\omega$ to $\omega_1^L$. In $L$, this partial order is not c.c.c., but in a universe where $\omega_1^L$ has become countable, then the poset is countable and hence c.c.c. And the partial order is abolutely definable.

• More generally, for any absolutely definable ordinal $\theta$, you may consider the partial order of all finite partial functions from $\omega$ to $\theta$. If $\theta$ is uncountable, then this partial order is not c.c.c., but if it is, then the whole partial order is countable and hence c.c.c. And it is absolutely definable.

Perhaps you will object that these examples are ad hoc in the sense you mention, but because these examples involve absolutely definable partial orders, I think you will have a hard time to cache out a robust concept of ad hoc that excludes them. The absolutely definable nature of these posets would seem to make them even more non-ad-hoc than the sample spaces you mention, which are not literally the same space in a model of set theory as in all its forcing extensions.

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No need to mess around with the partial orders: one can just do $\aleph_1^L$ equipped with the discrete topology and you've got another example of this sort, right? Still seems like cheating though... – Todd Eisworth Oct 6 '11 at 0:49
Yes, that's right, although the c.c.c. question in connection with partial orders is extremely natural from the forcing perspective, since the question of whether a poset is c.c.c. is an often-considered fundamental issue in forcing. If this is cheating, then one needs to refine the question, and it isn't clear how to ask the question in a way that would avoid these kinds of examples. – Joel David Hamkins Oct 6 '11 at 1:20
The Souslin tree example I gave is a little better than the $\aleph_1^L$ example (and the other examples I gave), since the question of whether it is ccc or not does not depend on collapsing the space to become countable. That is, one can kill the c.c.c. property of the tree $T$ while preserving all cardinals. So this makes it more attractive as an example of your phenomenon. – Joel David Hamkins Oct 6 '11 at 1:26
To me these examples feel very similar to Todd´s example (see comments after question), only that you use L to make them "canonical". In any case (although it is not what I was expecting) I like your Suslin-line example. – Ramiro de la Vega Oct 6 '11 at 11:37