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".
 A: 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.
