TITLE: Assuming Freiling's axiom of symmetry, can we define a specific set $X \subseteq \mathbb{R}$ such that $|\mathbb{N}|<|X|<|\mathbb{R}|$?
QUESTION [7 upvotes]: I learned here that GCH isn't very popular among set theorists, for a variety of reasons. I also learned about Freiling's axiom of symmetry, which (in the presence of the other ZFC axioms) is equivalent to the negation of the continuum hypothesis, which is equivalent to the statement that there exists a set $X$ such that $|\mathbb{N}|<|X|<|\mathbb{R}|.$
So I was wondering. Is it possible to write down a statement $P(x)$ in the language of set theory such that Freiling's axiom of symmetry proves that $X=\{x \in \mathbb{R}|P(x)\}$ satisfies $|\mathbb{N}|<|X|<|\mathbb{R}|?$

REPLY [3 votes]: It seems that you are asking for definable counterexamples to CH. It is consistent that there is one: If you add $\omega_2$ Cohen reals to $L$, then the set of constructible reals in the resulting model forms a set of size $\omega_1$ while continuum is $\omega_2$. I do not see how to get a model of ZFC + not CH + there is no such definable counterexample.
Addendum: A few weeks back I asked the question in the last sentence above from Arnold Miller and Kenneth Kunen and they both mentioned the following model to me: Levy collapse an inaccessible $\kappa$ to $\omega_1$ (this is Solovay's model) and add $\kappa^{+}$ Cohen reals. In the resulting model there is no set of reals ordinal definable (even) with a real parameter of size intermediate between $\omega$ and $2^{\omega} = \omega_2$.