Assuming Freiling's axiom of symmetry, can we define a specific set $X \subseteq \mathbb{R}$ such that $|\mathbb{N}|<|X|<|\mathbb{R}|$?

Question

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 $|\mathbb{N}|<|X|<|\mathbb{R}|?$

Answer 1

Much as I suspected, there is no specific set that we can write in explicit details which witnesses the failure of $\sf CH$ in the case the axiom of symmetry holds.

The following is taken from the following paper,

Galen Weitkamp, The $\Sigma^1_2$ Theory of Axioms of Symmetry. The Journal of Symbolic Logic, Vol. 54, No. 3 (Sep., 1989), pp. 727-734.

Instead of the real numbers, we use binary sequences, and since we can encode countable sequences as real numbers we can just think about each number as encoding a sequence (and therefore a set). Note that it is important that we encode sequences, because we will deal with models where $\{A\subseteq\Bbb R\mid |A|=\aleph_0\}$ has cardinality strictly greater than that of the continuum.

For every real number $x$, let $x_n$ be $n$-th number in the sequence coded by $x$.

Let $\Gamma$ be a pointclass (a collection of sets) over $2^\omega$, then $A(\Gamma)$ abbreviate the statement if $f\colon2^\omega\to2^\omega$ and the graph of $f$ is in $\Gamma$, then there are $x,y$ such that for every $n$, $$y\neq f(x)_n\land x\neq f(y)_n.$$ In other words, neither $x$ nor $y$ appear in the sequence encoded by the other.

The firt theorem Weitkamp proves is that in $\sf ZFC$ it is true that $A(\bf\Sigma^1_1)$ holds, where $\bf\Sigma^1_1$ is the class of analytic sets (continuous image of Borel sets).

He later proves the following theorem:

If $\Gamma$ is a "reasonable" pointclass such that $A\in\Gamma$ implies that $A$ is Lebesgue measurable, then in $\sf ZF+DC$ it is true that $A(\Gamma)$ holds.

If we assume the consistency of inaccessible cardinals, then two consequences are these:

1. It is consistent with $\sf ZFC$ that no "nicely describable set" forms a counterexample to $\sf CH$ even if the axiom of symmetry holds.
2. It is consistent with $\sf ZF+DC$ that all sets are Lebesgue measurable, and therefore the axiom of symmetry holds, but there is no set which has cardinality strictly between the integers and the real numbers.

Therefore we cannot really describe an explicit counterexample to $\sf CH$ even if the axiom of symmetry holds.

Answer 2

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 $2^{\omega} = \omega_2$.