Is there a set $A$ such that $|\mathbb Z|<|A|<|\mathbb R|$ is undecidable?

Question

CH guarantees that the statement $|\mathbb Z|<|A|<|\mathbb R|$ is false for all $A$, but since $\sf CH$ is undecidable it might still be possible that there exist a set $A$ for which the statement is undecidable without the CH or it's negation.

My intuition says that there should be such a set because in $\sf ZFC+\neg\sf CH$ there would exist a set $A$ such that $|\mathbb Z|<|A|<|\mathbb R|$, but in $\sf ZFC$ only the set $A$ could not be an counterexample of $\sf CH$ and it could not be countable or be as large as $\mathbb R$ because it then wouldn't fulfil it's property in $\sf ZFC+\neg\sf CH$.

If it's true that such a set exists, to what extent can such a set be described (in a concrete way)?

Answer 1

The problem is that any set which is not provably countable can be made countable in a forcing extension.

So for example $\Bbb R^L$ might be uncountable, in which case it always has size $\aleph_1$ and it might be a counterexample to $\sf CH$ (or not). But it is also possible that $\Bbb R^L$ is countable, in which case it is not a counterexample anymore.

You could ask about a definition of a set in whatever language, but again you can't say much. What you could say, however is that this set is not Borel or even analytic. It could be co-analytic, though.

Answer 2

I am not entirely sure what your question is but the CH is one of my favorite topics in mathematics due to the breadth of mathematical logic surrounding it, so all I can offer is some resources for more information.

The Wiki page for the CH is incredibly well written and contains a lot of information that may help you, particularly this..

"Solomon Feferman (2011) has made a complex philosophical argument that CH is not a definite mathematical problem. He conjectures that CH is not definite according to this notion, and proposes that CH should therefore be considered not to have a truth value. Peter Koellner (2011b) wrote a critical commentary on Feferman's article.

The Wiki page also has many links to important relevant information such as the axiom of constructibility.

Answer 3

: there exist a set A for which the statement is undecidable

This is a notoriously ambiguous claim because ZFC generally does not prove statements with concrete sets sets as parameters. One has to be careful with things like this as otherwise there could happen an absolute loss of sense. One of possible precise formulations of your para-question is this.

Is there a formula $F(x)$ of the language of ZFC such that

1) ZFC proves there is a unique $x$ st $F(x)$, and

2) ZFC proves that if CH fails then the $x$ of item 1 is a set of cardinality $\aleph_1$.

If this is what you ask then it answers in the positive by $F(x)$ being "$x=$ all countable ordinals" by obvious reasons. If you are looking for something else then pls specify it in unambiguous terms.