Are there any infinite sets that are not known to be either countable or uncountable?

Question

Are there any known examples of sets that are definitely infinite, but where we don't know whether or not they're countable? I haven't heard of anything like this before, but it seems that there should be some example of a set like this.

As mentioned in the comments, it's possible to construct examples of the form

$$ S = \left\{ \begin{array}{ll} \mathbb{N} & \quad \mbox{if } P \mbox{ holds} \\ \mathbb{R} & \quad \mbox{otherwise} \end{array} \right. $$

where $P$ is some unresolved conjecture in mathematics. Certainly these sets work, but I was hoping for an example that arose "naturally" in the course of mathematics.

Thanks!

Answer 1

Here's a wonderful open problem in set theory, which can be translated to a statement which you might be looking for.

Suppose that $\aleph_\omega$ is a strong limit cardinal. Is $2^{\aleph_\omega}<\aleph_{\omega_1}$?

We can prove that under the assumption that $\aleph_\omega$ is a strong limit cardinal, it is necessarily the case that $2^{\aleph_\omega}<\aleph_{\omega_4}$. This is one of Shelah's most celebrated results from PCF theory. And we can arrange for every countable successor ordinal $\alpha>\omega$, that is the case that $2^{\aleph_\omega}=\aleph_\alpha$ (well, under certain additional assumptions anyway).

So we can rephrase the question as follows:

Assuming that $\aleph_\omega$ is a strong limit cardinal. Is the set $\{\alpha\mid\aleph_\alpha<2^{\aleph_\omega}\}$ provably countable?

This set is always infinite, since every finite ordinal is necessarily in this set. But as the aforementioned problem shows, we do not know if it is always countable (even if we don't know exactly what is this set), or if it is possible for it to be uncountable.

Answer 2

A simple example that occurs naturally in set theory (although it is not as mysterious as the one in Asaf's answer and there are probably no more open problems about it) is the set $\mathbb{R} \cap L$ of all constructible reals. If $V = L$ then of course this set is uncountable, but the statement "$\mathbb{R} \cap L$ is countable" can be proved consistent relative to $\mathsf{ZFC}$ by forcing, and it also follows from large cardinal axioms (for example, from the existence of a measurable cardinal, or just from the existence of $0^\sharp$.)

The statement "$\mathbb{R} \cap L$ is countable" is equivalent to the statement "every $\Sigma^1_2$ set of reals with a $\Sigma^1_2$ well-ordering is countable," where $\Sigma^1_2$ denotes the pointclass of projections of co-analytic sets. There is an open problem along these lines about a pointclass much bigger then $\Sigma^1_2$:

Does any large cardinal axiom imply that every $(\Sigma^2_1)^{\text{uB}}$ set of reals with a $(\Sigma^2_1)^{\text{uB}}$ well-ordering is countable?

Here $\text{uB}$ denotes the pointclass of all universally Baire sets of reals, and a set of reals $A$ is called $(\Sigma^2_1)^{\text{uB}}$ if there is a formula $\varphi$ such that $x \in A \iff \exists B \in \text{uB}\,(\text{HC};\in , B) \models \varphi[x]$ for every real $x$. Essentially the question asks whether there is a large cardinal axiom that transcends the pointclass $(\Sigma^2_1)^{\text{uB}}$ in the way that $0^\sharp$ transcends the pointclass $\Sigma^1_2$.

For technical reasons, let's strengthen the notion of "$(\Sigma^2_1)^{\text{uB}}$ well-ordering of a $(\Sigma^2_1)^{\text{uB}}$ set of reals" to "$(\Sigma^2_1)^{\text{uB}}$-good well-ordering," by which we mean that the well-ordering has length $\le \omega_1$ and the set of reals coding its proper initial segments is a $(\Sigma^2_1)^{\text{uB}}$ set. It is consistent (as shown using forcing) that all such well-orderings are countable.

If there is a proper class of Woodin cardinals (a fairly mild large cardinal assumption,) then there is a canonical $(\Sigma^2_1)^{\text{uB}}$ set of reals that is maximal under inclusion among $(\Sigma^2_1)^{\text{uB}}$ sets of reals admitting $(\Sigma^2_1)^{\text{uB}}$-good well-orderings. This canonical $(\Sigma^2_1)^{\text{uB}}$ set of reals can be thought of as a higher-order analogue of $\mathbb{R} \cap L$, and the question asks whether any large cardinal axiom implies that it is countable. A "yes" answer would be a very important development for the field of inner model theory.