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Before asking my question I like to admit that i am not an expert in the foundation of mathematics and I am interested in this issue more from a philosophical perspective. So my question may be competently naive. If we assume $ZFC$ and the negation of the continuum hypothesis, what is know about the set $$\{|A|~|~A\subseteq \mathbb{R},~|\mathbb{N}|<|A|<|\mathbb{R}|\}.$$ Is in finite, countable or uncountable? Or are these claims independent of $ZFC+\neg CH$. As a Platonist in the philosophy of mathematics I suppose that the set is uncountable. Everything that possibly exists, exists. But what is the consistency strength of such a claim?

Thanks for the answers so far. Perhaps I should make my last question more precise. Is ZFC plus the assumption that the set above is uncountable consistent if ZFC is?

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    $\begingroup$ It is independent. It can take nearly any ordinal value. $\endgroup$
    – Monroe Eskew
    Commented Apr 27, 2019 at 18:14
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    $\begingroup$ If I told you that I can prove that $A\subseteq\Bbb R$ is either empty or it's not, and then I say "now assume it's not empty". What is the cardinality of $A$? $\endgroup$
    – Asaf Karagila ♦
    Commented Apr 27, 2019 at 18:41
  • $\begingroup$ Dear Monroe Eskew, where do i find such results? $\endgroup$
    – Jörg Neunhäuserer
    Commented Apr 27, 2019 at 19:49
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    $\begingroup$ @JörgNeunhäuserer Much of the general situation - much broader than merely $\mathbb{R}$ - is summarized by Easton's theorem. This isn't the end of the story, but in some sense it shows that many (if not most) of the "naive" questions about cardinality can't be resolved in ZFC alone. $\endgroup$
    – Noah Schweber
    Commented Apr 27, 2019 at 22:03
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    $\begingroup$ The answer to the added, more precise question is yes: If ZFC is consistent, it remains so when one adds that the set in the question is uncountable. $\endgroup$
    – Andreas Blass
    Commented Apr 28, 2019 at 0:11

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By a theorem of Solovay, $|\mathbb R|$ can consistently be $\aleph_\alpha$ for any ordinal number $\alpha>0$ that does not have countable cofinality. Then the set $\{|A|:A\subseteq\mathbb R, |\mathbb N|<|A|<|\mathbb R|$ in your question would have cardinality $\alpha-1$ if $\alpha$ is finite, and it would have cardinality $|\alpha|$ when $\alpha$ is infinite.

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    $\begingroup$ Thank you for recalling that theorem of Solovay. Does the question of consistency strength make sense ? Gerhard "Is Interested In Theory Building" Paseman, 2019.04.27. $\endgroup$
    – Gerhard Paseman
    Commented Apr 27, 2019 at 19:58
  • $\begingroup$ @Gerhard What question? The consistency strength of the statement that the set of intermediate cardinalities is uncountable? This is equiconsistent with ZFC. $\endgroup$
    – Andrés E. Caicedo
    Commented Apr 27, 2019 at 23:10
  • $\begingroup$ The original was the strength of "such a claim", where to me it was unclear precisely what the claim was (existence of everything? Uncountability?). Since then the question has been modified so that a new question of consistency arose, which you have answered. Gerhard "Time Is A Varying Concept" Paseman, 2019.04.27. $\endgroup$
    – Gerhard Paseman
    Commented Apr 27, 2019 at 23:35
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It depends. Do you care about naming or describing such sets?

In the system you propose, you may be able prove the existence of more than finitely many of them, but in a countable (non-infinitary, since you mention a small extension of ZFC) language, you will only be able to describe countable many of these sets or equivalence classes.

It may also be possible to show that there is no surjection from any given set with a cardinal (that you can describe or posit in this system) onto this class, but all that says is that the system you consider is too weak to analyze the class as fully as you want. Back at you: if you were given an answer, what would you hope to do with it?

I am not expert enough to answer the question about consistency strength. Based on the comments of Asaf and Monroe to the question, I suspect an assumption that the class you have is enumerated by a cardinal of type X is equivalent in strength to the assumption that a cardinal of type X exists.

Gerhard "Philosophers Don't Know What's Wanted" Paseman, 2019.04.27.

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    $\begingroup$ I am quite willing to accept (not challenge) a downvote for any well-expressed technical reason that the post above is flawed, for then the community learns. If the question and answer are too naive for MathOverflow, a reasoned statement accompanying the downvote would be appreciated as well. Otherwise the downvote just seems petty. Gerhard "Many Downvotes Just Seem Petty" Paseman, 2019.04.27. $\endgroup$
    – Gerhard Paseman
    Commented Apr 27, 2019 at 19:47
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    $\begingroup$ Gerhard, I didn't downvote your answer but it is not a good answer. The distinction between naming and describing isn't relevant to the question which was asked. The comments already answer the question (which is basic set theory, not research level). $\endgroup$
    – Nik Weaver
    Commented Apr 27, 2019 at 19:58
  • $\begingroup$ @Nik, I am unsure that the question is good. I don't care to distinguish between naming or describing. I am wondering about the motivation or purpose of the questioner. Whether a Platonist cares about naming does not matter to me; I care (for the purposes of understanding the motivation) whether it matters to Joerg. Do you know why he asks this question? (Thank you for your comment.) Gerhard "Wants To Make Question Good" Paseman, 2019.04.27. $\endgroup$
    – Gerhard Paseman
    Commented Apr 27, 2019 at 20:04
  • $\begingroup$ And as a standalone answer, I admit mine pales in comparison to Andreas's answer. However, my post will prove its worth if it engenders a good response from Joerg. He may not know enough foundations to clarify the question technically; I hope he knows enough philosophically that he can explain to us why the question is important to him or to his understanding. Gerhard "Feels There Is Underlying Question" Paseman, 2019.04.27. $\endgroup$
    – Gerhard Paseman
    Commented Apr 27, 2019 at 20:10

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