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Throughout this question we assume ZFC.

If CH holds, then the following is obvious:

(S) Every definable infinite subset of $\mathbb R$ has size either $\aleph_0$ or $2^{\aleph_0}$.

(It's true because every subset of reals satisfies this, in particular so does every definable)

Recently my friend has asked me whether the same is true if we assume CH fails. My answer was that this is independent of ZFC, because I'm fairly sure that results about pointwise definable models will gives us such model of ZFC + not CH (correct me if I'm wrong!) in which a set of reals of size $\aleph_1$ is definable, and I can recall seeing a result that (S) can hold, possibly under some large cardinal assumption. However, I failed to find a reference for that.

Hence my question is for a reference of the following result (if my memory isn't failing me and it's actually true):

It is consistent (relatively to large cardinals) that CH fails and (S) holds.

I also wanted to ask if this is true if CH fails badly:

Is it consistent for every cardinal $\kappa$ that (S) holds and $2^{\aleph_0}>\kappa$?

Thanks in advance.

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  • $\begingroup$ @VictoriaGitman That's not a problem, as I hope it's clarified now :) $\endgroup$ – Wojowu Sep 28 '15 at 19:21
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    $\begingroup$ Two possible models: (i) Start with GCH and force with finite partial functions from $\kappa$ to $\omega_1$ for a regular $\kappa \geq \omega_3$. (ii) Levy collapse an inaccessible $\kappa$ to $\omega_1$ (Solovay's model) and then add $\kappa^{+}$ Cohen reals. (i) works for ordinal definable sets of reals and (ii) also allows a real parameter. $\endgroup$ – Ashutosh Sep 28 '15 at 19:37
  • $\begingroup$ I don't have a published reference for this. These examples were suggested to me by Kunen and Arnie Miller when I was a grad student. $\endgroup$ – Ashutosh Sep 28 '15 at 19:50
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    $\begingroup$ Would I sound too naive if I ask what you exactly mean by "definable subsets of $\mathbb{R}$"? Sometimes people use that phrase for projective sets of reals, which have the perfect set property under suitable large cardinal assumptions. $\endgroup$ – Burak Sep 28 '15 at 19:51
  • $\begingroup$ @Burak Given a model $M$ of $ZFC$, I call set $S$ in that model definable if there is a formula $\phi(x)$ without parameters such that $\phi(x)\Leftrightarrow x\in S$. $\endgroup$ – Wojowu Sep 28 '15 at 19:53
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EDIT: This isn't really an answer - see Andres' comment - but you might find it interesting, and it is too long for a comment.

Your second question is vague, but has an affirmative answer in the following sense:

Let $V\models ZFC+V=L$, and let $\kappa$ be a cardinal in $V$. Then there is a forcing extension $V[G]$ of $V$ in which $2^{\aleph_0}>\kappa$ but there is a definable set of reals of cardinality $\aleph_1$.

Proof: Add $\kappa^+$-many Cohen reals; this bumps the continuum up to $\kappa^+$. Meanwhile, it does not collapse $\aleph_1$, so the set $V\cap\mathbb{R}$ of ground reals is still of size $\aleph_1$. And $V\cap\mathbb{R}=(\mathbb{R}^L)^V=(\mathbb{R}^L)^{V[G]}$, so is definable in $V[G]$ without parameters.


The first question seems harder. Any countable model of ZFC has a class forcing extension which is pointwise definable, but this class forcing extension doesn't seem to preserve large cardinal properties. Meanwhile, "big enough" large cardinals kill off all the obvious ways to try to define sets of reals of intermediate cardinality.

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  • $\begingroup$ This is not really addressing the question as given. The background theory here should be $\mathsf{ZFC}+\mathsf{LC}$ rather than just $\mathsf{ZFC}$, for $\mathsf{LC}$ a (significant) large cardinal assumption. That is, we want large cardinals in our universe, and still determinacy of definable games. This rules out models obtained by forcing over $L$ or thin inner models. $\endgroup$ – Andrés E. Caicedo Sep 28 '15 at 22:33
  • $\begingroup$ @AndresCaicedo That's true, but I wasn't sure if the OP was aware of constructions like this, so I thought it worth pointing out. I've edited to clarify. $\endgroup$ – Noah Schweber Sep 28 '15 at 22:53
  • $\begingroup$ Note that you can do the same trick as long as you have nicely absolute core models, so I think this style of argument is compatible with a decent amount of large cardinal structure. (Please correct me if I'm wrong.) $\endgroup$ – Noah Schweber Sep 28 '15 at 22:56
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    $\begingroup$ i dont know what Andres Caicedo's objection is but you produced a model of not(S) plus stuff while the op asked if there were models of (s) plus stuff. $\endgroup$ – Ashutosh Oct 1 '15 at 2:30

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