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This question was asked and bountied at MSE, without success.

My main question is whether, starting with a model of determinacy, a "generic $\mathbb{R}$" could be different in cardinality from any ground model set:

Is there a well-founded $V\models\mathsf{ZF+AD}$, a forcing $\mathbb{P}\in V$, and a $G$ which is $\mathbb{P}$-generic over $V$ such that $V[G]\models$ "There is no bijection between $\mathbb{R}^{V[G]}$ and any set in $V$"?

In case the answer is yes, there is a natural follow-up question - whether the above can happen "canonically:"

Are there $V\models\mathsf{ZF+AD}$, $\mathbb{P}\in V$, and $G,H$ mutually $\mathbb{P}$-generic over $V$ such that $V[G\times H]$ satisfies "$\mathbb{R}^{V[G]}$ and $\mathbb{R}^{V[H]}$ are in bijection with each other but are not in bijection with any set in $V$?"

If the answer to this question is yes, that would give a very surprising answer to this old question of mine. I suspect that the first question has an affirmative answer and strongly suspect that the second question has a negative answer, but I don't see how to prove either point.

Here are a couple quick comments:

  • Since this is only interesting if we add reals, $\mathsf{AD}$ will not be preserved. So determinacy doesn't give us a lot of "leverage" in the forcing extension that I can see.

  • To preempt worries about triviality, we can have $V[G]\models$ "There is some $a$ which is not in bijection with any $b\in V$." For example, Monro showed that we can have an amorphous set in $V[G]$ even if there are no amorphous sets in $V$; since amorphousness is downwards-absolute, any amorphous set in such a $V[G]$ is not (in $V[G]$, anyways) in bijection with any set in $V$. (This reference was pointed out to me by Asaf Karagila.)

  • That said, questions of this type are always trivial over $\mathsf{ZFC}$ since forcing preserves choice and adds no new ordinals.

  • But that said, the previous bulletpoint is quite fragile and I don't see that it gives any insight into my question - there's no obvious replacement for $Ord$ that I see here to serve the same role as something simultaneously "invariant" and "universal," especially since it would have to be "generically universal."

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  • $\begingroup$ What reference would you indicate for a beginner in Logic willing to understand what is being asked? =] $\endgroup$
    – Dry Bones
    Commented Aug 5, 2020 at 19:32
  • $\begingroup$ @DryBones Unfortunately this really isn't accessible without a significant amount of experience - specifically, comfort with forcing is the key ingredient (although you also need to be comfortable with the axiom of determinacy, that's somewhat a secondary concern - the question is still interesting if we replace $\mathsf{ZF+AD}$ with just $\mathsf{ZF}$). Forcing is treated for example in Kunen's old book "Set theory and independence proofs;" this and this are also good. $\endgroup$
    – Noah Schweber
    Commented Aug 5, 2020 at 19:36
  • $\begingroup$ Forcing, however, relies on a mastery of the basics of set theory and model theory. So it's quite a long road. $\endgroup$
    – Noah Schweber
    Commented Aug 5, 2020 at 19:36
  • $\begingroup$ Ok, I can imagine. Thanks, anyway! $\endgroup$
    – Dry Bones
    Commented Aug 5, 2020 at 19:38
  • $\begingroup$ @DryBones To be fair I might be overemphasizing the role of model theory. It's not really directly needed if you're willing to take a couple things on faith. However, the basic idea behind forcing is that we're building models of a certain very complicated theory (namely $\mathsf{ZF}$ or similar), and understanding basic properties of models will help dispel many (very reasonable) initial worries. So personally I wouldn't approach forcing without being comfortable with the downward Lowenheim-Skolem, compactness, completeness, and incompleteness theorems - the experience that confers will help. $\endgroup$
    – Noah Schweber
    Commented Aug 5, 2020 at 19:41

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