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."

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."