I'm broadly interested in notions of "generic presentability" - when a given object exists in *every* forcing extension of the universe by some fixed forcing, at least up to the appropriate notion of equivalence. Sometimes this is boring - [per Solovay](https://mathoverflow.net/questions/155915/who-proved-sets-in-every-generic-are-already-in-the-ground-model?noredirect=1&lq=1), the only "generically presentable sets up to equality" are those already in $V$ - but other times it can be more interesting. In particular, the appropriate notion of "generically presentable countable structure" is nontrivial ([1](https://www.cambridge.org/core/journals/journal-of-symbolic-logic/article/computable-structures-in-generic-extensions/A13DB0FC4DB8F0BD7F9F2D9FFC18DFCE), [2](https://onlinelibrary.wiley.com/doi/abs/10.1002/malq.201400094)).


I'd like to ask about an intermediate notion, that of **generically presentable cardinals** (or if one prefers, generically presentable sets up to *equipollence* instead of equality):



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> A generically presentable cardinal is a pair $(\nu,\mathbb{P})$ where $\nu$ is a $\mathbb{P}$-name and we have $$\Vdash\_{\mathbb{P}^2}\nu[G\_0]\equiv\nu[G\_1].$$
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Over $\mathsf{ZFC}$, the generically presentable cardinalities are boring in the sense that for every generically presentable cardinal $(\nu,\mathbb{P})$ there is some $a\in V$ such that $\Vdash\_\mathbb{P}\nu[G]\equiv\check{a}$. However, this uses choice in a crucial way (by choice we WLOG have $\Vdash\_\mathbb{P}\nu[G]\in Ord$, and now we just observe that forcing adds no new ordinals). So this leaves the following question open:



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> Is there a $V\models ZF$ containing a generically presentable cardinal $(\nu,\mathbb{P})$ such that $$\Vdash\_{\mathbb{P}^2}\forall a\in V(\nu[G\_0]\not\equiv a),$$ or at least such that $$\Vdash\_{\mathbb{P}}\forall a\in V(\nu[G\_0]\not\equiv a)?$$
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Note that per Laver/Woodin, this makes sense: the ground model is appropriately definable in its forcing extensions. (And the two questions are indeed distinct, since two non-equipollent sets can become equipollent after further forcing. So an affirmative answer to the former implies an affirmative answer to the latter, but not obviously conversely.)



A new cardinality living in every forcing extension?
Partial answer. There is a $V\models ZFA$ containing a generically presentable cardinal $(\nu,\mathbb{P})$ such that for all $a\in V$ we have $\Vdash\_{\mathbb{P}^2}\nu[G\_0]\not\equiv\check{a}.$ This should be equivalent to your first condition, assuming the first condition can even be stated i.e. the ground model is definable in the ZFA setting.


I am not sure about ZF models. (But I *don’t* see a reason why Pincus’ transfer theorems *don’t* apply, using the “surjectively boundable” condition for example. I think we only need to consider $a$ that don’t admit a surjection to $\omega\_1,$ because the below model should give $\Vdash\_{\mathbb{P}^2}\text{there is no surjection }\nu[G\_0]\to\omega\_1.$ And the forcing relation should be equivalent to some fairly tame sentence, surely?)
$
\DeclareMathOperator{dom}{dom}
\DeclareMathOperator{rng}{rng}
\DeclareMathOperator{supp}{supp}
$


Take $V$ to be the basic Fraenkel model, with set of atoms $A.$ Every set $x\in V$ has a minimal finite support $\supp(x)\subset A.$ So $x$ is fixed by permutations that fix each element of $\supp(x).$


Conditions $p\in \mathbb P$ are bijections such that $\dom p$ and $\rng p$ are disjoint finite subsets of $A.$ Graphically, imagine a finite set of vertex-disjoint directed edges using the vertex set $A.$ As usual $p\leq q$ if $p\supseteq q.$


Take $\nu$ to be the name for $\dom \bigcup G.$


I’ll make use of the dense subset $\mathbb Q\subset \mathbb P^2$ consisting of pairs $p=(p\_0,p\_1)$ such that $\dom p\_0 \cup \rng p\_0 = \dom p\_1 \cup \rng p\_1.$ Graphically, these conditions are a finite union of cycles (possibly 2-cycles) using the vertex set $A.$ In each cycle edges are alternately colored $0$ and $1,$ and every edge has a direction. (Note these aren’t “directed cycles”, just cycles with directed edges - the direction on different edges are unrelated.)


The interesting part is that these cycles have no orientation-reversing symmetry. Graphically, a reflection of a regular polygon would have to pass through the midpoint of an edge, which is ruled out by the directions, or pass through a vertex, which is ruled out by the coloring. Pick an isomorphism-invariant choice function for orientations of alternating-colored cycles with directed edges. This is kosher because we can define these using only pure sets, and in fact this is completely finitary. This gives a bijection $\nu[G\_0]\to \nu[G\_1]$ by identifying each set with edges of the appropriate color, and sending each edge in a cycle one step along the chosen orientation.


The rest of the argument is just the usual trick of making use of compatible conditions related by symmetries. Suppose for contradiction that there exists $z\in V$ and $p\in\mathbb Q$ and a $\mathbb Q$-name $\dot{f}$ such that $$p\Vdash\text{$\dot{f}$ is a bijection $\check{z} \to \nu[G\_0]$}$$
where $\Vdash=\Vdash\_{\mathbb Q}.$
By refining $p$ if necessary we can assume that the supports of $z$ and $\dot{f}$ are contained in $A\_0:=\dom p\cup \rng p.$ There exists $q\leq p$ and elements $x\_a\in z$ for $a\in A\_0$ such that $q\Vdash \bigwedge\_{a\in A\_0}\dot{f}(\check{x\_a})=\check{a}.$ Consider a permutation $\pi$ fixing $A\_0$ but moving every other atom used by $q,$ and every $x\_a,$ to unused atoms. Then $q$ and $\pi q$ are compatible, which shows that $\supp(\check{x\_a})\subseteq A\_0,$ so the choice of $q$ was irrelevant:


$$p\Vdash \bigwedge\_{a\in A\_0}\dot{f}(\check{x\_a})=\check{a}.$$


Pick an element $x\in z\setminus \{x\_a:a\in A\_0\}.$ Pick $r\leq p$ such that each element of $\supp(x)\setminus A\_0$ is in a distinct 2-cycle of $r.$ Let $A\_1$ be the set of elements of cycles in $r$ that use an element of $\supp(x)\setminus A\_0.$ Refine $r$ if necessary such that: for all $y=\pi x$ where $\pi$ fixes $A\_0$ and $\supp y\subseteq A\_1,$ there exists $a\_y\in A$ with $$r\Vdash \dot{f}(\check y)=\check{a\_y}$$


For each such $a=a\_y$ we have $a\not\in A\_0.$ Suppose for contradiction that $a\not\in A\_1.$ Pick $\pi$ fixing $A\_0\cup A\_1$ but sending all other atoms used by cycles in $r$ to unused atoms, and such that $a\neq \pi a.$ Then $\pi r$ and $r$ are compatible, which gives the contradiction $r\cup \pi r\Vdash \check a=\pi \check a.$ So $a\in A\_1.$


Each $a\_y$ lies in the set $\dom r\_0\cap A\_1$ of order $|A\_1|/2,$ but there are at least $\binom{|A\_1|}{|A\_1|/2}> |A\_1|/2$ choices of $y$ (with $\binom{0}{0}=1>0$) because sets with distinct supports are distinct.


I think this all generalizes to $\kappa$-closed conditions, using a generalization of the basic Fraenkel model to supports of order $<\kappa.$ However, the cycles can then be infinite, so the choice function for orientations now seems to genuinely need choice, for pairs of countable sets of reals.


