
> 
> Can we have a sequence of transitive sets $\langle\mathcal V\_0, \mathcal V\_1, \mathcal V\_2,...\rangle$, all modeling $\sf ZF$, such that $\mathcal P(V\_n) \subset \mathcal V\_{n+1}$, and the cardinality of each is inaccessible from the cardinalities of the sets preceding it, and where choice fails inside each one of them, yet each successor set $\mathcal V\_{n+1}$ proves choice externally over its predecessor $\mathcal V\_n$?
> 
> 
> 


Formal workup:


Add a primitive total unary function symbol $\mathcal V$ to the language of set theory; as a notation we shall write $\mathcal V(x)$ as $\mathcal V\_x$. To all axioms of $\sf ZF$ [in language $\{=,\in\}$], add the following axioms:


**Restriction:** $\forall x: x \not \in \omega \to \mathcal V\_x= \emptyset$


**Modeling:** if $\phi$ is an axiom of $\sf ZF\neg C$, then: $\forall n \in \omega: \phi^{\mathcal V\_n}$


**Transitivity:** $\forall n \in \omega: \mathcal V\_n \text { is transitive}$


**Power :** $\forall n \forall x: n \in \omega \land x \subseteq \mathcal V\_n \to x \in \mathcal V\_{n+1}$


**Inaccessibility:** $\forall n \in \omega: \operatorname {icc}( |\mathcal V\_n |)$


**Choice:** $\forall n \in \omega: \forall x \in \mathcal V\_n \exists f \in \mathcal V\_{n+1}: \\\operatorname {dom}(f)=x \setminus \{\emptyset\} \land \forall m \, (f(m) \in m ) $


Where for every $\Phi$ set of sentences, $[\Phi]^X$ is the relativization of all quantifiers in elements of $\Phi$ to $X$; and $``\operatorname {icc..}\!\!"$ stands for *".. is inaccessible"*, $``| \ |\!\!"$ stands for cardinality defined after Scott's. This means that the cardinality of each $\mathcal V\_n$ is not reachable by set unions and powering from below, i.e. it is strictly larger than the cardinality of any set union of a set of strictly smaller cardinality whose elements are of strictly smaller cardinalities, and it is strictly larger than the cardinality of the power set of any set of a strictly smaller cardinality.



Can we have this sequence where choice fails and returns?
Sure.


Start with countably many inaccessible cardinals, $\kappa\_n$, and now take the full support product adding $\kappa\_n^+$ subsets to each $\kappa\_n$. Then the $n$th model is the symmetric extension given by adding all the generics for the first $n-1$ coordinates, and violating choice in the $n$th one (e.g. a Cohen style model).


