6
$\begingroup$

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.

$\endgroup$

1 Answer 1

11
$\begingroup$

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).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.