The tail-flip hereditary-symmetric model

Work over a transitive $V\models \mathrm{Z}\mathrm{F}\mathrm{C}+V=L$ and force with $P=\mathrm{Add}(\omega ,\omega )$, the finite partial functions $p:\omega \times \omega \rightharpoonup 2$ from Cohen, collapse, and Lévy-collapse forcing orders. For a $V$-generic $G$, put

$S_n=\{k<\omega :(\bigcup G)(n,k)=1\}\mathrm{.}$

Let $\mathcal{G}=(2^{\omega \times \omega }{)}^V$ act on $P$ by bitwise addition modulo $2$: for $a\in \mathcal{G}$, the condition $a\cdot p$ has the same finite domain as $p$ and $(a\cdot p)(n,k)=p(n,k)+a(n,k)(\mathrm{m}\mathrm{o}\mathrm{d}2)$. Thus a ground-model set of bits, possibly infinite, may be flipped. For $m<\omega$ let

$H_m=\{a\in \mathcal{G}:a(n,k)=0\text{ whenever }n<m\}\mathrm{.}$

The group is abelian, the $H_m$ are normal and descending, and their upward closure is a normal filter $\mathcal{F}$. Hence $(P,\mathcal{G},\mathcal{F})$ is a symmetric system in the sense of Symmetric forcing systems, supports, and hereditarily symmetric names. Define the tail-flip hereditary-symmetric model by

${\mathsf{F}}_{\mathrm{t}\mathrm{f}}={\mathrm{H}\mathrm{S}}_{\mathcal{F}}^G\mathrm{.}$

Every $x\in {\mathsf{F}}_{\mathrm{t}\mathrm{f}}$ has an HS name $\dot{x}$ with one finite support bound: since $\mathrm{sym}(\dot{x})\in \mathcal{F}$, there is an $m<\omega$ such that

$H_m\subseteq \mathrm{sym}(\dot{x})\mathrm{.}$

This assertion is about the one name $\dot{x}$. It does not say that every name in the transitive closure of $\dot{x}$ is fixed by the same $H_m$, nor that $x$ and all its descendants lie in one hereditary definability class generated by $S_0,\dots ,S_{m-1}$.

The identifier of this item is retained for compatibility with the earlier draft, but ${\mathsf{F}}_{\mathrm{t}\mathrm{f}}$ is defined directly by hereditary symmetry. It is not defined as the literal union of the classes hereditarily definable from fixed finite tuples of the $S_n$. That literal union cannot be a ZF model: it contains each $S_n$ at some stage, but if its internal collection of all subsets of $\omega$ belonged to one fixed hereditary stage $m$, then every $S_n$, including $S_m$, would belong to that same stage. The coordinate-$m$ tail flip fixes its permitted predicates and ordinals while moving $S_m$, a contradiction. Thus the literal union fails Power Set.

Feferman's $M^{\ast }$ on printed pp. 340–341 instead comes from a transfinite ramified type, or ramified type-free, hierarchy. Each formula uses only finitely many predicate symbols, while the hierarchy can collect objects whose elements require unbounded finite supports. No equality between that ramified hierarchy, a fixed-stage hereditary-definability union, and ${\mathsf{F}}_{\mathrm{t}\mathrm{f}}$ is asserted here. Ground Choice enters through $V=L$; the symmetric-model definition itself assumes no Choice internally.