Permutation Models and Transfer to ZF: Examples and Counterexamples

Let $B\subseteq A$ be supported by finite $E$. If some $a\mathrm{\notin }E$ lies in $B$, then for every $b\mathrm{\notin }E$ the transposition $(ab)$ fixes $E$ and hence $B$, so $b\in B$. Thus $A\setminus E\subseteq B$ and $B$ is cofinite.

If no such $a$ exists, then $B\subseteq E$ and $B$ is finite. This includes the vacuous case $A\setminus E=\mathrm{\varnothing }$, although it cannot occur when $A$ is infinite and $E$ finite.

Write $P_n=\{a_n,b_n\}$. If $E$ is a finite atom support, choose the least $n$ with $P_n\cap E=\mathrm{\varnothing }$. The permutation $\pi$ interchanging $a_n,b_n$ and fixing all other atoms is allowed, fixes $E$, and satisfies $\pi P_k=P_k$ for every $k$.

If $c$ were supported by $E$ and $c(k)\in P_k$, then $\pi c=c$, so graph evaluation gives $c(n)=\pi (c(n))$. But $\pi$ has no fixed point in $P_n$, contradiction.

For the relation $R_{<}=\{(a,b):a<b\}$ and every allowed $\pi$, order preservation gives $(a,b)\in R_{<}$ iff $(\pi a,\pi b)\in R_{<}$, hence $\pi R_{<}=R_{<}$. Its stabilizer is the whole group, so $\mathrm{\varnothing }$ supports it.

In contrast, if a well-order had finite support $E$, an order automorphism fixing $E$ could move its least atom in $A\setminus E$, contradicting invariance. Thus “linearly ordered” and “well-orderable” separate in this model.

In a ZFA model with distinct atoms $a\mathrm{\ne }b$, both have no members. Unrestricted ZF Extensionality would conclude $a=b$, so the whole ZFA universe is not a ZF model.

The valid repairs are narrower: F1 says the pure kernel is a ZF model, and F2 transfers a prescribed power-iterate segment into a symmetric ZF model under its explicit ambient ZFA+AC premise. Neither identifies the entire atom universe with a ZF universe.