Symmetric Extensions and Basic Choice-Failure Models: Examples and Counterexamples

For every finite permutation $\pi$, $\pi A=\{\pi a_n:n\in \omega \}=\{a_{\pi (n)}:n\in \omega \}=A\mathrm{.}$ Thus the whole automorphism group stabilizes $A$, so $\mathrm{\varnothing }$ is a support.

Let $E\subseteq \omega$ be finite. Choose distinct $n,m\mathrm{\notin }E$ and let $\pi =(nm)$. Then $\pi \in \mathrm{fix}(E)$, but the pair $⟨n,a_n⟩$ is sent to $⟨n,a_m⟩$ in the action on the graph's value coordinate after the ground ordinal $n$ is fixed. Since $a_n\mathrm{\ne }{a}_m$, $\pi e\mathrm{\ne }e$. Hence no finite $E$ supports $e$.

The calculation separates an invariant unordered range from a non-symmetric ordering of that range. It does not claim merely that this particular graph is absent: F1 separately supplies the support argument excluding every enumeration. No Choice is used.

If $B\subseteq A$ is countably infinite, a displayed bijection $b:\omega \to B$ followed by inclusion is an injection $\omega \to A$. Conversely, the range of an injection $i:\omega \to A$ is a subset of $A$ bijective with $\omega$. These are explicit maps and require no simultaneous choices.

From an injection $i:\omega \to A$, define $h:A\to A\setminus \{i(0)\}$ by $h(i(n))=i(n+1)$ and $h(x)=x$ off $i[\omega ]$. The two pieces are disjoint, and the inverse sends $i(n+1)$ to $i(n)$ and fixes the complement, so $h$ is a bijection onto a proper subset.

Conversely, if $h:A\to B⊊A$ is a bijection, choose the single witness $x_0\in A\setminus B$ and recursively put $x_{n+1}=h(x_n)$. Injectivity of $h$ and the fact that $x_0$ is not in its range show by cancellation that the $x_n$ are distinct. Thus $n\mapsto x_n$ injects $\omega$ into $A$. This uses one existential witness and recursion, not Countable Choice.

F1 rules out the proper-subset bijection. By step 1.1, step 1.2, step 1.3 it therefore rules out an $\omega$-injection and a countably infinite subset as well. Both implications of every equivalence have been accounted for in ZF.

Suppose $p$ forces that a supported name $c$ is a choice function. Let $E$ support both $p$ and $c$. Pick $n$ outside the finitely many pair indices occurring in $E$, and strengthen to $q\le p$ deciding, after relabelling the two mates if necessary, $q⊩c(P_n)=R_{n,0}\mathrm{.}$ This is a single forcing decision, not a sequence of choices.

Because $q$ has finite domain, choose a natural-number coordinate cutoff beyond all coordinates of $q$ in the $n$th blocks. Define an automorphism $\pi$ which swaps the two $n$-blocks while translating the finitely used internal coordinates to unused coordinates, and fixes $E$. Then $q$ and $\pi q$ agree wherever both are defined, hence $q\cup \pi q$ is a condition.

The support of $c$ is fixed, so $\pi c=c$, while $\pi R_{n,0}=R_{n,1}$ and $\pi P_n=P_n$. By F2, equivariance transforms the decision in step 1.1 into $\pi q⊩c(P_n)=R_{n,1}\mathrm{.}$ Their common extension forces $R_{n,0}=R_{n,1}$, contradicting F3. Thus the fresh coordinate swap defeats the proposed choice.

Let $G$ be generic for the basic Cohen forcing over a ZFC ground. By F1 the ambient $M[G]$ satisfies AC. The hereditarily symmetric interpretation $N\subseteq M[G]$, however, contains the invariant set $A$ while every proposed enumeration is moved by a finite-support transposition. By F2, $N\models \mathrm{\neg }\mathrm{A}\mathrm{C}$. Thus $N$ directly refutes the universal statement.

The socks construction sharpens the same failure mode: the indexed pair family is invariant, while a swap outside a proposed choice name's finite support exchanges its selected mate. Hence the family lies in the symmetric model but no choice function does.

The failed inference is therefore $M[G]\models \mathrm{A}\mathrm{C}\Rightarrow N\models \mathrm{A}\mathrm{C}$. Transitivity and satisfaction of ZF pass to the symmetric interpretation by its separate model theorem; AC does not.