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A permutation model is a model of the axioms of set theory in which there is a non trivial automorphism of the set theoretic universe. Such models are used to show the consistency of the negation of the Axiom of Choice (AC).
A typical construction of a permutation model is done here. By $ZF^-$ we denote the axioms of $ZF$ minus the axiom of foundation. In particular we allow sets $a$ such that $a = \{a\}$ which we will call atoms. Let $A$ be an infinite set of atoms.
Define $V_\alpha(A)$ by induction on $\alpha$ as follows:
Finally define $V = \bigcup_{\alpha \in {ON}} V_\alpha(A)$ . Then we have $$ A = V_0(A) \subset V_1(A) \subset \cdots \subset V_\alpha(A) \cdots \subset V $$ For any $x \in V$ we can assign a rank, $$ \rank(x) = \text{ least } \alpha [ x \in V_{\alpha+1}(A)] $$ Let $G$ be the group of permutations of $A$ . For $\pi \in G$ we extend $\pi$ to a permutation of $V$ by induction on $\in$ by defining $$ \pi(x) = \{ \pi(y) :
y \in x \} $$ and letting $\pi(\emptyset) = \emptyset$ . Then $G$ permutes $V$ and fixes the well founded sets $WF \subset V$ .
Lemma 1 For all $x,y \in V$ and any $\pi \in G$ . $$ x \in y \iff \pi(x) \in \pi(y) $$
That is, $\pi$ is an $\in$ -automorphism of $V$ . From this we can prove that $\pi(\{X,Y\}) = \{\pi(X), \pi(Y)\}$ and so
Also by induction on $\alpha$ it is easy to show that $$ \rank(x) = \rank(\pi(x)) $$ for all $x \in V$ .
Let $a_1,\cdots,a_n \in A$ and define $$ [ a_1, \cdots, a_n ] = \{\pi \in G : \pi(a_i) = a_i,\text{ for } i = 1, \cdots, n \} $$ Call a set $X \in V$ symmetric if there exists $a_1,\cdots,a_n \in A$ such that $\pi(X) = X$ for all $\pi \in [a_1, \cdots, a_n]$ . Define the class $HS \subset V$ of hereditarily symmetric sets $$ HS = \{x \in V : x \text{ is symmetric and } x \subset HS \}
$$
Call a class $N$ transitive if $$ \forall x \in N [ x \subset N] $$ and call $N$ almost universal if (for sets S) $$ \forall S \subset N [ \exists Y \in N (S \subset Y) ] $$
$HS$ is transitive and almost universal.
To show that a class $N \models ZF^-$ is straightforward for most axioms of $ZF^-$ except for the axiom of Comprehension. To show $N$ is a model of Comprehension it suffices to show that $N$ is closed under Gödel Operations:
Theorem 1 ($ZF$) If $N$ is transitive, almost universal and closed under Gödel Operations, then $N \models ZF$ .
$HS$ is closed under Gödel operations and so $HS \models ZF^-$ . The class $HS$ is a permutation model. The set of atoms $A \in HS$ and furthermore:
Lemma 2 Let $f : \w \rightarrow A$ be a one to one function. Then $f \notin HS$ and so $A$ cannot be well ordered in $HS$ .
Which proves the theorem:
Theorem 2 $HS \models ZF^- + \neg AC$ .
which completes the proof that ${Con}(ZF^-) \implies {Con}(ZF^- + \neg AC)$ . In particular we have that $ZF^- \nproves AC$ .
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