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The Mostowski collapse theorem applies when $E$ is a well-founded relation and an extensional relation on the set $M$. Extensionality here means that different points have different sets of $E$-predecessors. There is then a unique transitive set $T$ and a unique bijection $\pi :M\to T$ such thatThe collapsing map is characterized by well-founded relation recursion:Thus $(M,E)$ is isomorphic to $(T,\in )$. Both well-foundedness and extensionality are hypotheses of the theorem; well-foundedness alone does not make the recursive map injective.

The epsilon-recursion theorem says that for every definable operation $G$ assigning a set $G(x,h)$ to each set $x$ and each function $h$ with domain $x$, there is a unique class function $F$ such thatfor every set $x$.

For uniqueness, suppose $F$ and $F^{\prime }$ both satisfy the recursion. If they agree on every $y\in x$, then their restrictions to $x$ agree and henceThe epsilon induction principle therefore gives $F=F^{\prime }$.

For existence, call a function $f$ coherent on a transitive set $T$ whenThe same epsilon-induction argument shows that any two coherent functions agree on the intersection of their transitive domains. Assuming coherent functions have been constructed below every member of $x$, their compatible union defines all values below $x$; adjoining the value prescribed by $G(x,\cdot )$ extends the construction to $x$. Epsilon induction establishes this construction for every $x$. The compatible local functions therefore unite to the required class function $F$.

A relation $r$ on $x$ is well-founded when every nonempty subset of $x$ has an element with no $r$-predecessor in that subset. It is extensional when

The Mostowski collapse theorem states that every well-founded extensional relation $(x,r)$ is isomorphic to membership on a unique transitive set $M$. Define recursivelyWell-founded recursion makes $\pi$ well defined. Epsilon-style induction along $r$, together with extensionality, shows that $\pi$ is injective. By definition,so $\pi$ is an isomorphism from $(x,r)$ onto $(M,\in )$, where $M=\mathrm{ran}\pi$. If $z\in \pi (a)\in M$, then $z=\pi (b)$ for some $bra$, so $z\in M$; hence $M$ is transitive. If another collapse ${\pi }^{\prime }$ has transitive range, induction gives ${\pi }^{\prime }(a)=\{{\pi }^{\prime }(b):bra\}=\pi (a)$, proving uniqueness.

Finally, suppose $x$ is finite. Its collapse $M$ is a finite transitive set, so every member of $M$ is hereditarily finite. Therefore every well-founded extensional relation on $x$ collapses to a transitive subset of $V_{\omega }$.

Conversely, let $x$ be infinite. Choose the initial infinite ordinal $\kappa$ equinumerous with $x$. The successor ordinal $\kappa +1$ has the same cardinality as $\kappa$. Pulling its membership relation back along a bijection $x\to \kappa +1$ gives a well-founded extensional relation on $x$ whose collapse is $\kappa +1$. But $\kappa \in \kappa +1$ is not hereditarily finite, so $\kappa +1⊈V_{\omega }$. Consequently,

A relation $r$ on a set $x$ is well-founded when every nonempty subset of $x$ has an $r$-minimal element. It is extensional when distinct elements have distinct predecessor sets:The Mostowski collapse theorem states that every well-founded extensional relation on a set is uniquely isomorphic to membership on a transitive set; the collapse is recursively

For the requested example, let $A=V_{\omega +1}$ and defineNo $x_n$ belongs to $A$, and thereforeThe map $x_n\mapsto n$ is an isomorphism from membership on $x$ to membership on the von Neumann ordinal $\omega$, so the Mostowski collapse is $\omega$. On the other hand, $x_0=V_{\omega +1}$ already has rank $\omega +1$, and hence the rank of $x$ is greater than $\omega$.

For statement (i), suppose $(x,r)$ is isomorphic to $(y,\in )$ with $y$ transitive. Membership is well-founded by the axiom of foundation. It is extensional on $y$ because transitivity ensures that the predecessors in $y$ of an element $a\in y$ are exactly the elements of $a$, and the axiom of extensionality distinguishes different sets. Both properties are preserved by isomorphism. Thus (i) is always true.

Write $[f]$ for the equivalence class of $f:\kappa \to V_{\lambda }$ in the ultrapowerand define its membership relation byThe kappa-complete filter property makes $E$ well-founded: an infinite descending $E$-chain would give countably many members of $U$ whose intersection belongs to $U$, and every index in that intersection would yield an infinite descending membership chain, contradicting the Axiom of foundation. The relation is extensional by Łoś's theorem.

The Mostowski collapse theorem therefore gives a unique isomorphism $\pi :(N,E)\to (M,\in )$ onto a transitive set $M$. Recursively, the notation missing from the printed formula may be defined byThe value is independent of the representative because it is defined on the ultrapower class $[f]$. Moreover $M\subseteq V_{\lambda }$: every $f:\kappa \to V_{\lambda }$ has its range contained in some $V_{\alpha }$ with $\alpha <\lambda$, since $\kappa <\lambda$ and the strongly inaccessible cardinal $\lambda$ is regular; induction on the resulting rank bound keeps $\pi ([f])$ inside $V_{\lambda }$.

Define the ultrapower embeddingThe constant-function map into $N$ is elementary by Łoś's theorem, and $\pi$ is an isomorphism, so their composite $j$ is elementary.