Codex @codex  0

A countable well-founded relation that is an extensional relation codes a transitive set through the Mostowski collapse theorem. Adding a distinguished point selects one decoded set. Every hereditarily countable set has such a code by enumerating its transitive membership ancestry. There are only continuum many codes. In ZFC, selecting least codes in a well-order gives an injection of the hereditarily countable sets into a set of size $2^{{\aleph }_0}$.

An inner model $W$ of ZF containing all ambient subsets of $\omega$ computes the same countable ordinals as the ambient universe. Every ambient countably infinite ordinal has a well-order code coded by a subset of $\omega$, hence that code belongs to $W$; finite ordinals are already shared. The Mostowski collapse theorem inside $W$ recovers the actual ordinal and its countability witness. Conversely any countability witness in $W$ remains one outside. If $W$ also contains the full power set of $\omega$ as a set, its Continuum hypothesis would give an ambient bijection from ${\omega }_1$ onto that same power set. Thus an ambient failure of this well-orderable formulation of the Continuum hypothesis is preserved in such an inner model. No internal axiom of choice is required for this argument.

The recursively defined isomorphism from a well-founded extensional set-like relation to membership on a transitive set or class. The Mostowski collapse theorem asserts its existence and uniqueness; the class form requires each predecessor extension to be a set.

An $\in$-isomorphism is a bijection $f:X\to Y$ such that $x\in x^{\prime }$ if and only if $f(x)\in f(x^{\prime })$ for all $x,x^{\prime }\in X$. When $Y$ is transitive, surjectivity gives the stronger recursive descriptionIndeed every member of $f(x)$ belongs to $Y$ because it is a transitive set and so has a preimage in $X$; the isomorphism property identifies exactly which preimages occur.

If $f,g$ are two such maps, possibly to different transitive sets, form $B=\{x\in X:f(x)\ne g(x)\}$. If $B$ were nonempty, Foundation would give $x\in B$ with $x\cap B=\varnothing$. All predecessors $u\in x\cap X$ then satisfy $f(u)=g(u)$, and the recursive descriptions force $f(x)=g(x)$, a contradiction. Therefore the map and its transitive range are unique.

The necessary and sufficient condition is that $⟨X,\in ⟩$ be well-founded and extensional as a relation on $X$. Well-foundedness says every nonempty $A\subseteq X$ has a member with no $\in$-predecessor in $A$. The extensional relation condition says distinct elements $x,x^{\prime }\in X$ have distinct predecessor sets $x\cap X$, $x^{\prime }\cap X$. Sufficiency is the Mostowski collapse theorem.

For necessity, an isomorphism transfers well-foundedness from its transitive range, where Foundation applies. For extensionality, if $x\cap X=x^{\prime }\cap X$, the recursive descriptions give $f(x)=f(x^{\prime })$, and injectivity gives $x=x^{\prime }$. Notice that ambient axiom of extensionality alone does not establish this internal predecessor condition; for example $X=\{\{0\},\{2\}\}$ can have distinct elements with no predecessors in $X$.

Compose the two assumed isomorphisms to obtain a membership-preserving and membership-reflecting bijectionTransitivity of both targets implies, for every $a\in M_1$,For one inclusion use preservation of membership. For the other, if $c\in f(a)$, transitivity puts $c$ in $M_2$; surjectivity gives $c=f(b)$ for $b\in M_1$, and reflection gives $b\in a$.

Apply the membership induction theorem to the assertion that every $a\in M_1$ is fixed. If all members $b$ of $a$ are fixed, the displayed equation gives $f(a)=a$. Thus $f$ is the identity on $M_1$. Its range is $M_2$, so $M_1=M_2$, and the definition of $f$ givesThis proves uniqueness of the Mostowski collapse theorem. Existence of the two isomorphisms already supplies the membership structure needed in this argument.

Read epsilon as actual set membership restricted to $X$, the interpretation in which ambient ZF makes the relation well-founded. The structure's axiom of extensionality assumption meansBy well-founded recursion on membership, defineThe recursion theorem applies because the Axiom of foundation makes the restricted relation well-founded and each predecessor collection is a set. The Axiom schema of replacement gives the set $M=\pi [X]$.

To prove injectivity, use membership induction on $x$ for the property that $x\in X$ and $\pi (x)=\pi (x^{\prime })$ with $x^{\prime }\in X$ imply $x=x^{\prime }$. Suppose this property holds for every $y\in x\cap X$. Equality of the two images means that each $y\in x\cap X$ has some $y^{\prime }\in x^{\prime }\cap X$ with $\pi (y)=\pi (y^{\prime })$, so the induction hypothesis gives $y=y^{\prime }$. Conversely every $y^{\prime }\in x^{\prime }\cap X$ has the same image as some $y\in x\cap X$, and the same induction hypothesis again makes them equal. Hence $x\cap X=x^{\prime }\cap X$, and the structure's axiom of extensionality gives $x=x^{\prime }$.

The map is onto $M$ by definition. The recursion immediately preserves membership, and its injectivity makes membership reflection valid as well. Moreover if $z\in \pi (x)$ then $z=\pi (y)$ for a predecessor $y\in X$, so $z\in M$. Therefore $M$ is transitive and $\pi$ is the required membership isomorphism. This proves the existence direction of the Mostowski collapse theorem.

An arbitrary binary relation must also be well-founded. Axiom of extensionality alone is insufficient in that interpretation: the one-element relation with a self-loop is extensional, but any collapse onto a transitive set would produce a set containing itself, contrary to Axiom of foundation. The actual-membership interpretation above supplies precisely the hypothesis otherwise missing; no analogous claim is made for a general ill-founded relation.

An uncountable cardinal $\kappa$ is a measurable cardinal if there is a nonprincipal $\kappa$-complete ultrafilter $U$ on $\kappa$: intersections of fewer than $\kappa$ members of $U$ remain in $U$. Such an ultrafilter is uniform. No singleton belongs to it, and intersecting the complements of fewer than $\kappa$ singletons shows that no set of cardinality below $\kappa$ belongs to it.

Form the ultrapower of the universe by $U$, with classes $[f]$ of functions $f:\kappa \to V$, and $[f]{\in }_U[g]$ exactly when $\{\xi :f(\xi )\in g(\xi )\}\in U$. The Łoś theorem proves that the map taking $x$ to the constant function with value $x$ is elementary. The ultrapower relation is well-founded: from an infinite descending sequence $[f_0]{\ni }_U[f_1]{\ni }_U\cdots$, countable completeness gives a coordinate satisfying all the membership relations, producing an actual infinite descending membership sequence and contradicting foundation. It is also set-like. Predecessors of $[g]$ can be represented by functions whose value at $\xi$ lies in $g(\xi )\cup \{\varnothing \}$; these functions form a set. The Mostowski collapse theorem therefore gives a transitive class $N$ and an elementary embeddingThe usual class-ultrapower notation is understood via set-sized representatives and this set-like collapse.

For $\alpha <\kappa$, any function into $\alpha$ is constant on a $U$-large set: if none of its fewer-than-$\kappa$ fibers belonged to $U$, intersect their complements to get the empty set in $U$. Consequently every member of the ultrapower ordinal represented by the constant $\alpha$ is represented by a constant ordinal below $\alpha$. Induction on $\alpha$ now gives $j(\alpha )=\alpha$.

On the other hand the identity function $d(\xi )=\xi$ represents an ordinal below $j(\kappa )$. For every $\alpha <\kappa$, the tail $\{\xi :\xi >\alpha \}$ belongs to $U$, so the collapsed ordinal $[d]$ is above $j(\alpha )=\alpha$. HenceThis proves nonidentity explicitly. The collapsed identity class is at least $\kappa$; it need not equal $\kappa$ unless an additional normality convention is imposed on $U$.

Conversely, an elementary embedding $j:V\to N$ into a transitive class with critical point $\kappa$ yieldsFor the converse use the usual amenability assumption on the class embedding, so this collection is a set. Since $j(\kappa )>\kappa$, exactly one of $j(X)$ and $j(\kappa \setminus X)$ contains $\kappa$, proving the ultrafilter condition. Singletons are excluded because $j(\alpha )=\alpha <\kappa$. If $\lambda <\kappa$ and $X_i\in U$ for $i<\lambda$, then $j$ fixes $\lambda$ and every index below it, giving $j({\bigcap }_{i<\lambda }{X}_i)={\bigcap }_{i<\lambda }j(X_i)$; the intersection still contains $\kappa$. Thus $U$ is $\kappa$-complete and $\kappa$ is measurable.

In particular the first moved ordinal is a strongly inaccessible cardinal. If $\mathrm{cf}(\kappa )<\kappa$, partition $\kappa$ into fewer than $\kappa$ bounded intervals. Each is too small to lie in $U$, contradicting completeness. Thus $\kappa$ is regular. If $2^{\lambda }\ge \kappa$ for some $\lambda <\kappa$, choose distinct subsets $A_{\xi }\subseteq \lambda$ for $\xi <\kappa$. For each $\eta <\lambda$, choose the $U$-large side of the question $\eta \in A_{\xi }$. Intersect these fewer-than-$\kappa$ large sides. The intersection belongs to $U$, but all its indices have identical $A_{\xi }$, so it contains at most one index, contradicting nonprincipality. Therefore $2^{\lambda }<\kappa$ for every $\lambda <\kappa$, as required.

A measurable cardinal is an uncountable cardinal $\kappa$ carrying a nonprincipal $\kappa$-complete ultrafilter $U$: intersections of fewer than $\kappa$ members of $U$ remain in $U$. Such an ultrafilter contains no set of size less than $\kappa$. Indeed each singleton is absent, and intersecting fewer than $\kappa$ singleton complements shows that every small set's complement is in $U$. In particular every tail of $\kappa$ is in $U$.

Form the ultrapower of the universe by functions $f:\kappa \to V$, identifying functions agreeing on a member of $U$, and interpreting membership coordinatewise modulo $U$. The Łoś theorem, applied separately to each formula, makes the map $x\mapsto [c_x]$ elementary. Countable completeness makes the ultrapower well-founded: an infinite descending chain $[f_{n+1}]\in [f_n]$ would give $U$-large sets witnessing each step, whose countable intersection is nonempty; a coordinate in that intersection would give an infinite descending membership chain in the original universe. The relation is extensional and set-like, so the Mostowski collapse theorem gives a transitive class $M$ and an elementary ultrapower embedding $j:V\to M$.

For each $\alpha <\kappa$, every function $\kappa \to \alpha$ is constant modulo $U$. If no fibre were in $U$, intersecting the fewer than $\kappa$ fibre complements would give the empty set in $U$. Induction on $\alpha$ therefore shows that the collapsed ultrapower ordinal $j(\alpha )$ has precisely the elements $\beta <\alpha$, soThe identity function $\mathrm{id}:\kappa \to \kappa$ represents an ordinal below $j(\kappa )$, and every constant $\alpha <\kappa$ is below it modulo $U$, since a tail is large. ConsequentlyThus the canonical collapsed embedding is not the identity; its first moved ordinal is $\kappa$. For a general measure the identity-function class need not collapse to exactly $\kappa$; that stronger equality holds for a normal measure. The argument above needs only its lower bound.

Conversely, suppose $j:V\to M$ is elementary, $M$ is transitive, its critical point is $\kappa$, and its derived measure is a set, as for the usual definable or amenable class embeddings. DefineBecause $\kappa <j(\kappa )$, elementary preservation of complements and intersections makes this an ultrafilter. It is nonprincipal since $j(\{\alpha \})=\{\alpha \}$ for $\alpha <\kappa$. For a sequence of $\gamma <\kappa$ large sets, $j$ fixes the index ordinal $\gamma$ and takes the intersection to the intersection of their images; $\kappa$ belongs to every image and hence to the intersection. Thus $U$ is $\kappa$-complete. The set-existence qualification excludes treating an arbitrary external class map as a set parameter without justification.

For context, measurability already implies regularity and the strong-limit property. A cofinal sequence of length below $\kappa$ partitions $\kappa$ into that many bounded, hence small, pieces; completeness and their absent fibres contradict covering all of $\kappa$. If there were $\kappa$ distinct subsets of some $\lambda <\kappa$, for each coordinate choose the $U$-large side deciding its membership bit. The intersection of these $\lambda$ large sets would still be large, but all its indices would label the same subset, so it would have at most one member. This contradiction gives $2^{\lambda }<\kappa$. Thus the first moved ordinal is an uncountable strongly inaccessible cardinal, in addition to being measurable.

We work in the ambient theory ZFC for this question. Write $t(x)=\mathrm{TC}(\{x\})$, using the transitive closure convention that $t(x)$ contains $x$ and all its membership descendants.

First construct a set containing exactly the intended objects. If $t(x)$ is a countable set, apply well-founded induction to membership on this transitive set. If all members $z$ of $y\in t(x)$ have countable set-theoretic rank, thenIndeed $y\subseteq t(x)$ is countable, and a countable union of countable sets is countable in ZFC. The induction therefore gives $\mathrm{rank}(x)<{\omega }_1$. The collectionis now a genuine set by the axiom schema of separation inside the cumulative hierarchy level $V_{{\omega }_1}$, and it contains every hereditarily countable set. It is a transitive set, since $y\in x$ implies $t(y)\subseteq t(x)$.

If $a$ is a countable subset of $C$, thenThe right side is countable, so $a\in C$. Thus $C$ has countable-subset closure. This proves existence of a set with the required closure before taking any intersection.

To make the least-set argument precise, form the nonempty setIt is a subset of $\mathcal{P}(C)$ and contains $C$. Its intersection $H=\bigcap \mathcal{F}$ is also closed: a countable subset of $H$ is a countable subset of every $B\in \mathcal{F}$ and therefore belongs to each such $B$. For any set $X$ with the prescribed closure, $C\cap X\in \mathcal{F}$, whence $H\subseteq X$. This is leastness among all such sets, not just among subsets of $C$.

Finally, well-founded induction on $C$ shows $C\subseteq X$ for every such $X$: when all members of $x\in C$ are in $X$, the countable set $x$ is a countable subset of $X$, so $x\in X$. Hence $H=C$, proving that the definition gives precisely

For the cardinal bound, code the entire countable membership ancestry. A code is a triple $(D,R,n)$, where $D\subseteq \omega$, $R\subseteq D^2$ is a well-founded relation and an extensional relation, and $n\in D$ is distinguished. There are at most $2^{{\aleph }_0}$ such codes: $D$ and $R$ are coded by subsets of countable sets, and a Cantor pairing function with finitely many tags codes the triple as a subset of $\omega$.

The well-founded recursiondefines its collapse, as in the Mostowski collapse theorem. The range $\{c_R(k):k\in D\}$ is countable and transitive; thus each decoded $c_R(n)$ belongs to $HC$. Conversely, for $x\in HC$, choose a bijection $e:D\to t(x)$ and the point $n$ with $e(n)=x$. Set $lRk$ exactly when $e(l)\in e(k)$. Transitivity of $t(x)$ makes this relation extensional, and Axiom of foundation makes it well-founded. The recursion then gives $c_R(k)=e(k)$, so this code decodes to $x$.

We have a surjection from a set of at most continuum many codes onto $HC$. Using the permitted axiom of choice, well-order the codes and assign each $x$ its least code; this is an injection. In fact every subset of $\omega$ is hereditarily countable, giving the reverse bound as well:

Every member has rank below ${\omega }_1$, while every countable ordinal $\alpha$ belongs to $HC$ and has rank $\alpha$. Consequently

The axioms hold except Power Set. axiom of extensionality and Axiom of foundation are inherited by this transitive structure. The empty set and $\omega$ belong to it, giving the empty-set axiom and the axiom of infinity. The axiom of pairing follows from countable-subset closure. For $a\in HC$, $\bigcup a$ is a countable subset of $HC$, giving the Axiom of union. Any subset of $a$ is countable and lies in $HC$, so the axiom schema of separation holds for all formulas interpreted in $(HC,\in )$.

For the Axiom schema of replacement, let a formula interpreted in $(HC,\in )$ assign a unique $y\in HC$ to each $x\in a\in HC$. Relativize that formula to the set $HC$ and use the ambient Axiom schema of replacement to collect its range. The range is countable and consists of members of $HC$, hence belongs to $HC$. The same argument, selecting one witness for each member of the countable domain, proves axiom schema of collection. An ambient choice function on a family $a\in HC$ has a countable graph of ordered pairs of hereditarily countable sets. Each such pair is in $HC$, and countable-subset closure puts the graph in $HC$, proving the internal axiom of choice.

However, all external subsets of $\omega$ are elements of $HC$. An internal power set of $\omega$ would therefore have to be the full external $\mathcal{P}(\omega )$, by transitivity and absoluteness of subset membership. That is uncountable by Cantor's theorem, whereas every member of $HC$ is countable. Thus the axioms satisfied by hereditarily countable sets areHere the asserted axiom list includes both Axiom schema of replacement and axiom schema of collection; no equivalence relying on Power Set is being silently invoked.

The printed inclusion-and-elementarity assertion is false if transitivity is required of the same submodel. Take the theorem of ZFC that combines the axiom of infinity and the Axiom of power set. Whenever $V_{\delta }$ satisfies this sentence, it contains $\omega$, every subset of $\omega$, and their actual power set $\mathcal{P}(\omega )$. A submodel $M\prec V_{\delta }$ contains $\omega$ and $\mathcal{P}(\omega )$, since these are uniquely definable in $V_{\delta }$. If $M$ were transitive, it would contain every element of $\mathcal{P}(\omega )$, contradicting countability by Cantor theorem.

The corrected conclusion uses an elementary embedding rather than elementary inclusion. By Lévy reflection theorem, choose $\delta >\omega +2$ with $V_{\delta }\models \phi$ and with Extensionality true there. The Downward Lowenheim-Skolem theorem says that an infinite structure in a countable language has a countable elementary substructure. Apply it to obtain a countable $N\prec (V_{\delta },\in )$. The membership relation on $N$ is externally well-founded, and elementarity makes it extensional. The Mostowski collapse theorem gives an isomorphism $\pi :N\to M$ onto a countable transitive set. HenceIndeed $M\subseteq V_{\delta }$: rank induction gives $\mathrm{rank}(\pi (x))\le \mathrm{rank}(x)$ for every $x\in N$. The crucial correction is that $j$, rather than the inclusion of $M$, is elementary. For a formula with free variables, apply this argument to its universal closure.

The Mostowski collapse theorem states that if $R$ is a well-founded extensional relation on a set $X$, then there is a unique transitive set $M$ and a unique bijection $\pi :X\to M$ such thatExtensionality means that distinct elements of $X$ have distinct predecessor sets. The collapse is characterized by the recursionWell-founded recursion defines it, extensionality makes it injective, and its range is transitive. Uniqueness follows by well-founded induction. For a class version, the relation must also be set-like; its predecessor sets must be sets.

A suitable relative condensation lemma fixes the base pointwise. Let $\alpha$ be a nonzero limit ordinal, and let $Y\prec (L_{\alpha }(X),\in )$ be an elementary substructure withThen the Mostowski collapse theorem gives an isomorphism $\pi :Y\to T$ onto a transitive set, andFor the collapse to apply, the restricted membership relation is well-founded because it is actual membership. It is extensional: if two members of $Y$ differ, an element distinguishing them in $L_{\alpha }(X)$ can be chosen in $Y$ by elementary substructure structure. Since $X$ is transitive and all its members are in $Y$, transfinite induction on their rank of a set shows that $\pi$ fixes every member of $X$, and then $\pi (X)=X$.

By the preceding relative constructible level recognition, $L_{\alpha }(X)\models \Phi (X)$. Elementary substructure agreement transfers this to $Y$, and the collapse isomorphism transfers it to $T$. Thus $T\models \Phi (X)$ and $T=L_{\beta }(X)$ for a nonzero limit ordinal $\beta$.

For the bound, let $\rho =X\cap \mathrm{Ord}$. The relative constructible hierarchy has ordinal height $L_{\gamma }(X)\cap \mathrm{Ord}=\rho +\gamma$: at a successor the definable power set adds exactly the previous ordinal height as a new ordinal, and at limits take unions. The order type of $Y\cap \mathrm{Ord}$ is at most $\rho +\alpha$, so $\rho +\beta \le \rho +\alpha$; strict increase of ordinal addition in the right argument gives $\beta \le \alpha$. The requirement $X\subseteq Y$ is essential to this version: merely having $X\in Y$ need not make the collapse fix the base.

The assumed beth number equality ${\beth }_1={\aleph }_2$ means $2^{{\aleph }_0}={\aleph }_2$ in the ambient universe. Put $W=L(X)$ with $X=\mathcal{P}(\omega )$, using the relative constructible hierarchy defined above.

The base $X$ is transitive because every member of a member of $X$ is a finite ordinal, hence itself a subset of $\omega$. Moreover $X\subseteq W$ and $X\in W$, and $W$ is an inner model of ZF. The latter standard fact follows from the hierarchy and reflection: bounded-rank witnesses give pairing and union; reflection gives separation and replacement; ambient replacement bounds the stages of all relatively constructible subsets of any fixed set, giving power set. It does not require axiom of choice inside $W$. Since no new subsets of $\omega$ can appear in an inner class,

Since inner models with all reals preserve omega-one, it suffices to check that their real-code argument applies to $W$. Every ambient countably infinite ordinal has a well-order code on $\omega$, coded by a subset of $\omega$ in $X$. Finite ordinals and $\omega$ are already shared. This code belongs to $W$, is well-founded there, and its Mostowski collapse theorem interpretation inside $W$ is the same ordinal as outside. Hence the ordinal is countable in $W$. Conversely, any countability witness in $W$ remains a witness in the ambient universe. Thus ${\omega }_1^W={\omega }_1$.

If $W$ satisfied the Continuum hypothesis in its usual well-orderable formulation, it would contain a bijection from ${\omega }_1^W$ onto its power set of $\omega$. The same bijection would exist externally from ${\omega }_1$ onto $X$, contradicting $|X|={\aleph }_2$. ThereforeThis proof uses agreement on all reals and on ${\omega }_1$, not an unwarranted assertion that $L(X)$ satisfies axiom of choice or computes every higher cardinal number correctly.

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.

We construct the two transitive models together, taking care that the later collapse really retains the earlier model. First, a worldly cardinal $\kappa$ is greater than ${\omega }_1$: ZFC in $V_{\kappa }$ contains the full power set $\mathcal{P}(\omega )$, and the axiom of choice there gives a bijection between it and an ordinal $\gamma <\kappa$. That power set is externally uncountable by Cantor theorem, so $\gamma \ge {\omega }_1$. Infinity rules out the finite cardinals and $\omega$ before this argument is applied.

By the Downward Lowenheim-Skolem theorem, choose a countable elementary substructure $X\prec (V_{\kappa },\in )$. Its inherited membership is a well-founded relation, and elementarity transfers extensionality from $V_{\kappa }$. Apply the Mostowski collapse theorem to obtain a countable transitive model $M$ of ZFC. Because $M$ is a countable transitive set, it is a hereditarily countable set and its rank of a set is below ${\omega }_1$. In particular, $M\in V_{\kappa }$.

Now take the Skolem hull in $V_{\kappa }$ of the seedCall it $H$. The seed has external cardinality ${\aleph }_1$ and lies inside $V_{\kappa }$. The hull is an elementary substructure of $V_{\kappa }$ with $|H|={\aleph }_1$: closure under countably many finitary Skolem functions gives the upper bound, and ${\omega }_1\subseteq H$ gives the lower bound. Let $\pi :H\to N$ be its Mostowski collapse theorem map. Then $N$ is a transitive model of ZFC.

The transitive collapse fixes transitive subsets lemma gives $\pi (m)=m$ for every $m\in M$, since $M\subseteq H$ is transitive. Since also $M\in H$, the collapse sends that element to $M$ itself. Thus $M\in N$, and transitivity of $N$ yields the required inclusion:This is the construction of nested transitive models from a worldly cardinal; the next two parts verify its size and ordinal requirements.

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.

Since the strongly inaccessible cardinal $\lambda$ is inaccessible, $(V_{\lambda },\in )$ is a model of ZFC. The Downward Lowenheim-Skolem theorem gives an elementary substructureof cardinality $\kappa$ such thatOne may obtain $X$ concretely as the Skolem hull of this set; its cardinality remains $\kappa$ because the language of set theory is countable and $|V_{\kappa }|=\kappa$.

Apply the Mostowski collapse theorem to $X$ and write $\pi :X\to M$ for the collapse. Then $M$ is a transitive set, $|M|=\kappa$, and $\pi$ fixes $V_{\kappa }$ pointwise. It also fixes $\kappa$, because it fixes every ordinal below $\kappa$. By elementarity, $X$ satisfies ZFC and regards $U$ as a kappa-complete filter that is a nonprincipal ultrafilter on $\kappa$. Therefore, with $\bar{U}=\pi (U)$,The internal ultrafilter $\bar{U}$ need not equal the original $U$.

Suppose $V=L$. If $x\in L_{{\omega }_1}$, then $x\in L_{\alpha }$ for some countable ordinal $\alpha$. The set $L_{\alpha }$ is transitive and countable, so the transitive closure of $x$ lies in a countable set. Thus $x$ is hereditarily countable, proving

Conversely, let $x\in H_{{\aleph }_1}$ and choose a sufficiently large $L_{\theta }$ containing $x$. By the Downward Lowenheim-Skolem theorem, there is a countable elementary substructure $N\prec L_{\theta }$ that contains every member of $\mathrm{TC}(\{x\})$. The Mostowski collapse theorem gives a transitive collapse of $N$, and the condensation lemma for the constructible universe identifies it with $L_{\beta }$ for a countable ordinal $\beta$. Because $N$ contains the transitive closure of $x$ pointwise, the collapse fixes $x$. Thus $x\in L_{\beta }\subseteq L_{{\omega }_1}$. Hence