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Fix the base convention $L_0(X)=X$. For a transitive set $X$, the relative constructible hierarchy isHere the definable power set isThe satisfaction for a set structure in this definition is a definable set-theoretic relation, obtained by finite syntax coding and recursion on a first-order formula. The transfinite recursion therefore defines a class in ZF.

Each level is a transitive set. For transitive $A$, every $a\in A$ is a subset of $A$ definable using the parameter $a$, so $A\subseteq \mathrm{Def}(A)$; also $A\in \mathrm{Def}(A)$ using the always-true first-order formula. Thus the levels increase and $X\in L_1(X)$. Starting instead with $X\cup \{X\}$ is another common indexing convention, but it is not the convention used here. The relative universe need not satisfy axiom of choice: arbitrary $X$ need not have an internally available well-order.

A definable continuous hierarchy is a definable class function $\alpha \mapsto H_{\alpha }$ on the ordinals, with set-valued levels, such that $H_{\alpha }\subseteq H_{\beta }$ for $\alpha <\beta$ andfor nonzero limit ordinals. Its union is a definable class in set theory $H$. Often the definition additionally requires every level to be a transitive set; the following statement also works without that requirement.

The reflection theorem for definable hierarchies says that for any finite collection $\mathcal{F}$ of first-order formulas, there is a closed unbounded class of ordinals $\alpha$ such that, for every $\phi \in \mathcal{F}$ and every parameter tuple from $H_{\alpha }$,A useful justification is the witness-closure proof. Close $\mathcal{F}$ under subformulas. At each level and for each existential subformula, bound the least level containing a witness for each parameter tuple for which a witness exists in $H$. Axiom schema of replacement bounds these indices. Iterating the finitely many bounds through $\omega$ produces a limit level containing all required witnesses. Induction on the first-order formulas, equivalently the finite-formula version of the Tarski-Vaught test, yields agreement. Continuity gives closedness of the reflecting class, and starting above any prescribed ordinal gives unboundedness. This is finite reflection, not a claim that all first-order formulas reflect simultaneously in an arbitrary hierarchy.

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 Rasiowa–Sikorski lemma constructs a generic filter $G_i$ over each $M_i$. Every resulting generic extension remains a countable transitive model of ZFC: there are only countably many ground-model names externally. The order $\mathbb{P}$ has the same elements and ordering at every stage, and atomless forcing order structure is absolute because all its quantifiers are bounded to $\mathbb{P}$. Hence the generic filter for an atomless order is new result gives $G_i\notin M_i$ for every $i$.

The increasing union $N={\bigcup }_{i<\omega }{M}_i$ is transitive and contains $\mathbb{P}$. If it satisfied Axiom of power set for $\mathbb{P}$, there would be a set $A\in N$ withChoose $i$ with $A\in M_i$. The next-stage generic filter $G_i$ belongs to $M_{i+1}\subseteq N$ and is an actual subset of $\mathbb{P}$. This subset assertion is absolute for the transitive set $N$, so $G_i\in A$. Transitivity of $M_i$ and $A\in M_i$ then give $G_i\in M_i$, a contradiction. Thus the power-set failure in an increasing union of generic extensions occurs already at the fixed ground-model order:The stages form an increasing chain, not an elementary chain, so the elementary chain theorem does not assert ZFC for their union.

Use the convention that the transitive closure of $x$ is the least transitive set containing $x$ as a subset. Define $x_0=x$ and $x_{n+1}=\bigcup x_n$. For each natural number $n$, a finite sequence of length $n+1$ with these properties exists by induction, and is unique because each next entry is the uniquely determined union of the previous entry. Thus the relation sending $n$ to its last entry is a definable function-class, not just a putative recursive specification. The Axiom schema of replacement applied to $\omega$ gives the set of all $x_n$. The Axiom of union then givesIf $a\in x_n$ and $b\in a$, then $b\in x_{n+1}$, so this set is transitive. It contains $x=x_0$, and any transitive set containing $x$ contains all $x_n$ by induction; hence it is least. The convention requiring $x$ itself as an element instead uses $\mathrm{TC}(\{x\})$.

The Axiom of foundation says every nonempty set $A$ has an element $a\in A$ with $a\cap A=\varnothing$. Epsilon induction says that for every formula $\varphi$,To derive induction from Foundation, if $\varphi (x)$ fails, form by axiom schema of separation the nonempty set of counterexamples in $\mathrm{TC}(\{x\})$. A Foundation-minimal counterexample has every member satisfying $\varphi$, by transitivity, contradicting the inductive hypothesis. Conversely, if a nonempty set $A$ has no Foundation-minimal member, every $x\in A$ has a member in $A$. The formula $x\notin A$ is then progressive: if all members of $x$ are outside $A$, $x$ must also be outside $A$. Epsilon induction makes $A$ empty, a contradiction. Thus the two principles are equivalent given the other axioms.

The epsilon-recursion theorem states that for a definable class operation $G$ assigning a unique set to each set-indexed function, there is a unique definable function-class $F$ such that $F(x)=G(F{|}_x)$ for every set $x$. A version with explicit $x$ in the argument is equivalent. Foundation gives well-foundedness and Replacement makes each restriction set-sized. Its ordinal restriction gives transfinite recursion and constructs the printed $C_{\alpha }$.

Here “countable” means at most countable, including finite sets and the empty set. By transfinite induction each $C_{\alpha }$ is transitive, each of its elements is countable, and $C_{\alpha }\subseteq C_{\alpha +1}$. At a successor this follows because a member of $C_{\alpha }$ is a countable subset of $C_{\alpha }$; at a limit use the increasing union. Let ${\omega }_1$ be the first uncountable ordinal. Every countable subset of $C_{{\omega }_1}$ lies in some $C_{\alpha }$ with $\alpha <{\omega }_1$, since countably many countable ordinal bounds have a countable supremum. It therefore belongs to $C_{\alpha +1}$. ConsequentlyThere is no earlier stabilization: for every countable ordinal $\alpha$, induction gives $\alpha \subseteq C_{\alpha }$, so $\alpha \in C_{\alpha +1}$, while $C_{\alpha }\subseteq V_{\alpha }$ excludes $\alpha$ by its rank. Thus ${\omega }_1$ is the first such stage. The fixed set is the set of hereditarily countable sets.

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.

We prove simultaneously that each $L_{\alpha }$ is a transitive set and that $L_{\alpha }\subseteq L_{\alpha +1}$. The key observation is that if $A$ is transitive and $x\in A$, then $x\subseteq A$ andThis is an internal definition with parameter $x$, so $x\in \mathrm{Def}(A,1)$. Hence $A\subseteq \mathrm{Def}(A,1)$. Every $z\in \mathrm{Def}(A,1)$ is a subset of $A$; if $y\in z$, then $y\in A\subseteq \mathrm{Def}(A,1)$. Thus the definable power set is transitive too.

Start with the empty set. At successor stages the observation proves transitivity and inclusion. At a limit stage the union of the earlier transitive sets is transitive, and the same observation then proves its inclusion in its own definable power set. This completes the transfinite induction and gives

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,

The Von Neumann hierarchy is defined by transfinite recursion:for every limit ordinal $\lambda$.

We prove simultaneously by transfinite induction that every $V_{\alpha }$ is a transitive set and that $V_{\alpha }\subseteq V_{\alpha +1}$. The claims are immediate for $V_0$. Suppose first that $V_{\alpha }$ is transitive. If $x\in y\in V_{\alpha +1}=\mathcal{P}(V_{\alpha })$, then $y\subseteq V_{\alpha }$, so $x\in V_{\alpha }$. Moreover, transitivity says that every $x\in V_{\alpha }$ satisfies $x\subseteq V_{\alpha }$, and henceThus $V_{\alpha +1}$ is transitive and $V_{\alpha }\subseteq V_{\alpha +1}$. At a limit $\lambda$, if $x\in y\in V_{\lambda }$, then $y\in V_{\beta }$ for some $\beta <\lambda$; transitivity of $V_{\beta }$ gives $x\in V_{\beta }\subseteq V_{\lambda }$. The inclusions at successor stages also imply

Finally, the rank of a set $x$ is the ordinalEvery $y\in x$ belongs to $V_{\mathrm{rank}(y)+1}\subseteq V_{\mathrm{rank}(x)}$, so $x\subseteq V_{\mathrm{rank}(x)}$. Thereforewhich proves that every set occurs in the hierarchy.

The Von Neumann hierarchy is defined by transfinite recursion:for a limit ordinal $\lambda$.

We prove by transfinite induction that every $V_{\alpha }$ is a transitive set. The claim is immediate for $V_0$. If $V_{\alpha }$ is transitive and $x\in y\in V_{\alpha +1}$, then $y\subseteq V_{\alpha }$, so $x\in V_{\alpha }$. Transitivity also gives $x\subseteq V_{\alpha }$, hence $x\in \mathcal{P}(V_{\alpha })=V_{\alpha +1}$. At a limit stage, if $x\in y\in V_{\lambda }$, then $y\in V_{\beta }$ for some $\beta <\lambda$, so $x\in V_{\beta }\subseteq V_{\lambda }$.

Transitivity implies $V_{\alpha }\subseteq \mathcal{P}(V_{\alpha })=V_{\alpha +1}$. Induction and unions at limit stages then give

By well-founded recursion on membership, define the rank of a setInduction on this rank gives $y\in V_{\mathrm{rank}(x)}$ for every $y\in x$. Thusand consequentlyEvery set therefore occurs at some level of the hierarchy.

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.

An strongly inaccessible cardinal $\kappa$ has the Keisler extension property when there is a proper transitive set $X⊋V_{\kappa }$ such that

Suppose $\kappa$ is strongly inaccessible and has this property. Because $X$ properly extends the transitive set $V_{\kappa }$, it contains $\kappa$. Strong inaccessibility of $\kappa$ is downward absolute from the ambient universe to the transitive set $X$: any internal witness that $\kappa$ is countable, singular, or not a strong limit would also be an ambient witness. Hencewith $\kappa$ as a witness. Since $V_{\kappa }\prec X$, the same sentence holds in $V_{\kappa }$. Its witness is an ordinal $\mu <\kappa$. The set $V_{\kappa }$ contains $V_{\mu +1}$, so it computes all subsets of cardinals below $\mu$ correctly; strong inaccessibility of $\mu$ is therefore absolute between $V_{\kappa }$ and the universe. Thus there is a strongly inaccessible $\mu <\kappa$, and $\kappa$ cannot be the least strongly inaccessible cardinal.

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 $U$.

Yes. Start with the model $M_0$ constructed in part a. If $M_0$ has no internally strongly inaccessible cardinal above $\kappa$, put $M=M_0$. Otherwise let $\delta$ be the least ordinal above $\kappa$ that $M_0$ regards as strongly inaccessible, and putIn the second case $M\models \mathrm{ZFC}$ because $M_0$ regards $\delta$ as inaccessible. The measure witnessing that $\kappa$ is measurable has rank below $\kappa +3<\delta$, so it still belongs to $M$. In both cases $M$ is a transitive set of cardinality $\kappa$, contains $V_{\kappa }$, and has no internally inaccessible ordinal strictly between $\kappa$ and its height.

We verify absoluteness for every ordinal $\alpha \in M$. If $\alpha <\kappa$, then $M$ and $V_{\lambda }$ both contain $V_{\alpha +1}$ and therefore compute all subsets and functions relevant to strong inaccessibility in the same way. At $\alpha =\kappa$, both models see a measurable cardinal and hence an inaccessible cardinal. Finally, if $\kappa <\alpha \in M$, then $M$ says that $\alpha$ is not inaccessible by construction. The larger model $V_{\lambda }$ cannot say that it is inaccessible, because strong inaccessibility is downward absolute to a transitive model of ZFC: any failure visible in the smaller model remains a failure in the larger one, while ambient inaccessibility would force internal inaccessibility. Hence “$\alpha$ is inaccessible” is absolute between $M$ and $V_{\lambda }$.

A condensation sentence for the constructible hierarchy may be obtained by taking a single conjunction $\sigma$ that expresses a sufficiently strong finite fragment of set theory, the assertion $V=L$, and that the ordinals have no largest member. The finite fragment is chosen strong enough to define the satisfaction relation needed for the $L$-construction and to prove its absoluteness for transitive sets.

If a transitive set $X$ satisfies $\sigma$, let $\lambda =X\cap \mathrm{Ord}$. The absence of a largest ordinal makes $\lambda$ a limit ordinal. Internal $V=L$ says every $x\in X$ belongs to some internally constructed $L_{\beta }$, while transitivity and absoluteness identify that level with the actual $L_{\beta }$. Conversely the finite closure axioms ensure that every $L_{\beta }$, $\beta <\lambda$, belongs to $X$. Hence $X=L_{\lambda }$.

For a transitive set $X$, use $L_0(X)=X$, $L_{\alpha +1}(X)=\mathrm{Def}(L_{\alpha }(X))$ and unions at nonzero limit ordinals. The successor operation is the definable power set. Each level is transitive and the levels increase; moreover $X\in L_1(X)$. Their union is $L(X)$, an inner model of ZF. The initial-set convention differs from a hierarchy using a predicate, or one starting at $X\cup \{X\}$. Axiom of choice need not hold for an arbitrary base without an internally available well-order.

For transitive sets $M,X$ with $X\in M$, a single first-order formula $\Phi (X)$ recognises the nonzero limit levels $L_{\alpha }(X)$. Require finite relation closure for set-theoretic coding, transitivity of $X$, that every element belongs to a coded relative constructible stage, and that every represented stage index has a larger represented index. Correct history codes have downward-closed indices with no maximum, hence form a nonzero limit ordinal $\alpha$. The exhaustion assertion and transitivity then give $M={\bigcup }_{\xi <\alpha }{L}_{\xi }(X)$. Conversely every limit level has the required closure, correct histories cofinal in its index, and exhaustion. This formulation does not assume ZF for an arbitrary input $M$ and does not replace finite coding closure with an unjustified internal constructibility assertion.