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For limit ordinals $\alpha \le \beta$, the structures $V_{\alpha }$ and $V_{\beta }$ agree on whether any shared set is a cardinal number. They agree on being an ordinal. If $\delta <\alpha$ is not a cardinal number, a bijection from some $\gamma <\delta$ onto $\delta$ has rank of a set at most $\delta +3$, hence belongs to $V_{\alpha }$. Its defining properties are bounded formulas in set theory. Both structures therefore see the same failure witness. This full agreement is stronger than downward absoluteness of cardinalhood between arbitrary transitive models.

The ordinal assigned recursively by $\mathrm{nrk}(\tau )=\sup \{\mathrm{nrk}(\sigma )+1:(\sigma ,p)\in \tau \}$. It measures dependency on subnames and justifies generic evaluation by recursion. The ordinary rank of a set also accounts for the coding of the ordered pairs and the forcing conditions; it need not equal this dependency rank of a set.

If $V_{\lambda }$, for an infinite cardinal number $\lambda$, satisfies Axiom schema of replacement for every external functional class, then an external cofinal function from an ordinal below $\lambda$ cannot exist: its range would have rank of a set $\lambda$ and would have to belong to $V_{\lambda }$. Similarly, a surjection $\mathcal{P}(\theta )\to \lambda$ with $\theta <\lambda$ is impossible. Infinity therefore makes $\lambda$ an uncountable regular cardinal and a strong limit cardinal. Full semantics for second-order logic, allowing arbitrary external functional relations, is essential.

This representation of an ordered pair by sets satisfies $⟨a,b⟩=⟨c,d⟩$ exactly when $a=c$ and $b=d$. Its intersection recovers $\{a\}$, and its union recovers $\{a,b\}$; if the latter is a singleton then $b=a$. Its rank of a set is $\max (\mathrm{rank}(a),\mathrm{rank}(b))+2$, which bounds the ranks of function graphs in absoluteness of cardinalhood in limit ranks.

Independence here is a relative consistency assertion: from a ground first-order model satisfying ZFC, construct a first-order model of each remaining theory in which the omitted axiom fails. The ground axiom of choice is a tool for constructing these countermodels, not an extra axiom imposed on the theories being tested. The constructible universe supplies the usual relative-consistency implication from ZF to ZFC, so this ground choice assumption requires no additional consistency strength. These constructions can be formalized inside an arbitrary ground model; they do not require the stronger assumption that a transitive model exists.

Union. Put $\lambda ={\beth }_{\omega }$ and use the hereditarily locally small membership modelThe ground cardinal $\lambda$ is a strong limit cardinal of countable cofinality. This transitive class contains the empty set and $\omega$. Pairing preserves the bound, as does taking a subset, so separation holds for every formula relativized to this class. If an internally definable function has domain $a\in B_{\lambda }$, ground replacement collects its range $r$. Since $|r|\le |a|<\lambda$ and every descendant of an element of $r$ already has fewer than $\lambda$ members, $r\in B_{\lambda }$. Thus the full functional Axiom schema of replacement holds. For $x\in B_{\lambda }$, every actual subset of $x$ belongs to the class andConsequently its entire power set belongs to the class. Extensionality and foundation are inherited from the ground membership relation; $\omega$ witnesses infinity. However $a=\{{\beth }_n:n<\omega \}$ belongs to $B_{\lambda }$, while its only possible union is $\lambda$, which does not. All the other ZF axioms hold, and union fails. Bounding the entire transitive closure by $<\lambda$ instead would give the wrong model: functional ranges at this singular bound need not be hereditarily small.

Power set. Use the transitive class $H_{{\omega }_1}$ of hereditarily countable sets. Pairing, union and separation preserve hereditary countability. A functional image of a countable domain is countable, and its members have countable transitive closures; their countable union is countable in the ground ZFC model. This proves replacement. The remaining structural axioms and infinity are inherited. Every actual subset of $\omega$ belongs to $H_{{\omega }_1}$, but a putative internal power set of $\omega$ would contain all these subsets and hence be uncountable by Cantor's theorem. Power set fails in $H_{{\omega }_1}$, with every other ZF axiom intact.

Replacement. Use $V_{\omega +\omega }$ in the cumulative hierarchy. At a limit ordinal cutoff, pairing, union, power set and every instance of separation remain below the cutoff; foundation, extensionality and infinity hold as well. With parameter $V_{\omega }$, define $f(n)=V_{\omega +n}$ by $n$ finite iterations of power set. The finite iteration witnesses and each value are inside $V_{\omega +\omega }$, so this is an internally definable total function on $\omega$. Its range has rank of a set $\omega +\omega$ and is absent from the model. This instance of replacement fails.

Extensionality. Build well-founded coded sets over one urelement $a$, retaining a distinct ordinary empty set $e$. Membership of an atom is always false, and membership of a set code is its coded extension. Thus $a\ne e$ but they have identical empty extensions. Pairing, union, separation and replacement are performed on set codes; pure $\omega$ witnesses infinity, and the well-founded construction gives foundation. There is a small point concerning the unmodified, untyped Axiom of power set: the atom is vacuously a subset of every object. Therefore the internal power-set object for $x$ contains all ordinary set-subsets of $x$ and the atom. It has exactly the required extension. All remaining ZF axioms hold, but extensionality fails. In the usual typed atom theory, the power-set axiom instead quantifies only over set-subsets.

Foundation. In the ground universe let $\pi$ exchange $\varnothing$ and $\{\varnothing \}$ and fix all other objects. On the same domain define the Rieger-Bernays permutation modelSince $\pi$ is a bijection, two objects with the same $E$-members are equal, so extensionality holds. Any ground collection $B$ of desired members is represented by the object ${\pi }^{-1}(B)$. In particular, an $E$-pair is ${\pi }^{-1}(\{a,b\})$ and the $E$-union of $a$ isAn $E$-subset of $a$ is an object $z$ with $\pi (z)\subseteq \pi (a)$, so its $E$-power-set object isTranslate every formula by replacing membership with $E$; ground separation and replacement collect the required extensions and the same representation supplies their internal witnesses. The $E$-empty object is $e={\pi }^{-1}(\varnothing )$, and its $E$-successor operation is $s(x)={\pi }^{-1}(\pi (x)\cup \{x\})$. Ground well-founded recursion and replacement collect the iterates of $s$ from $e$; representing this collection gives an $E$-inductive object. Thus infinity holds. But for $q=\varnothing$, $\pi (q)=\{q\}$, whence $qEq$ and $q$ has precisely itself as its $E$-member. Foundation fails.

Infinity. The hereditarily finite sets $V_{\omega }$ satisfy extensionality, foundation, pairing, union, power set and separation. Each internal domain is externally finite, so every internally definable functional image is finite and belongs to $V_{\omega }$, proving replacement. No finite object can contain the empty set and all its successive Finite von Neumann ordinals. Infinity fails, and all other ZF axioms hold.

Choice with atoms. For the negative model use the Basic Fraenkel permutation model. Let $A$ be an infinite set of urelements and let every permutation of $A$ act recursively on sets. Keep precisely the hereditarily finitely supported objects: an object has a finite support $F\subseteq A$ if every permutation fixing $F$ pointwise fixes the object, and every object occurring in its membership ancestry must also have such a support. The class of these objects contains the pure universe and $A$ itself.

Supports for a pair are united; union preserves a support; a formula defining a subset uses the finite union of the supports of its domain and parameters. An internally definable functional range has the same support, since its values are uniquely specified, and its elements are already hereditarily supported. This proves the full separation and replacement schemas. All hereditarily supported set-subsets of a fixed set form its symmetric internal power set; this collection is invariant under the pointwise stabilizer of a support of the original set. For the untyped version add all the atoms, as above. Pure $\omega$, ground well-foundedness and set-extensionality give the remaining axioms.

The family $[A{]}^2$ of unordered pairs of atoms is hereditarily supported. If it had a choice function $f$, let $F$ be a finite support of $f$. Choose distinct $a,b\in A\setminus F$. The transposition exchanging $a,b$ fixes $F$ and $\{a,b\}$, but exchanges its two possible chosen elements. Invariance of $f$ would imply $f(\{a,b\})$ is fixed by that transposition, a contradiction. Conversely, the full well-founded universe over the atoms, before restricting to finite supports, satisfies choice when the ground universe does. Both choice and its negation are relatively consistent with the stated weakening of extensionality.

Interpret the assertion in the usual well-founded, class-theoretic version of Zermelo set theory, including foundation, class comprehension and separation with class parameters. Classes are not extra sets. The axiom of limitation of size, in the form used here, says that every proper class is in bijection with the universe $V$.

First derive replacement. Let a functional class relation map a set $a$ onto its image class $B$. If $B$ were proper, the limitation axiom would give a bijection $h:B\to V$. Its composite with the original map would be a surjection $g:a\to V$. Class-parameter separation formsSurjectivity gives $d\in a$ with $g(d)=D$, whence $d\in D$ iff $d\notin D$, a contradiction. Thus $B$ is a set, proving each instance of replacement. The argument also gives class replacement when the functional class is a parameter.

The class $\mathrm{On}$ of ordinals is proper: were it a set, its union and successor would be ordinals larger than every ordinal in it. Limitation gives a bijection $e:\mathrm{On}\to V$. Ordering $V$ by the ordinal indices of $e$ gives a global well-order; each nonempty set has a member with least index. Choosing that member defines global choice.

Conversely assume replacement and global choice. Ordinary choice well-orders each rank segment, and global choice selects such a well-order uniformly, since the collection of well-orders of each fixed rank segment is a nonempty set. Order all sets first by their rank of a set and then by the selected order within that rank. The resulting global order is set-like: the predecessors of any element are contained in a rank segment and form a set. Restrict it to a proper class $A$. By transfinite recursion, enumerate $A$ in this order. It cannot be exhausted at a set ordinal, since replacement would then make $A$ a set. Every member appears, because its predecessors form a set, whose well-order has an ordinal order type. Hence $A$ has order type $\mathrm{On}$. The same applies to $V$, and composing the two enumerations gives a bijection $A\cong V$. Thus limitation of size is equivalent to replacement plus global choice in this class-theoretic setting.

A transitive set $x$ satisfies $z\in y\in x\Rightarrow z\in x$, equivalently $\bigcup x\subseteq x$. If $x$ is transitive and $z\in y\in \bigcup x$, choose $w\in x$ with $y\in w$. Transitivity gives $y\in x$, so $z\in \bigcup x$. Thus $\bigcup x$ is transitive. If $z\in y\in \mathcal{P}(x)$, then $z\in x$ and transitivity implies $z\subseteq x$, whence $z\in \mathcal{P}(x)$. Thus the power set is transitive as well.

The converse for the union is false: take $x=\{\{\varnothing \}\}$. Then $\bigcup x=\{\varnothing \}$ is transitive but $x$ is not. The power-set converse is true: if $z\in y\in x$, then $y\in \{y\}\in \mathcal{P}(x)$ implies $y\in \mathcal{P}(x)$, so $z\in x$.

Define $x_0=x$, $x_{n+1}=\bigcup x_n$ and $\mathrm{TC}(x)={\bigcup }_{n<\omega }{x}_n$. This is a set by replacement and union, contains $x$ as a subset, and is transitive because taking a member advances one stage. Induction shows it is contained in every transitive set containing $x$, proving the transitive closure property.

If the rank of a set $x$ is $\alpha$, thenEvery subset of $x$ has rank at most $\alpha$, and $x$ itself is an element of $\mathcal{P}(x)$, proving the first formula. The second is transitive closure preserves set-theoretic rank. Here the convention is $x\subseteq \mathrm{TC}(x)$, not $x\in \mathrm{TC}(x)$.

The cumulative hierarchy is defined by transfinite recursion:The rank of a set is $\mathrm{rk}(x)={\sup }_{y\in x}(\mathrm{rk}(y)+1)$, equivalently the least $\alpha$ such that $x\subseteq V_{\alpha }$. The standard hierarchy properties give $\mathrm{rk}(V_{\alpha })=\alpha$, so every ordinal occurs as a set rank; the ordinal $\alpha$ itself is another example of rank $\alpha$.

Let $x$ be transitive of rank $\alpha$, and fix $\beta <\alpha$. The supremum formula gives an element of $x$ whose rank is at least $\beta$. Choose such an element $y$ of least possible rank. If $\mathrm{rk}(y)>\beta$, the same supremum formula gives $z\in y$ with rank at least $\beta$. Transitivity gives $z\in x$, and membership strictly decreases rank, contradicting the choice of $y$. Thus $x$ contains an element of every rank below $\alpha$.

A nonempty finite set has rank the maximum of finitely many successor ordinals, hence a successor ordinal; the empty set has rank zero. Conversely, for any $\beta$, the singleton $\{V_{\beta }\}$ has rank $\beta +1$. ThusA finite transitive set of rank $\alpha$ contains, by the proved property, distinct elements of every rank $\beta <\alpha$. Finiteness therefore forces $\alpha <\omega$. Every finite ordinal is itself a finite transitive set of that rank. Consequently

The cumulative hierarchy is defined by transfinite recursion:Prove transitivity and nesting together by transfinite induction. The empty set is transitive. If $V_{\alpha }$ is transitive, each $x\in V_{\alpha }$ satisfies $x\subseteq V_{\alpha }$, so $V_{\alpha }\subseteq \mathcal{P}(V_{\alpha })$. If $y\in x\in \mathcal{P}(V_{\alpha })$, then $y\in V_{\alpha }\subseteq \mathcal{P}(V_{\alpha })$; thus the successor stage is transitive. At a limit, if $y\in x\in V_{\lambda }$, choose an earlier stage containing $x$ and apply its transitivity. The union is transitive and contains every earlier stage. These observations also give $V_{\alpha }\subseteq V_{\beta }$ for all $\alpha \le \beta$.

By Axiom of foundation and well-founded recursion, define the rank of a set by $\mathrm{rank}(x)=\sup \{\mathrm{rank}(y)+1:y\in x\}$. Axiom schema of replacement ensures this is an ordinal. Induction shows every element of $x$ lies in $V_{\mathrm{rank}(x)}$, soThus every set appears in the hierarchy; the argument uses foundation, not the existence of a set containing all sets.

If $\mathrm{rank}(x)=\beta$, every subset of $x$ has rank at most $\beta$, while $x$ itself belongs to $\mathcal{P}(x)$. ConsequentlyOnly successor ordinals occur. Conversely $\mathrm{rank}(V_{\beta })=\beta$, so choosing $x=V_{\beta }$ realizes every successor $\beta +1$. The exact answer is all successor ordinals, excluding zero and nonzero limit ordinals.

The axiom of infinity excludes $V_{\omega }$ and all smaller stages, so the cardinal $\lambda$ is uncountable. Full semantics for second-order logic is decisive: every externally specified functional relation whose ordered pairs lie in $V_{\lambda }$ is an allowed class parameter, even if the relation itself is not an element of $V_{\lambda }$.

If $\mathrm{cf}(\lambda )=\mu <\lambda$, choose an external cofinal function $h:\mu \to \lambda$. Its graph is an allowed class parameter: each input, output and ordered pair has rank of a set below the infinite cardinal $\lambda$, which is a limit ordinal. The domain $\mu$ belongs to $V_{\lambda }$. The second-order form of Axiom schema of replacement would give its range as an element of $V_{\lambda }$. But a cofinal subset of $\lambda$ has rank of a set $\lambda$ and cannot belong to $V_{\lambda }$. Therefore $\mathrm{cf}(\lambda )=\lambda$.

If $2^{\theta }\ge \lambda$ for a cardinal $\theta <\lambda$, choice supplies an external surjection $q:\mathcal{P}(\theta )\to \lambda$. The domain belongs to $V_{\lambda }$ and the ordered pairs of its graph have ranks below $\lambda$. Applying Axiom schema of replacement to this external functional class would put its range, the ordinal $\lambda$, in $V_{\lambda }$, again impossible. Hence $2^{\theta }<\lambda$ for every $\theta <\lambda$.

ConsequentlyThese are exactly the defining conditions for a strongly inaccessible cardinal. This full second-order replacement rank obstruction would not follow from Henkin semantics, where the available class relations may exclude the external functions just used.

The Axiom of foundation states that every nonempty set $A$ has a member $a$ with $a\cap A=\varnothing$. The principle of epsilon induction states that a formula $\varphi$ satisfying $\forall x[(\forall y\in x\varphi (y))\Rightarrow \varphi (x)]$ holds for every set $x$.

Assume Axiom of foundation and suppose a progressive formula has a counterexample $a$. In the transitive closure of $\{a\}$, use axiom schema of separation to collect the counterexamples into a nonempty set $B$. Foundation supplies $b\in B$ having no member in $B$. Since the closure is transitive, every member of $b$ satisfies $\varphi$, contradicting progressiveness. Conversely, if a nonempty set $A$ had no foundation-minimal member, every $x\in A$ would have a member $y\in x\cap A$. The formula $x\notin A$ is then progressive: if every member of $x$ lies outside $A$, $x$ cannot belong to $A$. Epsilon induction would make $A$ empty, a contradiction. This proves equivalence in the presence of the other Zermelo–Fraenkel set theory.

To obtain the cumulative hierarchy, perform epsilon induction on the assertion that $x$ belongs to some $V_{\alpha }$. If this holds for each member $y$, Axiom schema of replacement collects their least such ordinals into a set. Choose an ordinal $\beta$ above them all. Then $x\subseteq V_{\beta }$ and $x\in V_{\beta +1}$. Thus every set occurs in the hierarchy. Use the rank of a set convention $\mathrm{rk}(x)={\sup }_{y\in x}(\mathrm{rk}(y)+1)$, equivalently the least $\alpha$ with $x\subseteq V_{\alpha }$.

Use the Kuratowski ordered pair $(m,n)=\{\{m\},\{m,n\}\}$. Its rank of a set is $\max (m,n)+2$. Every pair in the Cartesian product has finite rank, but those ranks are unbounded in $\omega$. Therefore $\mathrm{rk}(\omega \times \omega )={\sup }_{m,n<\omega }(\max (m,n)+3)=\boxed{\omega }$.

An ordinal has rank equal to itself. Thus the rank of a set here is $\sup \{\omega +2,\omega +3,\omega +4\}=\boxed{\omega +4}$.

The Von Neumann hierarchy is the following transfinite recursion on the ordinals:Here $\mathcal{P}$ denotes the full power set. The levels are increasing transitive sets, and the construction is continuous at limit ordinals. Transfinite induction proves these properties: transitivity at a successor follows because a member of a subset of $V_{\alpha }$ already belongs to $V_{\alpha }\subseteq \mathcal{P}(V_{\alpha })$; at a limit it follows from the set union of increasing transitive sets. In ZF, Axiom of foundation and Axiom schema of replacement ensure that every set belongs to some $V_{\alpha }$. In terms of the rank of a set, $x\in V_{\alpha }$ exactly when $\mathrm{rank}(x)<\alpha$.

Use the standard convention that a strongly inaccessible cardinal is uncountable, regular and strong limit. The printed explanatory definition omits uncountability: literally it also admits $\kappa =\omega$, for which no finite cardinal is a worldly cardinal. Thus that omission makes the requested conclusion false under the literal abbreviated definition. The proof below applies to the usual, intended strongly inaccessible cardinal convention $\kappa >\omega$.

First $|V_{\gamma }|<\kappa$ for every $\gamma <\kappa$. At successor stages this follows from the strong limit cardinal property, and at limits from the regular cardinal property. Consequently $V_{\kappa }\models \mathsf{ZFC}$. In particular, for Axiom schema of replacement, a domain $a\in V_{\kappa }$ has size less than $\kappa$, so its functional image has fewer than $\kappa$ elements, so the supremum of the rank of a set over its elements is below $\kappa$ by regular cardinal structure. All other axioms, including axiom of choice, have their witnesses at bounded ranks below $\kappa$; uncountability supplies axiom of infinity.

Enumerate the first-order formulas as ${\phi }_0,{\phi }_1,\dots$. For each finite collection, reflection inside the set structure $V_{\kappa }$ gives a closed unbounded subset of $\kappa$ of agreeing ranks. To see why the witness bounds stay below $\kappa$, there are fewer than $\kappa$ parameter tuples at each $V_{\gamma }$, and the supremum of their least witness ranks remains below $\kappa$ by regular cardinal structure. Closing under these bounds and taking increasing countable limits gives the usual reflection theorem for definable hierarchies argument. The countable intersection of these closed unbounded subsets is still closed unbounded because $\kappa$ is regular and uncountable. At each resulting $\alpha$,The infinite cardinal numbers below $\kappa$ also form a closed unbounded subset: they are unbounded because ${\lambda }^{+}\le 2^{\lambda }<\kappa$ for infinite $\lambda <\kappa$, and a supremum of increasing cardinal numbers is a cardinal number. Intersect the two closed unbounded subsets. Every member of the intersection is a worldly cardinal, and an unbounded subset of a regular cardinal $\kappa$ has size $\kappa$. HenceIn fact this proves the stronger worldly cardinals below an inaccessible cardinal result that these worldly cardinals contain a closed unbounded subset of $\kappa$.

The essential fact is absoluteness of cardinalhood in limit ranks. Let $\alpha \le \beta$ be limit ordinals and $x\in V_{\alpha }$. Being an ordinal is absolute between these transitive sets, so a non-ordinal is a cardinal number in neither structure. For an ordinal $\delta \in V_{\alpha }$, we have $\delta <\alpha$.

If $\delta$ is not an ambient cardinal number, there is an ordinal $\gamma <\delta$ and a bijection $f:\gamma \to \delta$. With Kuratowski ordered pairs, its graph has rank of a set at most $\delta +3$. Since $\alpha$ is a limit ordinal, $\delta +3<\alpha$, so $f\in V_{\alpha }$. The assertion that $f$ is such a bijection is a bounded formula in set theory and is therefore evaluated correctly by each rank. Conversely, any internal bijection witnessing failure of cardinal number status is an actual one by the same set-theoretic absoluteness. ConsequentlyUnlike arbitrary transitive models, these rank-initial structures contain every sufficiently low-rank witness. That witness availability is what upgrades the usual downward absoluteness of cardinalhood to agreement in both directions.

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.

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 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.