The axiom of infinity fails. First prove $L_n=V_n$ for every finite ordinal $n$. The base stage is empty in both hierarchies. If $L_n=V_n$, that stage is a finite set. Every one of its subsets is definable over it with finitely many parameters, by a disjunction of equalities to the selected elements; the empty subset is defined by a false formula. HenceInduction and the limit-stage definitions givethe set of all hereditarily finite sets. Every member of this set is finite externally.

If this transitive structure satisfied the axiom of infinity, it would contain an inductive set $I$: zero belongs to $I$, and whenever $x\in I$, its successor $x\cup \{x\}$ belongs to $I$. These operations are the actual empty set and actual set successor, since they are absolute for this transitive structure and all relevant finite-rank sets lie in it. External induction would then place every finite ordinal in $I$, contradicting the finiteness of $I$.

Every individual finite ordinal does belong to $L_{\omega }$, but the set $\omega$ of all finite ordinals does not. The distinction between containing each finite ordinal and containing one inductive set is precisely the issue.

Use transfinite induction. The assertion is immediate at stage zero, and at a limit stage it follows by taking the increasing union of the earlier stages.

Assume $L_{\alpha }\cap \mathrm{On}=\alpha$. Any ordinal $\gamma \in L_{\alpha +1}$ is a subset of $L_{\alpha }$, since members of the definable power set are such subsets. Every element of $\gamma$ is an ordinal, sowhich implies $\gamma \le \alpha$. This gives the upper bound on the ordinals appearing at the successor stage.

For the reverse inclusion, the smaller ordinals are already in $L_{\alpha }$ and hence in $L_{\alpha +1}$. The remaining ordinal $\alpha$ itself is the definable subset $L_{\alpha }\cap \mathrm{On}$. The fact used here is absoluteness of ordinalhood between a transitive set and the universe: ordinalhood has a bounded definition, and bounded formulas are absolute for transitive sets. Thus the formula asserting that a member is an ordinal, interpreted over $L_{\alpha }$, defines exactly $\alpha$. Therefore $\alpha \in \mathrm{Def}(L_{\alpha })=L_{\alpha +1}$.

This provesIn particular each ordinal $\alpha$ is a member of $L_{\alpha +1}$, so $\boxed{\mathrm{On}\subseteq L}$.

For a set $A$, its definable power set consists of the subsets of $A$ definable by a first-order formula over $(A,\in )$ using finitely many parameters from $A$. The quantifiers range over $A$; this is generally much smaller than its full power set.

The constructible universe is defined by transfinite recursion:The last union denotes a definable proper class, not a set-sized union over a set of all ordinals. At a successor stage, only definitions over the previously constructed stage are permitted.

To prove that power-set equality is not upward absolute, use the standard constructible universe theorem: $L$ is a transitive inner model of ZF containing every ordinal, and set rank is absolute between $L$ and $V$. Since $V\ne L$, some rank stage contains a nonconstructible set. Let $\alpha$ be the least ordinal with $V_{\alpha }⊈L$. It is not zero; it cannot be a limit, since a union of earlier stages contained in $L$ is still contained in $L$. Thus $\alpha =\beta +1$ and $V_{\beta }\subseteq L$.

We need not merely this inclusion, but that $y=V_{\beta }$ itself belongs to $L$. Inside $L$, ZF constructs its own rank stage $(V_{\beta }{)}^L$, which is a set of $L$. By transitivity and rank absoluteness,Consequently $y\in L$. Let $x=(\mathcal{P}(y){)}^L$, the power set computed in $L$. Then $x\in L$ and $L\models x=\mathcal{P}(y)$.

By minimality there is $z\in V_{\beta +1}\setminus L$. Externally $z\subseteq V_{\beta }=y$, but $z\notin x$, because every member of the set $x\in L$ is in the transitive class $L$. Hence $V/\models x=\mathcal{P}(y)$. Taking $U_1=L$ and $U_2=V$ proves power-set equality is not upward absolute, even between transitive models of ZF.

There is no contradiction with absoluteness of $z\subseteq y$: that bounded relation agrees whenever both parameters are present in the smaller class. Power-set equality quantifies over all subsets, including a new $z$ absent from the smaller class. The step proving $V_{\beta }\in L$ is essential; inclusion in a class alone would not imply membership in it.

A sufficient syntactic condition is that $\varphi$ be a sigma-one formula in set theory, written ${\Sigma }_1$: an existential block $\exists z_1\cdots \exists z_m\theta$ with $\theta$ bounded, meaning that each of its quantifiers is of the form $\forall t\in a$ or $\exists t\in a$. A bounded formula is also called ${\Delta }_0$ and is absolute in both directions between transitive classes.

Prove the latter fact by structural induction. Atomic membership and equality have their actual meanings in both classes, and Boolean connectives preserve equivalence of truth. For a bounded quantifier with bound $a\in U_1$, transitivity ensures that every element of $a$ belongs to $U_1$; thus its range is exactly the same in both classes, and induction applies to its matrix. This proves the bounded case. If the smaller class satisfies a ${\Sigma }_1$ formula, take its existential witnesses in $U_1$. They remain available in $U_2$, and bounded absoluteness preserves the matrix. Hence ${\Sigma }_1$ formulas are upward absolute. An arbitrary unbounded universal quantifier has no such witness-preservation argument.

Let $U_1\subseteq U_2$ be transitive classes, both interpreted with actual membership. A formula $\varphi (\bar{x})$ is upward absolute from $U_1$ to $U_2$ if for every tuple of parameters $\bar{a}$ from $U_1$,For proper classes this means the usual relativization of the fixed formula's quantifiers. Full absoluteness requires the implication in both directions. The qualification on parameters matters: the smaller class must contain every object named in the comparison.

The generalized continuum hypothesis isEquivalently, for every ordinal $\alpha$, $2^{{\aleph }_{\alpha }}={\aleph }_{\alpha +1}$. Here $2^{\kappa }$ is the cardinality of the power set of a set of cardinality $\kappa$, and ${\kappa }^{+}$ is the least strictly larger cardinal. The hypothesis says that no intermediate cardinal lies between an infinite set and its power set.

König's theorem for cardinal numbers states that, for a set-indexed family with ${\kappa }_i<{\lambda }_i$ at every index,Here the sum is the cardinality of a disjoint union and the product is the cardinality of a Cartesian product. Work in ZFC. Choose disjoint sets $A_i$ of sizes ${\kappa }_i$ and sets $B_i$ of sizes ${\lambda }_i$. For any function $f:{⨆}_i{A}_i\to {\prod }_i{B}_i$, its $i$th coordinate values on $A_i$ form a subset of $B_i$ of cardinality at most ${\kappa }_i$. Choose $b_i\in B_i$ outside that subset, using choice. Then $b=(b_i{)}_i$ cannot equal $f(a)$ for any $a$: if $a\in A_j$, its $j$th coordinate was specifically excluded. Thus there is no surjection from the disjoint union onto the product.

Cardinal comparability under choice now implies the displayed strict inequality. Empty index set gives $0<1$ directly; empty individual $A_i$ cause no difficulty, since the corresponding coordinate excludes an empty image. This is the cardinal diagonal theorem, rather than the finitely branching tree result also bearing König's name.

To prove replacement failure at V omega plus omega, put $M=V_{\omega +\omega }$. The set $\omega$ belongs to $M$, and for every $n\in \omega$ the ordinal $\omega +n$ also belongs to $M$. The relation saying that $y$ is obtained by taking $n$ successive ordinal successors of $\omega$ is definable, with parameter $\omega$, and is single valued in $M$.

For clarity this does not assume Replacement inside $M$. For each finite $n$ its finite computation sequence exists in $M$: its entries have rank below $\omega +n+1$, and the finite coding overhead still has rank below $\omega +\omega$. The bounded formula verifying this finite successor chain is absolute between $M$ and the universe. The ordinary finite induction identifying its unique last entry uses the actual finite ordinals, which are the same in $M$. This justifies the internal definable function $n\mapsto \omega +n$ without invoking the axiom being tested.

If Replacement held there, its range $B=\{\omega +n:n\in \omega \}$ would be an element of $M$. But ordinal rank equals the ordinal, soEvery element of $V_{\omega +\omega }$ has strictly smaller rank. This contradiction proves Replacement fails in $V_{\omega +\omega }$. Each individual output is present; the missing object is the set collecting all outputs.

Take a nonzero limit ordinal $\lambda$ and $x\in V_{\lambda }$. Choose $\beta <\lambda$ with $x\in V_{\beta }$. Transitivity gives $x\subseteq V_{\beta }$. Hence every actual subset of $x$ belongs to $\mathcal{P}(V_{\beta })=V_{\beta +1}$, and the actual set $\mathcal{P}(x)$ belongs to $V_{\beta +2}$. Since $\lambda$ is a limit, $\beta +2<\lambda$, so $\mathcal{P}(x)\in V_{\lambda }$. Thus the internal power set contains every actual subset of $x$, and the power-set axiom holds in $V_{\lambda }$.

Here the absoluteness fact used is: every bounded formula is absolute between transitive sets or classes with the actual membership relation. In particular $z\subseteq x$, expressed by $\forall t\in z(t\in x)$, has the same meaning internally and externally. The argument supplies the actual power set as an internal witness. If zero is included in one's convention for limit ordinals, its empty stage has no nonempty first-order domain; the assertion is understood for nonzero limits.

If also $\lambda >\omega$, the preceding ordinal-membership result gives $\omega \in V_{\omega +1}\subseteq V_{\lambda }$. The actual set $\omega$ contains the empty set and is closed under $z\mapsto z\cup \{z\}$. These bounded membership and successor conditions are absolute in the transitive stage, so $\omega$ is an internal inductive set. Therefore Infinity holds as well. In contrast $V_{\omega }$ has only hereditarily finite sets and cannot use $\omega$ as an element.

The cumulative hierarchy is defined by transfinite recursion on the ordinals:The recursion theorem of ZF supplies each set stage. The whole universe is $V={\bigcup }_{\alpha \in \mathrm{Ord}}{V}_{\alpha }$, a proper-class union. Equivalently, $x\in V_{\alpha }$ exactly when the rank of $x$ is less than $\alpha$, where $\mathrm{rank}(x)=\sup \{\mathrm{rank}(z)+1:z\in x\}$.