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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.

Begin with bounded formulas in set theory. Atomic membership and equality have the same interpretation on shared elements of $U_1\subseteq U_2$, each a transitive class. Boolean connectives preserve agreement. For a bounded quantifier with bound $a\in U_1$, every member of $a$ lies in $U_1$ by transitivity. Its quantifier range is therefore exactly the same in both structures. Structural induction proves set-theoretic absoluteness for every bounded formula in set theory.

A Sigma-one formula in set theory is upward absolute: witnesses in $U_1$ are still witnesses in $U_2$, and the bounded matrix agrees. A Pi-one formula in set theory is downward absolute, by negation or by restricting its universal quantifiers.

For the given Delta-one formula modulo ZF $\phi$, select a ${\Sigma }_1$ equivalent $\sigma$ and a ${\Pi }_1$ equivalent $\pi$. Both structures satisfy ZF, so both equivalences hold in both structures. Upward absoluteness of $\sigma$ proves the forward implication, and downward absoluteness of $\pi$ proves the reverse implication. Thus Delta-one absoluteness gives

A standard membership model of set theory has the actual membership relation:When “standard model” is used with the transitive-model convention, one also requires $M$ to be transitive: $x\in y\in M$ implies $x\in M$. Under a convention where standard means only actual membership, this extra requirement defines a transitive model. The set-theoretic absoluteness arguments below use transitive models of ZFC, so that bounded quantification and ordinals have their intended meaning.

A syntactically bounded formula in set theory is built from the atomic first-order formulas $u=v$ and $u\in v$ using Boolean operations and bounded quantifiers $\exists u\in v$ and $\forall u\in v$. These abbreviate $\exists u(u\in v\wedge \cdots )$ and $\forall u(u\in v\Rightarrow \cdots )$, with $u$ different from the bounding variable $v$.

The superscript in the question allows provable equivalence: a ZF-equivalent bounded formula is any first-order formula $\phi (\vec{v})$ for which some syntactically bounded formula in set theory $\psi (\vec{v})$ satisfiesThus ${\Delta }_0^{\mathrm{ZF}}$ is the class of first-order formulas equivalent over ZF to a syntactic ${\Delta }_0$ first-order formula. If a convention uses ${\Delta }_0^{\mathrm{ZF}}$ just for syntactically bounded set-theoretic first-order formulas, it is the smaller syntactic class; the subsequent set-theoretic absoluteness proof works for both conventions.

For parameters $a_1,\dots ,a_n\in M$, set-theoretic absoluteness means agreement of the two membership first-order structures:This must hold for every such parameter tuple. Equivalently, formula relativization to a class produces two formulas ${\phi }^M$ and ${\phi }^N$ whose truth values agree on tuples from $M$. Agreement for parameters only in $N\setminus M$ is not part of the definition. For a proper class in set theory, this notation is understood separately for each fixed first-order formula, rather than as an unrestricted truth predicate for the entire universe.

First prove set-theoretic absoluteness for syntactically bounded formulas in set theory by induction on their construction. Atomic equality and membership use the actual equality and membership relations in both transitive classes. Boolean operations preserve agreement.

For the bounded existential step, take $a\in M$ and suppose $N\models \exists u\in a\psi (u,\vec{b})$. Its witness $c$ is an actual member of $a$. Since $M$ is a transitive class, $c\in M$; the induction hypothesis then gives $M\models \psi (c,\vec{b})$. The converse uses the same witness in $N$. The bounded universal step follows similarly because both structures quantify over exactly the actual members of $a$, or follows by negation from the existential step. HenceNo ZF axioms are needed for this syntactic statement beyond the ambient meaning of a transitive class. For a ZF-equivalent bounded formula $\phi$, choose a bounded equivalent $\psi$. The assumption that both structures model ZF gives ${\phi }^M\leftrightarrow {\psi }^M$ and ${\phi }^N\leftrightarrow {\psi }^N$. Combining these with the displayed equivalence proves the asserted set-theoretic absoluteness of ${\Delta }_0^{\mathrm{ZF}}$.

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.

For every tuple of parameters $\bar{a}\in M^n$, an upward absolute formula transfers truth from the smaller structure to the larger one:These definitions of directional set-theoretic absoluteness do not themselves require the structures to be transitive.

A formula has set-theoretic absoluteness between these structures when both directions hold for every $\bar{a}\in M^n$:Equivalently, it is both a downward absolute formula and an upward absolute formula between them.

Take the formula saying that a set is transitive:This is a bounded formula in set theory. If $x\in M$, transitivity of $M$ ensures that every $u\in x$ and every $v\in u$ is already in $M$, and the same holds in $N$. Both structures therefore run through exactly the same bounded witnesses and use the same actual membership relation. They agree on ${\phi }_1(x)$ for every $x\in M$, proving set-theoretic absoluteness.

Use the cardinalhood formula $C$ from part (b), and let a finite flag choose between it and its negation:Here $0,1$ abbreviate their parameter-free definitions as Finite von Neumann ordinals, and both belong to $M$. This is one first-order formula with two free variables.

For $\beta =({\omega }_1{)}^M$, part (b) proved $M\models C(\beta )$ and $N\models \neg C(\beta )$. The shared parameters $(\beta ,0)$ therefore make ${\phi }_3$ true in $M$ and false in $N$, disproving upward set-theoretic absoluteness. The shared parameters $(\beta ,1)$ make it false in $M$ and true in $N$, disproving downward set-theoretic absoluteness. Hence the same formula fails both directions, with explicit witnesses to each failure.

An uncountable cardinal number $\kappa$ is weakly compact when every $\kappa$-satisfiable theory in an infinitary language $L_{\kappa ,\kappa }$ with at most $\kappa$ nonlogical symbols is satisfiable. A cardinal is inaccessible when it is uncountable, regular, and a strong limit cardinal.

Two standard results supply the proof. First, every weakly compact cardinal is inaccessible. Second, every weakly compact $\kappa$ has the Keisler extension property: there is a transitive set $X⊋V_{\kappa }$ such thatand $\kappa \in X$. Since $\kappa$ is inaccessible, the relevant downward absoluteness makes $X\models$ “$\kappa$ is inaccessible”. Hence $X$ satisfies “there is an inaccessible cardinal”. By elementarity, $V_{\kappa }$ satisfies the same sentence, so it contains some inaccessible $\lambda$. Every ordinal in $V_{\kappa }$ is below $\kappa$, and inaccessibility is absolute here, giving

A property $P$ of $\kappa$ reflects below $\kappa$ whenis unbounded in $\kappa$. The standard elementary-embedding reflection argument starts with any $\gamma <\kappa$. Since $j(\gamma )=\gamma$ and $j(\kappa )>\kappa$, the target model can use $\kappa$ itself as a witness toElementarity then gives a witness $\mu$ with $\gamma <\mu <\kappa$ in the domain.

For a concrete example, the property of being a strongly inaccessible cardinal reflects below $\kappa$. A measurable cardinal is strongly inaccessible, and the assumed inclusion $V_{\kappa +1}\subseteq M$ makes this assertion about $\kappa$ absolute between $V_{\lambda }$ and $M$: both models have all subsets of every ordinal below $\kappa$. Thus $M$ sees that $\kappa$ is strongly inaccessible. Given $\gamma <\kappa$, it therefore satisfiesElementarity supplies a strongly inaccessible $\mu$ between $\gamma$ and $\kappa$ in $V_{\lambda }$. As $\gamma$ was arbitrary, the strongly inaccessible cardinals below $\kappa$ are unbounded.

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 $V_{\lambda }$.

Assume $V\ne L$. At the first cumulative stage containing a nonconstructible set, the index is a successor $\beta +1$. Rank absoluteness gives $y=V_{\beta }=(V_{\beta }{)}^L\in L$, not just $V_{\beta }\subseteq L$. Let $x$ be the power set of $y$ computed in $L$. It omits a nonconstructible subset of $y$ from the next external rank stage. Thus $L$ satisfies power-set equality for the parameters $x,y\in L$, while $V$ does not. Both are transitive models of ZF. Subset inclusion itself is bounded and absolute; quantification over all possible subsets causes the failure.

A first-order formula is ZF-equivalent bounded if ZF proves it equivalent, with the same free variables, to a syntactically bounded formula in set theory. It need not itself have only bounded quantifiers. Between transitive models of ZF, use the bounded equivalent and the axioms in each model to prove set-theoretic absoluteness. The syntactic bounded-formula result requires only transitivity; the provable-equivalence extension additionally requires the theory used for the equivalence.