Past exam of the analytic topology and axiomatic set theory course of the University of Oxford / 2009 / B / 6

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.

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.

Use the bounded formulaIt expresses $x\subseteq y$ with no unbounded quantifier. For parameters $x,y\in U_1$, transitivity puts every member of $x$ in $U_1$ already, and membership in $y$ has the same meaning in both classes. Thus the formula is ${\Delta }_0$ and is absolute in both directions, in particular upward absolute.

Equality with the union is expressed byThe first conjunct says that every member of $x$ lies in some member of $y$; the second gives the reverse inclusion. Every quantifier is bounded by a parameter or by an element of a previously bounded parameter. Therefore bounded absoluteness applies, proving full absoluteness and hence upward absoluteness of $x=\bigcup y$. This only compares the relation for existing parameters; it does not require either class to contain the union of every one of its sets.

A bounded expression for the successor relation isThe first two clauses put $y$ and all its elements into $x$, while the last forbids any additional member of $x$. Extensionality identifies this exactly with $x=y\cup \{y\}$. Since the formula is ${\Delta }_0$, it is absolute in both directions between the transitive classes and therefore upward absolute. This bounded description also justifies the successor computations used earlier in the rank-stage Replacement counterexample.

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.