Past exam of the mathematics course of the University of Cambridge / 2016 / iii / Paper 121 / 4

The Von Neumann hierarchy and the constructible hierarchy are defined by transfinite recursion:The union clauses apply to nonzero limit ordinals. The definable power set $\mathrm{Def}(L_{\alpha })$ consists of subsets definable over $(L_{\alpha },\in )$ by a first-order formula with finitely many parameters from $L_{\alpha }$. Definability over that structure is essential; it is not unrestricted definability in the ambient universe. The full constructible universe is $L={\bigcup }_{\alpha \in \mathrm{Ord}}{L}_{\alpha }$.

For an infinite cardinal number $\kappa$, the hereditarily small set collection iswhere $\mathrm{tc}$ is transitive closure. Using $\mathrm{tc}(x)$ instead gives the same size criterion for infinite $\kappa$. This bounds the whole membership ancestry, not just $|x|$.

One can take $\boxed{\alpha =\omega +1}$. Finite induction gives $L_n=V_n$ for every finite $n$: every subset of a finite level is definable using its finitely many elements as parameters. Hence $L_{\omega }=V_{\omega }$, the set of hereditarily finite sets.

Now reason inside the constructible universe $L$, a model of ZFC. The set $L_{\omega }$ is countable there, and there are countably many first-order formulas and finite parameter tuples. Thus $L_{\omega +1}=\mathrm{Def}(L_{\omega })$ is countable in $L$. But the constructible power set ${\mathcal{P}}^L(\omega )$ is uncountable in $L$ by the Cantor theorem. ChooseThis $r$ is constructible and a subset of $\omega$, so $r\subseteq V_{\omega }$ and $r\in V_{\omega +1}$. ThereforeAll comparisons were made inside $L$; this avoids assuming that the constructible reals are uncountable in the ambient universe. The witness remains valid externally because the two structures use the same $r$, the same $V_{\omega }$, and the same constructible level. Thus constructible sets of low rank can appear at later stages.

Use the Lévy hierarchy, with bounded formulas in set theory as the quantifier-free level for this purpose. Let $\mathrm{Ord}(\kappa )$ be the bounded ordinal predicate, and let $\mathrm{Bij}(f,\alpha ,\kappa )$ be the bounded predicate saying that $f$ is a graph of a bijection from $\alpha$ onto $\kappa$. Then cardinalhood is expressed byAll quantifiers in $\mathrm{Ord}$ and $\mathrm{Bij}$ are bounded by the displayed sets and the graph. Ordered-pair membership, graph functionality, injectivity and surjectivity have bounded expansions using the usual set coding of ordered pairs. The only unbounded quantifier shown is universal over $f$. The formula is therefore ${\Pi }_1$ and in particular a Pi-one formula modulo ZF.

In ZF, a cardinal number is an ordinal not in bijection with any smaller ordinal, so this formula has exactly the required meaning. Thus cardinalhood is Pi-one definable; the argument does not use the axiom of choice.

Let $\mathrm{Map}(f,\delta ,\kappa )$ be the bounded graph predicate for a function $\delta \to \kappa$. For an infinite cardinal number, regularity is expressed byThe predicate $\mathrm{InfCard}$ is the cardinal predicate in (i), conjoined with the bounded condition that $\kappa$ is a nonzero limit ordinal. For an infinite cardinal number, this latter condition holds automatically; it excludes the finite cardinals. The expression $f(\xi )<\beta$ abbreviates bounded graph membership with the output lying in $\beta$.

Every quantifier after $\forall f$ is bounded, and the cardinal predicate is already ${\Pi }_1$. Their conjunction can be placed in universal form, so this is a Pi-one formula modulo ZF. Its mathematical content is that no function from a smaller ordinal is cofinal in $\kappa$. By the definition of cofinality, that is equivalent to $\mathrm{cf}(\kappa )=\kappa$. Hence regular cardinalhood is Pi-one definable. This uses the standard convention that a regular cardinal is infinite.

A witness to failure of the Axiom of power set is $\boxed{{\omega }_1\in H_{{\omega }_2}}$. Every subset $A\subseteq {\omega }_1$ belongs to $H_{{\omega }_2}$, since its transitive closure has cardinality at most ${\aleph }_1$. If some $p\in H_{{\omega }_2}$ were the internal power set of ${\omega }_1$, transitivity and bounded subsethood would force $p$ to contain every ambient subset of ${\omega }_1$, and only those subsets. Thus it would be the full $\mathcal{P}({\omega }_1)$.

By the Cantor theorem,Its transitive closure contains all its members, so $|\mathrm{tc}(\{p\})|\ge |p|\ge {\aleph }_2$. This contradicts $p\in H_{{\omega }_2}$. ThereforeThe power-set failure in hereditarily small sets needs no assumption about the Continuum hypothesis.

Let $T$ be ZFC together with the assertion of unboundedly many inaccessible cardinals. Given a model of $T$, if it has no inaccessible limit of inaccessibles, it already witnesses the required nonimplication. Otherwise let $\iota$ be its least inaccessible limit of inaccessibles and pass to its rank segment $V_{\iota }$.

Because $\iota$ is an inaccessible cardinal, $V_{\iota }$ satisfies ZFC. Its ordinals are precisely those below $\iota$, and below $\iota$ the inaccessible cardinals are unbounded by the choice of $\iota$. Inaccessibility of any smaller ordinal is absolute to this segment: it has the same subsets and functions on all smaller ordinals, as in strong-inaccessibility absoluteness from rank agreement. Thus $V_{\iota }$ satisfies the unbounded-inaccessibles assertion.

No smaller ordinal can be an inaccessible limit of inaccessibles, by minimality of $\iota$ and the same absoluteness. Also $\iota$ is not an ordinal of $V_{\iota }$ itself. ConsequentlyThis rank cutoff at the first inaccessible limit of inaccessibles proves $\mathrm{Con}(T)\Rightarrow \mathrm{Con}(T+\neg \mathrm{ILI})$. The soundness theorem for first-order logic then gives $T/⊢\mathrm{ILI}$ if $T$ is consistent. This is a model-theoretic relative-consistency argument, and does not infer external transitivity from mere consistency.