Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 116 (Large Cardinals)

Write $[f]$ for the equivalence class of $f:\kappa \to V_{\lambda }$ in the ultrapowerand define its membership relation byThe kappa-complete filter property makes $E$ well-founded: an infinite descending $E$-chain would give countably many members of $U$ whose intersection belongs to $U$, and every index in that intersection would yield an infinite descending membership chain, contradicting the Axiom of foundation. The relation is extensional by Łoś's theorem.

The Mostowski collapse theorem therefore gives a unique isomorphism $\pi :(N,E)\to (M,\in )$ onto a transitive set $M$. Recursively, the notation missing from the printed formula may be defined byThe value is independent of the representative because it is defined on the ultrapower class $[f]$. Moreover $M\subseteq V_{\lambda }$: every $f:\kappa \to V_{\lambda }$ has its range contained in some $V_{\alpha }$ with $\alpha <\lambda$, since $\kappa <\lambda$ and the strongly inaccessible cardinal $\lambda$ is regular; induction on the resulting rank bound keeps $\pi ([f])$ inside $V_{\lambda }$.

Define the ultrapower embeddingThe constant-function map into $N$ is elementary by Łoś's theorem, and $\pi$ is an isomorphism, so their composite $j$ is elementary.

The critical point of an elementary embedding $j$ is the least ordinal $\alpha$ for which $j(\alpha )\ne \alpha$.

We first prove by transfinite induction that $j(\alpha )=\alpha$ for every $\alpha <\kappa$. Suppose this is known below $\alpha <\kappa$. If $[f]E[{\mathrm{const}}_{\alpha }]$, thenThe sets $A_{\beta }=\{\xi \in A:f(\xi )=\beta \}$ for $\beta <\alpha$ partition $A$ into fewer than $\kappa$ pieces. A kappa-complete filter that is an ultrafilter must contain one cell $A_{\beta }$: otherwise all their complements would belong to $U$, and their intersection would contradict $A\in U$. Hence $[f]=[{\mathrm{const}}_{\beta }]$. The predecessors of $j(\alpha )$ are consequently exactly the already-fixed ordinals below $\alpha$, so $j(\alpha )=\alpha$.

Now let $\mathrm{id}(\xi )=\xi$. An identical argument shows that the predecessors of $[\mathrm{id}]$ in the ultrapower are exactly $[{\mathrm{const}}_{\beta }]$ for $\beta <\kappa$, soSince $\{\xi <\kappa :\xi <\kappa \}=\kappa \in U$, one has $[\mathrm{id}]E[{\mathrm{const}}_{\kappa }]$. After collapsing, $\kappa \in j(\kappa )$, and therefore $j(\kappa )>\kappa$. Thus $\mathrm{crit}(j)=\kappa$.

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.

An strongly inaccessible cardinal $\kappa$ has the Keisler extension property when there is a proper transitive set $X⊋V_{\kappa }$ such that

Suppose $\kappa$ is strongly inaccessible and has this property. Because $X$ properly extends the transitive set $V_{\kappa }$, it contains $\kappa$. Strong inaccessibility of $\kappa$ is downward absolute from the ambient universe to the transitive set $X$: any internal witness that $\kappa$ is countable, singular, or not a strong limit would also be an ambient witness. Hencewith $\kappa$ as a witness. Since $V_{\kappa }\prec X$, the same sentence holds in $V_{\kappa }$. Its witness is an ordinal $\mu <\kappa$. The set $V_{\kappa }$ contains $V_{\mu +1}$, so it computes all subsets of cardinals below $\mu$ correctly; strong inaccessibility of $\mu$ is therefore absolute between $V_{\kappa }$ and the universe. Thus there is a strongly inaccessible $\mu <\kappa$, and $\kappa$ cannot be the least strongly inaccessible cardinal.

For a first-order theory $R$ extending ZFC, let $C_R$ be its set of formal consequences and let $\mathrm{Cons}$ denote the class of formal consistency statements $\mathrm{Con}(Q)$ for recursively axiomatized extensions $Q$ of ZFC. Using Gödel numbering to code proofs and theories, these objects and the following comparison are definable in the base theory ZFC.

The consistency-strength preorder isThus every consistency assertion provable in $T$ is also provable in $S$. Its strict part is

Let $\mathrm{IC}$ be the sentence asserting that a strongly inaccessible cardinal exists, and begin withDefine the iterated consistency progressionThe construction is effective, so every $T_n$ and $T_{\infty }$ is a recursively axiomatized first-order theory extending ZFC.

Because $T_{n+1}$ extends $T_n$, every theorem of $T_n$, including every formal consistency statement it proves, is a theorem of $T_{n+1}$; hence $T_n{\le }_{\mathrm{Cons}}{T}_{n+1}$. The theory $T_{n+1}$ proves $\mathrm{Con}(T_n)$ by construction, whereas a consistent $T_n$ cannot prove its own consistency by Gödel second incompleteness theorem. ThereforeLikewise $T_{\infty }$ extends every $T_n$ and contains $\mathrm{Con}(T_n)$ as an axiom already at stage $n+1$, while $T_n$ does not prove it. Consequentlyassuming the stated consistency hypotheses.

Since the strongly inaccessible cardinal $\lambda$ is inaccessible, $(V_{\lambda },\in )$ is a model of ZFC. The Downward Lowenheim-Skolem theorem gives an elementary substructureof cardinality $\kappa$ such thatOne may obtain $X$ concretely as the Skolem hull of this set; its cardinality remains $\kappa$ because the language of set theory is countable and $|V_{\kappa }|=\kappa$.

Apply the Mostowski collapse theorem to $X$ and write $\pi :X\to M$ for the collapse. Then $M$ is a transitive set, $|M|=\kappa$, and $\pi$ fixes $V_{\kappa }$ pointwise. It also fixes $\kappa$, because it fixes every ordinal below $\kappa$. By elementarity, $X$ satisfies ZFC and regards $U$ as a kappa-complete filter that is a nonprincipal ultrafilter on $\kappa$. Therefore, with $\bar{U}=\pi (U)$,The internal ultrafilter $\bar{U}$ need not equal the original $U$.

No. Letthe successor cardinal of $\kappa$ computed by $M$. Then $\alpha \in M$, and $M$ regards $\alpha$ as a cardinal number. Because $M$ is transitive, $\alpha \subseteq M$, so externallyOn the other hand $\kappa <\alpha$, hence $|\alpha |=\kappa$ in the ambient universe. A corresponding bijection belongs to $V_{\lambda }$ because its rank is below the inaccessible limit $\lambda$. Thus $V_{\lambda }$ regards $\alpha$ as equinumerous with $\kappa$ and therefore not as a cardinal. This is an instance of cardinal nonabsoluteness in a small transitive model.

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 $V_{\lambda }$.