Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 116 / 1

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.