Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 116 / 3

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 $V_{\lambda }$.