Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 116 / 2

Let $\lambda$ be inaccessible and let $j:V_{\lambda }\to M$ be an elementary embedding into a transitive set with critical point $\kappa$. It is $\beta$-strong whenA formula $\Phi (x,\kappa )$ is a beta-stable cardinal property when it is absolute between the universe and every transitive set containing $V_{\kappa +\beta }$.

To say that the embedding reflects such a property means that whenever $\Phi (\kappa )$ holds,is unbounded in $\kappa$: for every $\gamma <\kappa$ there is such a $\mu$ with $\gamma <\mu <\kappa$. Indeed, stability gives $M\models \Phi (\kappa )$, so $M$ sees the witness $\kappa$ between $\gamma$ and $j(\kappa )$. Elementarity reflects a witness between $\gamma$ and $\kappa$. This is reflection by a beta-strong embedding.

The standard closure lemma for the ultrapower embedding says that every $\kappa$-sequence of members of $M$ which belongs to $V_{\lambda }$ is itself in $M$. If $\kappa <\alpha <{\kappa }^{+}$, choose in $V_{\lambda }$ a surjectionAll ordinal values of $s$ belong to the transitive model $M$, so closure gives $s\in M$. Thereforeand $M$ does not regard $\alpha$ as a cardinal.

Every ordinal belowhas an ultrapower representative $f:\kappa \to {\kappa }^{+}$: by Łoś theorem, a representative below the successor of $j(\kappa )$ may be chosen below ${\kappa }^{+}$ on a set in the ultrafilter. Hence, in $V_{\lambda }$,The ultrapower is closed under $\kappa$-sequences, so it contains every subset of $\kappa$ and computes $2^{\kappa }$ correctly. By elementarity $M$ regards $j(\kappa )$ as measurable, hence strongly inaccessible by Question 1(b). ConsequentlyThus $V_{\lambda }$ has a set of cardinality at most $2^{\kappa }$ whose order type is $\eta >2^{\kappa }$, so

Letbe the supremum of the Kunen critical sequence, and putThe required choices areIndeed $X$ has rank $\hat{\kappa }$, so $X\in V_{\hat{\kappa }+1}$, while the Kunen lemma gives $X\notin M$ once the domain contains the omega-Jonsson function required by the proof, which is ensured by $\alpha \ge \hat{\kappa }+2$.

Let $\hat{\kappa }={\sup }_{n<\omega }{j}^n(\kappa )$. Every term of the critical sequence is below $\delta$. If $\mathrm{cf}(\delta )>{\aleph }_0$, then its countable supremum also satisfies $\hat{\kappa }<\delta$, and because $\delta$ is a limit ordinal,Apply the Kunen lemma to the restriction available inside $V_{\delta }$. It giveswhich is impossible because $V_{\hat{\kappa }+1}\subseteq V_{\delta }$. Hence $\mathrm{cf}(\delta )\le {\aleph }_0$. A nonzero limit ordinal has infinite cofinality, so