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

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 $\mathrm{crit}(j)=\kappa$.