Let
$$
\widehat\kappa=\sup_{n<\omega}j^n(\kappa)
$$
be the supremum of the <Kunen critical sequence>, and put
$$
X=j^{\prime\prime}\widehat\kappa
=\{j(\xi):\xi<\widehat\kappa\}.
$$
The required choices are
$$
\boxed{\beta=\widehat\kappa+2,\qquad
\gamma=\widehat\kappa+1,\qquad
X=j^{\prime\prime}\widehat\kappa.}
$$
Indeed $X$ has rank $\widehat\kappa$, so $X\in V_{\widehat\kappa+1}$, while the <Kunen lemma> gives $X\notin M$ once the domain contains the omega-Jonsson function required by the proof, which is ensured by $\alpha\geq\widehat\kappa+2$.

Solution

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

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2024/iii/paper-116/2/c/solution