For a measurable cardinal $κ$, show that $cf(γ)≠κ$ implies $j_U(γ)=\sup\{j_U(δ):δ<γ\}$ ($U$ $κ$-complete ultrafilter, $j_U$ associated embedding)

Question

For :

• $κ$ a measurable cardinal,

• $U$ a $κ$-complete ultrafilter over $κ$

• $j_U$ the elementary embedding of $V$ into the ultrapower of $V$ to $U$

How to show that : If $\operatorname{cf}(γ)≠κ$ Then $j_U(γ)= \sup \{j_U(δ)\mid δ<γ\}$ ?

Answer 1

Let us divide the cases. Let $\langle \gamma_\zeta\mid \zeta<\mu\rangle$ be a cofinal sequence converging to $\gamma$. (Here $\mu=\operatorname{cf}\gamma$.)

• $\operatorname{cf}\gamma<\kappa$: Let $[f]_U<j_U(\gamma)$. Without loss of generality we may assume that $f(\xi)<\gamma$ for all $\xi<\kappa$. (Why?) For each $\xi$, choose $\eta_\xi<\mu$ such that $f(\xi)<\gamma_{\eta_\xi}$. Then there is $\delta<\mu$ such that $\{\xi<\kappa\mid \eta_\xi=\delta\}\in U$. Can you see how to proceed from there?

• $\operatorname{cf}\gamma>\kappa$: Let $[f]_U<j_U(\gamma)$ again. Still, we may assume that $f(\xi)<\gamma$ for all $\xi<\kappa$. Since $\kappa<\mu$, $\delta:=\sup_{\xi<\kappa}f(\xi)<\gamma$. Then completing the proof is easy, so I will leave it to you.