Solution
= Solution
For regular $\kappa$, the <cobounded filter on a regular cardinal> is $\kappa$-complete. Part (c) extends it to a $\kappa$-complete ultrafilter $U$. Since $\kappa\setminus\{\alpha\}$ is cobounded for every $\alpha<\kappa$, no singleton belongs to $U$; hence $U$ is nonprincipal. Thus every strongly compact cardinal is <measurable cardinal>[measurable].