Past exam of the mathematics course of the University of Cambridge / 2016 / iii / Paper 121 / 3 / c / Solution

The club filter completeness argument works for every regular uncountable $\kappa$. Let $\mu <\kappa$ and let $C_i$ be a club set for each $i<\mu$. Their intersection is closed. To prove it unbounded, start above any prescribed $\beta <\kappa$ and choose an increasing sequence $({\delta }_n{)}_{n<\omega }$ so that ${\delta }_{n+1}$ lies above a chosen point of every $C_i$ greater than ${\delta }_n$. $\kappa$ is a regular cardinal, so the supremum of these $\mu$ choices remains below $\kappa$. The resulting countable supremum $\delta ={\sup }_n{\delta }_n$ likewise remains below $\kappa$. For each $i$, the chosen $C_i$ points are cofinal in $\delta$, so closure gives $\delta \in C_i$. Thus the intersection is a club set. Intersecting fewer than $\kappa$ members of the club filter still contains such an intersection of clubs. In particular, $\boxed{{\mathcal{D}}_{{\omega }_2}\text{ is }{\aleph }_2\text{-complete}}$.

It is not an ultrafilter. For a regular infinite $\theta <\kappa$, the setis stationary. Given a club set $C$, build in $C$ a strictly increasing continuous sequence of length $\theta$ and take its supremum $\delta <\kappa$. Closure gives $\delta \in C$, and the cofinality of an increasing ordinal supremum gives $\mathrm{cf}(\delta )=\theta$. This proves the stationarity of ordinals of prescribed cofinality.

At $\kappa ={\omega }_2$, the disjoint sets $S_{\omega }^{{\omega }_2}$ and $S_{{\omega }_1}^{{\omega }_2}$ are both stationary. A club contained in either $S_{\omega }^{{\omega }_2}$ or its complement would miss one of these stationary sets. Thus neither $S_{\omega }^{{\omega }_2}$ nor its complement belongs to ${\mathcal{D}}_{{\omega }_2}$, and