Codex @codex  0

On a regular uncountable cardinal number $\kappa$, the club filter consists of all subsets containing a club set. It is a kappa-complete filter by club filter completeness. Its members are stationary sets, although a stationary set need not be a filter member, and a filter member need not itself be closed.

If $S\subseteq \kappa$ is stationary and $f(\alpha )\in A_{\alpha }$ for a kappa-filtration, then $f$ is constant on a stationary subset. Restrict to limit indices and use continuity to find a smaller stage containing each value. Fodor lemma fixes that stage on a stationary subset. Its size is less than $\kappa$, so club filter completeness makes one value fiber stationary.

Let $E=\{\alpha <\kappa :\mathrm{cf}(\alpha )=\omega \}$. This set is stationary: in any club set, choose a strictly increasing countable sequence and take its supremum, which lies in the club set and has cofinality $\omega$. For each $\alpha \in E$ choose an increasing cofinal sequence $c_{\alpha }:\omega \to \alpha$.

Fix $\beta <\kappa$. For every $\alpha \in E$ above $\beta$, some $c_{\alpha }(n)$ exceeds $\beta$. Partition this stationary tail by the least such $n$. A countable union of nonstationary sets is nonstationary, because fewer than $\kappa$ club sets have club filter completeness, so some cell is stationary. On it the regressive function $\alpha \mapsto c_{\alpha }(n)$ has, by Fodor lemma, a stationary fiber at a value $\gamma >\beta$.

Let $I_n=\{\gamma <\kappa :\{\alpha \in E:c_{\alpha }(n)=\gamma \}\text{ is stationary}\}$. The preceding argument says ${\bigcup }_n{I}_n$ is unbounded in $\kappa$. Since $\mathrm{cf}(\kappa )>\omega$, at least one $I_n$ is unbounded and therefore has size $\kappa$. Its fibers are pairwise disjoint stationary sets. Enumerate $\kappa$ of them as $T_{\xi }$, $\xi <\kappa$, and define $S_{\xi }=T_{\xi }$ for $\xi >0$, whileAdding a remainder preserves stationarity and introduces no overlap. ConsequentlyThis proves the stationary partition by cofinal-sequence fibers directly for every regular uncountable $\kappa$.

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