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The club filter on a regular uncountable cardinal number $\kappa$ is $\kappa$-complete. Intersections of fewer than $\kappa$ club sets are closed. For unboundedness, repeatedly step past a point of each club and take a countable supremum; regularity keeps all suprema below $\kappa$, and closure puts the final supremum in every club.

Fix a regular uncountable cardinal number $\kappa$. A club set $C\subseteq \kappa$ is unbounded in $\kappa$ and closed under suprema below $\kappa$ of nonempty increasing sequences with no last term. Equivalently, every limit ordinal $\delta <\kappa$ with $\sup (C\cap \delta )=\delta$ belongs to $C$. The club filter isIt consists of all sets containing a club set, not only the club sets themselves.

An omega-measurable cardinal is a cardinal number carrying a nonprincipal ultrafilter closed under countable intersections, that is, a countably complete ultrafilter. Such a cardinal must be uncountable: on a countable set the intersection of the complements of all singletons would be empty.

A measurable cardinal is an uncountable cardinal number $\kappa$ carrying a nonprincipal $\kappa$-complete ultrafilter. A kappa-complete filter is closed under intersections of every family of size strictly below $\kappa$. Thus for an omega-measurable cardinal the completeness required is countable completeness, not merely closure under finitely many intersections.

A uniform example is $\boxed{A=\kappa \setminus \{\omega \}}$. It contains the final segment $[\omega +1,\kappa )$, a club set. Therefore $A$ belongs to the club filter. Every two club sets in a regular uncountable cardinal intersect: alternately choose larger points from each and take the countable supremum, which remains below $\kappa$ and lies in both by closure. Hence every member of the club filter is a stationary set, so $A$ is stationary.

However, every finite ordinal belongs to $A$, and their increasing supremum $\omega$ does not. Thus $A$ is not closed and is not a club set. This is a stationary nonclosed member of the club filter 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

For any nonzero limit ordinal $\delta$, a club set is an unbounded subset closed under limit points below $\delta$. A subset is stationary if it meets every such club set. Statements concerning the completeness of the club filter or Fodor lemma require regular uncountable height; they should not be transferred to arbitrary singular or countable limit ordinals without checking their hypotheses.

For a regular infinite $\theta <\kappa$, with $\kappa$ a regular uncountable cardinal number, $S_{\theta }^{\kappa }=\{\delta <\kappa :\mathrm{cf}(\delta )=\theta \}$ is stationary. Build a strictly increasing continuous $\theta$-sequence in any club set, and take its supremum. Closure places it in that club and the cofinality of an increasing ordinal supremum gives cofinality $\theta$. In particular the two disjoint sets $S_{\omega }^{{\omega }_2}$ and $S_{{\omega }_1}^{{\omega }_2}$ show that the club filter on ${\omega }_2$ is not an ultrafilter.

For any regular uncountable cardinal number $\kappa$, the set $\kappa \setminus \{\omega \}$ contains the final-segment club set $[\omega +1,\kappa )$, so belongs to the club filter and is a stationary set. It is not closed, since its finite ordinals have supremum $\omega$, which is missing.

A subset of a regular uncountable cardinal number $\kappa$ is stationary if it meets every club set in $\kappa$. Intersecting a stationary set with a club set preserves stationarity. Every member of the club filter is stationary. At a regular infinite $\theta <\kappa$, the stationarity of ordinals of prescribed cofinality supplies important examples.