Stationary sets and club filters

Fix a regular uncountable cardinal κ. A subset C is unbounded if above every ordinal below κ there is a larger member of C. It is closed if every nonzero limit δ<κ for which C∩δ is unbounded in δ belongs to C. A club satisfies both requirements. Closedness here concerns limits below κ; it does not require κ itself to belong to a subset of κ. Tails are clubs, and the nonzero limit ordinals below ω1 form a club.

A subset S of κ is stationary if it meets every club, and nonstationary if some club misses it. Every stationary set is unbounded because it must meet each tail, but an unbounded set need not be stationary. The successor ordinals below ω1 are unbounded and disjoint from the club of nonzero limits. The club filter consists of the subsets of κ containing a club, rather than only the clubs themselves. It is proper because a club cannot be empty under the unboundedness requirement.

The intersection of fewer than κ clubs is a club. Closedness is preserved directly. To obtain an element above a chosen starting point, repeatedly choose larger points meeting the required clubs, then take a limit still below κ. Regularity ensures that all suprema used in a construction of length smaller than κ remain below κ; uncountable cofinality lets the final countable approach stay below the boundary. This makes the club filter closed under intersections of fewer than κ members, which is often called κ-completeness.

An intersection of κ many clubs can be empty, as shown by taking increasingly late tails. The replacement is their diagonal intersection: δ belongs when it lies in every club whose index is below δ. This diagonal intersection is again club for a κ-long sequence of clubs. It asks each point to satisfy only its earlier tests, while ordinary intersection asks it to satisfy all tests at once. That distinction is the mechanism behind the pressing-down argument, not a harmless change in notation.

Fodor's lemma says that if S is stationary in κ and f maps S into κ with f(α)<α for every nonzero α in S, then f is constant on a stationary subset of S. Removing zero from S does not affect stationarity. Suppose every fiber were nonstationary and choose a club Cξ missing the ξ-fiber. A member δ of S in the diagonal intersection must lie in C at index f(δ), since that index is below δ. But δ also belongs to that very fiber, a contradiction.

The strict inequality in regressiveness and the stationarity of the domain both matter. An injective regressive function on the unbounded successor ordinals supplies no stationary constant fiber, because that domain is nonstationary. Likewise the identity function on a stationary set is not regressive. These examples distinguish the theorem from a simple claim that a large domain forces repetition. The proof uses the club filter's normal diagonal behavior, which depends on the regular uncountable setting and cannot be inferred from recurrence in a finite cycle.