Past exam of the mathematics course of the University of Cambridge / 2005 / iii / Paper 27 / 12 / Solution

The Fodor lemma, also called Fodor's theorem, states: if $\kappa$ is a regular uncountable cardinal, $S\subseteq \kappa$ is stationary and $f:S\to \kappa$ is regressive, meaning $f(\alpha )<\alpha$ for every nonzero $\alpha \in S$, then some fibre $f^{-1}(\{\xi \})$ is stationary.

Delete zero from $S$, which preserves stationarity. Suppose no fibre is stationary. For each $\xi <\kappa$, choose a club set $C_{\xi }$ avoiding that fibre. The diagonal intersectionis a club. To prove unboundedness, start above an arbitrary bound and choose a strictly increasing sequence ${\alpha }_n$ with ${\alpha }_{n+1}$ in every $C_{\xi }$ for $\xi \le {\alpha }_n$. The intersection of these fewer-than-$\kappa$ clubs is club by club filter completeness. Put $\alpha ={\sup }_n{\alpha }_n<\kappa$. For any $\xi <\alpha$, every sufficiently late ${\alpha }_n$ lies in $C_{\xi }$, so its closure gives $\alpha \in C_{\xi }$. Thus $\alpha \in C$. For closure of $C$, let $\alpha <\kappa$ be a limit point of $C$; for each $\xi <\alpha$, the members of $C$ above $\xi$ witnessing that limit lie in $C_{\xi }$, and closure gives $\alpha \in C_{\xi }$.

Stationarity supplies $\alpha \in S\cap C$. Since $f(\alpha )<\alpha$, we have $\alpha \in C_{f(\alpha )}$, but that club was chosen to avoid the fibre containing $\alpha$. This contradiction proves the theorem. For completeness, club filter completeness follows by a similar diagonal sequence: cycle through fewer than $\kappa$ clubs over $\omega$ rounds, each time choosing a larger member of the next club. Regularity keeps the supremum below $\kappa$, and every club is visited cofinally often in the construction, so its closure contains the supremum. Their intersection is also closed.