Fodor's lemma on singular cardinals

Question

Fodor's lemma asserts that if $\kappa$ is a regular and uncountable cardinal, then if $f(\alpha)<\alpha$ for a stationary subset of $\kappa$, then it is constant on stationary subset.

Suppose $\kappa>\operatorname{cf}(\kappa)=\mu>\omega$, that is $\kappa$ is singular with uncountable cofinality.

What sort of conditions do we have to have in order to ensure a regressive function is constant on a stationary subset? Or is there a counterexample easily to be found which I could not come up with so far?

(In particular I'm interested in the case that the function is defined on an end segment)

Answer 1

Assume that $\kappa$ is a cardinal of uncountable cofinality. A sufficient (but not even close to necessary) condition for a regressive function $f:\kappa\to\kappa$ to be constant on a stationary subset of $\kappa$ is that $\lim(f)<\operatorname{cf}(\kappa)$ where $$ \lim(f)=\min\{\alpha\leq\kappa:f^{-1}[0,\alpha)\text{ is stationary}\}. $$ Indeed, if $\lim(f)=\beta<\operatorname{cf}(\kappa)$ then the stationary set $f^{-1}[0,\beta)=\bigcup_{\alpha<\beta}f^{-1}\{\alpha\}$ is a union of fewer than $\operatorname{cf}(\kappa)$ sets $f^{-1}\{\alpha\},\alpha<\beta$. Because every union of fewer than $\operatorname{cf}(\kappa)$ nonstationary sets is nonstationary, there must be such an $\alpha<\beta$ that $f^{-1}\{\alpha\}$ is stationary. Hence $f$ is constant on a stationary set.

Answer 2

I’m pretty sure that all you can get without very strong extra conditions is the weak form of the pressing-down lemma that says that the function must be bounded on a stationary set. For example, let $\kappa = \omega_{\omega_1}$, $\mu = \omega_1$, and $C = \{\omega_\xi:\xi\in\omega_1\}$, and define $\varphi:\kappa \to \kappa$ by setting $\varphi(\alpha) = \xi$ for $\omega_\xi \le \alpha < \omega_{\xi+1}$; clearly $\varphi$ is regressive, but it’s not constant on any stationary subset of $\kappa$. It’s not clear to me that any reasonable conditions would avoid examples like this, though of course what’s reasonable depends on what you want the result for.

To prove the weak form, suppose that $\kappa > \operatorname{cf}\kappa = \mu > \omega$; then $\kappa$ has a cub $C = \{\alpha_\xi:\xi \in \mu\}$, where the enumeration is continuous. Let $S$ be a stationary subset of $\kappa$; then $S_C = \{\xi \in \mu:\alpha_\xi \in S\}$ is a stationary subset of $\mu$. Let $\varphi$ be a regressive function on $\kappa$. Define $$\varphi_C:S_C \to \mu:\xi \mapsto \min\{\eta\in\mu:\varphi(\alpha_\xi) \le \alpha_\eta\};$$

clearly $\varphi_C(\xi)\le\xi$. Suppose that $\varphi_C(\xi)=\xi$; then for every $\eta<\xi$ we have $\alpha_\eta<\varphi(\alpha_\xi)<\alpha_\xi$, and since $C$ is closed, $\xi$ must be a successor ordinal. Thus, $\varphi_C$ is regressive on $\{\xi\in S_C:\operatorname{cf}\xi\ge\omega\}$, which is a stationary subset of $\mu$, and therefore $\varphi_C$ is constant on some stationary subset $T$ of $S_C$, say with value $\eta_0$. Then for each $\xi\in T$, $\varphi_C(\xi)=\eta_0$, so $\varphi(\alpha_\xi)\le\alpha_{\eta_0}$. Since $\{\alpha_\xi:\xi \in T\}$ is stationary in $\kappa$, we’re done: $\varphi$ is bounded on the stationary set $\{\alpha_\xi:\xi \in T\}$.

We actually do get a little more. For every $\eta<\eta_0$, $\varphi(\alpha_\xi)>\alpha_\eta$, so $\varphi(\xi)\ge\sup\{\alpha_\eta:\eta<\eta_0\}=$ $\alpha_{\sup\{\eta:\eta<\eta_0\}}$; thus, $\varphi$ is constant on a stationary set if $\eta_0$ happens to be a limit ordinal. If $\eta_0=\gamma+1$, however, all we can say is that $\varphi$ maps some stationary set into $[\alpha_\gamma,\alpha_{\gamma+1}]$.