I'm stuck at exercise 9a of chapter 2.2 of Introduction to cardinal arithmetic by Holz, Steffens and Weitz. It is as follows:
Assume that $\kappa > \omega$ is a regular cardinal, $I$ is the nonstationary ideal of $\kappa$, $\Phi: \kappa \to \kappa$ is a function, $+$ is the ordinal sum and $||\Phi||_I$ is the Galvin-Hajnal rank of $\Phi$ with respect to the ideal $I$. Prove: If the set $\{\xi < \kappa : \Phi(\xi) \leq \xi + \xi\}$ is stationary in $\kappa$, then $||\Phi||_I \leq \kappa + \kappa$. (Hint: Use transfinite induction, Fodor's theorem and Lemma 1.8.4 (this Lemma deals with certain properties of stationary and nonstationary sets))
To try to solve the problem: Assume $\Psi: \kappa \to \kappa$ is a function with $\Psi <_I \Phi$ (which means $ \{ \xi < \kappa : \Psi(\xi) \geq \Phi(\xi) \} \in I$ ) and aim to show that $||\Psi||_I+1 \leq \kappa + \kappa$ which is the same as $||\Psi||_I \leq \kappa + \alpha$ for some $\alpha < \kappa$. Because $\Psi <_I \Phi$ and $\{\xi < \kappa : \Phi(\xi) \leq \xi + \xi\}$ is stationary, the set $\{\xi < \kappa : \Psi(\xi) < \xi + \xi\}$ is stationary. Now I would like to find $\alpha < \kappa$ such that the set $\{\xi < \kappa : \Psi(\xi) \leq \xi + \alpha\}$ is stationary. It would imply that $||\Psi||_I \leq \kappa + \alpha$, but does there necessarily exist such $\alpha$? Or is this altogether a wrong approach to the problem?
This led me ponder another related question. Again, let $\kappa > \omega$ be a regular cardinal. We say that a function $f: \kappa \to \kappa$ eventually dominates another function $g: \kappa \to \kappa$, if there is $\alpha < \kappa$ such that $f(\xi) > g(\xi)$ for all $\xi$ between $\alpha$ and $\kappa$. Let $f_\alpha, g: \kappa \to \kappa$ be the functions $f_\alpha(\xi) = \xi + \alpha$ and $g(\xi)=\xi + \xi$. Is there a function $\kappa \to \kappa$ which eventually dominates all the functions $f_\alpha,\alpha < \kappa$ and is eventually dominated by the function $g$? In view of the exercise the answer should be negative but is there an easy way to see this?