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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?

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up vote 3 down vote accepted

Suppose that for each $\alpha \in \kappa$ there is a cub $C_\alpha$ on which $\Psi(\xi) > \xi + \alpha$. Let $D$ be the diagonal intersection of these cubs, i.e., $D = \{\xi \in \kappa:\xi \in \bigcap \limits_{\alpha < \xi} C_\alpha\}$. Then for any $\xi \in D$ we have $\Psi(\xi) \ge \sup \limits_{\alpha < \xi}(\xi+\alpha) = \xi+\xi$, contradicting the stationarity of $\{\xi \in \kappa:\Psi(\xi) < \xi+\xi\}$.

I don't see any substantially simpler argument for the related question, though the result can be proved directly in a similar fashion. Suppose that $f$ is such a function. Fix $\alpha \in \kappa$ such that $f(\xi) < \xi+\xi$ on $\kappa \setminus \alpha$. For each $\beta \in \kappa$ there is an $\alpha_\beta \in \kappa$ such that $f(\xi) > \xi+\beta$ on $\kappa \setminus \alpha_\beta$. Now let $D$ be the diagonal intersection of the tails $\kappa \setminus \alpha_\beta$; for any $\xi \in D$, $\xi+\xi > f(\xi) \ge \sup\limits_{\beta < \xi}(\xi+\beta) = \xi+\xi$, which is absurd. (I actually did this first; it gave me the idea for the general argument.)

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Neat solution. Thank you for help. As your argument essentially depends on the fact that the diagonal intersection of cubs is a cub (actually only nonempty is what is needed) and hence on regularity of $\kappa$, I have to wonder what happens in the case $\kappa$ is a singular cardinal of uncountable cofinality. – LostInMath Jun 26 '11 at 22:23
@LostInMath: Let $\lambda=\text{cf }\kappa$, and let $C=\{\gamma_\alpha:\alpha\in\lambda\}$ be a cub in $\kappa$ s.t. $\gamma_\alpha<\gamma_\beta$ iff $\alpha<\beta$; $C$ is a cofinal copy of $\lambda$ in $\kappa$. For $\alpha\in\lambda$ let $C_\alpha$ be a cub in $C$ on which $\Psi(\xi)>\xi+\gamma_\alpha$. Now you can take the diagonal intersection of these cub subsets of $C$ and finish pretty much as before. – Brian M. Scott Jun 27 '11 at 3:52
Right. Thank you again! – LostInMath Jun 27 '11 at 16:42

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