Existence of scales with special properties

Question

Let $\kappa$ be a singular cardinal, and let $\langle \kappa_i \mid i<\mathrm{cf}(\kappa) \rangle$ be an increasing sequence of regular cardinals cofinal in $\kappa$. Recall that a scale on $\Pi_{i<\mathrm{cf}(\kappa)} \kappa_i$ is a sequence $\langle f_\alpha \mid \alpha < \kappa^+ \rangle$ such that:

1. For every $\alpha < \kappa^+$, $f_\alpha \in \Pi_{i<\mathrm{cf}(\kappa)} \kappa_i$.
2. For every $\alpha < \beta < \kappa^+$, there is $i < \mathrm{cf}(\kappa)$ such that $f_\alpha <_i f_\beta$, i.e. for every $j\geq i$, $f_\alpha(j) < f_\beta(j)$.
3. For every $g\in \Pi_{i<\mathrm{cf}(\kappa)} \kappa_i$, there is $\alpha < \kappa^+$ and $i < \mathrm{cf}(\kappa)$ such that $g <_i f_\alpha$.

Question: Is it consistent that there is a scale on $\Pi_{i<\mathrm{cf}(\kappa)} \kappa_i$ such that, for every $\beta < \kappa^+$ and every $i<\mathrm{cf}(\kappa)$, $\left|{\{\alpha < \beta \mid f_\alpha <_i f_\beta\}}\right| < \kappa$ ?

My intuition is that the answer should be no, but I haven't been able to find a proof.

Answer 1

I have a negative answer assuming some mild cardinal arithmetic assumptions. Namely, if $(\kappa_i)^i < \kappa$ for every $i<\mathrm{cf}(\kappa)$, then there can be no scale with the desired property. This is true, for example, whenever $\mathrm{cf}(\kappa) = \omega$ or $\kappa$ is strong limit. We also make the harmless assumption that $\mathrm{cf}(\kappa) < \kappa_0$.

Assume for sake of contradiction that $\langle f_\alpha \mid \alpha < \kappa^+ \rangle$ is such a scale. For $j<\mathrm{cf}(\kappa)$, define $g_j \in \Pi_{i<\mathrm{cf}(\kappa)}\kappa_i$ as follows: Using the fact that $(\kappa_j)^j < \kappa$, fix $B_j \subseteq \kappa^+$ and $f \in \Pi_{i\leq j}\kappa_i$ such that $\left|{B_j}\right|=\kappa_j$ and, for every $\alpha \in B_j$ and $i\leq j$, $f_\alpha(i)=f(i)$. For $i\leq j$, let $g_j(i)=f(i)+1$. For $i>j$, let $g_j(i)=\sup(\{f_\alpha(i)+1 \mid \alpha \in B_j \})$. Now define $g \in \Pi_{i<\mathrm{cf}(\kappa)}\kappa_i$ by letting $g(i)=\sup(\{g_j(i) \mid j<\mathrm{cf}(\kappa) \})$. Finally, find $\beta < \kappa^+$ and $i<\mathrm{cf}(\kappa)$ such that $g <_i f_\beta$. Letting $B = \bigcup_{j<\mathrm{cf}(\kappa)}B_j$, we have that $\left|{B}\right| = \kappa$ and $f_\alpha <_i f_\beta$ for every $\alpha \in B$. Contradiction.

Answer 2

Shelah's Dichotmoy theorem (see link text) says more or less that both options are valid. Every increasing sequence can either:

a) Have an exact upper bound (in which case your condition fails.)

b) Or have an "interleaved cofinal sequence" i.e. another (very small length) sequence is "interleaved" with the original one, in which case your condition must hold.