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.



Existence of scales with special properties
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.


