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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.

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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.

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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.

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Isn't it the trichotomy theorem? – Asaf Karagila Jan 22 '13 at 23:26
I'm not entirely sure what you're saying here. The entire scale certainly does have an exact upper bound, namely the function $g$ with $g(i)=\kappa_i$. On the other hand, I don't see how an initial segment $\langle f_\alpha \mid \alpha < \beta \rangle$ for $\beta < \kappa^+$ of the scale having an e.u.b. (and it will for stationarily many $\beta$) implies that my condition fails or that having a small cofinally interleaved sequence implies that it holds. Also, the Dichotomy theorem is about functions increasing modulo an ultrafilter, not modulo the bounded ideal. Please elaborate. – Chris Lambie-Hanson Jan 23 '13 at 4:00
- Filters vs. ideals - a similar theorem (Trichotomy) exists for ideal increasing sequences of functions. - Re. e.u.b - you are right. In case of a scale each least upper bound of a sequence modulo an ideal, is also an exact upper bound (removing 3. in your question is more interesting). In that case let me suggest the following: order the scale on a tree, using the $\lt$ order. Thus assuming your condition, each level has size $\lt \kappa$. The fact that the scale is $\kappa^+$-long entails the $\kappa^+$-Aronszajn property, which we know is independent. – Eran Jan 23 '13 at 11:56
Re. "having a small cofinally interleaved sequence implies that it holds" - if you have such a (short - usually $2^{cf(κ)}$) interleaved sequence $S$, then you cannot have a $\kappa$-size set of $\lt_i f_\beta$ for any $\beta$ as you will have to have at least $\kappa$ elements in the short sequence $S$ - impossible. – Eran Jan 23 '13 at 12:12
Even in the Trichotomy theorem, the small cofinally interleaved family of functions is only cofinally interleaved modulo an ultrafilter, not necessarily the bounded ideal. Also, the scale ordered by $<$ is not necessarily a tree - it is quite possible that the $<$-predecessors of a given $f_\alpha$ are not linearly ordered. Even if it were a tree, my condition would not imply that it had levels of size $<\kappa$. In fact, the tree would have to have height $<\kappa$. – Chris Lambie-Hanson Jan 23 '13 at 17:06

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