Returning now to our main project, we extract a lemma from Shelah’s argument and examine it in isolation.

Let ${\mathfrak{M}=\langle M_\alpha:\alpha<\mu^+\rangle}$ and ${M^*}$ be as assumed, and consider the structure

$\displaystyle \left( \prod (\mu\cap{\sf Reg}), <_J\right), \ \ \ \ \ (1)$

where ${J}$ denotes the ideal of bounded subsets of ${\mu}$. It’s pretty easy to see that this structure is ${\mu^+}$-directed, i.e., that every collection of fewer than ${\mu^+}$ functions in ${\prod(\mu\cap{\sf Reg})}$ has a ${<_J}$-upper bound.

Thus, we can define a sequence of functions ${\langle f_\alpha:\alpha<\mu^+\rangle}$ in ${\prod(\mu\cap{\sf Reg})}$ by letting ${f_\alpha}$ be the ${<_\chi}$-least function in ${\prod(\mu\cap{\sf Reg})}$ serving as a ${<_J}$-upper bound for ${M_\alpha\cap\prod(\mu\cap{\sf Reg})}$. Using this definition, we achieve

$\displaystyle \langle f_\beta:\beta\leq\alpha\rangle\in M_{\alpha+1} \ \ \ \ \ (2)$

for each ${\alpha}$, and so

$\displaystyle \alpha<\beta<\mu^+\Longrightarrow f_\alpha<_J f_\beta \ \ \ \ \ (3)$

as well.

Lemma 1 Let ${\mathfrak{a}}$ be a set of regular cardinals satisfying

• ${\sup(\mathfrak{a})=\mu}$,
• ${\mathfrak{a}}$ is progressive, and
• ${\mathfrak{a}\in M^*}$.

Then ${\langle f_\alpha\upharpoonright\mathfrak{a}:\alpha<\mu^+\rangle}$ is a scale in ${\prod\mathfrak{a}}$ modulo the ideal ${J^{\rm bd}[\mathfrak{a}]}$ of bounded subsets of ${\mathfrak{a}}$.

Proof: Let ${\mathfrak{a}}$ satisfy the assumptions of the proposition. Since ${{\rm pp}(\mu)=\mu^+}$, we know that ${\prod\mathfrak{a}}$ carries a scale of length ${\mu^+}$ modulo the ideal ${J^{\rm bd}[\mathfrak{a}]}$. Since ${M^*}$ is an elementary model of ${\mathfrak{A}}$ and ${\mathfrak{a}\in M^*}$, it follows that there is such a scale living in ${M^*}$, say ${\bar{f}^{\mathfrak{a}}=\langle f^{\mathfrak{a}}_i:i<\mu^+\rangle}$. Note that each ${f^{\mathfrak{a}}_i}$ is in ${M^*}$ as ${\mu^+\subseteq M^*}$.

Now consider ${\langle f_\alpha\upharpoonright\mathfrak{a}:\alpha<\mu^+\rangle}$. Since ${\mathfrak{a}}$ is unbounded in ${\mu}$, it follows easily that this sequence is increasing modulo ${J^{\rm bd}[\mathfrak{a}]}$ in ${\prod\mathfrak{a}}$. Given ${i<\mu^+}$, there is an ${\alpha<\mu^+}$ with ${f^{\mathfrak{a}}_i\in M_\alpha}$. But clearly this implies

$\displaystyle f^{\mathfrak{a}}_i< f_\alpha\upharpoonright\mathfrak{a}\mod J^{\rm bd}[\mathfrak{a}]. \ \ \ \ \ (4)$

Thus, the sequence ${\langle f_\alpha\upharpoonright\mathfrak{a}:\alpha<\mu^+\rangle}$ is also cofinal in ${\prod\mathfrak{a}}$ modulo ${J^{\rm bd}[\mathfrak{a}]}$, and we are done. $\Box$

Notice that the above lemma gives us a way to smuggle information into ${M^*}$: given ${\mathfrak{a}}$ as in the proposition, it follows that ANY function ${g\in\prod\mathfrak{a}}$, whether in ${M^*}$ or not, is bounded modulo ${J^{\rm bd}[\mathfrak{a}]}$ by a function of the form ${f_\alpha\upharpoonright\mathfrak{a}}$ from ${M^*}$. Formulating this as a corollary, we obtain:

Corollary 2 Let ${\mathfrak{a}}$ be a set of regular cardinals as in the preceding lemma. If ${g\in\prod\mathfrak{a}}$, then there is an ${\alpha<\mu^+}$ such that

$\displaystyle g