## Continuation on pp

The main result of this post is natural, but unsurprising. Oddly enough, though, I can’t find this exact version published anywhere. I mentioned something in this direction in my Handbook of Set Theory argument; I wish I had thought about it a bit more back then, because the proof is certainly easy enough. Anyway, the result is the following:

Theorem 1 The following statements are equivalent for a singular cardinal ${\mu}$:

1. ${{\rm pp}(\mu)=\mu^+}$
2. If ${\mathfrak{a}\subseteq\mu}$ is a cofinal progressive set of regular cardinals, then ${\prod\mathfrak{a}}$ carries a scale of length ${\mu^+}$ modulo the ideal ${J^{\rm bd}[\mathfrak{a}]}$ of bounded subsets of ${\mathfrak{a}}$.

Proof:

We prove first that (2) implies (1), as this implication is almost immediate. Suppose ${\mathfrak{a}\subseteq\mu}$ is a progressive set of regular cardinals cofinal in ${\mu}$ and of cardinality ${{\rm cf}(\mu)}$. Our assumptions imply that there is a sequence ${\langle f_\alpha:\alpha<\mu^+\rangle}$ of functions from ${\prod\mathfrak{a}}$ such that

• ${\alpha<\beta\rightarrow f_\alpha, and
• if ${g\in\prod\mathfrak{a}}$, then there is an ${\alpha<\mu^+}$ such that ${g.

If ${D}$ is any ultrafilter on ${\mathfrak{a}}$ disjoint to ${J^{\rm bd}[\mathfrak{a}]}$, then the two above statements continue to hold modulo ${D}$, and hence ${\prod\mathfrak{a}/D}$ has cofinality at most ${\mu^+}$. But clearly this cofinality must be at least ${\mu^+}$ as ${D}$ extends the co-bounded filter on ${\mathfrak{a}}$. The statement ${{\rm pp}(\mu)=\mu^+}$ now follows from the definitions involved.

For the other direction, suppose ${{\rm pp}(\mu)=\mu^+}$, let ${\mathfrak{a}}$ satisfy the assumptions of statement (2), and let ${I}$ denote the ideal ${J^{\rm bd}[\mathfrak{a}]}$. It is clear that

$\displaystyle \prod\mathfrak{a}/I\text{ is }\mu^+\text{-directed} \ \ \ \ \ (1)$

(see Claim 1.3(4) on page 4 of Cardinal Arithmetic), that is, given a collection of functions in ${\prod\mathfrak{a}}$ of cardinality at most ${\mu}$, there is a single function in ${\prod\mathfrak{a}}$ above them all modulo the ideal ${I}$.

By Conclusion 3.2 on page 62 of Cardinal Arithmetic, ${{\rm cf}\left(\prod\mathfrak{a}/I\right)}$ is equal to the supremum of all cardinals of the form ${{\rm cf}\left(\prod\mathfrak{a}/D\right)}$, where ${D}$ is an ultrafilter on ${\mathfrak{a}}$ disjoint to ${I}$. This latter cardinal is at most ${{\rm pp}(\mu)=\mu^+}$, and so it follows that

$\displaystyle {\rm cf}\left(\prod\mathfrak{a}/I\right)=\mu^+. \ \ \ \ \ (2)$

Given (1) and (2), it is straightforward to construct the desired scale. $\Box$