## Conclusion 3.2A (part 4)

So let us now return to the proof of Conclusion 3.2A. We’ll keep notation the same; what remains to be shown is that if ${\mu}$ is singular of cardinality ${\kappa}$ and ${\mu<\aleph_\mu}$, then there is a ${\tau<\mu}$ for which

$\displaystyle {\rm cf}_\kappa\left(\prod(\tau,\mu)\cap{\sf Reg}\right)\leq{\rm pp}(\mu). \ \ \ \ \ (1)$

We have assumed ${\mu<\aleph_\mu}$, so for all sufficiently large ${\tau<\mu}$ we know ${(\tau,\mu)\cap{\sf Reg}}$ is progressive. Given such a ${\tau}$, we can apply Claim 3.2 (June 15, 2011) to conclude

$\displaystyle {\rm cf}_\kappa\left(\prod(\tau,\mu)\cap{\sf Reg}\right)=\sup{\rm pcf}_{\Gamma(\kappa^+,\aleph_0)}((\tau,\mu)\cap{\sf Reg}). \ \ \ \ \ (2)$

So we fix can fix ${\tau<\mu}$ large enough so that

• ${\mathfrak{a}:=(\tau,\mu)\cap{\sf Reg}}$ is progressive,
• ${{\rm cf}_\kappa\left(\prod\mathfrak{a}\right)=\sup{\rm pcf}_{\Gamma(\kappa^+,\aleph_0)}(\mathfrak{a})}$, and
• ${\sup\{\max{\rm pcf}(\mathfrak{b}):\mathfrak{b}\subseteq\mathfrak{a}, |\mathfrak{b}|=\kappa\}\leq{\rm pp}(\mu)}$,

where for the last statement we use the result of the previous posting.

We will get what we need by comparing the last two bullet points:

Suppose ${\nu\in{\rm pcf}_{\Gamma(\kappa^+,\aleph_0)}(\mathfrak{a})}$. This means that we there is an ultrafilter ${D}$ and a set ${\mathfrak{b}\subseteq \mathfrak{a}}$ of cardinality ${\leq\kappa}$ such that

• ${\mathfrak{b}\in D}$, and
• ${{\rm cf}(\prod\mathfrak{a}/D)=\nu}$.

Clearly we may assume that ${\mathfrak{b}}$ has cardinality exactly ${\kappa}$, and the second bullet implies

$\displaystyle {\rm cf}(\prod\mathfrak{b}/ D)=\nu \ \ \ \ \ (3)$

as well, where we view ${D}$ as an ultrafilter on ${\mathfrak{b}}$. Chaining things together, we have

$\displaystyle \nu={\rm cf}\left(\prod\mathfrak{a}/D\right)={\rm cf}\left(\prod\mathfrak{b}/D\right)\leq\max{\rm pcf}(\mathfrak{b})\leq{\rm pp}(\mu). \ \ \ \ \ (4)$

So for our choice of ${\tau}$, we have

$\displaystyle \sup{\rm pcf}_{\Gamma(\kappa^+,\aleph_0)}((\tau,\mu)\cap{\sf Reg})\leq{\rm pp}(\mu), \ \ \ \ \ (5)$

and so (1) follows from (2).