## Conclusion 3.2A (part 3)

This post is an immediate continuation of the previous one, so all notation etc. carries over from there. I noticed last night that Conclusion 3.2A is really two separate results which have been smashed together. The main part actually looks like it holds without the assumption that ${\mu}$ is not a fixed point, so I’m going to work in a slightly more general context today. If things work out, I’ll formulate “the real theorem” underlying things.

Lemma 1 Let ${\mu}$ be a singular cardinal of cofinality ${\kappa}$. Then there is a ${\tau<\mu}$ such that

$\displaystyle \sup\{{\rm cf}\left(\prod\mathfrak{b}, <\right):\mathfrak{b}\subseteq (\tau,\mu)\cap{\sf Reg}, |\mathfrak{b}|=\kappa\}\leq{\rm pp}(\mu). \ \ \ \ \ (1)$

Proof: First, note that if we restrict ourselves to ${\tau}$ with ${{\rm cf}(\tau)<\tau<\mu}$, then any ${\mathfrak{b}}$ relevant to computing the left-hand side of the above equation is necessarily progressive, and hence the cofinality of ${\prod\mathfrak{b}}$ will just be ${\max{\rm pcf}(\mathfrak{b})}$. Thus, we need only prove that there is a ${\tau}$ in the interval ${(\kappa,\mu)}$ for which

$\displaystyle \sup\{\max{\rm pcf}(\mathfrak{b}):\mathfrak{b}\subseteq (\tau,\mu)\cap{\sf Reg}, |\mathfrak{b}|=\kappa\}\leq{\rm pp}(\mu). \ \ \ \ \ (2)$

Suppose this is not the case, so for any ${\tau<\mu}$ there is a ${\mathfrak{b}\subseteq (\tau,\mu)\cap{\sf Reg}}$ of cardinality ${\kappa}$ with ${{\rm pp}(\mu)<\max{\rm pcf}(\mathfrak{b})}$. The following proposition takes this hypothesis and moves us closer to a contradiction:

Proposition 2 Under the above assumptions, for every ${\tau<\mu}$ there is a ${\mathfrak{b}}$ such that

• ${\mathfrak{b}}$ is a progressive subset of ${(\tau,\mu)\cap{\sf Reg}}$,
• ${|\mathfrak{b}|\leq\kappa}$,
• ${{\rm pp}(\mu)<\max{\rm pcf}(\mathfrak{b})}$, and
• ${\mathfrak{b}}$ is a bounded subset of ${\mu}$.

Proof: Without loss of generality, we assume ${\kappa<\tau}$ and so we can find (progressive) ${\mathfrak{b}\subseteq (\tau,\mu)\cap{\sf Reg}}$ of cardinality ${\kappa}$ such that ${{\rm pp}(\mu)<\max{\rm pcf}(\mathfrak{b})}$. If ${\mathfrak{b}}$ is bounded below ${\mu}$ then we are done, so assume ${\mathfrak{b}}$ is unbounded.

Let ${D}$ be an ultrafilter on ${\mathfrak{b}}$ such that

$\displaystyle {\rm pp}(\mu)<{\rm cf}\left(\prod\mathfrak{b}/D\right)=\max{\rm pcf}(\mathfrak{b}). \ \ \ \ \ (3)$

The above equation, taken together with the fact that ${\mathfrak{b}}$ is unbounded in ${\mathfrak{a}}$ of cardinality ${\kappa}$, tells us (by way of the definition of ${{\rm pp}(\mu)}$) that ${D}$ cannot be disjoint to the bounded subsets of ${\mathfrak{b}}$, i.e,. that there is a ${\sigma<\mu}$ such that

$\displaystyle \mathfrak{b}^*:=\mathfrak{b}\cap\sigma \in D. \ \ \ \ \ (4)$

Now this ${\mathfrak{b}^*}$ has all of the required properties. $\Box$

Armed with the above proposition, we can construct sequences ${\langle \mathfrak{b}_i:i<\kappa\rangle}$ and ${\langle \mu_i:i<\kappa\rangle}$ such that

• ${\langle \mu_i:i<\kappa\rangle}$ is a strictly increasing sequence of cardinals cofinal in ${\mu}$
• ${\mu_0=\kappa}$
• ${\mathfrak{b}_i\subseteq (\mu_i, \mu_{i+1})\cap{\sf Reg}}$
• ${|\mathfrak{b}_i|\leq\kappa}$ (hence ${\mathfrak{b}_i}$ is progressive)
• ${{\rm pp}(\mu)<\max{\rm pcf}(\mathfrak{b}_i)}$ for each ${i<\kappa}$.

Now define

$\displaystyle \mathfrak{b}=\bigcup_{i<\kappa}\mathfrak{b}_i. \ \ \ \ \ (5)$

By construction, ${\mathfrak{b}}$ is a progressive set of regular cardinals cofinal in ${\mu}$ of cardinality ${{\rm cf}(\mu)}$, and so the definition of ${{\rm pp}(\mu)}$ tells us that if ${D}$ is an ultrafilter on ${\mathfrak{b}}$ disjoint to the bounded ideal, then

$\displaystyle {\rm cf}\left(\prod\mathfrak{b}/D\right)\leq{\rm pp}(\mu). \ \ \ \ \ (6)$

Let ${\sigma={\rm pp}(\mu)^+}$, and consider the ideal ${J}$ on ${\mathfrak{b}}$ generated by ${J_{<\sigma}[\mathfrak{b}]}$ and the bounded subsets of ${\mathfrak{b}}$. It is easy to show that ${J}$ must be a proper ideal on ${\mathfrak{b}}$: if not, then there would be a ${\nu<\mu}$ such that

$\displaystyle \mathfrak{b}\cap (\nu,\mu)\in J_{<\sigma}[\mathfrak{a}], \ \ \ \ \ (7)$

but this is impossible as

$\displaystyle {\rm pp}(\mu)<{\rm pp}(\mu)^+=\sigma\leq\max{\rm pcf}(\mathfrak{b}\cap(\nu,\mu)) \ \ \ \ \ (8)$

by our construction of ${\mathfrak{b}}$. If ${D}$ is an ultrafilter on ${\mathfrak{b}}$ disjoint to ${J}$, then

$\displaystyle {\rm pp}(\mu)<\sigma\leq{\rm cf}\left(\prod\mathfrak{b}/D\right), \ \ \ \ \ (9)$

and we have a contradiction of (6). $\Box$