\(

\DeclareMathOperator{\pp}{pp}

\DeclareMathOperator{\tcf}{tcf}

\DeclareMathOperator{\pcf}{pcf}

\DeclareMathOperator{\cov}{cov}

\def\cf{\rm{cf}}

\def\REG{\sf {REG}}

\def\restr{\upharpoonright}

\def\bd{\rm{bd}}

\def\subs{\subseteq}

\def\cof{\rm{cof}}

\def\ran{\rm{ran}}

\DeclareMathOperator{\ch}{Ch}

\DeclareMathOperator{\PP}{pp}

\DeclareMathOperator{\Sk}{Sk}

\)

Assumptions for the next few posts are:

- $\kappa$ weakly compact
- $\mu>2^\kappa$ is singular of cofinality $\kappa$
- $A\subseteq\mu$ is a set of regular cardinals
- $A$ is unbounded in $\mu$ of cardinality $\kappa$
- $J$ is a $\kappa$-complete ideal on $A$ extending the bounded ideal
- $\lambda=\tcf(\prod A/ J)$

Since $J$ extends the bounded ideal, we may as well assume

- $(2^\kappa)^+<\min(A)$

Of course this implies $|A|<\min(A)$, and so without loss of generality $A$ is progressive.

Let $D$ be an ultrafilter on $A$ disjoint to $J$. Then $\cf(\prod A/D)=\lambda$ and hence $\lambda\in \pcf(A)$. Furthermore, if $B_\lambda[A]$ is a pcf generator for $\lambda$ then $B_\lambda[A]$ must be in $D$ and therefore unbounded in $A$.

It follows that $\tcf(\prod B_\lambda[A]/J)=\lambda$ as well, and so we may assume (by passing to $B_\lambda[A]$ if necessary) that

- $\lambda=\max\pcf(A)$

Our assumption that $2^\kappa<\min(A)$ tells us that

(1) $\pcf(\pcf(A))=\pcf(A)$, and

(2) $|\pcf(A)|\leq 2^\kappa<\min(A)=\min(\pcf(A))$

and so $\pcf(A)$ has a transitive set of generators, that is, a sequence $\langle B_\theta:\theta\in\pcf(A)\rangle$ such that

- $B_\theta\subseteq \pcf(A)$

- $B_\theta$ is a generator for $\theta$ in $\pcf(\pcf(A))=\pcf(A)$, and

- $\tau\in B_\theta\Longrightarrow B_\tau\subseteq B_\theta$.

## No comments:

Post a Comment