## Wednesday, July 20, 2016

### On a theorem of Gitik and Shelah II: Blanket Assumptions

$\DeclareMathOperator{\pp}{pp} \DeclareMathOperator{\tcf}{tcf} \DeclareMathOperator{\pcf}{pcf} \DeclareMathOperator{\cov}{cov} \def\cf{\rm{cf}} \def\REG{\sf {REG}} \def\restr{\upharpoonright} \def\bd{\rm{bd}} \def\subs{\subseteq} \def\cof{\rm{cof}} \def\ran{\rm{ran}} \DeclareMathOperator{\ch}{Ch} \DeclareMathOperator{\PP}{pp} \DeclareMathOperator{\Sk}{Sk}$

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