The quarter is ending here, so I should have a little more time for maintaining this blog. I mentioned before that I figured out how to fix the proof of Claim 3.5, which then gives a fairly transparent proof of the cov vs. pp theorem. I need to clarify this a little, because what I’ve got doesn’t recover the full result originally claimed by Shelah. Here’s what I can show:

Theorem 1 Suppose ${\mu}$ is singular, and ${\aleph_0<\sigma\leq{\rm cf}(\mu)<\theta<\mu}$ with ${\sigma}$ and ${\theta}$ regular. Then the following two statements are equivalent for a regular cardinal ${\kappa}$:

1. ${{\rm pp}_{\Gamma(\theta,\sigma)}(\mu)<\kappa}$.
2. ${{\rm cov}(\mu,\mu,\theta,\sigma)<\kappa}$.

This is enough to deduce Shelah’s cov vs. pp theorem (Theorem 5.4 of Chapter II in The Book), which states:

Theorem 2 If ${\sigma}$ is a regular uncountable cardinal, and ${\sigma<\theta\leq\kappa\leq\lambda}$, then

$\displaystyle {\rm cov}(\lambda,\kappa,\theta,\sigma)+\lambda =\sup\{{\rm pp}_{\Gamma(\theta,\sigma)}(\mu): \kappa\leq\mu\leq\lambda\wedge \sigma\leq{\rm cf}(\mu)<\theta\}+\lambda \ \ \ \ \ (1)$

Now the issue left unresolved concerns the attainment of suprema in one particular case: if ${\mu}$ is singular and ${{\rm pp}_{\Gamma(\theta,\sigma)}(\mu)}$ is a weakly inaccessible cardinal ${\kappa}$, must there exist a cofinal subset of ${\mu\cap{\sf Reg}}$ of cardinal less than ${\theta}$ and a ${\sigma}$-complete ideal ${J}$ on ${\mathfrak{a}}$ containing the bounded subsets of ${\mathfrak{a}}$ such that

$\displaystyle \kappa={\rm tcf}\left(\prod\mathfrak{a}/J\right). \ \ \ \ \ (2)$

Said another way, if ${{\rm pp}_{\Gamma(\theta,\sigma)}(\mu)}$ is regular, then must the supremum in the definition of ${{\rm pp}_{\Gamma(\theta,\sigma)}(\mu)}$ be attained?

PS: I’m still very sleep-deprived…it’s all part of being the father of young children.