November 14, 2011 at 17:09 | Posted in Uncategorized | Leave a comment

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.

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