## First steps in the hard direction

May 10, 2011 at 16:01 | Posted in Uncategorized | Leave a comment

Now what about the “hard direction”? Given cardinals ${\sigma}$, ${\theta}$, and ${\mu}$ such that ${\sigma}$ is regular and ${\aleph_0<\sigma\leq{\rm cf}(\mu)<\theta\leq\mu}$, our aim is to prove

$\displaystyle {\rm cov}(\mu,\mu,\theta,\sigma)\leq{\rm pp}_{\Gamma(\theta,\sigma)}(\mu), \ \ \ \ \ (1)$

and moreover, if ${{\rm cov}(\mu,\mu,\theta,\sigma)}$ is regular, then there exist ${\mathfrak{a}}$ and ${I}$ such that

1. ${\mathfrak{a}}$ is a progressive set of regular cardinals with ${\sup(\mathfrak{a})=\mu}$ and ${|\mathfrak{a}|<\theta}$,
2. ${I}$ is a ${\sigma}$-complete ideal on ${\mathfrak{a}}$ extending the bounded ideal,
3. the structure ${(\prod\mathfrak{a}, <_I)}$ has a true cofinality, and
4. ${{\rm cov}(\mu, \mu, \theta, \sigma)\leq {\rm tcf}(\prod\mathfrak{a}, <_I)}$.

Notice that by the preceding post, condition 4 is equivalent to the statement

$\displaystyle {\rm cov}(\mu, \mu, \theta,\sigma)={\rm tcf}(\prod\mathfrak{a}, <_I), \ \ \ \ \ (2)$

and so the equality

$\displaystyle {\rm pp}_{\Gamma(\theta,\sigma)}(\mu)={\rm cov}(\mu,\mu,\theta,\sigma) \ \ \ \ \ (3)$

holds (at least in the case where ${\aleph_0<\sigma}$) in a strong sense. Note that the underlying issue here is that the pp number is defined as the supremum of a certain set of regular cardinals; we will show that if this supremum itself is a regular cardinal (and hence weakly inaccessible), then it actually occurs as the true cofinality of ${(\prod\mathfrak{a}, <_I)}$ for some relevant ${\mathfrak{a}}$ and ${I}$.

We’ll do this by proving the following:

Theorem 1 Suppose ${\lambda}$ is a regular cardinal such that ${{\rm tcf}(\prod\mathfrak{a}, <_I)<\lambda}$ whenever ${\mathfrak{a}}$ and ${I}$ satisfy conditions 1, 2, and 3 above. Then ${{\rm cov}(\mu,\mu,\theta,\sigma)<\lambda}$ as well.

The usual caveats apply here: it is closer to the truth to say that I will attempt to prove the above theorem. We’ll see what happens…