## The easy direction

In this post, we’ll prove a more nuanced version of the result we discussed in our January 31 post. This is the “easy direction” of the theorem we’re aiming to prove, and the argument goes through without any special assumptions on the cofinality of the singular cardinal ${\mu}$.

Theorem 1 Suppose ${\sigma}$, ${\theta}$, and ${\mu}$ are cardinals satisfying ${\sigma\leq{\rm cf}(\mu)<\theta\leq\mu}$. Then

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

Proof: Suppose ${\mathfrak{a}\subseteq\mu}$ is a progressive set of regular cardinals such that ${|\mathfrak{a}|<\theta}$ and ${\sup(\mathfrak{a})=\mu}$. Further assume that ${I}$ is a ${\sigma}$-complete ideal on ${\mathfrak{a}}$ extending the bounded ideal such that ${\prod\mathfrak{a}/I}$ has true cofinality, and let

$\displaystyle \kappa:={\rm tcf}(\prod\mathfrak{a}, <_I). \ \ \ \ \ (2)$

We must show

$\displaystyle \kappa\leq{\rm cov}(\mu, \mu, \theta,\sigma). \ \ \ \ \ (3)$

Suppose this fails, and let ${\mathcal{P}}$ witness ${{\rm cov}(\mu,\mu,\theta,\sigma)<\kappa}$. By our choice of ${\kappa}$, there is a sequence ${\langle f_\alpha:\alpha<\kappa\rangle}$ such that

• each ${f_\alpha}$ is in ${\prod\mathfrak{a}}$,
• ${\alpha<\beta\Longrightarrow f_\alpha<_I f_\beta}$, and
• if ${g\in\prod\mathfrak{a}}$, then ${g<_I f_\alpha}$ for some ${\alpha<\kappa}$.

Since ${|\mathfrak{a}|<\theta}$, it follows that for each ${\alpha<\kappa}$ the range of ${f_\alpha}$ is covered by a union of fewer than ${\sigma}$ sets from ${\mathcal{P}}$. Since ${I}$ is ${\sigma}$-complete, we conclude that for each ${\alpha<\kappa}$ there is a set ${A_\alpha\in\mathcal{P}}$ such that

$\displaystyle \{\eta\in\mathfrak{a}:f_\alpha(\eta)\in A_\alpha\}\notin I. \ \ \ \ \ (4)$

Now ${\kappa}$ is regular and ${|\mathcal{P}|<\kappa}$, so we can assume that there is a single set ${A\in\mathcal{P}}$ with the property that

$\displaystyle B_\alpha:=\{\eta\in\mathfrak{a}:f_\alpha(\eta)\in A\}\notin I \ \ \ \ \ (5)$

for each ${\alpha<\kappa}$.

Now define a function ${g\in\prod\mathfrak{a}}$ as follows:

$\displaystyle g(\eta)= \begin{cases} \sup(A\cap\eta) &\text{if this is less than }\eta,\\ 0 &\text{otherwise.} \end{cases} \ \ \ \ \ (6)$

Since ${|A|<\theta\leq\mu}$, we know that ${A\cap\eta}$ is bounded in ${\eta}$ for all sufficiently large ${\eta\in\mathfrak{a}}$. Since ${I}$ contains the bounded subsets of ${\mathfrak{a}}$, it follows that for each ${\alpha<\kappa}$,

$\displaystyle \{\eta\in\mathfrak{a}: f_\alpha(\eta)\leq g(\eta)\}\in I^+. \ \ \ \ \ (7)$

But this is a contradiction, as there must exist an ${\alpha<\kappa}$ such that ${g<_I f_\alpha}$. $\Box$