## Characterize the sup of pcf sigma-complete

This post should be considered a warmup. To start, I want to recall one of the standard results of pcf theory, namely that if ${\mathfrak{a}}$ is a progressive set of regular cardinal, then

$\displaystyle \max{\rm pcf}(\mathfrak{a})={\rm cf}(\prod\mathfrak{a}, <), \ \ \ \ \ (1)$

where the ordering on ${\prod\mathfrak{a}}$ is given by

$\displaystyle f

My aim here is to prove an analogous result characterizing ${\sup{\rm pcf}_{\sigma{\rm-complete}}(\mathfrak{a})}$. For those new to this, a cardinal ${\kappa}$ is in ${{\rm pcf}_{\sigma{\rm-complete}}(\mathfrak{a})}$ if there is a ${\sigma}$-complete ideal ${J}$ on ${\mathfrak{a}}$ with

$\displaystyle \kappa={\rm tcf}(\prod\mathfrak{a}, <_J), \ \ \ \ \ (3)$

which means that there is a ${<_J}$-increasing ${\kappa}$-sequence of functions in ${\prod\mathfrak{a}}$ cofinal in ${\prod\mathfrak{a}}$. In other words, (3) means that ${(\prod\mathfrak{a}, <_J)}$ has cofinality ${\kappa}$, AND every subset of ${\prod\mathfrak{a}}$ of cardinality less than ${\kappa}$ has an upper bound in ${(\prod\mathfrak{a}, <_J)}$.

The characterization we give involves the following notion:

Definition 1 Let ${\sigma}$ be a regular cardinal, and suppose ${\mathfrak{a}}$ is a set of regular cardinals. We say that a family ${F\subseteq\prod\mathfrak{a}}$ is ${<\sigma}$-cofinal in ${\prod\mathfrak{a}}$ if for every ${f\in\prod\mathfrak{a}}$, there is a family ${F_0\subseteq F}$ of size less than ${\sigma}$ such that ${f<\sup F_0}$, i.e., such that

$\displaystyle f(\eta)<\sup\{g(\eta):g\in F_0\} \ \ \ \ \ (4)$

for all ${\eta\in \mathfrak{a}}$. The ${<\sigma}$-cofinality of ${\mathfrak{a}}$ is defined to be the minimum cardinality of a ${<\sigma}$-cofinal subset of ${\mathfrak{a}}$. We will denote this cardinal by ${{\rm cf}^{\sigma}\prod(\mathfrak{a})}$.

So now we come to the promised characterization:

Theorem 2 If ${\mathfrak{a}}$ is a progressive set of regular cardinals, and ${\sigma}$ is a regular cardinal with ${\sigma\leq|\mathfrak{a}|}$, then

$\displaystyle {\rm cf}^{\sigma}(\prod\mathfrak{a}) = \sup{\rm pcf}_{\sigma{\rm -complete}}(\mathfrak{a}). \ \ \ \ \ (5)$

Proof: We prove the ${\geq}$ direction first, so assume ${F}$ is a ${<\sigma}$-cofinal family in ${\prod\mathfrak{a}}$, and ${\kappa\in{\rm pcf}_{\sigma{\rm-complete}}(\mathfrak{a})}$. Let ${J}$ be a ${\sigma}$-complete ideal on ${\mathfrak{a}}$ such that

$\displaystyle {\rm tcf}(\prod\mathfrak{a}, <_J)=\kappa, \ \ \ \ \ (6)$

and assume by way of contradiction that ${|F|<\kappa}$. Given this, (15) implies that ${F}$ is bounded modulo ${J}$ in ${\prod\mathfrak{a}}$, so there is a ${g\in\prod \mathfrak{a}}$ with

$\displaystyle f<_J g\text{ for all }f\in F. \ \ \ \ \ (7)$

We have assumed, however, that ${F}$ is ${<\sigma}$-cofinal in ${\prod\mathfrak{a}}$, so we can fix ${\alpha<\sigma}$ and ${F_0=\{f_\beta:\beta<\alpha\}\subseteq F}$ with

$\displaystyle g<\sup F_0. \ \ \ \ \ (8)$

Given ${\beta<\alpha}$, define

$\displaystyle A_\beta=\{\eta\in\mathfrak{a}: g(\eta)

By our choice of ${F_0}$, we have

$\displaystyle \mathfrak{a}=\bigcup_{\beta<\alpha}A_\beta. \ \ \ \ \ (10)$

Since ${J}$ is ${\sigma}$-complete and ${\alpha<\sigma}$, there is a ${\beta<\alpha}$ with ${A_\beta\in J^+}$. But this is a contradiction as ${g}$ was an upper bound for ${F}$ in ${(\prod\mathfrak{a}, <_J)}$. Thus, ${\kappa\leq |F|}$ and the inequality “${\geq}$” has been established.

For the other direction, we are going to need a little pcf theory. In general, for each ${\kappa\in{\rm pcf}(\mathfrak{a})}$ we have an associated ideal ${J_{<\kappa}[\mathfrak{a}]}$ and generator ${B_{\kappa}[\mathfrak{a}]}$, and

$\displaystyle {\rm tcf}(\prod\mathfrak{a}, <_{J_{<\kappa}[\mathfrak{a}]+(\mathfrak{a}\setminus B_\kappa[\mathfrak{a}])})=\kappa. \ \ \ \ \ (11)$

This entails the existence of a sequence of functions ${\langle f^\kappa_\alpha:\alpha<\kappa\rangle}$ witnessing the true cofinality of the partial order, and we let ${F_\kappa}$ denote the collection ${\{f^\kappa_\alpha:\alpha<\kappa\}}$.

Let us define

$\displaystyle F:=\bigcup\{F_\kappa: \kappa\in{\rm pcf}_{\sigma{\rm-complete}}(\mathfrak{a})\}. \ \ \ \ \ (12)$

Clearly ${|F|=\sup{\rm pcf}_{\sigma{\rm-complete}}(\mathfrak{a})}$. We claim that ${F}$ is ${<\sigma}$-cofinal in ${\prod\mathfrak{a}}$.

Suppose this is not the case. Then there is a function ${g\in\prod\mathfrak{a}}$ that cannot be covered by fewer than ${\sigma}$ functions from ${F}$. We define an collection ${J}$ of subsets of ${\mathfrak{a}}$ by putting ${A}$ into ${J}$ if and only if there is a family ${F_0\subseteq F}$ of cardinality ${<\sigma}$ such that

$\displaystyle \eta\in A\Longrightarrow g(\eta)<\sup\{f(\eta):f\in F_0\}. \ \ \ \ \ (13)$

It is easy to check that ${J}$ is a ${\sigma}$-complete ideal on ${\mathfrak{a}}$, and this ideal is proper because of our choice of ${g}$.

Let ${\kappa\in{\rm pcf}(\mathfrak{a})}$ be minimal with ${B_\kappa[\mathfrak{a}]\notin J}$, and define

$\displaystyle I= J + (\mathfrak{a}\setminus B_\kappa[\mathfrak{a}]). \ \ \ \ \ (14)$

Since ${J}$ is ${\sigma}$-complete, it follows easily that ${I}$ is as well, and our choice of ${\kappa}$ guarantees that ${I}$ is a proper ideal on ${\mathfrak{a}}$. Our choice of ${\kappa}$ also guarantees

$\displaystyle J_{<\kappa}[\mathfrak{a}]\subseteq J, \ \ \ \ \ (15)$

and hence

$\displaystyle J_{<\kappa}[\mathfrak{a}]+(\mathfrak{a}\setminus B_\kappa[\mathfrak{a}])\subseteq I. \ \ \ \ \ (16)$

Thus, the functions ${\langle f^\kappa_\alpha:\alpha<\kappa\rangle}$ show us

$\displaystyle \kappa={\rm tcf}(\prod\mathfrak{a}, <_I), \ \ \ \ \ (17)$

and therefore

$\displaystyle \kappa\in{\rm pcf}_{\sigma{\rm-complete}}(\mathfrak{a}). \ \ \ \ \ (18)$

We begin to close in on a contradiction. We know that ${J}$ is a proper ideal on ${\mathfrak{a}}$ and therefore ${B_\kappa}$ (we’ll drop the “${[\mathfrak{a}]}$”) from now on) is not in ${J}$; we will finish the proof by demonstrating that in fact ${B_\kappa}$ MUST be in ${J}$.

Given the definition of ${F_\kappa}$, we know there is an ${f\in F_\kappa}$ for which

$\displaystyle g< f\mod J_{<\kappa}[\mathfrak{a}]+(\mathfrak{a}\setminus B_\kappa). \ \ \ \ \ (19)$

Said another way,

$\displaystyle \{\eta\in B_\kappa: f(\eta)\leq g(\eta)\}\in J_{<\kappa}[\mathfrak{a}]\subseteq J, \ \ \ \ \ (20)$

where the last inclusion is from (15).

But since ${F_\kappa\subseteq F}$, we know

$\displaystyle \{\eta\in B_{\kappa}: g(\eta)

as well, by the very definition of ${J}$.

Thus, ${B_\kappa}$ is the union of two sets in ${J}$, hence an element of ${J}$. Contradiction.

$\Box$