## Claim 3.2 page 377

Last time, we left off with the statement of the following theorem:

Theorem 1 Suppose ${\mathfrak{a}}$ is a progressive set of regular cardinals, and

$\displaystyle \aleph_0\leq{\rm cf}(\sigma)=\sigma<\theta\leq|\mathfrak{a}|. \ \ \ \ \ (1)$

Then

$\displaystyle {\rm cf}^\sigma_{<\theta}(\prod\mathfrak{a})=\sup{\rm pcf}_{\Gamma(\theta,\sigma)}(\mathfrak{a}). \ \ \ \ \ (2)$

The above is just Claim 3.2 on page 377 of The Book with the notation changed a little. Today I want to give a proof of this theorem. If you’ve already read through our June 10th post, then today will be simple because it’s essentially the same proof.

Proof: Start by letting ${F\subseteq\prod\mathfrak{a}}$ be ${<\sigma}$-cofinal in ${\prod\mathfrak{b}}$ for any ${\mathfrak{b}\subseteq\mathfrak{a}}$ of cardinality ${<\theta}$, and suppose ${\kappa\in{\rm pcf}_{\Gamma(\theta,\sigma)}(\mathfrak{a})}$. We will show that ${\kappa\leq|F|}$, thereby establishing the “${\geq}$” part of the theorem.

Assume by way of contradiction that ${|F|<\kappa}$. Let ${J}$ and ${\mathfrak{b}}$ witness that ${\kappa}$ is in ${{\rm pcf}_{\Gamma(\theta,\sigma)}(\mathfrak{a})}$, i.e.,

• ${J}$ is a ${\sigma}$-complete ideal on ${\mathfrak{a}}$,
• ${(\prod\mathfrak{a}, <_J)}$ has true cofinality ${\kappa}$,
• ${|\mathfrak{b}|<\theta}$, and
• ${\mathfrak{a}\setminus\mathfrak{b}\in J}$.

Since ${|F|<\kappa}$, there is a function ${g\in\prod\mathfrak{a}}$ such that

$\displaystyle f<_J g\text{ for all}f\in F. \ \ \ \ \ (3)$

Since ${|\mathfrak{b}|<\theta}$, we can find ${F_0\subseteq F}$ of cardinality ${<\sigma}$ such that

$\displaystyle g(\eta)<\sup\{f(\eta):f\in F_0\}\text{ for all }\eta\in\mathfrak{b}. \ \ \ \ \ (4)$

Given ${f\in F_0}$, we define

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

Then (4) tells us

$\displaystyle \mathfrak{b}\subseteq\bigcup\{A_f:f\in F_0\}. \ \ \ \ \ (6)$

A couple of things now come to our rescue: since ${\mathfrak{a}\setminus\mathfrak{b}\in J}$ and ${J}$ is ${\sigma}$-complete, there must be an ${f^*\in F_0}$ such that ${A_f}$ is not in ${J}$, and this tells us

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

This is absurd, of course, for ${g}$ was an upper bound of ${F}$ in ${(\prod\mathfrak{a}, <_J)}$. Thus, ${\kappa\leq |F|}$ and the “${\geq}$” part of the theorem has been established.

For the other direction, we once again mirror our June 10th post. Given ${\kappa\in{\rm pcf}(\mathfrak{a})}$, let ${J_{<\kappa}}$ and ${B_{\kappa}}$ be the associated ideal and generating set. Basic pcf theory tells us

$\displaystyle {\rm tcf}\left(\prod\mathfrak{a}, <_{J_{<\kappa}+(\mathfrak{a}\setminus B_\kappa)}\right)=\kappa, \ \ \ \ \ (8)$

and we let ${F_\kappa=\{ f^\kappa_\alpha:\alpha<\kappa\}}$ serve to witness this.

Now define

$\displaystyle F:=\bigcup\{F_\kappa:\kappa\in{\rm pcf}_{\Gamma(\theta,\sigma)}(\mathfrak{a})\}. \ \ \ \ \ (9)$

Clearly ${|F|=\sup{\rm pcf}_{\Gamma(\theta,\sigma)}(\mathfrak{a})}$, so we will be done if we can show ${F}$ is ${<\sigma}$-cofinal in ${\prod\mathfrak{b}}$ for every ${\mathfrak{b}\subseteq\mathfrak{a}}$ of cardinality ${<\theta}$.

Suppose this is not the case. Then we can find a set ${\mathfrak{b}\subseteq\mathfrak{a}}$ of cardinality ${<\theta}$ and a function ${g\in\prod\mathfrak{a}}$ such that the restriction of ${g}$ to ${\mathfrak{b}}$ cannot be covered by fewer than ${\sigma}$ functions in ${F}$.

Let us now define a 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

$\displaystyle \eta\in A\cap\mathfrak{b}\Longrightarrow g(\eta)<\sup\{f(\eta):f\in F_0\}. \ \ \ \ \ (10)$

It’s easy to check that ${J}$ is an ideal, and ${\mathfrak{a}\setminus\mathfrak{b}\in J}$ for trivial reasons. Our choice of ${g}$ guarantees that ${J}$ is a ${\sigma}$-complete proper ideal as well.

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

$\displaystyle I=J+(\mathfrak{a}\setminus B_\kappa). \ \ \ \ \ (11)$

The ideal ${I}$ is essentially ${J}$ restricted to ${B_\kappa}$; it is easy to see that ${I}$ is a proper (by choice of ${\kappa}$) ${\sigma}$-complete (as ${J}$ is) ideal on ${\mathfrak{a}}$. Clearly ${\mathfrak{a}\setminus\mathfrak{b}\in I}$ as well, since this set is in ${J}$.

The minimality of ${\kappa}$ ensures

$\displaystyle J_{<\kappa}\subseteq J, \ \ \ \ \ (12)$

which immediately implies

$\displaystyle J_{<\kappa}+(\mathfrak{a}\setminus B_\kappa)\subseteq I. \ \ \ \ \ (13)$

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

$\displaystyle \kappa={\rm tcf}\left(\prod\mathfrak{a}, <_I\right). \ \ \ \ \ (14)$

Putting all of the above together, we see

$\displaystyle \kappa\in{\rm pcf}_{\Gamma(\theta,\sigma)}(\mathfrak{a}) \ \ \ \ \ (15)$

by way of our ideal ${I}$.

Our goal now is to reach a contradiction exactly as in the June 10th post by demonstrating that ${B_\kappa\in J}$, and ${B_\kappa\notin J}$.

Part of this is trivial, as ${\kappa}$ was selected so that ${B_\kappa\notin J}$. For the other part, we know that given our choice of ${F_\kappa}$, there is an ${f\in F_\kappa}$ such that

$\displaystyle g

which means,

$\displaystyle \{\eta\in B_\kappa: f(\eta)\leq g(\eta)\}\in J_{<\kappa}\subseteq J. \ \ \ \ \ (17)$

But since ${F_\kappa\subseteq F}$, the definition of ${J}$ tells us

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

Since ${\mathfrak{a}\setminus\mathfrak{b}\in J}$, we have

$\displaystyle B_\kappa\setminus\mathfrak{b}\subseteq \mathfrak{a}\setminus\mathfrak{b}\in J, \ \ \ \ \ (19)$

and now the conjunction of (17), (18), and (19) together with the fact that ${J}$ is an ideal shows us that ${B_\kappa\in J}$. Contradiction. $\Box$