## Traveling

I’ll be doing a lot of traveling over the next two weeks; I’ll pick up with things when I’m back from Barcelona.

## The real theorem

This post is a bit of a detour, but I wanted to point out that Conclusion 3.2A is really just Claim 3.2 grafted onto a nice characterization of ${{\rm pp}(\mu)}$. Shelah implicitly uses this characterization in his proof of Conclusion 3.2A, but I can’t find a formulation of it anywhere in The Book.

Theorem 1 Let ${\mu}$ be a singular cardinal. Then the following three cardinals are equal:

1. ${{\rm pp}(\mu)}$
2. ${\min_{\tau<\mu}\sup\{{\rm cf}(\prod\mathfrak{b}, <):\mathfrak{b}\subseteq (\tau,\mu)\cap{\sf Reg}\wedge|\mathfrak{b}|\leq{\rm cf}(\mu)\}}$
3. ${\min_{\tau<\mu}\sup\{\max{\rm pcf}(\mathfrak{b}):\mathfrak{b}\subseteq(\tau,\mu)\cap{\sf Reg}\wedge |\mathfrak{b}|\leq{\rm cf}(\mu)<\min(\mathfrak{b})\}}$.

Notice that in 2 and 3, the suprema involved can only decrease as ${\tau}$ increases, and this tells us a couple of things. First, the cardinals in 2 and 3 are easily seen to be equal, as without loss of generality ${{\rm cf}(\mu)<\tau}$, hence the collection of relevant sets ${\mathfrak{b}}$ is exactly the same in each case, and we finish using the equality

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

for progressive ${\mathfrak{b}}$.

Second, we know that as ${\tau}$ approaches ${\mu}$, the suprema involve eventually stabilize and hence the minimum value is actually attained by some ${\tau<\mu}$. What the theorem shows is that this stable value is exactly ${{\rm pp}(\mu)}$. Notice as well that although ${{\rm pp}(\mu)}$ is computed using sets cofinal in ${\mu}$, we do not assume this for the ${\mathfrak{b}}$ involved in 2 and 3.

We turn now to the proof:

Proof: We have already noted that 2 and 3 are equal for easy reasons. The fact that 1 is less than or equal to 2 is also trivial from the definition of ${{\rm pp}(\mu)}$:

Suppose ${\eta<{\rm pp}(\mu)}$. Then there is a progressive ${\mathfrak{b}\subseteq \mu\cap{\sf Reg}}$, a cardinal ${\kappa}$, and an ultrafilter ${D}$ on ${\mathfrak{b}}$ such that

• ${|\mathfrak{b}|={\rm cf}(\kappa)}$
• ${D}$ is disjoint to the bounded ideal on ${\mathfrak{b}}$,
• ${\kappa={\rm cf}(\prod\mathfrak{b}/D)}$, and
• ${\eta<\kappa}$.

Given ${\tau<\mu}$, we know ${\kappa\in{\rm pcf}(\mathfrak{b}\cap(\tau,\mu))}$ as ${\mathfrak{b}\cap (\tau,\mu)\in D}$, hence

$\displaystyle \tau<\kappa\leq\max{\rm pcf}(\mathfrak{b})\cap(\tau,\mu). \ \ \ \ \ (2)$

Thus,

$\displaystyle \eta<\kappa\leq\sup\{\max{\rm pcf}(\mathfrak{b}):\mathfrak{b}\subseteq(\tau,\mu)\cap{\sf Reg}\wedge |\mathfrak{b}|\leq{\rm cf}(\mu)<\min(\mathfrak{b})\} \ \ \ \ \ (3)$

for each ${\tau<\mu}$ and the desired inequality follows immediately.

What about the other inequality? This is exactly Lemma 1 from our June 17, 2011 post, which was written with this application in mind.

$\Box$

## Conclusion 3.2A (part 4)

So let us now return to the proof of Conclusion 3.2A. We’ll keep notation the same; what remains to be shown is that if ${\mu}$ is singular of cardinality ${\kappa}$ and ${\mu<\aleph_\mu}$, then there is a ${\tau<\mu}$ for which

$\displaystyle {\rm cf}_\kappa\left(\prod(\tau,\mu)\cap{\sf Reg}\right)\leq{\rm pp}(\mu). \ \ \ \ \ (1)$

We have assumed ${\mu<\aleph_\mu}$, so for all sufficiently large ${\tau<\mu}$ we know ${(\tau,\mu)\cap{\sf Reg}}$ is progressive. Given such a ${\tau}$, we can apply Claim 3.2 (June 15, 2011) to conclude

$\displaystyle {\rm cf}_\kappa\left(\prod(\tau,\mu)\cap{\sf Reg}\right)=\sup{\rm pcf}_{\Gamma(\kappa^+,\aleph_0)}((\tau,\mu)\cap{\sf Reg}). \ \ \ \ \ (2)$

So we fix can fix ${\tau<\mu}$ large enough so that

• ${\mathfrak{a}:=(\tau,\mu)\cap{\sf Reg}}$ is progressive,
• ${{\rm cf}_\kappa\left(\prod\mathfrak{a}\right)=\sup{\rm pcf}_{\Gamma(\kappa^+,\aleph_0)}(\mathfrak{a})}$, and
• ${\sup\{\max{\rm pcf}(\mathfrak{b}):\mathfrak{b}\subseteq\mathfrak{a}, |\mathfrak{b}|=\kappa\}\leq{\rm pp}(\mu)}$,

where for the last statement we use the result of the previous posting.

We will get what we need by comparing the last two bullet points:

Suppose ${\nu\in{\rm pcf}_{\Gamma(\kappa^+,\aleph_0)}(\mathfrak{a})}$. This means that we there is an ultrafilter ${D}$ and a set ${\mathfrak{b}\subseteq \mathfrak{a}}$ of cardinality ${\leq\kappa}$ such that

• ${\mathfrak{b}\in D}$, and
• ${{\rm cf}(\prod\mathfrak{a}/D)=\nu}$.

Clearly we may assume that ${\mathfrak{b}}$ has cardinality exactly ${\kappa}$, and the second bullet implies

$\displaystyle {\rm cf}(\prod\mathfrak{b}/ D)=\nu \ \ \ \ \ (3)$

as well, where we view ${D}$ as an ultrafilter on ${\mathfrak{b}}$. Chaining things together, we have

$\displaystyle \nu={\rm cf}\left(\prod\mathfrak{a}/D\right)={\rm cf}\left(\prod\mathfrak{b}/D\right)\leq\max{\rm pcf}(\mathfrak{b})\leq{\rm pp}(\mu). \ \ \ \ \ (4)$

So for our choice of ${\tau}$, we have

$\displaystyle \sup{\rm pcf}_{\Gamma(\kappa^+,\aleph_0)}((\tau,\mu)\cap{\sf Reg})\leq{\rm pp}(\mu), \ \ \ \ \ (5)$

and so (1) follows from (2).

## Conclusion 3.2A (part 3)

This post is an immediate continuation of the previous one, so all notation etc. carries over from there. I noticed last night that Conclusion 3.2A is really two separate results which have been smashed together. The main part actually looks like it holds without the assumption that ${\mu}$ is not a fixed point, so I’m going to work in a slightly more general context today. If things work out, I’ll formulate “the real theorem” underlying things.

Lemma 1 Let ${\mu}$ be a singular cardinal of cofinality ${\kappa}$. Then there is a ${\tau<\mu}$ such that

$\displaystyle \sup\{{\rm cf}\left(\prod\mathfrak{b}, <\right):\mathfrak{b}\subseteq (\tau,\mu)\cap{\sf Reg}, |\mathfrak{b}|=\kappa\}\leq{\rm pp}(\mu). \ \ \ \ \ (1)$

Proof: First, note that if we restrict ourselves to ${\tau}$ with ${{\rm cf}(\tau)<\tau<\mu}$, then any ${\mathfrak{b}}$ relevant to computing the left-hand side of the above equation is necessarily progressive, and hence the cofinality of ${\prod\mathfrak{b}}$ will just be ${\max{\rm pcf}(\mathfrak{b})}$. Thus, we need only prove that there is a ${\tau}$ in the interval ${(\kappa,\mu)}$ for which

$\displaystyle \sup\{\max{\rm pcf}(\mathfrak{b}):\mathfrak{b}\subseteq (\tau,\mu)\cap{\sf Reg}, |\mathfrak{b}|=\kappa\}\leq{\rm pp}(\mu). \ \ \ \ \ (2)$

Suppose this is not the case, so for any ${\tau<\mu}$ there is a ${\mathfrak{b}\subseteq (\tau,\mu)\cap{\sf Reg}}$ of cardinality ${\kappa}$ with ${{\rm pp}(\mu)<\max{\rm pcf}(\mathfrak{b})}$. The following proposition takes this hypothesis and moves us closer to a contradiction:

Proposition 2 Under the above assumptions, for every ${\tau<\mu}$ there is a ${\mathfrak{b}}$ such that

• ${\mathfrak{b}}$ is a progressive subset of ${(\tau,\mu)\cap{\sf Reg}}$,
• ${|\mathfrak{b}|\leq\kappa}$,
• ${{\rm pp}(\mu)<\max{\rm pcf}(\mathfrak{b})}$, and
• ${\mathfrak{b}}$ is a bounded subset of ${\mu}$.

Proof: Without loss of generality, we assume ${\kappa<\tau}$ and so we can find (progressive) ${\mathfrak{b}\subseteq (\tau,\mu)\cap{\sf Reg}}$ of cardinality ${\kappa}$ such that ${{\rm pp}(\mu)<\max{\rm pcf}(\mathfrak{b})}$. If ${\mathfrak{b}}$ is bounded below ${\mu}$ then we are done, so assume ${\mathfrak{b}}$ is unbounded.

Let ${D}$ be an ultrafilter on ${\mathfrak{b}}$ such that

$\displaystyle {\rm pp}(\mu)<{\rm cf}\left(\prod\mathfrak{b}/D\right)=\max{\rm pcf}(\mathfrak{b}). \ \ \ \ \ (3)$

The above equation, taken together with the fact that ${\mathfrak{b}}$ is unbounded in ${\mathfrak{a}}$ of cardinality ${\kappa}$, tells us (by way of the definition of ${{\rm pp}(\mu)}$) that ${D}$ cannot be disjoint to the bounded subsets of ${\mathfrak{b}}$, i.e,. that there is a ${\sigma<\mu}$ such that

$\displaystyle \mathfrak{b}^*:=\mathfrak{b}\cap\sigma \in D. \ \ \ \ \ (4)$

Now this ${\mathfrak{b}^*}$ has all of the required properties. $\Box$

Armed with the above proposition, we can construct sequences ${\langle \mathfrak{b}_i:i<\kappa\rangle}$ and ${\langle \mu_i:i<\kappa\rangle}$ such that

• ${\langle \mu_i:i<\kappa\rangle}$ is a strictly increasing sequence of cardinals cofinal in ${\mu}$
• ${\mu_0=\kappa}$
• ${\mathfrak{b}_i\subseteq (\mu_i, \mu_{i+1})\cap{\sf Reg}}$
• ${|\mathfrak{b}_i|\leq\kappa}$ (hence ${\mathfrak{b}_i}$ is progressive)
• ${{\rm pp}(\mu)<\max{\rm pcf}(\mathfrak{b}_i)}$ for each ${i<\kappa}$.

Now define

$\displaystyle \mathfrak{b}=\bigcup_{i<\kappa}\mathfrak{b}_i. \ \ \ \ \ (5)$

By construction, ${\mathfrak{b}}$ is a progressive set of regular cardinals cofinal in ${\mu}$ of cardinality ${{\rm cf}(\mu)}$, and so the definition of ${{\rm pp}(\mu)}$ tells us that if ${D}$ is an ultrafilter on ${\mathfrak{b}}$ disjoint to the bounded ideal, then

$\displaystyle {\rm cf}\left(\prod\mathfrak{b}/D\right)\leq{\rm pp}(\mu). \ \ \ \ \ (6)$

Let ${\sigma={\rm pp}(\mu)^+}$, and consider the ideal ${J}$ on ${\mathfrak{b}}$ generated by ${J_{<\sigma}[\mathfrak{b}]}$ and the bounded subsets of ${\mathfrak{b}}$. It is easy to show that ${J}$ must be a proper ideal on ${\mathfrak{b}}$: if not, then there would be a ${\nu<\mu}$ such that

$\displaystyle \mathfrak{b}\cap (\nu,\mu)\in J_{<\sigma}[\mathfrak{a}], \ \ \ \ \ (7)$

but this is impossible as

$\displaystyle {\rm pp}(\mu)<{\rm pp}(\mu)^+=\sigma\leq\max{\rm pcf}(\mathfrak{b}\cap(\nu,\mu)) \ \ \ \ \ (8)$

by our construction of ${\mathfrak{b}}$. If ${D}$ is an ultrafilter on ${\mathfrak{b}}$ disjoint to ${J}$, then

$\displaystyle {\rm pp}(\mu)<\sigma\leq{\rm cf}\left(\prod\mathfrak{b}/D\right), \ \ \ \ \ (9)$

and we have a contradiction of (6). $\Box$

## Conclusion 3.2A (part 2)

Recall from the last post that I’ve committed myself to providing a proof of the following:

Theorem 1 (Conclusion 3.2A (i) page 378, reformulated) Suppose ${\mu}$ is a singular cardinal of cofinality ${\kappa}$ with ${\mu<\aleph_\mu}$. Then for some ${\tau<\mu}$,

$\displaystyle {\rm pp}(\mu)=^+{\rm cf}_{\kappa}\left(\prod(\tau,\mu)\cap{\sf Reg}\right). \ \ \ \ \ (1)$

We’ve already discussed what the notation means, so I’m going to plunge right in. I’ll probably split this up into a few posts; the result is true, but the proof given in The Book has the unfortunate property that at a key moment, the reader is given an incorrect reference as explanation, and it might take me a post or two to fill in the necessary details.

Proof: We’ll start with the easy stuff and prove first the inequality

$\displaystyle {\rm pp}(\mu)\leq {\rm cf}_{\kappa}\left(\prod(\tau,\mu)\cap{\sf Reg}\right) \ \ \ \ \ (2)$

whenever ${\tau<\mu}$ is such that ${(\tau,\mu)\cap{\sf Reg}}$ is progressive.

Let ${\tau}$ be given (such ${\tau}$ exist because ${\mu<\aleph_\mu}$ — see posts of June 15), and define

$\displaystyle \mathfrak{a}:=(\tau,\mu)\cap{\sf Reg}. \ \ \ \ \ (3)$

By Claim 3.2 (June 15, 2011), we know

$\displaystyle {\rm cf}_{\kappa}\left(\prod(\tau,\mu)\cap{\sf Reg}\right)={\rm cf}^{\aleph_0}_{<\kappa^+}(\prod\mathfrak{a})=\sup{\rm pcf}_{\Gamma(\kappa^+,\aleph_0)}(\mathfrak{a}). \ \ \ \ \ (4)$

So let us assume that ${\mathfrak{b}}$ is a progressive set of regular cardinals cofinal in ${\mu}$, ${J}$ is an ideal on ${\mathfrak{b}}$ containing the bounded subsets of ${\mathfrak{b}}$, and

$\displaystyle \nu = {\rm tcf}\left(\prod\mathfrak{b}, <_J\right). \ \ \ \ \ (5)$

To establish our inequality, we must show

$\displaystyle \nu\in{\rm pcf}_{\Gamma(\kappa^+,\aleph_0)}(\mathfrak{a}). \ \ \ \ \ (6)$

This is actually quite easy. Let us define an ideal ${I}$ on ${\mathfrak{a}}$ by

$\displaystyle A\in I\Longleftrightarrow A\cap \mathfrak{b}\in J. \ \ \ \ \ (7)$

Set ${\mathfrak{c}:=\mathfrak{b}\cap (\tau,\mu)}$, and note

• ${|\mathfrak{c}|\leq\kappa<\kappa^+}$ (as ${\mathfrak{c}\subseteq\mathfrak{b}}$), and
• ${\mathfrak{a}\setminus\mathfrak{c}\in I}$ (as ${J}$ contains the set ${\mathfrak{b}\cap\tau^+}$).

Thus, to finish this part of the proof we need only check

$\displaystyle \nu = {\rm tcf}\left(\prod\mathfrak{a}, <_I\right), \ \ \ \ \ (8)$

but this follows easily from (5).

I’ll end this post here, and pick up with establishing the other inequality (and the “${+}$” part) tomorrow.

$\Box$

## Conclusion 3.2A (part 1)

In this post, we want to look at the next result in The Book: Conclusion 3.2A on page 378. What I will do is state this exactly as it appears in the book, and then reformulate it in conjunction with providing some commentary. I’ll attempt to prove it in a future post.

Theorem 1 (Conclusion 3.2A page 378) If ${\lambda=\aleph_{\xi(*)+\zeta(*)}}$, ${\zeta(*)<\aleph_{\xi(*)}}$ a limit ordinal, then:

1. for ${i_0<\zeta(*)}$ large enough,

$\displaystyle {\rm pp}(\lambda)=^+{\rm cf}_{{\rm cf}[\zeta(*)]}\left(\prod\{\aleph_{\xi(*)+i+1}: i_0

2. if ${{\rm cf}[\zeta(*)]\leq\kappa\leq |\zeta(*)|^+}$ then for ${i_0<\zeta(*)}$ large enough

$\displaystyle {\rm pp}_\kappa(\lambda)=^+{\rm cf}_\kappa\left(\{\aleph_{\xi(*)+i+1}:i_0

I’m going to concentrate on the first part of the above; here’s my attempt at reformulating it.

Theorem 2 (Conclusion 3.2A (i), reformulated) Suppose ${\mu}$ is a singular cardinal of cofinality ${\kappa}$ with ${\mu<\aleph_\mu}$. Then for some ${\tau<\mu}$,

$\displaystyle {\rm pp}(\mu)=^+{\rm cf}_{\kappa}\left(\prod(\tau,\mu)\cap{\sf Reg}\right). \ \ \ \ \ (3)$

The right-hand side of the above is not something we’ve explicitly defined yet, but it’s a reasonable variant of the ${{\rm cf}^\sigma_{<\theta}(\prod\mathfrak{a})}$ notation we’ve been working with. The notation

$\displaystyle {\rm cf}_\kappa\left(\prod\mathfrak{a}\right) \ \ \ \ \ (4)$

refers to the minimum cardinality of a family ${F\subseteq\prod\mathfrak{a}}$ with the property that for any ${\mathfrak{b}\subseteq\mathfrak{a}}$ of cardinality ${\kappa}$, the collection

$\displaystyle \{f\upharpoonright\mathfrak{b}:f\in F\} \ \ \ \ \ (5)$

is cofinal in ${(\prod\mathfrak{b}, <)}$.

This definition is robust under small modifications: it is also the minimum cardinality of a family ${F\subseteq\prod\mathfrak{a}}$ with the property that for any ${\mathfrak{b}\subseteq\mathfrak{a}}$ of cardinality at most ${\kappa}$ and any ${g\in\prod\mathfrak{a}}$, there is a finite set ${F_0\subseteq F}$ such that

$\displaystyle g(\eta)<\max\{f(\eta):f\in F_0\}\text{ for all }\eta\in\mathfrak{b}. \ \ \ \ \ (6)$

In other words,

$\displaystyle {\rm cf}_\kappa\left(\prod\mathfrak{a}\right)={\rm cf}_{<\kappa^+}^{\aleph_0}\left(\prod\mathfrak{a}\right), \ \ \ \ \ (7)$

and we are back with the sorts of cardinals we’ve been considering of late.

There is something else implicit in this theorem that I want to bring out: it shows us that if ${\mu}$ is a singular cardinal that isn’t a fixed point, then ${{\rm pp}(\mu)}$ can be calculated just by looking at structure

$\displaystyle \left(\prod(\mu\cap{\sf Reg}), <\right). \ \ \ \ \ (8)$

In this situation ${{\rm pp}(\mu)}$ is the minimum cardinality of a family of functions in ${\prod(\mu\cap{\sf Reg})}$ with the property that for some ${\tau<\mu}$, for any ${\mathfrak{b}\subseteq(\tau,\mu)\cap{\sf Reg}}$ of cardinality ${{\rm cf}(\mu)}$ the family

$\displaystyle \{f\upharpoonright\mathfrak{b}:f\in F\} \ \ \ \ \ (9)$

is cofinal in ${\prod\mathfrak{b}}$. Roughly speaking, if ${\mu}$ is singular and not a fixed point, then ${{\rm pp}(\mu)}$ is the minimum cardinality of a family of functions in ${\prod\mu\cap{\sf Reg}}$ which can cover all products of size ${{\rm cf(\mu)}}$ drawn from some fixed tail of ${\mu\cap{\sf Reg}}$. That seems like it is worth knowing…

## A non fixed point by any other name…

This is a silly post, but looking ahead to Conclusion 3.2A reminded me of one of the frustrations I had when first trying to read The Book. One of the major themes of pcf theory is that it tells us quite a lot about cardinal arithmetic at singular cardinals ${\mu}$ that are not fixed points of the ${\aleph}$ function. The problem is that this is equivalent to lots of other conditions, and very often Shelah will use a different formulation of this property in the statement of a theorem, so it isn’t clear right away that what he’s really talking about is a cardinal that isn’t a fixed point. This little proposition is just an elementary exercise in set theory, but it’s useful to keep in mind when reading The Book. (And let’s hope I haven’t flubbed it since I’m typing from home while babysitting…)

Proposition 1 Let ${\mu}$ be a singular cardinal. Then the following are equivalent:

1. ${\mu}$ is not a fixed point, i.e., ${\mu<\aleph_\mu}$.
2. ${\mu=\aleph_{\xi+\zeta}}$ where ${\zeta}$ is a limit ordinal ${<\aleph_\xi}$.
3. There is a ${\tau<\mu}$ such that ${(\tau,\mu)\cap{\sf Reg}}$ is progressive.

Proof: Suppose ${\mu<\aleph_\mu}$, and fix ${\delta<\mu}$ such that ${\mu=\aleph_\delta}$. Since ${\mu}$ is a limit cardinal, we know ${|\delta|^+<\mu}$, so choose ${\xi<\delta}$ such that

$\displaystyle |\delta|^+=\aleph_\xi. \ \ \ \ \ (4)$

Now fix ${\zeta}$ such that

$\displaystyle \delta=\xi+\zeta. \ \ \ \ \ (5)$

It is clear that ${\zeta}$ must be a limit (as ${\delta}$ is), and the fact that

$\displaystyle \mu=\aleph_\delta=\aleph_{\xi+\zeta} \ \ \ \ \ (6)$

is completely trivial. For the last bit, we know

$\displaystyle |\zeta|\leq |\delta|<|\delta|^+=\aleph_\xi, \ \ \ \ \ (7)$

and we are done with proving that (1) implies (2)

Now assume (2) holds, set ${\tau=\aleph_\xi}$, and define

$\displaystyle \mathfrak{a}=(\tau,\mu)\cap{\sf Reg}. \ \ \ \ \ (8)$

Now note

$\displaystyle |\mathfrak{a}|\leq |(\tau,\mu)|=|\zeta|<\tau<\tau^+=\min(\mathfrak{a}), \ \ \ \ \ (9)$

and we have (3).

Moving on to the last implication, let us suppose ${\tau<\mu}$ and ${\mathfrak{a}=(\tau,\mu)\cap{\sf Reg}}$ is progressive, i.e., ${|\mathfrak{a}|\leq\tau}$. But then ${|(\tau,\mu)\cap{\sf Card}|\leq\tau}$ as well, and hence

$\displaystyle |\mu\cap{\sf Card}|\leq\tau. \ \ \ \ \ (10)$

Thus, we have

$\displaystyle \mu\leq\aleph_{\tau^+}<\aleph_\mu, \ \ \ \ \ (11)$

as required.

$\Box$

## 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. $\Box$

## Some terminology

Having set the mood with our last post, we now return to “The Book” and examine “covering all small products”. I want to use this post to try out some terminology to see if it’s possible to make the material a little more palatable.

Definition 1 Let ${\mathfrak{a}}$ be a set of regular cardinals, let ${\sigma}$ be a regular cardinal, and let ${F\subseteq\prod\mathfrak{a}}$. Given a set ${\mathfrak{b}\subseteq\mathfrak{a}}$, we say that ${F}$ is ${<\sigma}$-cofinal in ${\prod\mathfrak{b}}$ if for any function ${f\in\prod\mathfrak{a}}$, there is a family ${F_0}$ of fewer than ${\sigma}$ elements of ${F}$ such that

$\displaystyle f(\eta)<\sup\{g(\eta):g\in F_0\}\text{ for all }\eta\in \mathfrak{b}. \ \ \ \ \ (1)$

In the last post, we characterized the minimum cardinality of a ${<\sigma}$-cofinal subset of ${\prod\mathfrak{a}}$ in the case where ${\mathfrak{a}}$ is progressive and ${\sigma<|\mathfrak{a}|}$. Today, I want to consider a related question:

What is the minimum cardinality of a family ${F\subseteq\prod\mathfrak{a}}$ such that ${F}$ is ${<\sigma}$-cofinal in ${\mathfrak{b}}$ for every ${\mathfrak{b}\subseteq\mathfrak{a}}$ of cardinality ${<\theta}$?

Let us agree to denote the above cardinal by ${{\rm cf}^{\sigma}_{<\theta}(\prod\mathfrak{a})}$ (again, don’t blame me for the notation!).

Ok, so why on earth would I want to consider such a cardinal? Well, we’re going to see that this is actually quite relevant for cardinal arithmetic — we will show that such cardinals are intimately tied up with Shelah’s covering numbers. Our first task, though, will be to characterize such cardinals in terms of pcf theory. The characterization will require a little terminology.

Definition 2 Suppose ${\mathfrak{a}}$ is a progressive set of regular cardinals, and let ${\sigma}$ and ${\theta}$ be cardinals such that

$\displaystyle \aleph_0\leq{\rm cf}(\sigma)=\sigma<\theta\leq|\mathfrak{a}|. \ \ \ \ \ (2)$

We define the set ${{\rm pcf}_{\Gamma(\theta,\sigma)}(\mathfrak{a})}$ to consist of all (regular) cardinals ${\kappa}$ for which there is an ideal ${J}$ on ${\mathfrak{a}}$ such that

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

Notice that the above is very close to the definition of ${{\rm pcf}_{\sigma{\rm-complete}}(\mathfrak{a})}$; the difference is that the last clause further restricts the acceptable ideals to those which also “reduce” ${\prod\mathfrak{a}}$ to something of the form ${\prod\mathfrak{b}}$ where ${|\mathfrak{b}|<\theta}$.

We’ll end this post with the statement of a theorem; we’ll prove the theorem in the next post.

Theorem 3 Suppose ${\mathfrak{a}}$ is a progressive set of regular cardinals, and

$\displaystyle \aleph_0\leq{\rm cf}(\sigma)=\sigma<\theta\leq|\mathfrak{a}|. \ \ \ \ \ (3)$

Then

$\displaystyle {\rm cf}^\sigma_{<\theta}(\prod\mathfrak{a})=\sup{\rm pcf}_{\Gamma(\theta,\sigma)}(\mathfrak{a}). \ \ \ \ \ (4)$

## 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.

$\Box$

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