## Main Proof (part 2)

Proposition 1 Suppose ${\langle\alpha_n:n<\omega\rangle}$ is increasing in ${{\rm nacc}(C_\delta)}$ and ${\langle i_n:n<\omega\rangle}$ is increasing in ${\kappa}$. Then

$\displaystyle \bigcup_{n<\omega}\mathcal{N}(\alpha_n,i_n)\in M^*. \ \ \ \ \ (1)$

Proof: Define

• ${\alpha^*:=\sup\{\alpha_n:n<\omega\}}$,
• ${C^*=\bigcup_{n<\omega}C_{\alpha_n}}$, and
• ${i^*:=\sup\{i_n:n<\omega\}}$.

Here is where the choice of ${\langle C_\alpha:\alpha<\mu^+\rangle}$ comes to our rescue. Since ${\alpha_n\in{\rm nacc}(C_\delta)}$ for each ${n}$, it follows that

$\displaystyle C_{\alpha_n}=C_\delta\cap\alpha_n \ \ \ \ \ (2)$

for each ${n}$, and therefore

$\displaystyle C_{\alpha_n} = C_{\alpha_{n+1}}\cap\alpha_n\subseteq C_{\alpha_{n+1}}, \ \ \ \ \ (3)$

and

$\displaystyle C^*=C_\delta\cap\alpha^*. \ \ \ \ \ (4)$

Now it follows easily that

$\displaystyle \bigcup_{n<\omega}\mathcal{N}(\alpha_n, i_n)={\rm Sk}_{\mathfrak{A}}(\mu_{i^*}\cup\{f_\xi:\xi\in C^*\}). \ \ \ \ \ (5)$

Next, we define

$\displaystyle \epsilon:=\min({\rm nacc}(C_\delta)\setminus\alpha^*). \ \ \ \ \ (6)$

Given our definition of ${\langle M_\alpha:\alpha<\mu^+\rangle}$, it is not hard to see

• ${\epsilon\cup\mu\subseteq M_\epsilon}$, and
• every proper initial segment of ${\langle f_\xi:\xi<\epsilon\rangle}$ is in ${M_\epsilon}$.

In particular,

$\displaystyle \mu_{i^*}\cup\{f_\xi:\xi\in C^*\}\subseteq M_\epsilon. \ \ \ \ \ (7)$

Note that ${C^*\in M_{\epsilon+1}}$, as it is definable from ${\bar{C}}$, ${\epsilon}$, and ${\alpha^*}$:

$\displaystyle C^*= C_\epsilon\setminus \{\alpha^*\}, \ \ \ \ \ (8)$

and so

$\displaystyle \bigcup_{n<\omega}\mathcal{N}(\alpha_n, i_n)\in M_{\epsilon+1}\subseteq M^*, \ \ \ \ \ (9)$

as it can be computed in ${M_{\epsilon+1}}$ by taking the Skolem hull of ${\mu_{i^*}\cup\{f_\xi:\xi\in C^*\}}$ in the model ${M_\epsilon}$.

$\Box$

## The main proof, part 1.

General results on the approachability ideal ${I[\mu^+]}$ tell us there is a sequence

$\displaystyle \bar{C}=\langle C_\alpha:\alpha<\mu^+\rangle \ \ \ \ \ (1)$

such that

• ${C_\alpha}$ is a closed (not necessarily unbounded!) subset of ${\alpha}$,
• ${C_\alpha}$ has order-type at most ${\kappa^+}$, where ${\kappa={\rm cf}(\mu)}$,
• ${\beta\in{\rm nacc}(C_\alpha)\Longrightarrow C_\beta= C_\alpha\cap\beta}$, and
• the set of ${\delta<\mu^+}$ of cofinality ${\kappa^+}$ with ${\delta=\sup(C_\delta)}$ is stationary.

The above is a result from [Sh:420]; it says that there is a stationary subset of ${S^{\mu^+}_{\kappa^+}}$ in the ideal ${I[\mu^+]}$. The details of this are worked out in my Handbook article — see Theorem 3.18 and Theorem 3.7.

\bigskip

Without loss of generality, the sequence ${\bar{C}}$ lies in the model ${M_0}$, hence in each ${M_\alpha}$ and ${M^*}$ as well. Let ${\langle \mu_i:i<\kappa\rangle}$ be the ${<_\chi}$-least continuous and increasing sequence of cardinals cofinal in ${\mu}$ satisfying ${\kappa<\mu_0}$; note this means that the sequence is going to be in every elementary submodel of the structure ${\mathfrak{A}}$.

Moving back to the main proof, let us fix ${a\in [\mu^+]^{{\rm cf }\mu}}$; recall that we want to show ${a}$ is a subset of some element ${M^*\cap [\mu]^{<\mu}}$, and the preceding post gives us our strategy.

For each ${\alpha<\mu^+}$ and ${i<\kappa}$, we define

$\displaystyle \mathcal{N}(\alpha,i)={\rm Sk}_{\mathfrak{A}}(\mu_i\cup \{f_\xi:\xi\in C_\alpha\}), \ \ \ \ \ (2)$

and

$\displaystyle \mathcal{N}^+(\alpha, i)={\rm Sk}_{\mathfrak{A}}(\mu_i\cup \{f_\xi:\xi\in C_\alpha\}\cup a). \ \ \ \ \ (3)$

We note the following:

• ${\mathcal{N}(\alpha,i)\subseteq\mathcal{N}^+(\alpha, i)\subseteq M_{\alpha+1}\subseteq M^*}$,
• ${\mathcal{N}(\alpha, i)\in M_{\alpha+1}}$, and
• both ${\mathcal{N}(\alpha,i)}$ and ${\mathcal{N}^+(\alpha,i)}$ have cardinality ${\mu_i}$.

Next, we define

$\displaystyle \mathfrak{a}_{\alpha, i}=\mathcal{N}(\alpha,i)\cap\mu\cap{\sf Reg}\setminus{\mu_i^{++}}, \ \ \ \ \ (4)$

and let ${g_{\alpha, i}}$ be the characteristic function of ${\mathcal{N}^+(\alpha, i)}$ (note the “+”!) in ${\mathfrak{a}_{\alpha, i}}$, i.e., the domain of ${g_{\alpha, i}}$ is ${\mathfrak{a}_{\alpha, i}}$, and for ${\theta\in\mathfrak{a}_{\alpha, i}}$, we have

$\displaystyle g_{\alpha, i}(\theta)=\sup(\mathcal{N}^+(\alpha,i)\cap\theta)<\theta. \ \ \ \ \ (5)$

The function ${g_{\alpha, i}}$ lies in ${\prod\mathfrak{a}_{\alpha, i}}$, but there is no reason to believe it is an element of ${M^*}$. Note, however, that ${\mathfrak{a}_{\alpha, i}}$ IS an element of ${M^*}$. It is also a progressive set of regular cardinals cofinal in ${\mu}$, and therefore by earlier work, we know

$\displaystyle \langle f_\beta\upharpoonright\mathfrak{a}_{\alpha, i}:\beta<\mu^+\rangle \ \ \ \ \ (6)$

is a scale in ${\mathfrak{a}_{\alpha, i}}$ modulo the ideal of bounded subsets of ${\mathfrak{a}_{\alpha, i}}$.

Since ${\prod(\mu\cap{\sf Reg})}$ is ${\mu^+}$-directed modulo the ideal of bounded subsets of ${\mu}$, it follows that for each ${\alpha<\mu^+}$, there exists a ${\gamma(\alpha)<\mu^+}$ such that

$\displaystyle (\forall i<\kappa)\left(g_{\alpha, i}< f_{\gamma(\alpha)}\upharpoonright\mathfrak{a}_{\alpha, i}\mod J^{\rm bd}[\mathfrak{a}_{\alpha, i}]\right) \ \ \ \ \ (7)$

The function mapping ${\alpha}$ to ${\gamma(\alpha)}$ need not be in ${M^*}$. Nevertheless, the set of ${\delta<\mu^+}$ is closed unbounded in ${\mu^+}$, and so we can find a ${\delta}$ of cofinality ${\kappa^+}$ such that

• ${\delta}$ is closed under the map ${\alpha\mapsto\gamma(\alpha)}$, and
• ${C_\delta}$ is closed unbounded in ${\delta}$ (of order-type ${\kappa^+}$).

We’ll end this post on this cliff-hanger…

## The Elementary Submodel argument

In this post, we examine another ingredient we’ll need in the course of the proof: the relationship between elementary submodels of ${\mathfrak{A}}$ and their “characteristic functions”. There are many variants of this lemma scattered throughout the book and allied literature, but we’ll just worry about what is needed for our current project. Remember that our conventions regarding elementary submodels were laid out in an earlier post.

Lemma 1 Let ${N\subseteq M}$ be elementary submodels of ${\mathfrak{A}}$, and assume

• ${N\cap\mu}$ is unbounded in ${\mu}$, and
• for any ${\theta\in N\cap\mu\cap{\sf Reg}}$ we have ${\sup(M\cap\theta)=\sup(N\cap\theta)}$.

Then ${M\cap\mu= N\cap\mu}$.

Proof: Suppose not. Define

$\displaystyle \alpha=\min(M\cap\mu\setminus N), \ \ \ \ \ (1)$

and

$\displaystyle \beta=\min(N\cap\mu\setminus\alpha). \ \ \ \ \ (2)$

Notice that ${\beta}$ exists by our assumptions, which also imply that ${\beta}$ cannot be a regular cardinal, as ${\beta\in N\cap\mu}$ and

$\displaystyle \sup(N\cap\beta)\leq\alpha<\alpha+1\leq\sup(M\cap\beta). \ \ \ \ \ (3)$

The proof will be complete if we manage to prove to the contrary that ${\beta}$ MUST be a regular cardinal.

Clearly ${\beta}$ must be a limit ordinal because ${N}$ is closed under the predecessor function. If ${\beta}$ is not a regular cardinal, then ${\gamma={\rm cf}(\beta)}$ is a regular cardinal in ${N}$ (and ${M}$), and therefore ${N}$ contains a function ${f}$ mapping ${\gamma}$ onto a cofinal subset of ${\beta}$.

Since ${\gamma\in N}$, it follows that ${\gamma<\alpha}$ and therefore ${M\cap\gamma=N\cap\gamma}$. In the model ${M}$, there must be an ${\epsilon<\gamma}$ with ${\alpha. But since ${\epsilon}$ must be in ${N}$ as well, it follows that

$\displaystyle \alpha

contradicting the definition of ${\beta}$. $\Box$

How is this going to be used? Let us agree for a moment to call models ${M}$ and ${N}$ as above “${\mu}$-twins”. In the context of the main project we have an elementary submodel ${M^*}$ obtained as the union of a ${\mu^+}$-approximating sequence ${\langle M_\alpha:\alpha<\mu^+\rangle}$. The goal is to show that any ${a\in [\mu]^{{\rm cf}(\mu)}}$ is a subset of some element of ${M^*\cap [\mu]^{<\mu}}$. We do this by producing an elementary submodel ${N}$ satisfying

• ${|N|<\mu}$,
• ${N\in M^*}$ (so ${N\cap\mu\in M^*\cap[\mu]^{<\mu}}$), and
• ${N}$ and ${M:={\rm Sk}_\mathfrak{A}(N\cup a)}$ are ${\mu}$-twins.

The conclusion of the lemma tells us that

$\displaystyle a\subseteq M\cap\mu = N\cap \mu\in M^*\cap [\mu]^{<\mu} \ \ \ \ \ (5)$

which is exactly what we need.

Returning now to our main project, we extract a lemma from Shelah’s argument and examine it in isolation.

Let ${\mathfrak{M}=\langle M_\alpha:\alpha<\mu^+\rangle}$ and ${M^*}$ be as assumed, and consider the structure

$\displaystyle \left( \prod (\mu\cap{\sf Reg}), <_J\right), \ \ \ \ \ (1)$

where ${J}$ denotes the ideal of bounded subsets of ${\mu}$. It’s pretty easy to see that this structure is ${\mu^+}$-directed, i.e., that every collection of fewer than ${\mu^+}$ functions in ${\prod(\mu\cap{\sf Reg})}$ has a ${<_J}$-upper bound.

Thus, we can define a sequence of functions ${\langle f_\alpha:\alpha<\mu^+\rangle}$ in ${\prod(\mu\cap{\sf Reg})}$ by letting ${f_\alpha}$ be the ${<_\chi}$-least function in ${\prod(\mu\cap{\sf Reg})}$ serving as a ${<_J}$-upper bound for ${M_\alpha\cap\prod(\mu\cap{\sf Reg})}$. Using this definition, we achieve

$\displaystyle \langle f_\beta:\beta\leq\alpha\rangle\in M_{\alpha+1} \ \ \ \ \ (2)$

for each ${\alpha}$, and so

$\displaystyle \alpha<\beta<\mu^+\Longrightarrow f_\alpha<_J f_\beta \ \ \ \ \ (3)$

as well.

Lemma 1 Let ${\mathfrak{a}}$ be a set of regular cardinals satisfying

• ${\sup(\mathfrak{a})=\mu}$,
• ${\mathfrak{a}}$ is progressive, and
• ${\mathfrak{a}\in M^*}$.

Then ${\langle f_\alpha\upharpoonright\mathfrak{a}:\alpha<\mu^+\rangle}$ is a scale in ${\prod\mathfrak{a}}$ modulo the ideal ${J^{\rm bd}[\mathfrak{a}]}$ of bounded subsets of ${\mathfrak{a}}$.

Proof: Let ${\mathfrak{a}}$ satisfy the assumptions of the proposition. Since ${{\rm pp}(\mu)=\mu^+}$, we know that ${\prod\mathfrak{a}}$ carries a scale of length ${\mu^+}$ modulo the ideal ${J^{\rm bd}[\mathfrak{a}]}$. Since ${M^*}$ is an elementary model of ${\mathfrak{A}}$ and ${\mathfrak{a}\in M^*}$, it follows that there is such a scale living in ${M^*}$, say ${\bar{f}^{\mathfrak{a}}=\langle f^{\mathfrak{a}}_i:i<\mu^+\rangle}$. Note that each ${f^{\mathfrak{a}}_i}$ is in ${M^*}$ as ${\mu^+\subseteq M^*}$.

Now consider ${\langle f_\alpha\upharpoonright\mathfrak{a}:\alpha<\mu^+\rangle}$. Since ${\mathfrak{a}}$ is unbounded in ${\mu}$, it follows easily that this sequence is increasing modulo ${J^{\rm bd}[\mathfrak{a}]}$ in ${\prod\mathfrak{a}}$. Given ${i<\mu^+}$, there is an ${\alpha<\mu^+}$ with ${f^{\mathfrak{a}}_i\in M_\alpha}$. But clearly this implies

$\displaystyle f^{\mathfrak{a}}_i< f_\alpha\upharpoonright\mathfrak{a}\mod J^{\rm bd}[\mathfrak{a}]. \ \ \ \ \ (4)$

Thus, the sequence ${\langle f_\alpha\upharpoonright\mathfrak{a}:\alpha<\mu^+\rangle}$ is also cofinal in ${\prod\mathfrak{a}}$ modulo ${J^{\rm bd}[\mathfrak{a}]}$, and we are done. $\Box$

Notice that the above lemma gives us a way to smuggle information into ${M^*}$: given ${\mathfrak{a}}$ as in the proposition, it follows that ANY function ${g\in\prod\mathfrak{a}}$, whether in ${M^*}$ or not, is bounded modulo ${J^{\rm bd}[\mathfrak{a}]}$ by a function of the form ${f_\alpha\upharpoonright\mathfrak{a}}$ from ${M^*}$. Formulating this as a corollary, we obtain:

Corollary 2 Let ${\mathfrak{a}}$ be a set of regular cardinals as in the preceding lemma. If ${g\in\prod\mathfrak{a}}$, then there is an ${\alpha<\mu^+}$ such that

$\displaystyle g

## Continuation on pp

The main result of this post is natural, but unsurprising. Oddly enough, though, I can’t find this exact version published anywhere. I mentioned something in this direction in my Handbook of Set Theory argument; I wish I had thought about it a bit more back then, because the proof is certainly easy enough. Anyway, the result is the following:

Theorem 1 The following statements are equivalent for a singular cardinal ${\mu}$:

1. ${{\rm pp}(\mu)=\mu^+}$
2. If ${\mathfrak{a}\subseteq\mu}$ is a cofinal progressive set of regular cardinals, then ${\prod\mathfrak{a}}$ carries a scale of length ${\mu^+}$ modulo the ideal ${J^{\rm bd}[\mathfrak{a}]}$ of bounded subsets of ${\mathfrak{a}}$.

Proof:

We prove first that (2) implies (1), as this implication is almost immediate. Suppose ${\mathfrak{a}\subseteq\mu}$ is a progressive set of regular cardinals cofinal in ${\mu}$ and of cardinality ${{\rm cf}(\mu)}$. Our assumptions imply that there is a sequence ${\langle f_\alpha:\alpha<\mu^+\rangle}$ of functions from ${\prod\mathfrak{a}}$ such that

• ${\alpha<\beta\rightarrow f_\alpha, and
• if ${g\in\prod\mathfrak{a}}$, then there is an ${\alpha<\mu^+}$ such that ${g.

If ${D}$ is any ultrafilter on ${\mathfrak{a}}$ disjoint to ${J^{\rm bd}[\mathfrak{a}]}$, then the two above statements continue to hold modulo ${D}$, and hence ${\prod\mathfrak{a}/D}$ has cofinality at most ${\mu^+}$. But clearly this cofinality must be at least ${\mu^+}$ as ${D}$ extends the co-bounded filter on ${\mathfrak{a}}$. The statement ${{\rm pp}(\mu)=\mu^+}$ now follows from the definitions involved.

For the other direction, suppose ${{\rm pp}(\mu)=\mu^+}$, let ${\mathfrak{a}}$ satisfy the assumptions of statement (2), and let ${I}$ denote the ideal ${J^{\rm bd}[\mathfrak{a}]}$. It is clear that

$\displaystyle \prod\mathfrak{a}/I\text{ is }\mu^+\text{-directed} \ \ \ \ \ (1)$

(see Claim 1.3(4) on page 4 of Cardinal Arithmetic), that is, given a collection of functions in ${\prod\mathfrak{a}}$ of cardinality at most ${\mu}$, there is a single function in ${\prod\mathfrak{a}}$ above them all modulo the ideal ${I}$.

By Conclusion 3.2 on page 62 of Cardinal Arithmetic, ${{\rm cf}\left(\prod\mathfrak{a}/I\right)}$ is equal to the supremum of all cardinals of the form ${{\rm cf}\left(\prod\mathfrak{a}/D\right)}$, where ${D}$ is an ultrafilter on ${\mathfrak{a}}$ disjoint to ${I}$. This latter cardinal is at most ${{\rm pp}(\mu)=\mu^+}$, and so it follows that

$\displaystyle {\rm cf}\left(\prod\mathfrak{a}/I\right)=\mu^+. \ \ \ \ \ (2)$

Given (1) and (2), it is straightforward to construct the desired scale. $\Box$

## What is pp?

We may as well make a short post dealing with the definition of ${\rm{pp}(\mu)}$ for ${\mu}$ singular; this will also let me test the latex2wp setup at home.

Definition 1 If ${\mu}$ is a singular cardinal, then ${{\rm pp}(\mu)}$ is the supremum of all cardinals of the form

$\displaystyle {\rm cf}\left(\prod\mathfrak{a}, <_D\right) \ \ \ \ \ (1)$

where

• ${\mathfrak{a}}$ is a set of regular cardinals cofinal in ${\mu}$,
• ${|\mathfrak{a}|={\rm cf}(\mu)}$, and
• ${D}$ is an ultrafilter on ${\mathfrak{a}}$ extending the co-bounded filter.

This is not the “official” definition given as Definition 1.1 on page 41 of Cardinal Arithmetic, but it is equivalent. The official definition is designed with a nod towards generalization and refinement.

Note that the above definition is not the same as looking at

$\displaystyle \sup\{\theta:\theta\in{\rm pcf}\mathfrak{a}\text{ for some }\mathfrak{a}\text{ as above}\}. \ \ \ \ \ (2)$

The reason is that in the definition of ${{\rm pcf}(\mathfrak{a})}$ we do not require the ultrafilters under consideration to extend the co-bounded filter on ${\mathfrak{a}}$, and this can make a difference if ${{\rm otp}(\mathfrak{a})>\omega}$: you can potentially get a very large ${\theta}$ in ${{\rm pcf}(\mathfrak{a})}$ by virtue of an ultrafilter concentrating on an initial segment of ${\mathfrak{a}}$. In this situation, though, the fact that ${\theta}$ is in ${{\rm pcf}(\mathfrak{a})}$ isn’t really anything to do with ${\mu}$, and so the definition of ${{\rm pp}(\mu)}$ is designed to ignore this phenomenon.