# Notes on Sh506 (part I)

In pcf theory, we look at reduced products $\prod A/I$ where $A$ is a set of regular cardinals with singular limit $\mu$ satisfying $min(A)>|A|$ (such sets are referred to as progressive in the literature). Of course, if $\mu$ is an $\aleph$-fixed point, we cannot ask that $A$ is a progressive interval of regular cardinals, which is also a required hypothesis of many theorems of pcf theory. Part of the point of Sh 506 is to allow us to do pcf theory at $\aleph$-fixed points, which interests me because this may be applicable to the problem of finding a Jónsson successor of a singular. So, I’m going to work through that paper using this blog. We first need a few definitions.

If $I$ is an ideal over a set $A$ and $\theta$ is a cardinal, then we say that $I$ is $\theta$-weakly saturated if $X$ cannot be partitioned into $\theta$-many $I$ positive sets. We say that $wsat(I)$ is the smallest cardinal $\theta$ for which $I$ is $\theta$-weakly saturated. Throughout, $\mu$ will denote a fixed singular cardinal with $A=\langle \mu_i : i<\mu\rangle$ a sequence of cardinals with $sup A=\mu$, and $I$ an ideal over $\mu$.

Definition:
$pcf_I(A)=\{\lambda=cf(\prod A/D): D$ is an ultrafilter disjoint from $I\}$.

$J_{<\lambda}^I[A]=\{B\subseteq \mu :$ if $D$ is an ultrafilter over $\mu$ extending $I$ with $B\in D$, then $cf(\prod A/D)<\lambda\}\cup I$.

For the sake of simplicity, we will just use $J_{<\lambda}$ to refer to $J_{<\lambda}^I[A]$ when $I$ and $A$ are clear from the context. One of the main results of pcf theory is the existence of generators for these pcf ideals provided that $A$ is a progressive set of regular cardinals, and instrumental in that is the fact that $\prod A/J_{<\lambda}$ is $\lambda$-directed. It turns out that we can weaken the assumption that $A$ is progressive.

Theorem(Lemma 1.5 of Sh 506): Assume that $A$ is a sequence of regular cardinals with the property that $min(A)>wsat(I)$, then if $\lambda\geq wsat (I)$ is a cardinal with $J_{<\lambda}^I[A]$ proper, then $\prod A/J_{<\lambda}$ is $\lambda$-directed.

Proof: We will inductively show that, by induction on $\lambda_0 <\lambda$ that $\prod A/J_{<\lambda}$ is $\lambda_0^+$ directed. If $|F|\leq wsat(I), then we let $g$ be defined by $g(i)=sup\{f(i) : f\in F\}$. Then since each $\mu_i$ is regular, it follows that $g\in \prod A$ and $f\leq g$ everywhere.

By way of induction, assume we have shown, for $wsat(I)<\lambda_0<\lambda$ a cardinal, that $\prod A/J_{<\lambda}$ is $\lambda_0$-directed, and let $F\subseteq \prod A$ of size $\lambda_0$ be given. We first assume that $\lambda_0$ is singular. In this case, we can write $F=\bigcup_{\alpha such that $|F_\alpha|<\lambda_0$. Then by assumption, we can bound each $F_\alpha$ by some $g_\alpha$, and then bound the set $\{ g_\alpha : \alpha by some $g\in \prod A$. We then have that $f\leq g$ modulo $J_{<\lambda}$ for each $f\in F$.

So assume that $\lambda_0$ is regular. We begin by replacing $F=\{h_i : i<\lambda_0\}$ with a $\leq_{J_{<\lambda}}$-increasing sequence $\vec{f}=\langle f_i : i<\lambda_0\rangle$. We just let $f_i$ be a $\leq_{J_{<\lambda}}$-upper bound for $\{h_j : j\leq i\}\cup \{f_j : j. By construction, if we can find a $g\in \prod A$ such that $f_i \leq g$ modulo $J_{<\lambda}$ for each $i<\lambda_0$, then we will be done. At this point, we will proceed by induction on $\alpha and attempt to construct a $\leq_{J_{<\lambda}}$-increasing sequence of candidates for bounds of $\vec{f}$. As usual, we will show that this construction must terminate at some point, or we will be able to generate a contradiction.

By induction on $\alpha , we will define functions $g_\alpha$, ordinals $\xi(\alpha)$, and sequences $\langle B^\alpha_\xi : \xi <\lambda_0\rangle$ with the following properties:

1. $g_\alpha\in \prod A$ and for all $\beta<\alpha$, we have that $g_\beta\leq g_\alpha$;
2. $B_\xi^\alpha:=\{i<\mu : f_\xi(i)>g_\alpha (i)\}$;
3. For each $\alpha , and every $\xi\in[\xi(\alpha+1),\lambda_0)$, we have that $B^\alpha_\xi\neq B^{\alpha+1}_\xi$ modulo $J_{<\lambda}$.

The construction proceeds as follows. At stage $\alpha=0$, we simply let $g_0=f_0$, and set $\xi(\alpha)=0$ (note that $\xi(\alpha)$ only matters when $\alpha=\beta+1$ for some ordinal $\beta). At limit stages, assume that $g_\beta$ has been defined for each $\beta<\alpha$, and define $g_\alpha$ by setting $g_\alpha(i)=sup_{\beta<\alpha} f_\beta(i)$. Note since $\alpha and each $\mu_i$ is regular, that $g_\alpha\in\prod A$.

At successor stages, let $\alpha=\beta+1$, and suppose that $g_\beta$ has been defined. If $g_\beta$ is a $\leq_{J_{<\lambda}}$-upper bound for $\vec{f}$, then we’re done and we can terminate the induction. Otherwise, note that the sequence $\langle B^\beta_\xi : \xi <\lambda_0\rangle$ is $\subseteq_{J_{<\lambda}}$-increasing and so there is a minimum $\xi(\alpha)$ for which every $\xi\in[\xi(\alpha),\lambda_0)$ has the property that $B_\xi^\beta\notin J_{<\lambda}$ (else if there is no such $\xi(\alpha)$, then $g_\beta$ was indeed the desired bound). By definition, that means we can find some ultrafilter $D$, disjoint from $J_{<\lambda}$ such that $B_{\xi(\alpha)}^\beta\in D$ and $cf(\prod A/D)\geq\lambda$. Thus is follows that $\vec{f}$ must have a $<_D$-upper bound in $\prod A$, say $h_\alpha$. We then define $g_\alpha\in \prod A$ by $g_\alpha (i)=max\{g_\beta(i), h_\alpha(i)\}$.

Note that for each $\xi\in[\xi(\beta+1),\lambda_0)$, we have that $B_\xi^\beta\in D$. On the other hand, our definition of $g_\alpha$ gives us that $B_\xi^{\beta+1}\notin D$ since $g_\alpha$ is at least $h_\alpha$ everywhere. Thus, condition 3 is satisfied, as are 1 and 2 trivially by construction.

We claim that this process must have terminated at some stage. Otherwise, we let $\xi(*)=\sup\{\xi(\alpha) : \alpha, and note that each $B^\alpha_{\xi(*)}\notin J_{<\lambda}$ since the induction never terminated. Next, we note that condition 1) gives us that for $\alpha\leq \beta$, we have $B_{\xi(*)}^\beta\subseteq B_{\xi(*)}^\alpha$ and so $B_{\xi(*)}^\alpha\setminus B_{\xi(*)}^{\alpha+1}\notin J_{<\lambda}$. Therefore, for $\alpha<\beta$, we have that $B_{\xi(*)}^\beta\subseteq B_{\xi(*)}^{\alpha+1}$ and so the sets $B_{\xi(*)}^\alpha\setminus B_{\xi(*)}^{\alpha+1}$ and $B_{\xi(*)}^\beta\setminus B_{\xi(*)}^{\beta+1}$ are disjoint $I$-positive sets (since $J_{<\lambda}$ extends $I$). But then we have a partition $\{B_{\xi(*)}^\alpha\setminus B_{\xi(*)}^{\alpha+1} : \alpha < wsat(I)\}$ of $\mu$ into $wsat(I)$-many disjoint $I$-positive sets, which is a contradiction. Therefore the process terminated at some point and $\vec{f}$ (hence $F$) has a $\leq_{J_{<\lambda}}$-upper bound. This completes the induction and the proof.

A couple of observations:

Since $\lambda$-directedness seems to be the key ingredient in obtaining universal cofinal sequences, I suspect that we can recover such sequences for each $\lambda$ in $pcf_I(A)$. Further, the proof has a lot of the flavor of Shelah’s trichotomy theorem, especially the bit where we used witnessing ultrafilters to generate upper bounds for $\vec{f}$. So, I again suspect that one can prove a variant of the Trichotomy theorem for these pseudo-progressive sets of regular cardinals when we have a sufficiently nice ideal laying about. I unfortunately haven’t had time to work through the details and write this stuff down, but hopefully I’ll get around to it in the near future.

Also, it’s relatively easy to see that we get the usual characterization that for an ultrafilter $D$ over $A$ disjoint from $I$, we have $cf(\prod A/D)\geq \lambda$ if and only if $J_{<\lambda}^I[A]\cap D=\emptyset$. From there, we also get a natural correspondence between $\lambda\in pcf_I(A)$ and the existence of a set $X_\lambda\in J_{<\lambda^+}\setminus J_{<\lambda}$. We can also use this to show that $pcf_I(A)$ has a maximum element. So, we recover a lot of nice information about the structure of $pcf_I(A)$ from just this lemma. On the other hand, we don’t necessarily get that $pcf_I(A)$ must be an interval of regular cardinals since that requires what Abraham and Magidor call the “no holes argument”. The only proof of that theorem that I know heavily utilizes the theory of exact upper bounds, which would need the trichotomy theorem (or something equivalent to it).