Let $pcf(a)$ denote the set of regular cardinals such that $J\_{\leq \lambda} - J\_{<\lambda} \neq \emptyset$ and let $maxpcf(a)$ denote the maximum of $pcf(a)$. The $J\_{\leq \lambda}$ are the usual ideals built up inductively in which you throw in all set on which you can have a scale mod $J\_{<\lambda}$


There is a fact I want to prove: Let $a$ be a set of regular cardinals such that $a < min(a)$. Then there exists a family $\Im \subseteq \prod a$ with |$\Im$|=$maxpcf(a)$ such that $\forall g\in \prod a$ $\exists f\in \Im$ with $g < f$ (pointwisely). we say that such a family dominates.


I know the proof which uses generators of the $J\_{<\lambda}$ ideals, however, since proving that such generators exists is another thing itself, I would like to be able to prove the fact without using generators. I know there is a proof using Hull and Substructures arguments but it is the first time I am using these objects.


Can someone help me please finish the proof? Here is the beginning:


We do an induction on $maxpcf(a)$. When we have $b\subseteq a$ with $maxpcf(b) < maxpcf(a)$ assume for the induction that there exists a family $\Im\_b \subseteq \prod b$ with $g< f$. 


Let $\delta$ be strictly bigger than $maxpcf(a)$. We need to constuct a continuous chain of elementary substructures $M\_{\alpha}$ of $V\_\delta$, for $\alpha < |a|^+$, such that |$M\_\alpha$|$\leq maxpcf(a)$ and $\forall \alpha<|a|^+, <{M\_\beta :\beta < \alpha}> \in M\_{\alpha+1}$. If \alpha is a limit then we just take unions: $M\_{\alpha+1}= \bigcup\_{\beta < \alpha+1} M\_\beta$. 


If $M\_\alpha \bigcap \prod a$ is dominating, we are done. If not then $\exists g\_\alpha \in \prod a$ such that $\neg g\_\alpha < f, \forall f \in M\_\alpha \bigcap \prod a$. One can assume that for $\alpha'<\alpha$ then $g\_\alpha' < g\_\alpha$ everywhere. This entails that $g\_\alpha \in M\_{\alpha+1}$. 


Now I am stuck here, I want to arrive at a contradiction using my induction hypothesis.



Some Pcf Theory
Hi. I know this is over a year late, but I the proof you're looking at matches that given on page 61-62 of "Cardinal Arithmetic". The argument can be finished along the following lines:


(1) First, we may as well assume $|\mathfrak{a}|^+<\min(\mathfrak{a})$, as we can derive the result you want if we get it in this more restricted situation.


(2) You know that $\prod\mathfrak{a}$ has true cofinality $\lambda$ modulo the ideal $J\_{<\lambda}[\mathfrak{a}]$ by way of Lemma 1.8 on page 9 of Cardinal Arithmetic. Let $\langle f\_\alpha:\alpha<\lambda\rangle$ witness this. Without loss of generality, this sequence of functions is an element of $N\_0$, and each $f\_\alpha\in N\_0$ as well. (Here we are using $|N\_0|=\lambda$)


(3) Given one of your $g\_i$, we know there is an $\alpha<\lambda$ with $g\_i < f\_\alpha$ modulo $J\_{<\lambda}[\mathfrak{a}]$.  

By the usual sorts of argument, there is a single $\alpha<\lambda$ such that $g\_i<f\_\alpha$ modulo $J\_{<\lambda}[\mathfrak{a}]$ for all $i<|\mathfrak{a}|^+$.


(4) Now look at the collection $\langle\mathfrak{c}\_i:i<|\mathfrak{a}|^+\rangle$, where $\mathfrak{c}\_i$ is defined to be the set of $\theta\in\mathfrak{a}$ for which $f\_\alpha(\theta)\leq g\_i(\theta)$. You've arranged that $g\_i<g\_j$ (everywhere) whenever $i<j$, so the sequence $\langle \mathfrak{c}\_i:i<|\mathfrak{a}|^+\rangle$ is an increasing sequence of subsets of $\mathfrak{a}$. Such a sequence of length $|\mathfrak{a}|^+$ must be eventually constant, say with value $\mathfrak{c}$.


(5) This set $\mathfrak{c}$ has three important properties:


(5a) $\mathfrak{c}$ is in $J\_{<\lambda}[\mathfrak{a}]$ (as all $\mathfrak{c}\_i$ are in this ideal), 


(5b) $g\_i(\theta)<f\_\alpha(\theta)$ whenever $i(\*)\leq i<|\mathfrak{a}|^+$ and $\theta\in \mathfrak{a}\setminus\mathfrak{c}$, and 


(5c) $|\mathfrak{c}|\in N\_{i(\*)+1}$ as it is definable from $g\_{i(\*)}$ and $f\_\alpha$.


(6) we want to now apply our induction hypothesis to $\mathfrak{c}$ in the model 
$N\_{i(\*)+1}$. This model has cardinality greater than maxpcf $\mathfrak{c}$, and so $N\_{i(\*)+1}$ is going to contain every member of a dominating family for $\prod\mathfrak{c}$. 


(7) Consider now the function $g\_{i(\*)+1}$. It is not an element of $N\_{i(\*)+1}$; in fact, it is not even bounded by an element of $N\_{i(\*)+1}$. However, there is a function $g$ in $N\_{i(\*)+1}\cap\prod\mathfrak{c}$ dominating $g\_{i(\*)+1}$ on $\mathfrak{c}$. And the function $f\_\alpha$ dominates $g\_{i(\*)+1}$ on the set $\mathfrak{a}\setminus\mathfrak{c}$. By pasting $g$ and $f\_\alpha$ together, we get a function in $N\_{i(\*)+1}$ which dominates $g\_{i(\*)+1}$, and this is a contradiction.


Best,


Todd


PS: This is only the 2nd time I've attempted a post here; I'm not sure if the LaTeX looks right but I'll try to edit things so it looks OK.


PPS: Well, that took longer than I thought to edit, but maybe it looks OK now.


