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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.

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up vote 3 down vote accepted

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.



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.

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The proof I know is different as it uses smaller elementary substructures: |Nα|<κ for α<κ where κ is a regular cardinal with |A|<κ < min(A).

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Thanks Péter. Is it easier to use smaller elementary substructure? Also the proof with elementary substructure is susceptible to have a lot of valuable information since we're not using generators, it would be nice to compare this "new" information. – Carlo Von Schnitzel Jun 30 '10 at 17:52

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