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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 $J_{<\lambda}[\mathfrak{a}]$. <\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), and (5b) $g_i(\theta)<f_\alpha(\theta)$ whenever$i(*)\leq i<|\mathfrak{a}|^+$and$\theta\in \mathfrak{a}\setminus\mathfrak{c}$. mathfrak{a}\setminus\mathfrak{c}$, and

(5c) $|\mathfrak{c}|= |\mathfrak{c}|\in N_{i(*)+1}$ 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 $N_{i(*)+1$. 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}$. It is not an element of$M_{i(*)+1}$; N_{i(*)+1}$; in fact, it is not even bounded by an element of $M_{i(*)+1}$. N_{i(*)+1}$. However, there is a function$g$in$M_{i(*)+1}\cap\prod\mathfrak{c}$N_{i(*)+1}\cap\prod\mathfrak{c}$ dominating $g_{i(*)}+1$ 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$together, we get a function in$M_{i(*)+1}$N_{i(*)+1}$ which dominates $g_{i(*)+1}$, and this is a contradiction.

Now I'm sure there's an easier proof of this not involving generators; I'll see if I can work it out for you later.

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.

2 deleted 7 characters in body; added 2 characters in body; added 7 characters in body; added 9 characters in body

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 (4) Now look at the collection $\langle \mathfrak{c}_i:i<|\mathfrak{a}|^+\rangle$, \langle\mathfrak{c}_i:i<|\mathfrak{a}|^+\rangle$, where $\mathfrak{c}i$ \mathfrak{c}_i$ is defined to be the collection set of $\theta\in\mathfrak{a}$ for which $f\alpha(\theta)\leq f_\alpha(\theta)\leq g_i(\theta)$. You've arranged that $g_i<g_j$ (everywhere) whenever$i

(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), and

(5b) $g_i(\theta)<f_\alpha(\theta)$ whenever $i(*)\leq i<|\mathfrak{a}|^+$ and $\theta\in \mathfrak{a}\setminus\mathfrak{c}$.

(5c) $|\mathfrak{c}|= N_{i(*)+1}$ 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 $M_{i(*)+1}$; in fact, it is not even bounded by an element of $M_{i(*)+1}$. However, there is a function $g$ in $M_{i(*)+1}\cap\prod\mathfrak{c}$ 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 $M_{i(*)+1}$ which dominates $g_{i(*)+1}$, and this is a contradiction.

Now I'm sure there's an easier proof of this not involving generators; I'll see if I can work it out for you later.

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.

1

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 $g_i (4) Now look at the collection$\langle \mathfrak{c}_i:i<|\mathfrak{a}|^+\rangle$, where$\mathfrak{c}i$is defined to be the collection of$\theta\in\mathfrak{a}$for which$f\alpha(\theta)\leq g_i(\theta)$. You've arranged that$g_i

(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), and

(5b) $g_i(\theta) (5c)$|\mathfrak{c}|= 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}$. It is not an element of$M_{i(*)+1}$; in fact, it is not even bounded by an element of$M_{i(*)+1}$. However, there is a function$g$in$M_{i(*)+1}\cap\prod\mathfrak{c}$dominating$g_{i(*)}+1\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$together, we get a function in$M_{i(*)+1}$which dominates$g_{i(*)+1}\$, and this is a contradiction.

Now I'm sure there's an easier proof of this not involving generators; I'll see if I can work it out for you later.

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.