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Often in theorems of pcf theory one has the assumption that the length of sequences of functions has to respect some bound so things can be well-defined. For instance, let $a=[\aleph_2,...,\aleph_n,...:n<\omega]$ be a set of regular cardinals, say you have a sequence $f_\beta$ in $\prod a$ of length at most $|a|^+$. Then $sup_\beta f_\beta \in \prod a$ since $|a|^+ < min(a)$. But why is this true? If you have for example an $\omega_2$ sequence of functions $f:\kappa \rightarrow \kappa$ such that $f(\kappa)\in \kappa$, $\kappa$ some $\aleph_n$, $n$ not 0 and not 1,then why is $f_\beta$ for $\beta=\omega_2$ outside of the product, as far as we know, we don't know if $2^{\aleph_0}= \aleph_2 $ since $a$ is a countable set of regular cardinals (say the set of $\aleph_n$'s)? Thanks

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There are several typos in your question, could you please read it over and edit appropriately. –  François G. Dorais Aug 21 '10 at 20:07
    
I hope it is fixed now. –  Carlo Von Schnitzel Aug 21 '10 at 20:21
    
Not quite: You seem to be asking why $f_\beta \in \prod a$ when $f_\beta \in \prod a$. Also $|a|^+ < \min(a)$ is false when $a = \{\aleph_1,\aleph_2,\dots\}$. –  François G. Dorais Aug 21 '10 at 20:28
    
Oh yes sorry, got confused, since $|a|^+$ is in this case is $\aleph_1$. Let me start $a$ at $\aleph_2$, and let me consider an $\aleph_2$ sequence of functions. I am editing this. –  Carlo Von Schnitzel Aug 21 '10 at 20:56
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Maybe you mean that $\sup_\beta f_\beta \in \prod a$? –  François G. Dorais Aug 21 '10 at 23:33

1 Answer 1

up vote 3 down vote accepted

Let $a$ be a set of regular cardinals. An element of $\prod a$ is a function $f:a \to \sup a$ such that $f(\kappa) < \kappa$ for every $\kappa \in a$. Suppose we are given a family $f_i$, $i \in I$, of elements of $\prod a$. In order to ensure that $\sup_{i \in I} f_i \in \prod a$ we need to make sure that $\sup_{i \in I} f_i(\kappa) < \kappa$ for every $\kappa \in a$. A sufficient condition for this is that $|I| < \min a$. Indeed, since $|I| < \mathrm{cf}(\kappa) = \kappa$, we then have $\sup_{i \in I} f_i(\kappa) < \kappa$ for every $\kappa \in a$. Thus the assumption $|a| < \min a$ ensures that the supremum of any sequence of elements of $\prod a$ with length less than $|a|^+$ has a supremum in $\prod a$.

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Thank you François, but I am really disappointed as I should have figured this by myself. Somehow I had dropped the requirement that $f(\kappa)<\kappa$ and was being silly. I am kind of mad right now. It just hold by regularity of all the cardinals in $a$... –  Carlo Von Schnitzel Aug 22 '10 at 18:22

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