# An assumption in pcf theory

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
Maybe you mean that $\sup_\beta f_\beta \in \prod a$? –  François G. Dorais Aug 21 '10 at 23:33

## 1 Answer

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