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 welldefined. 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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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 $\prod a$. 


$a = \{\aleph_1,\aleph_2,\dots\}$
. – François G. Dorais♦ Aug 21 '10 at 20:28