I'm asking a question following the reading of Set Theory from Jech (page 126).


For the record, a $\kappa$-sequence of functions $\langle f\_\alpha : \alpha < \kappa\rangle$ is called
a $\kappa$-scale if $f\_\beta < f\_\alpha$ whenever $\beta < \alpha$, and if for every $g\colon \omega \rightarrow \omega$ there exists an $\alpha$ such that $g < f\_\alpha$.


Suppose that $\aleph\_1 = 2^{\aleph\_0}$. A scale $\langle f\_\alpha : \alpha < \omega\_1 \rangle$ is constructed by transfinite induction to $\omega\_1$:


Let $\{g\_\alpha : \alpha < \omega\_1\}$ enumerate all functions from $\omega$ to $\omega$. At stage $\alpha$, we
construct, by diagonalization, a function $f\_\alpha$ such that for all $\beta < \alpha$, $f\_\alpha > f\_\beta$
and $f\_\alpha > g\_\beta$. Then $\langle f\_\alpha : \alpha < \omega\_1 \rangle$ is an $\omega\_1$-scale.


**I don't see how to manage the transfinite induction in order to avoid for limit ordinals that $f\_\alpha$ becomes infinite.**


Thanks for your support!



Building a scale by diagonalization and transfinite induction
The inequality relation between functions $\omega\to\omega$ that is written $f<g$ in the question must mean that $f$ is *eventually* (not everywhere) below $g$, i.e., that $f(n)<g(n)$ holds for all sufficiently large $n$. (If it meant "for *all* $n$, then the induction would indeed break down at the first limit ordinal.) Now to construct an upper bound for countably many functions $f\_\alpha$, first re-enumerate them in an $\omega$-sequence, say as $g\_n$, and then use the function $k\mapsto 1+\max\{g\_n(k):n<k\}$. 


By the way, the usual notation for the "eventually $<$" relation is not $<$ but $<^\*$.


