In Jech's Set Theory he defines a $\kappa$-scale as a family of functions $\langle f_\alpha\colon\omega\to\omega | \alpha < \kappa \rangle$ for which:

- $f_\alpha < f_\beta$ except maybe for a finite set
- For any $g\colon\omega\to\omega$ there is some $f_\alpha>g$ (again for all but perhaps a finite set)

(the definition can be found in chapter 10 which deals with measurable cardinals)

Clearly if a $\kappa$-scale does exist then $\kappa \le 2^{\aleph_0}$ (as there are only that many functions to begin with).

If we look at the quasi-order on $\omega^\omega$ defined by as above, that is $f < g$ if $f(n) < g(n)$ for all but a finite number of $n\in\omega$, then we can immediately say that:

- There are no maximal elements (for any $f$ we can define $g(n) = f(n) +1$ and clearly $f< g$)
- For any given $f,g$ there exists an upper-bound for both (namely $\max (f(n), g(n))$)
- If there exists a $\kappa$-scale then there is a cofinal subset of the quasi-order of cardinality $\le \kappa$ (cofinal in the sense that for every $f$ you can find some $g$ in the cofinal subset for which $f< g$)

My question is, if so, assuming $\kappa$ is the least cardinal number for which a $\kappa$-scale exists, and $A\subseteq \omega^\omega$ some $ < $-cofinal subset. Does it follow that $|A|\ge\kappa$? Does the assumption that $2^{\aleph_0} = \lambda$ holds some property (e.g. regularity) affects the existence of such $\kappa$ or $A$?