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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:

  1. $f_\alpha < f_\beta$ except maybe for a finite set
  2. 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:

  1. There are no maximal elements (for any $f$ we can define $g(n) = f(n) +1$ and clearly $f< g$)
  2. For any given $f,g$ there exists an upper-bound for both (namely $\max (f(n), g(n))$)
  3. 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 $A$?

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This is a question regarding two famous cardinal characteristics:

  • $\mathfrak{b}$ the minimal cardinality of an unbounded family in $(\omega^\omega,{<^*})$.

  • $\mathfrak{d}$ the minimal cardinality of a cofinal family in $(\omega^\omega,{<^*})$.

It is not difficult to show that a scale exists if and only if $\mathfrak{b} = \mathfrak{d}$, and then the minimal length of a scale is the common value of these two cardinal characteristics (which is also the cofinality of any scale). However, this equality need not hold. In fact, the only inequalities that must hold are $$\aleph_1 \leq \mathfrak{b} = cf(\mathfrak{b}) \leq cf(\mathfrak{d}) \leq \mathfrak{d} \leq \mathfrak{c}.$$ Using Hechler's Theorem (see this answer of mine) any pattern of cardinals consistent with the above is possible to attain by a forcing. See Andreas Blass's handbook of set theory chapter (available here) for more on this topic.

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