Editing Beth number (section)

==Generalization==
The more general symbol <math>\beth_\alpha(\kappa)</math>, for ordinals <math>\alpha</math> and cardinals <math>\kappa</math>, is occasionally used. It is defined by:
:<math>\beth_0(\kappa)=\kappa,</math>
:<math>\beth_{\alpha+1}(\kappa)=2^{\beth_\alpha(\kappa)},</math>
:<math>\beth_\lambda(\kappa)=\sup\{ \beth_\alpha(\kappa):\alpha<\lambda \}</math> if ''λ'' is a limit ordinal.

So
:<math>\beth_\alpha=\beth_\alpha(\aleph_0).</math>

In [[Zermelo–Fraenkel set theory]] (ZF), for any cardinals <math>\kappa</math> and <math>\mu</math>, there is an ordinal <math>\alpha</math> such that:

:<math>\kappa \le \beth_\alpha(\mu).</math>

And in ZF, for any cardinal <math>\kappa</math> and ordinals <math>\alpha</math> and <math>\beta</math>:

:<math>\beth_\beta(\beth_\alpha(\kappa)) = \beth_{\alpha+\beta}(\kappa).</math>

Consequently, in ZF absent [[ur-element]]s, with or without the [[axiom of choice]], for any cardinals <math>\kappa</math> and <math>\mu</math>, the equality

:<math>\beth_\beta(\kappa) = \beth_\beta(\mu)</math>

holds for all sufficiently large ordinals <math>\beta</math>. That is, there is an ordinal <math>\alpha</math> such that the equality holds for every ordinal <math>\beta \geq \alpha</math>.

This also holds in Zermelo–Fraenkel set theory with ur-elements (with or without the axiom of choice), provided that the ur-elements form a set which is equinumerous with a [[pure set]] (a set whose [[transitive set#Transitive closure|transitive closure]] contains no ur-elements). If the axiom of choice holds, then any set of ur-elements is equinumerous with a pure set.