2
$\begingroup$

How to construct an ordinal with uncountable cofinality? All the very "large" ordinals I can think of, such as $\omega_\omega^{\omega_\omega}$, still seem to have countable cofinality. I need a better intuitive sense of what such a large ordinal can be.

Relevant links: http://en.wikipedia.org/wiki/Cofinality#Cofinality_of_ordinals_and_other_well-ordered_sets http://en.wikipedia.org/wiki/Ordinal_number

  • 5
    It's difficult to really get a sense for uncountable cofinalities, because by definition there's no (integer-)indexed sequence 'leading up to' that ordinal, but why not simply $\omega_1$, the ordinal of all countable ordinals ordered by set inclusion?2012-09-26
  • 0
    Pretty much every ordinal that you can "construct" is countable and hence has countable cofinality.2012-09-26
  • 1
    @LevonHaykazyan: It's hard for me to conceive a sensible notion of constructibility in which $\omega_\omega^{\omega_\omega}$ is constructible, but $\omega_1$ isn't...2012-09-26
  • 0
    @StevenStadnicki I think ${\omega_1}$ has countable cofinality, despite being uncountable. http://en.wikipedia.org/wiki/Cofinality#Cofinality_of_ordinals_and_other_well-ordered_sets2012-09-26
  • 1
    @tom4everitt Assuming the Axiom of Choice, $\omega_1$ has cofinality precisely $\omega_1$; its cardinal $\aleph_1$ is a so-called _regular_ cardinal. See http://en.wikipedia.org/wiki/Regular_cardinal for the basics of regular cardinals, and why $\omega_1$ can't have countable cofinality assuming AC.2012-09-26
  • 0
    @StevenStadnicki yes, you're right, just realized. thanks2012-09-26

1 Answers 1