Prove that there exist countable ordinal $\xi$ such that $\xi=\omega^\xi$. The $\xi= \sup \{b^i \mid b_1=w, b_{i+1}=w^{b^i}\}$ should work. But how to prove that $\xi$ is countable?

  • 2
    $\begingroup$ If you use the usual representation of ordinals as sets, you can write the $\sup$ as a union, and use that countable unions of countable sets are countable, I think. $\endgroup$ – fgp Mar 8 '14 at 15:36
  • $\begingroup$ I think countable union of countable sets is countable, if you assume countable Axiom of Choise $\endgroup$ – Nazar Serdyuk Mar 8 '14 at 17:45
  • $\begingroup$ Are you assuming the axiom of choice? $\endgroup$ – Andrés E. Caicedo Mar 8 '14 at 18:17
  • $\begingroup$ The reason I ask is that, yes, if you assume the axiom of choice, the statement is straightforward, as $\epsilon_0$ is the countable union of countable ordinals. The point here is that you should be able to show easily that $\alpha^\beta$ is countable if both $\alpha$ and $\beta$ are countable, using for instance this characterization of ordinal exponentiation. $\endgroup$ – Andrés E. Caicedo Mar 8 '14 at 18:25
  • 2
    $\begingroup$ (There is also a lazy way of avoiding choice. In the constructible universe $L$, choice holds, so $\epsilon_0$ is countable there. Check that ordinal exponentiation is absolute, so $\epsilon_0$ is computed the same in $V$ and in $L$, and of course the enumeration in $L$ still verifies that $\epsilon_0$ is countable in $V$.) $\endgroup$ – Andrés E. Caicedo Mar 8 '14 at 18:30

Browse other questions tagged or ask your own question.