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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?

Michael Hardy
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Nazar Serdyuk
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    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. – fgp Mar 08 '14 at 15:36
  • I think countable union of countable sets is countable, if you assume countable Axiom of Choise – Nazar Serdyuk Mar 08 '14 at 17:45
  • Are you assuming the axiom of choice? – Andrés E. Caicedo Mar 08 '14 at 18:17
  • 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. – Andrés E. Caicedo Mar 08 '14 at 18:25
  • There was no mention about AC so I suppose we cannot assume it. – Nazar Serdyuk Mar 08 '14 at 18:27
  • However, if you do not assume the axiom of choice, then one needs to work a bit harder, and verify that we get (uniformly) enumerations for all of $\omega^\omega,\omega^{\omega^{\omega}},\dots$, simultaneously, so we can implement the prof that countable unions of countable sets are countable, but without invoking the axiom of choice. – Andrés E. Caicedo Mar 08 '14 at 18:28
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    (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 $V$.) – Andrés E. Caicedo Mar 08 '14 at 18:30
  • OK. This question has been asked here a few times before. Here is a link to one of those instances. I suggest you study tomasz's answer, as it does not require knowledge of absoluteness or constructibility, and explicitly indicates how to produce the uniform enumerations I alluded to above. – Andrés E. Caicedo Mar 08 '14 at 18:35

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