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This is a question in ordinal arithmetic. (If anyone has read 'Classic Set Theory' by Derek Goldrei, this question comes from page 252.)

$\epsilon_0$ = sup {$\omega$, $\omega^\omega$, ... } and $\omega$ is the smallest infinite ordinal.

$\omega_1$ is the least uncountable ordinal so that $\alpha < \omega_1$ if and only if $\alpha$ is a countable ordinal.

I will take the definition of an ordinal to be: a set $\alpha$ is an ordinal if (i) $\alpha$ is well-ordered by $\in$ and (ii) if $\beta \in \alpha$ then $\beta$ is a subset of $\alpha$. (or $\alpha$ is $\in$-transitive).

I am also not assuming that $\epsilon_0$ is countable at this point.

C.K
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  • I'm not sure I understand. I can see that if $\epsilon_0$ < $\omega_1$, then $\epsilon_0$ must be countable since $\omega_1$ is the least uncountable ordinal but I'm not sure how to explicitly prove that without assuming that $\epsilon_0$ is countable. – C.K Feb 13 '14 at 13:28
  • In your question http://math.stackexchange.com/questions/650792/prove-that-omega-omega-1-omega-cdot-omega-1-omega-omega-1-ome I gave you a hint which suffices how to prove that. – Asaf Karagila Feb 13 '14 at 13:30
  • Also, YOU asked this before: http://math.stackexchange.com/questions/649957/how-to-prove-epsilon-0-is-countable – Asaf Karagila Feb 13 '14 at 13:31
  • I think I've confused myself now, sorry. I didn't get very far with math.stackexchange.com/questions/649957/… because I'm aware my question was badly formed. In that question I took for granted that $\epsilon_0 < \omega_1$. But now I was wanting a way to prove that $\epsilon_0 < \omega_1$ without assuming first that $\epsilon_0$ is countable. – C.K Feb 13 '14 at 13:59

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As each $\omega^{\omega^\cdots}<\omega_1$, $\epsilon_0\le\omega_1$. Can't be equal because the cofinality of $\epsilon_0$ is obviously $\omega$ and ${\rm cf}(\omega_1)=\cdots$

Martín-Blas Pérez Pinilla
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Each of $\omega, \omega^{\omega}, \dots$ is countable, as you can easily prove by induction; then $\epsilon_0 = \bigcup \{ \omega, \omega^{\omega}, \dots \}$ is a union of countably many countable sets, so it must be countable, hence $<\omega_1$.

Clive Newstead
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