Is there any ordinal $\alpha$ such that $\omega ^ {\omega ^ \alpha} = \alpha$?
Could you please suggest me how to even try to solve this?
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If $2^\alpha>\alpha$ (Cantor), then also $\omega^\alpha>\alpha$ and $\omega^{x}\geq x$ holds as well. – Peter Franek Jun 28 '14 at 07:43
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1@Peter: That's true for cardinal arithmetic; not for ordinal arithmetic. – Asaf Karagila Jun 28 '14 at 07:45
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@AsafKaragila: I see; I don't understand it completely at this moment, so sorry for the confusion. – Peter Franek Jun 28 '14 at 07:48
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@AsafKaragila: doesn't these identities hold for ordinal numbers? Can you have an $\alpha$ with $2^\alpha=\alpha$? – Peter Franek Jun 28 '14 at 07:53
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2@Peter: In ordinal arithmetic $2^\omega=\omega$. This is not the same exponentiation as cardinal exponentiation. See this post. – Asaf Karagila Jun 28 '14 at 07:56
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@AsafKaragila so the right answer is that it is the supremum of ${\omega^{\ldots ^{\omega^\omega}}}$? – Peter Franek Jun 28 '14 at 08:10
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@Peter: The right answer to which question? – Asaf Karagila Jun 28 '14 at 08:13
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Let us continue this discussion in chat. – Peter Franek Jun 28 '14 at 08:13
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@Peter: Sorry, I have a great dislike towards the chat system; and I have to leave my desk for a while now too. – Asaf Karagila Jun 28 '14 at 08:21
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@AsafKaragila: no problem and thanks! – Peter Franek Jun 28 '14 at 08:22
2 Answers
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Ordinals such that $\omega^{\alpha}=\alpha$ are called $\epsilon$-ordinals. The first such, $\epsilon$ zero is a tower of exponents, $$\epsilon_0=\omega^{\omega^{\omega^{\ddots}}}$$ (well I dont know how to make the diagonal dots go in the other direction)
It can be defined as follows $$\epsilon_0=\sup \beta_n$$
where $\beta_n$ is defined as
$$\beta_0=\omega \qquad \beta_{n+1}=\omega^{\beta_n}.$$
The epsilon ordinals $\epsilon_{\nu}$ form a closed unbounded set.
Rene Schipperus
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Thanks for your answer! I've got the idea but I don't know how to prove that $sup \beta_n$ is an ordinal and that $\omega ^ \epsilon = \epsilon$ – Alex Jun 28 '14 at 12:50
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1Well supremum of a set of ordinals is again an ordinal, this is a standard result. for the equality use the fact that $\omega^{\alpha}$ is continous. – Rene Schipperus Jun 28 '14 at 12:55
