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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 at 7:43
    
@Peter: That's true for cardinal arithmetic; not for ordinal arithmetic. –  Asaf Karagila Jun 28 at 7:45
    
@AsafKaragila: I see; I don't understand it completely at this moment, so sorry for the confusion. –  Peter Franek Jun 28 at 7:48
    
@AsafKaragila: doesn't these identities hold for ordinal numbers? Can you have an $\alpha$ with $2^\alpha=\alpha$? –  Peter Franek Jun 28 at 7:53
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@Peter: In ordinal arithmetic $2^\omega=\omega$. This is not the same exponentiation as cardinal exponentiation. See this post. –  Asaf Karagila Jun 28 at 7:56

2 Answers 2

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.

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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 at 12:50
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Well 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 at 12:55

HINT: What happens if $\alpha=\omega^\alpha$? Can we even have that?

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