Find ordinals $\alpha$ such that (a) $n^{\alpha}=\alpha\; $ (b) $\omega_1^{\alpha}=\alpha$

On (a) I could verify that all ordinals of the form $\omega, \omega^{\omega},\omega^{\omega^\omega},\cdots$ satisfy the equation, but are they all of them? Does it work for $\alpha=\omega_1$? For all $\omega_{\alpha}$? Well, $n^{\omega_1}=\sup_{\delta<\omega_1} n^{\delta}$ is this equal to $\omega_1 ?$ I know certainly that it is $\geq$ but couldn't prove $\leq$. And if it does holds I think I could repeat this argument for all ordinals $\omega_{\alpha}$.

On (b) I couldn't get anything. Does it holds for all cardinals? I thought following the same argument on (a) but it doesn't seem to be very trustful.

Could you help me?

  • $\begingroup$ What is $n$? Do we know that $n\gt1$? Is $n\lt\omega$? $\endgroup$
    – bof
    Jul 6 '20 at 11:51
  • $\begingroup$ Is your task to find all solutions of those equations, or to find some solution, or the least solution? $\endgroup$
    – bof
    Jul 6 '20 at 11:53
  • $\begingroup$ For (b), since $\alpha=0$ doesn't work, we need $\alpha\ge1$. But then $\alpha=\omega_1^\alpha\ge\omega_1^1=\omega_1$, so $\alpha\ge\omega_1$. But then $\alpha=\omega_1^\alpha\ge\omega_1^{\omega_1}$, so $\alpha\ge\omega_1^{\omega_1}$. $\endgroup$
    – bof
    Jul 6 '20 at 12:00
  • $\begingroup$ So how about defining $\alpha_0=1$ and recursively $\alpha_{n+1}=\omega_1^{\alpha_n}$, and then $\alpha=\lim_{n\to\omega}\alpha_n$? $\endgroup$
    – bof
    Jul 6 '20 at 12:06

Let $\epsilon_\alpha$ be the $\alpha$th solution of the equation $\alpha=\omega^\alpha$. They are known as epsilon numbers. Then we have

  1. $\alpha=\omega$ is the simplest solution. Moreover, $\omega$ is the only possible solution that are less than $\omega^2$. (Just put $\alpha=\omega\cdot k+l$ into the equation $n^\alpha=\alpha$.) Now assume that $\alpha\ge\omega^2$. I claim that $\alpha=\epsilon_\beta$ for some $\beta$.

    If $n\ge 2$, then $n^{\epsilon_\beta}=\epsilon_\beta$. This follows from some simple ordinal inequalities.

    On the other hand, we can see that $n^\alpha=\alpha$ implies $\omega^\alpha=\alpha$: $\alpha\ge\omega^2$ implies $\alpha=\omega+\alpha$. From this, we can show that $\omega\cdot\alpha=n^\omega n^\alpha=n^{\omega+\alpha}=\alpha$ and $\omega^\alpha = (n^\omega)^\alpha = n^\alpha=\alpha$. (I use the equality $n^\omega=\omega$.)

(The previous solution does not consider the case $\alpha<\omega^2$. Thanks to @Simply Beautiful Art for pointing it out.)

  1. Clearly, every ordinal which satisfies $\omega_1^\beta=\beta$ is an epsilon number. However, not every epsilon number satisfies $\beta=\omega_1^\beta$: You can see that $\omega_1$ is an epsilon number, and $\omega_1^{\omega_1}>\omega_1$.

    However, if $\beta$ is an epsilon number greater than $\omega_1^\omega$, then we have $\omega_1^\beta=(2^{\omega_1})^\beta = 2^\beta\le \beta$. @bof has already shown that if $\omega_1^\beta=\beta$, then $\beta>\omega_1^\omega$, so we have all possible solutions.

  • $\begingroup$ Your first point is unclear to me. You state that $n^\alpha=\alpha$ implies $\omega^\alpha=\alpha$, yet also say that you use the equality $n^\omega=\omega$. Putting them together, it seems you are claiming that $\omega^\omega=\omega$, which is clearly false. And as per the second point, those are simply $\alpha=\varepsilon_{\omega_1+\beta}$ for $\beta\ge1$ since $\omega_1=\varepsilon_{\omega_1}$. $\endgroup$ Aug 4 '20 at 12:48
  • $\begingroup$ @SimplyBeautifulArt My proof of the first part has an error, thank you for pointing it out. (I am fixing it now.) $\endgroup$
    – Hanul Jeon
    Aug 4 '20 at 13:11
  • $\begingroup$ @SimplyBeautifulArt I fixed it. I hope my proof is not incorrect any further. $\endgroup$
    – Hanul Jeon
    Aug 4 '20 at 13:33

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