# Find ordinals $\alpha,\beta$ such that $n^{\alpha}=\alpha$ and $\omega_1^{\beta}=\beta$

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?

• What is $n$? Do we know that $n\gt1$? Is $n\lt\omega$?
– bof
Jul 6 '20 at 11:51
• Is your task to find all solutions of those equations, or to find some solution, or the least solution?
– bof
Jul 6 '20 at 11:53
• 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 $\alpha\ge\omega_1^{\omega_1}$.
– bof
Jul 6 '20 at 12:00
• So how about defining $\alpha_0=1$ and recursively $\alpha_{n+1}=\omega_1^{\alpha_n}$, and then $\alpha=\lim_{n\to\omega}\alpha_n$?
– 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.

• 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 $\omega_1=\varepsilon_{\omega_1}$. Aug 4 '20 at 12:48
• @SimplyBeautifulArt My proof of the first part has an error, thank you for pointing it out. (I am fixing it now.) Aug 4 '20 at 13:11
• @SimplyBeautifulArt I fixed it. I hope my proof is not incorrect any further. Aug 4 '20 at 13:33