# Finite ordinal to the power a = a

Let $\alpha>\omega$ be an ordinal such that $2^\alpha$ = $\alpha$.

Then $\alpha$ is an epsilon number?

I have tried many different ways, but i can only work with the left side of $\alpha$(e.g, I have proved such ordinals satisfy $\omega$$\alpha$ = $\alpha$ and etc), but i think it's critical to work with right side of $\alpha$ in proof and i cant handle this.. Help

Plus i want to know even when the base is not 2 but finite, whether $\alpha$ is an epsilon number

-
So far, i have proved that $\alpha$ is a limit ordinal. –  Katlus Jun 4 '12 at 8:08
A slightly more complete answer: Let $E = \{\omega\} \cup \{\epsilon_\alpha | \alpha$ an ordinal}. Then, for any $\beta$, the ordinals $\alpha$ such that $\beta^\alpha = \alpha$ is precisely $\{\alpha | \alpha \in E, \alpha > \beta\}$. –  Deedlit Sep 17 '12 at 9:29

Lemma: $2^{\omega\alpha}=\omega^\alpha.$

Proof: Given $2^{\omega\alpha}=\omega^\alpha,$ we have $2^{\omega(\alpha+1)}=2^{\omega\alpha+\omega}=\omega^{\alpha+1},$ where the first equality is by definition of the function $\omega x$ and the second is by the hypothesis and calculation of the limit of $2^{\omega\alpha+n}.$

At limit ordinals, it's true because the composition of continuous functions is continuous.$\square$

Writing $\alpha$ in Cantor normal form to base $\omega,$ if $\alpha$ has any $\omega^n$ terms for finite $n,$ then $2^\alpha>\alpha$ by normality of $2^x$ and inspection:

Since $2^x$ is a normal function, we know $2^\alpha\geq\alpha.$ Addition of 1 in the exponent is multiplication by 2; addition of $\omega$ in the exponent is multiplication by $\omega;$ addition of $\omega^2$ yields multiplication by $\omega^\omega$ (all of these are meant as on the right). All of these produce larger ordinals.

If $\alpha$ does not have any $\omega^n$ terms, then $\alpha=\omega\alpha$ and so if $2^{\alpha}=\alpha$ then $\alpha=\omega^\alpha$ by the lemma; this is the defining property of an $\epsilon$-number.

Note that the same applies to any finite base.

-
Nice. Thank you :) –  Katlus Jun 4 '12 at 9:55