Does $\alpha^{\omega}$ have a countable fixed point $\alpha>1$?

Question

In ordinal arithmetic, does the function $f(\alpha)=\alpha^{\omega}$ have a fixed point which is countable and greater than 1? I think the answer is "no", as I've tried quite hard to construct examples, but I do not understand how to use the countability. Really, I would like to prove $\alpha$ countable implies $\alpha < \alpha^{\omega}$ for $\alpha >1$. I can do this via an induction but I don't see how to treat the case where $\alpha$ is a limit.

I would appreciate guidance in how to show that functions defined by arithmetic operations on ordinals have no countable fixed point more generally.

Answer 1

You are correct. There are none. To see why, note that there is a monotonicity here: $$\alpha<\alpha\cdot\alpha=\alpha^2<\dots<\alpha^\omega,$$ for any $\alpha>1$.

To prove this, simply note that $\alpha+1<\alpha^2$ as well. And therefore $\alpha$ is a proper initial segment of $\alpha^2$, and thus of $\alpha^\omega$.

As a final remark, see that the above doesn't even use the fact that $\alpha$ is countable.

Answer 2

For ordinals $a,b$ let $S(a,b)$ be the set of functions $f:b\to a$ such that $\{x\in b:f(x)\ne 0\}$ is finite. Give $S(a,b)$ the reverse-lexicographic order: If $f,g\in S(a,b)$ and $y=\max \{x\in b: f(x)\ne g(x)\}$ then $f<g$ iff $f(y)<g(y).$ This is a well-order.

$a^b$ is defined to be the unique ordinal that is order-isomorphic to this order on $S(a,b).$

It is straightforward to show that if $a>1$ and $b_1<b_2$ then $S(a,b_1)$ is a proper initial segment of $S(a,b_2),$ and therefore $a^{b_1}<a^{b_2}.$

For example, if $a>1$ then $a<a^2<a^3<....<a^{\omega}.$