Is there a canonical mapping between countable transfinite ordinals and $\omega$? What about recursive ordinals?

Question

Consider $\omega^2$. We can build a simple bijection between the ordinal and $\omega$ similarly to how the bijection between $\mathbb{Q}$ and $\mathbb{N}$ can be built.

I was wondering if there is a way to get such a bijection between all countable transfinite ordinals and $\omega$. In the title I used the word canonical. Perhaps a way of defining this is: can we build a function $F$ which, given a transfinite ordinal $\tau$, returns a bijection $f$ with $\omega$, without using the axiom of choice?

I'm also satisfied if this is true only for recursive ordinals.

Thank you.

Answer 1

It is consistent with $\mathsf{ZF}$ that no such $F$ exists (this will happen in any model where $\omega_1$ is singular), so the axiom of choice is necessary here. This is basically the same as this answer of mine to an MSE question of yours from a while back.

On the other hand, we can definitely get such an $F$ for the computable (infinite) ordinals: for $\alpha$ computable let $n_\alpha$ be the smallest index of a Turing machine building a copy of $\alpha$ with domain $\mathbb{N}$, and then just use the (unique) isomorphism between $\alpha$ and this copy.

(More generally, here's a method that works in $\mathsf{ZF}$ alone for all ordinals $<\omega_1^L$ - which is vastly bigger than $\omega_1^{CK}$, and indeed is $\mathsf{ZFC}$-consistently equal to $\omega_1$. Given $\alpha<\omega_1^L$, let $f$ be the $\le_L$-least bijection between $\alpha$ and $\omega$. Here $\le_L$ is the canonical well-ordering of $L$; on reals, this is $\Delta^1_2$. Note that if no $F$ as in your question exists in $V$, then this process still works up to $\omega_1^L$ but we have $\omega_1^L<\omega_1$ in this case.)