necessity of AC

Question

Let $X = \bigcup\lbrace\alpha_i \mid i\in\mathbb{N}\rbrace,$ where each $\alpha_i$ is a countable ordinal. My understanding is that in order to show that $X$ itself is a countable ordinal one must assume a weak form of the Axiom of Choice -- namely, that the union of countably-many countable sets is countable. This confuses me, however. We know that, for each $\alpha_i,$ there is a bijection between it and $\mathbb{N}$. Is the problem that we can't "point" to these so as to then construct the bijection between $X$ and $\mathbb{N}$? Are there no occasions when we know that such a bijection exists despite being unable to "point" to it -- even in the context of ZF alone? Or is this precisely what we lack without AC? Thanks!

Answer 1

Yes, what you describe is correct. There are models of ZF in which the axiom of choice fails in such a way that $\omega_1,$ which is the least uncountable ordinal, can be expressed as a union just as you wrote.

The issue is that although for each $i\in\mathbb{N}$ there is a bijection between $\alpha_i$ and $\mathbb{N},$ there isn't (in such a model) a function $f$ such that for each $i\lt\omega,$ $f(i)$ is a bijection between $\alpha_i$ and $\mathbb{N}.$

Answer 2

If you know that there is a bijection between $X=\bigcup\{X_i\mid i\in\Bbb N\}$ and $\Bbb N$, then by restricting this bijection to each $X_i$, we can uniformly identify a bijection $f_i\colon X_i\to\Bbb N$.

Therefore, if you are in a situation where $X$ is uncountable, then it means exactly that we cannot uniformly choose bijections.

(And indeed, Feferman and Levy showed that it is possible that the countable union of countable ordinals is in fact $\omega_1$; and that even the real numbers can be written as a countable union of countable sets.)