Say $\omega_1$ is the first uncountable ordinal. The reason I care about $\omega_1$ is
- Any countable subset of $\omega_1$ is bounded (or if you prefer, there is no countable cofinal subset).
This is obvious because
- A countable union of countable sets is countable.
Now, I'm told that we cannot prove (2) without at least some weak version of AC. Question: Does (1) also require AC?
Seems possible that some ordinalistic magic lets us be explicit about the relevant sequence of bijections?
Yes, it requires the axiom of choice.
It was shown by Feferman and Levy, as one of the first uses of forcing, that it is consistent that $\omega_1$ is the countable union of countable sets. This means that there is a countable subset of $\omega_1$ which is unbounded. This is just assuming the consistency of $\sf ZFC$. (The idea is simple, for each $n$ add many bijections between $\omega$ and $\omega_n$; if you are careful enough, you can ensure that these are all the new bijections you add, and therefore there is no bijection between $\omega$ and the "original" $\omega_\omega$, which makes it the least uncountable ordinal of the model.)
This result was later generalized by Gitik who showed that assuming additional axioms are consistent, we can construct a model where every limit ordinal is the countable union of smaller ordinals. It was shown that at least some additional axioms are needed for such a result, although it seems that Gitik's original assumptions are probably stronger than needed.