I can't understand a simple thing again, though found the similar questions here. Given the set $S$ of all countably infinite ordinals, how can we find its order type, or at least its cardinality? Do we need the axiom of choice or continuum hypothesis here? I've just managed to show $\omega_0\leq Ord(S)\leq\omega_1$.
P.S. I'm trying not to use the Neumann's approach (where ordinals are sets with the certain properties), considering the separate class of objects. It's also given that $\omega_1$ is the set of all countable (not necessarily infinite) ordinals.
Best Answer
We don't need the AC or CH.
We can show that the collection of all well-orderings of $\mathbb N$ is a set (this makes essential use of power set). Each of these well-ordered sets corresponds to an ordinal - its order type (this makes essential use of replacement), and so (by replacement again) there is a set of all countable ordinals. This is a well-ordered set under the usual ordering $\in$, so it has an order type. We call this ordinal $\omega_1$ by definition.
$\omega_1$ is not countable since it is an ordinal, and by its definition any countable ordinal is strictly less.
So that's a complete answer to the question, and no choice was used. Where choice might get involved is when we want to say $\omega_1$ is the smallest uncountable cardinality. It's clearly the smallest uncountable ordinal, and so it can be embedded into any uncountable well-orderable set, so it is of size less than or equal to it. But we can't say every set is well-orderable without AC, and for instance, it is consistent with ZF that $\omega_1$ does not have an injection into the reals. (However it is provable in ZF that there is a surjection of the reals onto $\omega_1$.) It's also consistent that there are infinite sets that $\omega$ does not even inject into, so no AC also wrecks $\omega$'s status as the smallest infinite cardinality.