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As stated in the title, which axiom in ZF shows that a class of all countable (or any cardinal number) ordinals is a set?

Not sure which axiom, that's all.

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Why should there be an axiom, most things usually require several axioms to hold. That's what makes mathematics interesting! :-) – Asaf Karagila Sep 17 '12 at 16:09
Related. – Cameron Buie Jan 26 '13 at 18:17
up vote 6 down vote accepted

Each countable (= countably infinite) ordinal can be realized as a well-ordering of $\omega$ in at least one (actually uncountably many, but that is irrelevant) way, so the set of countable ordinals can be made by applying Replacement to $\mathcal P(\omega\times\omega)$ and a formula that produces the order type of its input if the input is a well-ordering and $\omega$ otherwise.

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One question: How does the power set of $(\omega \times \omega)$ results in the set of all countable ordinals? – Transfinite Lover Sep 18 '12 at 0:06
A well-ordering of $\omega$ is a special case of a ordering of $\omega$, which is a special case of a binary relation in $\omega$, which is a set of ordered pairs, which is an element of $\mathcal P(\omega\times\omega)$. Each well-ordering gives rise to a corresponding induction principle which you can use to construct the ordinal number for each of its initial segments one by one, and eventually the ordinal number corresponding to the entire well-ordering. – Henning Makholm Sep 18 '12 at 0:20

Every countable ordinal can be order embedded in $\mathbb{R}$. So the set of all countable ordinals can be identified with a subset of $2^{\mathbb{R}\times\mathbb{R}}$. So you can construct it from $2^{\mathbb{R}\times\mathbb{R}}$ using comprehension and replacement.

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Of course the approach of Henning makes much more sense. – Michael Greinecker Sep 17 '12 at 15:06

As Asaf pointed out in the comments, the "which axiom" part of the question does not have any obvious formal meaning, but I think it does have some kind of meaning so I will try to answer it. I think it makes sense to attribute the theorem "the class of countable ordinals is a set" primarily to the axioms of power set and replacement even though some other axioms are necessarily involved in the proof. This is because to me the most natural models of fragments of ZFC where the theorem does not hold are $V_{\omega+\omega}$ (sets of rank $<\omega+\omega$) and $H(\omega_1)$ (hereditarily countable sets) and these models satisfy all of the ZFC axioms except for the replacement axiom and the power set axiom respectively.

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A combination of them, actually. The proof I've seen requires power set and replacement, and I think union or pairing, too, but I can't recall off the top of my head. I can post an outline of that proof, if you like.

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