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I'm trying to prove that there is an uncountable ordinal all which members are countable ordinal. This is fairly easy if I can state that the class of all countable ordinals is a set and then take the union on of that set (which I already know is an ordinal). However I'm not sure how to justify the fact that it is indeed a set and not a proper class.

Secondly, I was wondering if a union of a proper class of ordinals is necessarily an ordinal in the same way a union of a set of ordinals is?

Looking forward for your answers. Thanks!

Serpahimz
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2 Answers2

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First note that every countable ordinal is in bijection with $\omega$, therefore we can view it simply as a relation on $\omega$ which well-orders it by this order type.

Therefore we may define an equivalence relation on $\mathcal P(\omega\times\omega)$ in which every two well-orders are equivalent if and only if they are isomorphic (and everything which is not a well-ordering of $\omega$ is lumped into one big equivalence class).

Now you can show that this too makes a set, it is a definable collection of $\mathcal P(\mathcal P(\omega\times\omega))$, and since every well-ordered relation is isomorphic to a unique ordinal there is a definable map from that set onto the class of countable ordinals. Therefore countable ordinals make a set.

The union of this set is again a set, by the axiom of union, and it is an ordinal by showing it is transitive and $\in$ well-orders it.

As for the second question the answer is obviously no. If the union of a proper class of ordinals would be an ordinal then it would have to be an ordinal which is a set, which has a proper class of members. The union of any proper class of ordinals, if so, is the entire class of ordinals.

Also related: How do we know an $ \aleph_1 $ exists at all?

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Asaf Karagila
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  • Thanks Assaf, I had a feeling I could expect a swift reply from you. As always you've been most helpful! Could you elaborate please on how to formally show that a union of every proper class of ordinal is the entire class of ordinals? – Serpahimz Jan 12 '13 at 16:19
  • @Serpahimz: Every set of ordinals is a subset of an ordinal. So if we have a proper class of ordinals it has to be unbounded in the ordinals themselves, and so every ordinal is an element of a member of this class. – Asaf Karagila Jan 12 '13 at 16:23
  • @Serpahimz: Also, I should point out that "Assaf" and "Asaf" both write the same in Hebrew, but in the myriad of set theorists named like this (and I know at least three) I am "Asaf" with one "s". :-) – Asaf Karagila Jan 12 '13 at 16:43
  • Heh, I will remember it for next time :) It's also rather surprising there are so many set theorists that share the name, maybe you should consider forming a society :) – Serpahimz Jan 12 '13 at 20:39
  • There is one, but I already said too much... Watch out for any cluster of three or more people walking with cellphones at the same time. They know you know now. – Asaf Karagila Jan 12 '13 at 20:44
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The union of any proper class of ordinals is the class of all ordinals, actually, which is not, itself, an ordinal. If this were not so, then there would be a least ordinal $\alpha$ not in the union. Then $\alpha$ would necessarily be an upper bound on the ordinals in the class, so the class is contained in the set $\alpha\cup\{\alpha\}$, and so not proper, at all.

For your first question, consider the set $$\mathcal{W}=\bigl\{\langle X,R\rangle\in\mathcal{P}(\omega)\times\mathcal{P}(\omega\times\omega):R\text{ well-orders }X\bigr\}.$$ Every countable well-order type is represented by at least one of the members of this set, so by replacement, the collection of countable well-order types (ordinals) is a set.

Cameron Buie
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