# How to prove that every infinite cardinal is equal to $\omega_\alpha$ for some $\alpha$?

How to prove that every infinite cardinal is equal to $\omega_\alpha$ for some $\alpha$ in Kunen's book, I 10.19?

I will appreciate any help on this question. Thanks ahead.

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Are you asking for the definition of the $\aleph$ numbers? Are you asking to prove the well ordering theorem? Have you tried something for yourself? – Asaf Karagila Jan 13 '12 at 5:30
I've tried. However it is a litte difficult for a beginner. – Paul Jan 13 '12 at 5:43
Could you give me some encouragement, but not criticism if you don't intend to help to solve this question? – Paul Jan 13 '12 at 5:49
I should tell you that I am probably willing to go above and beyond in helping people understand set theory. I just want to see them trying on their own as well. While I do believe that you tried prior to coming here, what is it that you tried? Why do you think that the claim is true? Where do you feel that you cannot approach the question? The more to give me to work with, the better I can help you. – Asaf Karagila Jan 13 '12 at 6:03
@John: what is $\omega_\alpha$ and what is your definition of cardinal? – Giorgio Mossa Jan 13 '12 at 9:11

I took the trouble to read through Kunen in order to understand the problem, as well the definitions which you can use for this.

1. Cardinal is defined to be an ordinal $\kappa$ that there is no $\beta<\kappa$ and a bijection between $\kappa$ and $\beta$.

2. The successor cardinal $\kappa^+$ is the least cardinal which is strictly larger than $\kappa$.

3. $\aleph_\alpha=\omega_\alpha$ defined recursively, as the usual definitions go: $\aleph_0=\omega$; $\aleph_{\alpha+1}=\omega_{\alpha+1}=\omega_\alpha^+$; at limit points $\aleph_\beta=\omega_\beta=\sup\{\omega_\alpha\mid\alpha<\beta\}$.

Now we want to show that:

Every cardinal is an $\omega_\alpha$ for some $\alpha$.

Your question concentrates on the second part of the lemma.

Suppose $\kappa$ is an infinite cardinal. If $\kappa=\omega$ we are done. Otherwise let $\beta=\sup\{\alpha+1\mid\omega_\alpha<\kappa\}$. I claim that $\kappa=\omega_\beta$.

Now suppose that $\omega_\beta<\kappa$ then we reach a contradiction since this means that $\beta<\sup\{\alpha+1\mid\omega_\alpha<\kappa\}=\beta$ (since $\beta$ is in this set, then $\beta<\beta+1\le\sup{\cdots}=\beta$).

If so, $\kappa\le\omega_\beta$. If $\beta=\alpha+1$ then $\omega_\alpha<\kappa\le\omega_\beta$ and by the definition of a successor cardinal we have equality. Otherwise $\beta$ is a limit cardinal and we have that $\omega_\alpha<\kappa$ for every $\alpha<\beta$, then by the definition of a supremum we have that $\omega_\beta\le\kappa$ and again we have equality.

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You wouldn’t know unless you were familiar with the book, but Lemma I.10.19 in Ken’s Set Theory doesn’t require AC: he defines a cardinal to be an ordinal $\alpha$ such that $\alpha=|\alpha|$, where for a well-orderable set $A$, $|A|$ is defined to be the least ordinal $\alpha$ such that there’s a bijection between $A$ and $\alpha$. Thus, the problem is merely to show that if $\omega\le\alpha\in\mathbf{ON}$, and there is no bijection between $\alpha$ and any smaller ordinal, then $\alpha=\omega_\xi$ for some $xi$. – Brian M. Scott Jan 14 '12 at 19:34
@Brian: Indeed I am less familiar with this book than with Jech's book; however to have that $|A|$ is an ordinal for every set $A$? Well, if that is not the well ordering theorem, I'm not sure what is :-) I personally very dislike the way cardinality is defined only for well orderable sets. This produces comments like that one someone left here somewhere "without the axiom of choice there is not much sense to the cardinality of the continuum." which is completely wrong. – Asaf Karagila Jan 14 '12 at 19:38
I agree that to have $|A|$ an ordinal for every set $A$ is the well-ordering theorem, but as I said, at that point he defines cardinality only for well-orderable sets. That’s really all that he needs, since ‘[t]his book is concerned mainly with set theory with AC’ (p. 15), and it makes the introductory material a bit simpler. – Brian M. Scott Jan 14 '12 at 20:29
@Brian: After reading most of this section I can give my approval for his method. He does a good job, mentioning what AC gives us and such. – Asaf Karagila Jan 14 '12 at 22:13
@Asaf: Nice. That's just what I want to know:) – Paul Jan 15 '12 at 4:45