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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?2012-01-13
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    I've tried. However it is a litte difficult for a beginner.2012-01-13
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    Could you give me some encouragement, but not criticism if you don't intend to help to solve this question?2012-01-13
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    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.2012-01-13
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    @John: what is $\omega_\alpha$ and what is your definition of cardinal?2012-01-13
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    John: After the comments by Brian I went and read half the section in order to properly answer your question. This is not how it should be, you should supply the definitions and the relevant lemmas. You should not rely on the people coming to answer spending a good deal of time trying to locate the resources you have at hand.2012-01-14
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    @Asaf: I see. Sorry for just posting a question but not with relevant defiitions and lemmas.2012-01-14
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    @ineff: The definitions can be seen in the chapter I of Kunen's book:)2012-01-14

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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 $xi$.2012-01-14
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    @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.2012-01-14
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    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.2012-01-14
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    @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.2012-01-14
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    @Asaf: Nice. That's just what I want to know:)2012-01-15
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    @AsafKaragila Where did you use AC in you answer?2012-01-16
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    @Matt: I did not really use AC. I *thought* that the question was about the well ordering theorem; however the discussion with Brian sent me reading in Kunen and after some reading I realized what the question was about. Kunen simply defines *cardinals* as aleph numbers (he does a proper job introducing cardinality in general). This requires no real use of AC, it's just a proof that every aleph number is an initial ordinal.2012-01-16
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    @AsafKaragila I should've read the editing history! Thanks for the comment, Asaf.2012-01-16