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Here is the proof from p. 48 of the Millennium Edition, corrected 4th printing 2006: Proof of Thm 5.1 from Jech's Set Theory Millennium Edition

My question: how do we know that there is any ordinal $\theta$ such that $A=\{a_\xi\,\colon\xi<\alpha\}$? Presumably we need to use the fact that $A$ is a set because if $A$ is a proper class I don't think it's true. On the other hand, I don't see how we've used it. Am I missing something? And if I'm not missing something, what is the easiest/most elegant way to patch the hole? If the process keeps going indefinitely, $A$ would have to be a proper class, but I am not clear on how to stick the landing on that approach.

Thank you!

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    $\begingroup$ See Jech's coverage of the Hartogs function in the proof of Lemma 3.4, bottom of p. 29. $\endgroup$
    – blargoner
    Commented Jul 7 at 22:07
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    $\begingroup$ en.wikipedia.org/wiki/Hartogs_number might be helpful $\endgroup$
    – Lucenaposition
    Commented Jul 7 at 22:28
  • $\begingroup$ Thank you, @blargoner and @Lucenaposition! So it looks to me like there is something missing in the proof, and that this is the way to patch it. $\endgroup$
    – Nat Kuhn
    Commented Jul 8 at 0:48
  • $\begingroup$ @NatKuhn I'd hesitate to say it's missing, because that makes it sound like there's a problematic gap or hole. It's well within the realm of details Jech expects the reader to fill in using techniques he's illustrated. If that proof has a gap, then so does almost every other textbook proof. $\endgroup$
    – blargoner
    Commented Jul 8 at 1:09

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If there is no such $\theta$, then there is a map from $A$ onto the class of ordinals, that's impossible, since $A$ is a set.

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  • $\begingroup$ This was my intuition but I was not sure how to make it rigorous, but I see now that the axiom of replacement would imply that the class of all ordinals is a set. $\endgroup$
    – Nat Kuhn
    Commented Jul 8 at 1:07

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