Can it be proved in ZFC - Pow (ZFC excluding the Axiom of Power Set) that Well Ordering Theorem holds? I have seen several proofs of Well Ordering Theorem in ZFC (including Zermelo's original one in 1904, Zermelo's in 1908), all involving choosing a choice function defined on the power set of the given set which we are going to well-order (To be more precise, the proof given by Wikipedia does not involve such a choice function, but it's deduced from Zorn's Lemma, and anyway it still requires the use of power sets.). Therefore, these proofs rely on the Axiom of Power Set, which makes me wonder whether such a proof is still possible without Power Set. Can anybody give such a proof (if possible) or otherwise construct a model of ZFC - Pow in which the Well Ordering Theorem fails?

  • $\begingroup$ In ZF there are many statements that are demonstrably equivalent to AC.. I think that what you are asking is whether, in ZF-P, it can proven that Zorn's Lemma implies that every set can be well-ordered. $\endgroup$ – DanielWainfleet Jul 22 '17 at 0:48
  • $\begingroup$ @Daniel: No, the point is the proof of the well ordering theorem in ZFC uses the power set axiom in a significant way. $\endgroup$ – Asaf Karagila Jul 22 '17 at 7:39
  • $\begingroup$ @Daniel: I mean, yes, the question can be phrased as an implication in ZF-P, but you're talking about the wrong implication. And I think it is instructive to think about this as a question in ZFC-P instead. $\endgroup$ – Asaf Karagila Jul 22 '17 at 8:23
  • $\begingroup$ @AsafKaragila . In that case the Q depends on which "version" of AC is specified. For example in Kunen's "Set Theory: An Introduction To Independence Proofs", he takes "Every set can be well-ordered" as AC... In this Q, it appears that AC is taken as the existence of choice-functions. $\endgroup$ – DanielWainfleet Jul 22 '17 at 17:23
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    $\begingroup$ @Daniel: I am a very big believer in the statement that AC always means the existence of choice functions. Once you start talking about various formulations and so on, you sort of lose some of the meaning. I mean, why not just take "Every vector space has a basis" and that's it? $\endgroup$ – Asaf Karagila Jul 22 '17 at 17:35

No, it cannot be proved that every set can be well-ordered.

You can find details, and much much more, in the following paper.

Gitman, Victoria; Hamkins, Joel David; Johnstone, Thomas A., What is the theory $\mathsf{ZFC}$ without power set?, ZBL06642525, arXiv.


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