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I'm new to set theory and ordinals and such things and currently have a question.

I have a proposition which says that the axiom of choice is equivalent to this: "the power set of every well-ordered set is well-ordered."

It's trivial that the statement is true by well-ordering (which is equivalent to axiom of choice). but I can't find the proof in the opposite way. I also have a proof which I can't understand. It seems that the author tried to prove well-ordering (for any sets) using the mentioned statement. but I can't get it.

Note that I need to prove well-ordering using the mentioned statement not proving the statement with or without AC!

This is the proof: Click here

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marked as duplicate by Asaf Karagila axiom-of-choice Jul 27 '18 at 13:57

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  • $\begingroup$ The key problem is that the Axiom of Regularity (or Foundation) plays a significant role in this proof. So it's often not easy to explain from a naive point of view. $\endgroup$ – Asaf Karagila Jul 27 '18 at 13:59
  • $\begingroup$ The proof you link to seems to have some typos, though. $\endgroup$ – Asaf Karagila Jul 27 '18 at 14:00