## a question about the Axiom of choice [closed]

how to proof the following conclusion: for any infinite set S,there exists a bijection f:S--->S*S implies the Axiom of choice.

Can you give a proof without the theory of ordinal numbers.

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This is not a research-level question. – Chris Eagle Aug 9 2011 at 8:12
I already gave a full answer to this question on math.SE, math.stackexchange.com/questions/56466 – Asaf Karagila Aug 9 2011 at 9:58
@Asaf: Isn't the OP asking for a proof WITHOUT using ordinals? – Emil Jeřábek Aug 9 2011 at 10:44
@Emil: I'm not sure if such proof was given, nor if it would be any close to the elegance and simplicity. I would much rather wait for the comment of the OP (preferably, of course, on math.SE) on whether or not he find this proof satisfying. Despite the relatively heavy reliance on Hartogs number in the proof, I do not see any great use of the "theory of ordinals" in the proof other than the fact ordinals are well ordered and a least ordinal can be chosen in the second case of the lemma. – Asaf Karagila Aug 9 2011 at 10:57
I just posted an answer on MSE. It is a proof that can be carried out in Z (ZF - Replacement), which is the best interpretation of "without the theory of ordinal numbers" that I could think of. – François G. Dorais Aug 9 2011 at 18:18
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