Well-ordering of the Reals

  • Thread starter dmuthuk
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Main Question or Discussion Point

We all know that the axiom of choice is equivalent to the existence of a well-ordering for any set. And, this of course implies that [tex]\mathbb{R}[/tex] can be well-ordered, in particular. However, how do we know that the axiom of choice is actually needed in the case of the reals? That is, if we remove the axiom of choice, do the reals become a set that cannot be well-ordered? Furthermore, is the axiom of choice needed for every uncountable set?
 

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CRGreathouse
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You're essentially asking if ZF + there exists a well-ordering of the reals is weaker than ZFC, right?
 
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You're essentially asking if ZF + there exists a well-ordering of the reals is weaker than ZFC, right?
Yes, I believe I am. So, I guess what I wanted to know is if there exists a proof that the reals can be well-ordered without AC.
 
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Yes, I believe I am. So, I guess what I wanted to know is if there exists a proof that the reals can be well-ordered without AC.
No there isn't. When Cohen proved the independence of AC he used a model in which there was no well-ordering of the reals.
 
CRGreathouse
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Does that actually prove what dmuthuk asked? I know that ZF + "there is no well-ordering of the reals" is consistent*, but what about ZF + ¬C + "there is a well-ordering of the reals"?


* By "consistent", I mean "equiconsistent with ZFC".
 

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