Is the Axiom of Choice Necessary for Well-Ordering the Reals?

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SUMMARY

The discussion centers on the necessity of the Axiom of Choice (AC) for the well-ordering of the real numbers (\mathbb{R}). It is established that the Axiom of Choice is equivalent to the existence of a well-ordering for any set, including \mathbb{R}. However, there is no proof that \mathbb{R} can be well-ordered without AC, as demonstrated by Cohen's work on the independence of AC, which showed a model where \mathbb{R} cannot be well-ordered. The conversation also touches on the consistency of ZF + "there is no well-ordering of the reals" being equiconsistent with ZFC.

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  • Understanding of Zermelo-Fraenkel set theory (ZF)
  • Familiarity with the Axiom of Choice (AC)
  • Knowledge of well-ordering principles
  • Basic concepts of model theory and independence proofs
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  • Research the implications of Cohen's independence proofs on set theory
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  • Explore well-ordering theorems and their applications
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Mathematicians, logicians, and students of set theory who are exploring the foundations of mathematics and the implications of the Axiom of Choice on well-ordering and real numbers.

dmuthuk
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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 \mathbb{R} 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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You're essentially asking if ZF + there exists a well-ordering of the reals is weaker than ZFC, right?
 
CRGreathouse said:
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.
 
dmuthuk said:
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.
 
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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