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

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The discussion centers on the necessity of the Axiom of Choice (AC) for well-ordering the real numbers. It is established that while AC implies the existence of a well-ordering for any set, including the reals, the question remains whether the reals can be well-ordered without AC. A key point raised is that Cohen's proof of AC's independence demonstrated a model where the reals cannot be well-ordered. The conversation also touches on the consistency of ZF combined with the assertion that there is no well-ordering of the reals, but questions remain about the consistency of ZF without AC while still asserting a well-ordering exists. Ultimately, the necessity of AC for well-ordering the reals remains unresolved.
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".
 
I was reading documentation about the soundness and completeness of logic formal systems. Consider the following $$\vdash_S \phi$$ where ##S## is the proof-system making part the formal system and ##\phi## is a wff (well formed formula) of the formal language. Note the blank on left of the turnstile symbol ##\vdash_S##, as far as I can tell it actually represents the empty set. So what does it mean ? I guess it actually means ##\phi## is a theorem of the formal system, i.e. there is a...

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