# Well-ordering of the Reals

1. Apr 22, 2008

### dmuthuk

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?

2. Apr 22, 2008

### CRGreathouse

You're essentially asking if ZF + there exists a well-ordering of the reals is weaker than ZFC, right?

3. Apr 28, 2008

### dmuthuk

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.

4. May 1, 2008

### Hubert

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.

5. May 2, 2008

### CRGreathouse

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".