Is the existence of a well-ordering on R independent of ZF? - MathOverflow

Is the existence of a well-ordering on R independent of ZF?

2009-11-11T22:48:53Z

I am reasonably certain this is the case, but can't find a reference that actually states this, although the Wikipedia article states something close.

Answer by Ori Gurel-Gurevich for Is the existence of a well-ordering on R independent of ZF?

2009-11-11T23:02:07Z

It is possible to have all the subsets of R be measurable (Solovay, Robert M. (1970). "A model of set-theory in which every set of reals is Lebesgue measurable". Annals of Mathematics. Second Series 92: 1–56.) which implies the nonexistence of a well ordering of R.

Answer by Richard Dore for Is the existence of a well-ordering on R independent of ZF?

2009-11-12T00:02:04Z

Yes. Here's a sketched example:

Start in L. Let P be the forcing which adds ω₁ many Cohen reals, and let G be an L-generic filter for P. Then L(ℝ)^L[G] will model ZF, but will have no well ordering of the reals. The point is that if σ is an automorphism of P, then σ can be extended to an elementary map from L[G] to L[σ[G]], and this extension will fix L(ℝ)^L[G]. So if there was a well ordering of ℝ in L(ℝ)^L[G], it would give a well ordering of G which was fixed by σ. But σ can reorder the elements of G because of the homogeneity of P.