Is the existence of a well-ordering on R independent of ZF? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T07:15:19Z http://mathoverflow.net/feeds/question/5116 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/5116/is-the-existence-of-a-well-ordering-on-r-independent-of-zf Is the existence of a well-ordering on R independent of ZF? Qiaochu Yuan 2009-11-11T22:48:53Z 2009-11-12T00:13:08Z <p>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. </p> http://mathoverflow.net/questions/5116/is-the-existence-of-a-well-ordering-on-r-independent-of-zf/5119#5119 Answer by Ori Gurel-Gurevich for Is the existence of a well-ordering on R independent of ZF? Ori Gurel-Gurevich 2009-11-11T23:02:07Z 2009-11-11T23:02:07Z <p>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.</p> http://mathoverflow.net/questions/5116/is-the-existence-of-a-well-ordering-on-r-independent-of-zf/5127#5127 Answer by Richard Dore for Is the existence of a well-ordering on R independent of ZF? Richard Dore 2009-11-12T00:02:04Z 2009-11-12T00:13:08Z <p>Yes. Here's a sketched example:</p> <p>Start in L. Let P be the forcing which adds &omega;<sub>1</sub> many Cohen reals, and let G be an L-generic filter for P. Then L(&#8477;)<sup>L[G]</sup> will model ZF, but will have no well ordering of the reals. The point is that if &sigma; is an automorphism of P, then &sigma; can be extended to an elementary map from L[G] to L[&sigma;[G]], and this extension will fix L(&#8477;)<sup>L[G]</sup>. So if there was a well ordering of &#8477; in L(&#8477;)<sup>L[G]</sup>, it would give a well ordering of G which was fixed by &sigma;. But &sigma; can reorder the elements of G because of the homogeneity of P.</p>