Timeline for Why can't an explicit well-ordering of the reals be ruled out in ZF?
Current License: CC BY-SA 3.0
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Sep 6, 2012 at 21:42 | comment | added | Andreas Blass | Joel, you're right. In fact, it seems that the second formulation of B is actually equivalent (in PA or appropriate weaker theories) to $\neg$Con(ZF). | |
Sep 6, 2012 at 20:27 | comment | added | Joel David Hamkins | I think your second formulation actually is consistent with ZF, since it is consistent with ZF that $\neg\text{Con}(\text{ZF})$, and in such a model, proofs from ZF are easy to come by. That is, if there are any models of ZF at all, then there are models of ZF that have a definable well-ordering of the reals such that, in that model, there is a proof from ZF that this definition does define a well-ordering of the reals. | |
Sep 6, 2012 at 19:54 | history | answered | Andreas Blass | CC BY-SA 3.0 |