Re: [math-fun] Fermat's Last Theorem question
The controversy arises from Wiles' use of Grothendieck's development of algebraic geometry, and in particular of so-called "cohomological number theory". As a matter of technical convenience, Grothendieck assumes the existence of a "universe", which is, essentially, a set that can model ZFC. The existence of such a set implies the consistency of ZFC, so (by Godel Incompleteness) it is an assumption strictly stronger than ZFC.
By the Law of Excluded Middle, either ZFC is consistent (in which case it admits a model, and Wiles' proof works) or inconsistent (in which case, by the Principle of Explosion, we can derive any statement, including FLT). So, ZFC --> FLT. Sincerely, Adam P. Goucher
On Wed, Feb 20, 2013 at 2:00 PM, Adam P. Goucher <apgoucher@gmx.com> wrote:
The controversy arises from Wiles' use of Grothendieck's development of algebraic geometry, and in particular of so-called "cohomological number theory". As a matter of technical convenience, Grothendieck assumes the existence of a "universe", which is, essentially, a set that can model ZFC. The existence of such a set implies the consistency of ZFC, so (by Godel Incompleteness) it is an assumption strictly stronger than ZFC.
By the Law of Excluded Middle, either ZFC is consistent (in which case it admits a model, and Wiles' proof works) or inconsistent (in which case, by the Principle of Explosion, we can derive any statement, including FLT). So, ZFC --> FLT.
Careful! remember that since Consis(ZF) is not provable in ZF, then if ZF is consistent, then ZF + (Not Consis(ZF)) is also consistent, and hence has a model. "Having a model" is different from "has as a model the thing I normally think of as sets and membership". Assuming ZF is consistent, there are things that are provable from ZF + Consis(ZF) that are not provable from ZF (trivially; Consis(ZF) is one). It's possible that FLT is one. Andy
Sincerely,
Adam P. Goucher
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