On 12/10/2012 2:43 PM, Virgil wrote: > In article > <firstname.lastname@example.org>, > WM <email@example.com> wrote: > >> On 10 Dez., 01:02, fom <fomJ...@nyms.net> wrote: >>> On 12/9/2012 12:30 PM, WM wrote: >>> >>>> On 9 Dez., 17:24, fom <fomJ...@nyms.net> wrote: >>>>> On 12/9/2012 3:20 AM, WM wrote: >>> >>> <snip> >>> >>>>> So, why is there no global axiom of choice? >>> >>>> As far as I am informed, *the* axiom of choice is global. There is no >>>> exception. Zermelo proved: Every set can be well-ordered. >>> >>> The axiom of choice only applies to sets within >>> a given model. >> >> Zermelo proved that every set can be well-ordered - without mentioning >> any model. My interest is solely the set of real numbers. It is >> covered by Zermelo's proof. > > Does WM claim that Zermelo's "proof" must hold in every model, that it > is somehow universal?
Well, it certainly holds whenever the theory being modeled has the axoim of choice.
I wonder how the claim holds when the axiom of determinacy is in force and the axiom of choice is inconsistent.
I suppose, that the claim is interpretable along the lines of finitism. Completeness is of no issue. What can be proved using a sound deductive system is what is true. Then the only real numbers are the constructive real numbers.