```Date: Jan 24, 2013 7:50 PM
Author: Jesse F. Hughes
Subject: Re: ZFC and God

WM <mueckenh@rz.fh-augsburg.de> writes:> On 24 Jan., 14:46, "Jesse F. Hughes" <je...@phiwumbda.org> wrote:>>> It would be swell if you could write it in more or less set-theoretic>> terms, since, after all, you are allegedly providing a proof in ZF.>>>> Thanks much.>> A last approach to support your understanding:> Define the set of all terminating decimals 0 =< x =< 1 in ZF.> Do all that you want to do (with respect to diagonalization).> Stop as soon as youÂ encounter a non-terminating decimalI've no idea what you're talking about.Let t:N -> [0, 1) be the usual list of non-terminating decimals.I define  d(j) = 7 if t_j(j) != 7      = 6 elseThere.  Done.  Just as soon as I specified what d is, it is obviouslynon-terminating.  There's no process here to stop.  The variable d wasundefined and then it was defined and once defined, it is clearly anon-terminating decimal.  (I even hate to use temporal talk like "oncedefined", but hopefully my meaning is clear enough.)I patiently await a proof in ZF that d is not non-terminating.  Funfact: the axioms of ZF don't talk about time or stopping or anythinglike that.  So, you'll have to rework this "argument" so that it is avalid argument in ZF.Let's start with the theorem you want to prove, shall we?  Whattheorem is it precisely?  Is it this:  Let d be defined as above.  Then d is a terminating decimal.Is that your claim?  I.e.,  Let d be defined as above.  Then there is an i in N such that for  all j > i, d(j) = 0.Or do you plan on proving something else?  Thanks much.  Eagerly awaiting further enlightenment, etc.-- "Being in the ring of algebraic integers is just kind of being in aweird place, but it's no different than if you are in an Elk's Lodgewith weird made up rules versus just being out in regular society."                                      -- James S. Harris, teacher
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