```Date: Jan 24, 2013 7:39 PM
Author: Jesse F. Hughes
Subject: Re: ZFC and God

WM <mueckenh@rz.fh-augsburg.de> writes:> On 24 Jan., 14:16, "Jesse F. Hughes" <je...@phiwumbda.org> wrote:>> WM <mueck...@rz.fh-augsburg.de> writes:>> > You will have have recognized that here the diagonal argument is>> > applied. It is obvious that up to every line = column the list is a>> > square.>>>> It is clear that, for all j, d(j) != t_j(j) and hence d != t_j.  If>> that's what you mean by the diagonal argument, great!>>>> Once again, however, you say something that has no clear meaning to>> me.  Can you clarify "It is obvious that up to every line = column the>> list is a square?"  I've no clue what it means.>> Then ponder a while about the following sequence>> d>> d1> 2d>> d11> 2d2> 33d>> and so on. In every square there are as many d's as lines. The same> could be shown for the columns.Yes, in this sequence of three squares, what you say is true.But none of this is relevant, because we've explicitly defined theanti-diagonal d and it is a triviality to see that it is an infinitesequence of non-zero and non-nine digits.  And this fact really hasnothing at all to do with limits of sequences of squares.  It is allperfectly explicit.>> > Neither the set of t_i does have a largest element. Nevertheless there>> > is no t_i of actually infinite length.>>>> Correct, to both claims.  So what?  The digit d(j) is defined for>> every j, and is neither 0 nor 9, so d is a real number which has no>> terminating decimal representation.>>>> Do you agree with that (obvious) claim or not?>> No. Pleeze try to understand: Presently we work in the system of> terminating decimals, by definition. There is no non-terminating> decimal. If we were so unlucky to met any non-terminating decimal we> would not know what to do. It would not even be defined. If the> diagonal gets non-terminating, we have to stop and cut it before it> becomes non-terminating.Golly.Perhaps we should start more slowly.Do you agree that (by presumption) t_i is defined for every i in N?>>>>> Sorry, I could've sworn you were talking about what ZF proves.  Can>> you state this in the language of ZF and prove it?  Much thanks!>> Can't you imagine to define in ZF the set of all terminating decimals?> Is it too hard? Just take the above set that you couldn't recover in> our arguing, namely the set of all FIS (1, 2, ..., n) with n in n as> indices of the digits.I don't want to imagine what you are thinking, because I will riskgetting it wrong.  I'd prefer that you explicitly give an argument inZF so that we can determine whether it is valid or not.>>> > In particular, what would be changed in the length of d if we admitted>> > also non-terminating t_i (of infinite length)?>>>> Nothing would change.  So what?>> Look, presently we work in the system of terminating decimals - by> definition. If nothing changes when we switch to the system of non-> terminating decimals, do we switch then at all? How could we recognize> that we have switched?I don't have any idea what these questions mean and will let thempass.  Let's instead see a valid proof of a contradiction in ZF.-- Jesse F. Hughes"It's not really winning if you don't get to where you want to go."                   -- An inspirational slogan from James S. Harris
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