Date: Jan 24, 2013 7:39 PM
Author: Jesse F. Hughes
Subject: Re: ZFC and God
WM <email@example.com> 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
> 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 the
anti-diagonal d and it is a triviality to see that it is an infinite
sequence of non-zero and non-nine digits. And this fact really has
nothing at all to do with limits of sequences of squares. It is all
>> > 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.
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 risk
getting it wrong. I'd prefer that you explicitly give an argument in
ZF 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 them
pass. 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