Date: Jan 26, 2013 5:04 PM
Author: mueckenh@rz.fh-augsburg.de
Subject: Re: ZFC and God

On 26 Jan., 16:06, "Jesse F. Hughes" <je...@phiwumbda.org> wrote:
> WM <mueck...@rz.fh-augsburg.de> writes:
> > On 26 Jan., 02:50, "Jesse F. Hughes" <je...@phiwumbda.org> wrote:

> >> I asked how you define terminating decimal representation.  How is
> >> that meaningless?

>
> > Sorry, where did you ask?
>
> You've snipped the question three times, in the thread directly
> preceding this post.


Please excuse me, but there are very many text that I have to read.
Sometimes I overlook something. Nevertheless, I answered it:

> > The latter is not quite correct, because a terminating decimal
> > representation has nothing behind its last digit d_n, neither zeros
> > nor any other digits. (But of course, we can expand every terminating
> > decimal by a finite set of further decimals d_j = 0 for every j with n
> > <j <m, m in N.)

>
> I don't know why you want to avoid using the usual convention that
> 0.1 = 0.1000...., but okay.  It makes no difference.


Sorry, it makes a difference. 0.1000... is an infinite path in the
Binary Tree. But in the original question I distingusihed the Binary
Tree constructed by all finite initial segments of infinite paths and
the Binary Tree constructed by (all) infinite paths. Of course it
makes not a difference with respect to nodes. But, according to the
belief of matheologians, it makes a big difference with respect to
paths.
>
> Let's state the definition explicitly then:
>
>   Let x be a real number in [0,1].  We say that x has a terminating
>   decimal representation iff there is a natural number k and a
>   function f:{1,...,k} -> {0,...,9} such that
>
>    x = sum_i=1^k f(i) * 10^-i.
>
> Right?


Right.
>
> Now, let {t_i} be a list of all the finite decimal representations of
> reals, that is, each t_i is a finite decimal representation, and every
> finite decimal representation is in the list.  For each t_i, let k_i
> be the "length" of t_i.
>
> And we define a sequence d_j so that
>
>   d_j = 7 if j > k or t_j(j) != 7
>   d_j = 6 if j <= k and t_j(j) = 7.
>
> As before, we can notice the following facts:
>
>   d_j is defined for every j in N.
>   d_j = 7 or d_j = 6 for every j in N.
>
> Clearly, d_j is *NOT* a finite sequence.  Moreover, since the sequence
> d_j does not end in trailing 0s or 9s, the real number d defined by
>
>   d = sum_i=1^oo d_i & 10^-i
>
> has no finite decimal representation.
>
> Now, please tell me what is unclear about these obvious facts?


It is unclear why you apparently are unable to understand, that we are
working in the set of terminating decimals. Therefore the diagonal
cannot be actually infinite, although there is no last digit.

Can't you understand that the Binary Tree constructed by all
terminating paths has no last level and is nevertheless not actually
infinite because it does not contain any actually infinite path like
that of 1/9, 1/7, 1/3, and many, many more?

You have to distinguish these both cases, because the most important
argument of matheologians is the following: The tree constructed by
means of all finite paths does not contain the actually infinite
paths. Therefore I restrict the discussion to all finite paths. And
obviously the diagonal cannot be longer, because its digits consist
only of digits of the finite paths. It the diagonal nevertheless
appears infinite to you, then this fact only shows that the most
important argument is wrong.

Regards, WM