Date: Jan 22, 2013 4:23 PM
Author: Virgil
Subject: Re: ZFC and God

In article 
<e3d3bd8d-5831-46aa-9f98-cd13ef3d09fe@t5g2000vba.googlegroups.com>,
WM <mueckenh@rz.fh-augsburg.de> wrote:

> On 22 Jan., 15:49, "Jesse F. Hughes" <je...@phiwumbda.org> wrote:
>
> FIS: Finite Initial Segment
> FISON: Finite Initial Segment Of Naturals (or indices)
>

> > Well, it's not a "union" in the usual sense, but let's let it pass.
>
> It is a union in that sense that every FISON {1, 2, ..., n+1} added
> contains all smaler FISONs {1, 2, ..., n}. This does never change. In
> particular the set is always finite. If you add with always doubling
> frequence, you can add all FISONs in finite time - given that an "all"
> is meaningfull here. But at the end, you think, we cannot follow so
> quickly, and abracadabra we get something larger than every FISON? Not
> in mathematics!


One can in ZFC have the set of all FISONs, and thus also the union of
the set of fall FISONs, say |N, and then, since there is no maximal
FISON, That union cannot be a FISON.
> >
> > Here's the problem. Let d(n) be the n'th digit of d, and similarly
> > b_k(n) the n'th digit of b_k. Let d_k be the finite initial segment
> > of d consisting of the first k digits.
> >
> > We can prove, as you say, that
> >
> > (An)(d(n) != b_n(n)).

>
> From this "for every n" you claim "for all". And that is wrong for
> infinite sets.


It is true in ZFC for members of arbitrary sets, even if not in
WMytheology.
> >
> > From this, you claim that it follows that
> >
> > (Ek)(An)(d_k != b_n) (*)

>
> No I do not confuse quantifiers.


Actually quantifier dyslexia is one of WM's known fao=ults.

> I simply know that for FIS of the
> diagonal there exist a FIS in the list such that:
>
> (An)(Ek) (d_1,d_2, ..., d_n) = (b_k1, b_k2, ..., b_kn)


The above does not make any sense, as stated.
> >
> > I simply do not see how that follows.

>
> It follows from the purported completeness of the list (or decimal
> tree) that contains all possible FIS of decimal representations of
> reals.


Such a claim needs far more proof than WM's handwaving.

Note that while each finite initial segment of the "diagonal" will
appear somewhere in WM's list, that finite initial segment of length n
will NOT appear in line n (because the diagonal is contructed to prevent
it from happening.), and it is irrelevant in any other line.

This can be shown by adding one more "letter" to the alphabet used and
filling in all the blanks of the finite sequences with that new letter.
Now all the sequences, including the diagonal, are equally infinite, and
it is trivial to see that the diagonal, which does not contain that new
letter, differs from every sequence that does contain it, which every
listed line.





>
> And all these sets are finite. In particular no FIS contains more
> natural indices than every FIS. So the union cannot contain more
> either.


Nor less, so must contain each and every of more than any finite number
of them.
>
> > > By the way, every FISON, which stands for finite initial segment of
> > > the naturals, is finite. This does not change by unioning as many as
> > > are available. The union never gets larger than every FISON. Can you
> > > understand that?

> >
> > Of course that claim is simply nonsense, but let's focus on the matter
> > at hand. Are you claiming that the "reason" that some d_k is missing
> > from the list of b_n's is that

>
> There is no FIS (d_1,d_2, ..., d_n) of the diagonal missing, since
> the list is complete!


But as the diagonal contains at least one more than any finite listing,
it differs from each finite listing.
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