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Topic: ZFC and God
Replies: 45   Last Post: Apr 18, 2013 3:47 AM

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 mueckenh@rz.fh-augsburg.de Posts: 18,076 Registered: 1/29/05
Re: ZFC and God
Posted: Jan 22, 2013 11:06 AM

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!
>
> 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.
>
> From this, you claim that it follows that
>
> (Ek)(An)(d_k != b_n) (*)

No I do not confuse quantifiers. 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)
>
> 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. Up to any finite n, such a FIS (b_k1, b_k2, ..., b_kn) is
present in a complete list. And we know that there is no infinite n.

> > Do you agree that the actually infinite path {1, 2, ...} does not
> > differ from B, such that "behind" every natural nothing can happen? In
> > particular the inclusion of {1, 2, ...} does not change anything in B?
> > Therefore {1, 2, ...} cannot differ from the union B?

>
> I'm not at all sure what your question means, but I agree (of course)
> that the identity stated in (B) is correct. The set N is the union of
> all the sets {1,2,...,n} (using the convention that N's least element
> is 1, of course).

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

> > 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!

Regards, WM