Date: Mar 2, 2013 5:55 PM
Author: Virgil
Subject: Re: Matheology � 222 Back to the roots
In article

<f050fc72-6c3d-4360-87e3-cdc5293a2cf7@x15g2000vbj.googlegroups.com>,

WM <mueckenh@rz.fh-augsburg.de> wrote:

> On 2 Mrz., 19:24, William Hughes <wpihug...@gmail.com> wrote:

> > On Mar 2, 6:27 pm, WM <mueck...@rz.fh-augsburg.de> wrote:

> >

> > We both agree that there is a natural number

> > valued function of time, m(t), such that

> > at any time t, m(t) is the index of an existing

> > line which contains all existing FIS of d.

> > We each believe that our m(t) is not constant.

> >

> > We also agree that there does not exist

> > (in the sense of not able to find) a

> > natural number n such that the

> > nth line of L is coFIS with the

> > diagonal.

> >

> > I find your characterization of this

> > situation as "there is a natural

> > number m such that the mth line

> > of L is coFIS with the diagonal"

>

> since there do not exist more than m FIS of the diagonal.

But m(t) is not merely a value but is a function with no (finite) upper

bound.

>

> > to be silly.

>

> Because you do not yet fully understand potential infinity: There do

> not exist more than m FIS of the diagonal.

But m(t) is not merely a value but is a function with no (finite) upper

bound.

>

> Question: Do you find your characterization of the situation in

> finished infinity not silly?

A good deal less silly that WM's claim of a natural with no successor.

> Don't you see a mathematical

> contradiction of the sentence: There are all FIS of d in the list but

> not in one single line?

Nope!

Create the list of von Neumann naturals taken in order, so that each

line is both a member of and a FIS of the next line, then the set of all

FISs is just the set of all such naturals, which cannot, by definition,

be contained in any one line.

Note that WM violently dislikes the von Neumann naturals because they

are so useful in poking holes in WM's phony arguments.

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