Date: Feb 24, 2013 3:04 PM
Author: William Hughes
Subject: Re: Matheology § 222 Back to the roots

On Feb 24, 8:32 pm, WM <mueck...@rz.fh-augsburg.de> wrote:
> On 24 Feb., 15:56, William Hughes <wpihug...@gmail.com> wrote:
>

> > So, when WM says that a natural number m does not
> > exist, he may mean that you can prove it exists
> > but you cannot find it.

>
> > Suppose that P is a predicate such that
> > for every natural number m, P(m) is true.

>
> Example: Every line of the list L
>
> 1
> 1, 2
> 1, 2, 3
> ...
>
> contains all its predecessors.
>
>
>
>
>
>
>
>
>
>
>

> > Let x be a natural number such that
> > P(x) is false. According to WM you cannot
> > prove that x does not exist.  (WM
> > rejects the obvious proof by contradiction:

>
> >      Assume a natural number, x, such that P(x)
> >      is false exists.
> >      call it k
> >      Then P(k) is both true and false.
> >      Contradiction,  Thus the original assumption
> >      is false and no natural number, x, such
> >      that P(x) is false exists)

>
> > We will say that x is an unfindable natural
> > number.

>
> > It is interesting to note that WM agrees with
> > the usual results if you insert the term findable.

>
> > E.g.
>
> > There is no findable last element of the potentially
> > infinite set |N.

>
> > There is no findable index to a line of L that
> > contains d.

>
> > There is no ball with a findable index in the vase.
>
> > etc.
>
> > It does not really matter if nonfindable natural
> > numbers exist or not. They have no effect.

>
> > I suggest we give WM a teddy bear marked unfindable.
>
> I suggest, William keeps abd comforts it until he can find the first
> line of L that is not capable of containing everthing that its
> predecessors contain.

Every line of L is capable of containing everything that
its predecessors contain. However, according to WM, this
does not prove that there is not a line that is not
capable of containing everything that its predecessors
contain. In WM's world, just because you can't
find something does not mean it does not exists.

In Wolkenmuekenheim, we cannot prove that there is a
no line of L that contains every FIS of d. However,
since any such line is not findable questions about
its existence are moot.