```Date: Feb 24, 2013 5:20 PM
Author: Virgil
Subject: Re: Matheology � 222 Back to the roots

In article <a07faa13-25cf-43de-8a4e-499f8e839339@l9g2000yqp.googlegroups.com>, WM <mueckenh@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.> As long as one has d, which DOES contain every line of L that is capable of containing everthing that its predecessors contain, one does not need such an L.In standard math, d is just a sort of union of all L's and its existence is required in such standard set theories as ZF and NBG.Model: in ZF with the von Neumann naturals each of WM's "lines" is merely modeled by the identity mapping on a nonempty natural, so every line is a superset of any prior line. But in ZF, the union of any such well-defined set of sets is itself a set so that union is d, the identity function on N.This works perfectly in ZF, so until WM can disprove ZF, he loses.--
```