Date: Feb 24, 2013 5:20 PM
Subject: Re: Matheology � 222 Back to the roots
WM <email@example.com> 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, 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.