In article <firstname.lastname@example.org>, 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 > 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. --