Virgil
Posts:
4,482
Registered:
1/6/11
|
|
Re: Matheology � 203
Posted:
Jan 29, 2013 7:23 PM
|
|
In article <2c0d6d3b-4b27-488a-a006-7db8b50e685a@u1g2000yql.googlegroups.com>, WM <mueckenh@rz.fh-augsburg.de> wrote:
> On 29 Jan., 10:18, William Hughes <wpihug...@gmail.com> wrote: > > On Jan 29, 10:09 am, WM <mueck...@rz.fh-augsburg.de> wrote: > > > > > > > > > > > > > On 29 Jan., 09:54, William Hughes <wpihug...@gmail.com> wrote: > > > > > > On Jan 29, 9:33 am, WM <mueck...@rz.fh-augsburg.de> wrote: > > > > > > > "All" and "every" in impredicative statements about infinite sets. > > > > > > > Consider the following statements: > > > > > > > A) For every natural number n, P(n) is true. > > > > > B) There does not exist a natural number n such that P(n) is false. > > > > > C) For all natural numbers P is true. > > > > > > > A implies B but A does not imply C. > > > > > > Which is the point. Even though A > > > > does not imply C we still have > > > > A implies B. > > > > > > Let L be a list > > > > d the antidiagonal of L > > > > P(n), d does not equal the nth line of L > > > > > > We have (A) > > > > > > For every natural number n, P(n) is true. > > > > > > This implies (B) > > > > > > There does not exist a natural number n > > > > such that P(n) is false. > > > > > > In other words, there is no line of L that > > > > is equal to d. > > > > > And how can C be correct nevertheless? Because "For all" is > > > contradictory. > > > > B: There is no line of L that is equal to d > > > > does not imply > > > > C: For all n, line n is not equal to d. > > > > B correct does not mean "C correct nevertheless"- > > But we know of cases where C is correct nevertheless. I quoted four of > them in the § 203. Or do you disagree to one of them? > > In case you have forgotten the old discussion concerning the > configurations of the Binary Tree construction, here it is repeated: > > The complete infinite binary tree is the limit of the sequence of its > initial segments B_k:
How is this any different from |N being the the "limit' (or union) of all its finite initial segments but then having properties that none of its finite initial segments have, like infinitely many elements.? > > The structure of the Binary Tree excludes that there are any two > initial segments, B_k and B_(k+1), such that B_(k+1) contains two > complete infinite paths both of which are not contained in B_k.
In fact, no finite initial segment can contains any "Path" at all.
> Nevertheless the limit of all B_k is the complete binary tree > including all (uncountably many) infinite paths. Contradiction. There > cannot exist more than countably many infinite paths.
WRONG!
There can be uncountably many infinite binary sequences representing different real numbers in the unit interval, [0,1], even allowing for the countably many duplicate representations of binary rationals in that interval, and an easy bijection between them and the infinite paths in an infinite binary tree. --
|
|