
Re: UNCOUNTABILITY
Posted:
Dec 22, 2012 7:26 AM


On Dec 22, 11:09 am, Virgil <vir...@ligriv.com> wrote: > In article > <c30be87331904c3da11a67660228c...@v7g2000yqv.googlegroups.com>, > > > > > > > > > > Zuhair <zaljo...@gmail.com> wrote: > > On Dec 21, 8:44 pm, WM <mueck...@rz.fhaugsburg.de> wrote: > > > On 21 Dez., 17:36, Zuhair <zaljo...@gmail.com> wrote: > > > > Note finally: Every Cantor diagonal r differs from any other real > > > number by a finite initial segment n(r) of its string of digits. That > > > is not possible with the Binary Tree. A diagonal does not differ from > > > all finite paths, i.e., for every initial segment n(r) of every real > > > number r there exists a finite path of the Binary Tree that is n(r). > > > You may consider actual infinity as well as uncountable languages, but > > > that does not change the fact that Cantor's argument does not apply. > > > NO, Cantor's diagonal argument construct a diagonal r that differs > > from every element of a COUNTABLE set of reals by a finite initial > > segment n(r). You are just not getting Cantor's argument. Cantor's > > argument is not about diagonalizing the set of ALL reals since that is > > clearly not possible, Cantor's argument is about diagonalizing any > > COUNTABLE set of reals. So again for any *countable* set S of reals > > there is a diagonal r that differs from each real in S by a finite > > initial segment n(r). > > This has been PROVED by Cantor. This logically entails that the set R > > of ALL reals cannot be countable, since otherwise this leads to the > > obvious contradiction that R is missing a member of it which cannot > > be. > > > The complete infinite binary tree have paths that can represent any > > real! So no diagonal path can be defined over the whole set, this is > > clear. But again also using Cantor's argument we can prove that for > > Any subtree T of the complete infinite binary tree if T has countably > > many paths then we can define a diagonal path that is missing from T, > > i.e. a diagonal path that belongs to the complete infinite binary tree > > but yet missing from T. Thus the infinite binary tree itself cannot be > > countable! > > > I reviewed your writings about the infinite binary tree. You want to > > prove that if we assume completed infinity (which you don't believe it > > to be a consistent assumption, so you don't believe even in the > > possible existence of such objects that are completed infinite sets) > > then we will arrive at a contradiction, this contradiction is the > > CantorWM contradiction, that is: > > > If we assume that the sets N and R of all naturals and reals > > respectively are completed infinite sets, then it follows that > > > (1) R is strictly bigger than N by Cantor's proof > > (2) R is as big as N by the WM infinite binary tree proof > > I think you meant to say > > (2) R is no bigger than N by the WM infinite binary tree proof' > What you actually said would, at least in standard English, allow R and > N still to satisfy (1). >
Note 'strictly' in (1).
Zuhair

