On 14 Jun., 22:47, William Hughes <wpihug...@hotmail.com> wrote: > On Jun 14, 2:35 pm, WM <mueck...@rz.fh-augsburg.de> wrote: > > > On 14 Jun., 16:49, William Hughes <wpihug...@hotmail.com> wrote: > > Your claim is that "no possibility exists to construct or to > distinguish by one or many or infinitely many nodes > of the tree another path." > > > There are two statements: > > 1) Path p can be distinguished from every path of P. > > 2) Path p cannot be distinguished from every path of P. > > > The first is assumed to be correct before P was used to construct the > > tree. > > Nope. No assumption. You agreed that statement 1 > is correct before P was used to construct the tree.
I agreed under the assumption that actual infinity exists, in order to contradict this assumption. > > You agreed that constucting the tree does not change > path p or any element of P
That is correct. It is a basic of mathematics. > > You are now trying to claim statement 2 is correct > after P is used to construct the tree.
I can prove that statement 2 is correct. It is proved by your inability to distinguish p from all paths of the tree. > > (Note P is used to construct the nodes of the > tree, the nodes of the tree are used to construct > the paths of the tree. The set of paths in P > and the set of paths in the tree are not the same.
There is nothing added. So you are trying to invent magic.
> The fact that p cannot be distinguished from > all paths in the tree does not mean that p cannot > be distinguished from all paths in P)
It does. > > > The second statement can be proved to be correct, because in fact you > > are not able to distinguish p from P (by means of digits). > > P contains every digit in p, so you are not able to distinguish > p from P by digits. P does not contain every subset of digits > in p, so you are able to distinguish p from P by subsets of digits.
Cantor's list argument, applied to sequences of digits, is applied to single digits only. If this is not justified, then his proof is false.
