In article <firstname.lastname@example.org>, WM <email@example.com> wrote:
> 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.
WRONG! Unless WM can also prove that every path in the constructed tree WA already there in his set P, he lies. > > > > (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.
Wile there are no nodes added, there now exist all sorts of sets of nodes that were not present in any member of P.
WM's inability to distinguish between nodes and sets of nodes, among a number of other important distinctions that he seems incapable of making, repeatedly destroys the validity of his arguments. > > > 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.
Not outside of WM's world of MathUnrealism.
If P is a countable set of paths(as sets of nodes), even though the union of P is the set of all nodes, there are sets of nodes which are maximal totally ordered subsets under the 'ancestor of" partial ordering of the set of all nodes ordering which are not in P.
> > Cantor's list argument, applied to sequences of digits, is applied to > single digits only.
WM cannot have read the same proof I read then.
In the proof I read, Cantor is presented with a list of binary digit sequences, say s_0, s_1, s_2, ..., and creates an entire sequence t_0 such that for all n, t_0(n) = 1 - s_n(n). But One can as easily create for each k a sequence of sequences t_k such that for all k and all n, t_k(n) = 1- s_n(n+k).
We now have at least as many nonmembers of that list as we have members.
And this holds for any such list of lists of binary digits.
Thus if any of WM's outre claims should turn out to be true, it is clear that WM is not competent to prove them. And many of them admit to valid counterproofs.