In article <email@example.com>, WM <firstname.lastname@example.org> wrote:
> On 14 Jun., 14:49, William Hughes <wpihug...@hotmail.com> wrote: > > On Jun 14, 8:26 am, WM <mueck...@rz.fh-augsburg.de> wrote: > > > > > On 14 Jun., 14:01, 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." > > > > > > > > > > Assuming infinite paths exist: > > > > > > The path p is in the tree > > > > > > The path p can be distinguished from > > > > every element of P. > > > > > For every path p_n of P there is a digit such that p differs from p_n. > > > But there is no digit of p that differs from every path p_n of P > > > because, if p exists as binary representation, then p is the union of > > > all p_n and as such cannot differ from the union. > > > > The statement says nothing about p being distinguished from the union > > of all p_n, the statment says that p can be distinguished from every > > *element* > > of the union of all p_n. Do you wish to repudiate your repeated > > agreement to > > > > The path p can be distinguished > > from every element of P > > > > ? > > No. The union of paths is not larger than every element - unless we > agree to magic.
WM again loses track of what is going on. A union of paths, such as WM mentions, would be a set of nodes which is not, in general, a path at all. Hughes talks about unions of sets of paths, which would produce a new set of paths, each being a special sort of set of nodes, not merely a set of nodes which is not a path at all.
> > In particular p cannot be distinguished from the paths of P, used to > construct the tree, if I had included p in P.
But there is no way to include all potential paths in such a construction as WM describes, so whatever ones he uses, others appear in his final tree.
> I agree that p was not > included. But that does not play a role unless you are able to obtain > that from the complete tree. But you are not!
Maybe WM is not able to do so. He seems self-crippled in so many ways, but those less handicapped can. > > > > > > > > > > > > > > > > > > You refuse to agree to > > > > > > Assuming infinite paths exist: > > > > > > There is a path p in the tree that can > > > > be distinguished from every element of P. > > > > > > Until you explain this, no other argument as > > > > to why you refuse to accept this statment > > > > will be considered. > > > > > It is obviously impossible to distinguish p from all paths p_n that > > > are in the binary tree after construction has been completed. > > > > Indeed. However, the statement is not about > > distinguishing p from paths in the > > tree, it is about distinguishing p from paths in P. > > If this was possible before, then it would also be possible post > festum, because only P was used to construct the tree.
The point is that if only countably many paths are used in that construction, there will inevitably appear in the constructed tree other paths. At least so long as paths are defined as maximal totally ordered sets of nodes under the "ancestor" partial ordering of nodes.
> But it is not.
> Therefore it turns out that the original assumption "p can be > distuinguished from every path of P" is contradicted. Wrong!
> A classical proof by conradiction.
Except that it is not a proof because it relies on assumptions contrary to fact.
> By the way, one of the finest proofs that > exists in mathematics.
WM will break his arm patting himself on the back so hard.
But WM is doubly mistaken in his ego trip.
His argument is neither a proof nor is it in mathematics.