In article <138f41db-b767-4ea6-87d7-3c37918ed63e@r16g2000vbn.googlegroups.com>, WM <mueckenh@rz.fh-augsburg.de> wrote:
> On 20 Jun., 13:19, William Hughes <wpihug...@hotmail.com> wrote: > > On Jun 20, 4:25 am, WM <mueck...@rz.fh-augsburg.de> 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." > > > > A: actually infinite paths exist, > > B: the infinite tree contains a path p that can be > > distinguished from every path of P. > > > > You agree > > > > A ==> B > > > > So if A is true then B is true > > and your claim is false. > > > > You want to show > > > > ~A [Follows from ( A ==> B, ~B) ==> ~A ] > > > > by proving ~B > > > > (Note that assuming ~A is circular) > > > > > I prove ~B, for instance by your inability to distinguish t from T, > > > with no regard to the truth of A. > > > > Nope. You need to distinguish the subset of nodes t from every > > element of P, not from the tree, > > The paths of the tree are the same as the subsets of P.
Not in the maximal infinite binary tree which one can so easily build from theactually infinite set of all 1-origin naturals by defining the left child of n to be 2*n+0 and the right child of n to be 2*n+1.
> There is no > reasonable argument that leads to a difference.
Cantor's argument re the incompleteness of any list of binary sequences is reasonable argument which leads to a difference. It is WM who has no reasonable argument prohibiting such a difference.
> Further we can derive > from the axioms that the union of sets does not contain any element, > that is not contained in at least one of the sets united.
It has been already granted that a countable set of paths can have every member of a countable set of nodes in at least one path.
But such a fact does not in any way require that such a countable set of paths contain each possible path of a maximal infinite binary tree, and Cantor proved that it did not.
We unite the > singlets {p_n} where p_n is in P. T = U{p_n}. There is no other path > in the tree. Every path is a countable set of nodes, all of which are members of members of P. But a family of sets whose union is a given set need not contain all subsets of that union, which is what WM's argument would require. So most paths of the tree are not in WM's set of paths, P. > > > > and you have > > repeatedly agreed that you can distinguish t from every element of > > P if A is true. > > You erroneously believed so.
It is hardly erroneous to believe Cantor's diagonal proof. So given any COUNTABLE set of paths, P, in a maximal infinite binary tree, there are "more" paths in the tree but not in P than actually in P.
> I could not contradict your error.
But you did contradict his truth.
> Therefore I let it go. Now you can see that you were wrong.
On the contrary, we all still see where WM is still wrong.
As soon as any actually infinite set exists, and one is necessary for any construction of a maximal infinite binary tree, then WM is wrong about such trees.
And unless there is at least one such actually infinite set, there are no such trees for WM to pontificate about.
> > >You have never shown that you cannot distinguish > > t from every element of P without regard to the truth of A. > > I cannot show that because I cannot disprove a dream unless the sleper > is awaked. This is accomplished by the tree.
Then it is WM who is doing all the dreaming, as his sort of maximal infinite binary tree can not ever exist anywhwere. Such trees can only exist in a world containing an actually infinite set. > > > > We are left we the result that you > > cannot prove A or ~A. > > We are left with the result ~B.
WM is left with the result that his sort of tree cannot exist at all.
In WM's world, a set of paths, P, with each path limited to finitely many right child nodes must contain a path containing infinitely many right child nodes.
That is not allowed to happen in mathematical set theories.
