In article <556f71b5-452d-41d1-a00a-b6aef151bd36@l8g2000vbp.googlegroups.com>, WM <mueckenh@rz.fh-augsburg.de> wrote:
> On 18 Jun., 14:44, 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." > > > > 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) > > > > <snip evasion> > > > > Please answer yes or no > > > > t is not an element of P > > > > The actually infinite path t is not an element of P and not a path of > T, because t does not exist.
WE have an actually infinite set of naturals N = {1, 2, 3, ...}. WE can make that set into the set of nodes of a maximal infinite binary tree by assigning to each natural N a left child, 2+n + 0 ,and a right child, 2*n+1.
Note that we have, a priori, no paths at all but, a postiori, uncountably many of them
> Proof by the fact that t cannot be distinguished from the set P after > its elements have been ordered as paths of the tree.
In my tree, if P is only countably infinite, as a set of subsets of N, then there are more paths in the tree not in P than in it, even if the union of P is the set of all nodes.
One of WM's models of P would be, in my tree as defined above, the set of paths having at most finitely many odd naturals as nodes.
And WM tries to claim that among a family sets, P, with each set limited to finitely many odd naturals there must be a set containing infinitely many odd naturals.
