In article <ac5850a7-9c4c-4cd2-9ff7-cb56e419bcf3@y17g2000yqn.googlegroups.com>, WM <mueckenh@rz.fh-augsburg.de> wrote:
> On 17 Jun., 20:19, 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. > > So it is. > > > > > What do you understand by "all nodes"? > > > > "all nodes of t are in the tree" > > There is no node that is in t but is not in the tree > > That is correct. > > > > "all nodes of t are in P" > > There is no node that is in t but is not in an element of P > > Also correct. Every node of t and all predecessor nodes are in one > path of P.
But unless every chain of successor nodes to any node in t is in P, there are paths not in P. > > > > WM: the list of paths P is the same as the tree
Nope: A set of sets of nodes is not a set of nodes. > > > > Nope. You have agreed that the tree contains a subset of > > nodes that is not contained in one element of the list of paths P. > > All subsets of nodes that are in the tree also are in the list of > paths.
It is not even true that all totally-ordered-by-'ancestor of' sets of nodes are in P, at least if P is countable. And no set that is not thus totally ordered can possibly be in P.
