On 11 Jun., 21:33, William Hughes <wpihug...@hotmail.com> wrote: > On Jun 11, 2:51 pm, WM <mueck...@rz.fh-augsburg.de> wrote: > > > > > > > On 11 Jun., 20:02, William Hughes <wpihug...@hotmail.com> wrote: > > > > On Jun 11, 1:21 pm, 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. > > > > > If you have a tree that is (the nodes of which are) completely covered > > > > by a set of paths then you cannot show any node that will distinguish > > > > a further path from that given set of paths. > > > > You can, however, show a subset of nodes that will distinguish > > > a further path from that given set of paths. > > > But only before those paths have occupied the tree. > > Nope. Let the given set of paths be P. > > Please acknowledge > > There is a subset of nodes that cannot be found in a single > element of P. > > A subset of nodes is distinguished from an element > p of P if and only if it is not contained in p. > > A subset of nodes is distinguished from every > element of P if and only if it is not contained in > a single element of P. > > There is a subset of nodes > which forms a path that is not contained in a single element of P. > This subset of nodes is distinguished from every element > of P. > > The subset of nodes can be found in the tree, however > it cannot be found in a single element of P.
All right as long as P is not used to construct the binary tree. Afterwards, however, we can see that every p is present in the tree.
Either sneaking in at _some_ (which?) stage, based upon unmathematical superstition, or recognizing that there are no actually infinite paths.
