In article <9b068051-20e6-40e7-8fa5-8842974228e4@k38g2000yqh.googlegroups.com>, WM <mueckenh@rz.fh-augsburg.de> wrote:
> 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.
Are they like an invading army? How do paths "occupy" a tree.
If you mean that the set of nodes forming the path is subset of the set of nodes forming the tree, there must be a better way of expressing it. But if that is what you mean, Then every path comes into being as soon as the set of nodes of that tree comes into being. > > > > 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.
The completion of the set of nodes of a tree automatically and immediately creates each subset of the set of nodes, thus creates each path in that tree.
There is no deferred creation of such paths possible.
