In article <email@example.com>, WM <firstname.lastname@example.org> wrote:
> On 11 Jun., 13:00, William Hughes <wpihug...@hotmail.com> wrote: > > On Jun 11, 6:21 am, WM <mueck...@rz.fh-augsburg.de> wrote: > > > > > On 11 Jun., 00:12, William Hughes <wpihug...@hotmail.com> wrote: > > > > > > - There is more that one path through "that node" > > > > > > - There are countably many nodes below "that node" > > > > We are discussing WM's statement: > > > > Let us choose one of them and map it one that node. > > For the others we will find other nodes, below that one chosen. > > > > WM has agreed > > > > - There is more that one path through "that node" > > > > - There are countably many nodes below "that node" > > > > We are not at > > > > > > - Assuming that "For the others we > > > > will find other nodes, below that one chosen" > > > > is equivalent to assuming that there are only > > > > countably many paths through "that node" > > > > > No. It is not assuming this but *proving* this. > > > > Nope, "assuming" is not "proving". You give a putative > > proof that starts out by assuming what you want > > to prove. This is circular. > > > Not at all. I show that there are no nodes left over, after a > countable number of infinite paths have been constructed.
Since infinitely many paths use each node, and you have only eliminated one per node, you would then need to show that infinity minus one is zero for your argument to hold.
> My only > assumption is that in a tree, the nodes of which are completely > covered by paths, no further path can be constructed.
But that is not the case for a MAXIMAL infinite binary tree. For example, WM's own construction of a tree in which no path has more than finitely many 1 branchings and infinitely many 0 branchings By his own argument, correct for once, there is a path through every node of even a maximal infinite binary tree.
But such a tree is far from maximal as there are more paths missing than present.
> > My proof rests upon the fact that after the set of all finite paths of > the form 0.1, 0.11, 0.111, ... has been constructed, there is no > chance to construct the path 0.111... in addition.
Nevertheless, if a maximal tree that path must be present, so that WM's construction does not produce the desired maximal infinite binary tree. > > Regards, WM > > > > > Because paths cannot be distinguished without nodes. > > > > Proves nothing. We know that a countable > > number of elements can distinguish an uncountable > > number of subsets. A countable number of nodes can > > distinguish an uncountable number of paths. > > > > - William Hughes