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. My only assumption is that in a tree, the nodes of which are completely covered by paths, no further path can be constructed.
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.
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