In article <4223f1bf-a04e-4e14-8513-c642212d4827@21g2000vbk.googlegroups.com>, WM <mueckenh@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" > > > > - 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.
WM has not the least notion of how to distinguish valid proofs from invalid attempts, so that his claims of "proof" are all nonsense.
> Because paths cannot > be distinguished without nodes.
Non Sequitur. > > Equivalently: Assuming that that, in Cantor's proof, every line > contains a digit that differs from the corresponding digit of the anti > diagonal would be assuming that the anti diagonal is not in the list.
There is a rule by which just such a situation can be made to occur.
And until WM can show why that rule does not achieve the claimed result, which he has often tried but universally failed to do, the proof remains valid. > > > > - You are trying to prove that there are countably many > > paths through the root node. > > > > - This is equivalent to proving that there are countably many > > paths through "that node" > > This is equivalent to proving that there are only countably many > possibilities of distinguishing paths below "that node".
Through through node in a maximal infinite binary tree there are as many paths as from the root node, and more than there are nodes, at least in cardinality.
And that will remain the case until someone can show a surjection from the set of nodes to the set of paths.
