In article <firstname.lastname@example.org>, WM <email@example.com> wrote:
> On 10 Dez., 06:32, Zuhair <zaljo...@gmail.com> wrote: > > On Dec 10, 12:21 am, WM <mueck...@rz.fh-augsburg.de> wrote: > >
> > If so what is the proof that ALL reals belong to that tree? > > There is no proof because the Binary Tree contains all reals between 0 > and 1 by definition.
What definition is that? I know of no such definition.
While it is certainly possible to surject the set of paths of a truly complete binary tree, of the sort that is forbidden to grow in WM's garden, none of those reals really IS a path.
> Every binary sequence that is not in the BinaryTree
Only in Wolkenmuekenheim are there any. Elsewhere Every binary sequence is in any complete infinite binary tree.
> is not the representation of a real number between 0 and 1. > Cantor's proof (concerning the binaries with bits w and m) shows that > not all that are in the tree can be in the list.
Right for once! > > > > > And what is the proof that the number of paths in that tree is > > countable? > > This proof is given by constructing the whole Binary Tree by countably > many actually infinite paths (i.e. finite paths with infinite endings > like 010101... or 000... or 111... or 001001001... or any desired > ending that I do not publish). My proof rests upon your inability to > find out what paths are missing.
And Cantor's counterproof says that no such list contains all paths and provides an unambiguous way of determining from any such list ( which exists as a consequence of WM's claim of countability) any finite number of the uncountably many paths which are missing. > > > I can see that the number of paths in any FINITE binary > > tree is less than the number of its nodes? but can that feature > > survive at infinite level? and what is the proof? I do see that the > > number of nodes in your tree is countable. But would it follow that > > the number of paths must be so at infinite level? > > Find a path that I have not used!
List the ones you have used (and until you have listed them , or at least prove you can list them, your claim of countability is notstablished) and then it is trivial to find others. > > > > How I see matters is that if I assume that there is a bijection > > between N and the set of all infinite paths in your tree, then I can > > easily construct a diagonal using Cantor's argument, and this diagonal > > would provably be a path that is not in that Tree. So either your tree > > must have uncountably many paths (with countably many nodes) or your > > tree has countably many paths but is incomplete, i.e. there are > > infinite binary paths that are not paths of it. > > > Or the idea of countable and uncountable sets is humbug. Why do you > refuse to take into account this possibility?
In large part because those like WM who argue against it here are so obviously unable to make their case.
> Because you cannot > believe that many thousands of mathematicians have behaved like fools > in the last hundred years? No reason to be ashamed. I have been among > them myself for a long time.