In article <firstname.lastname@example.org>, WM <email@example.com> wrote:
> On 12 Jun., 09:17, Rainer Rosenthal <r.rosent...@web.de> wrote: > > WM schrieb: > > > > > On 11 Jun., 14:13, Rainer Rosenthal <r.rosent...@web.de> wrote: > > >> WM wrote: > > >>> 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. > > >> Why not start with path 0.111... and then add those other paths > > >> 0.1, 0.11, 0.111, ...? > > > > > [Many lines snipped, because I felt they didn't get to the point I > > > was trying to ask] > > > However, your contribution to this problem is of high value and I > > > appreciate it very much. > > > > Thank you, but it seems as if I wasn't clear in my question. > > Didn't you say: "If I am going to enumerate the real numbers > > in such and such a way then there will be some of them not > > in my list"? > > It is not enumerating all real numbers. > What I do is to use actually infinite paths (at the end we will see > that they do not exist, but I assume their existence) for constructing > the complete binary tree. As a simple example I use only such paths > that have tails of the form 000..., i.e., "terminating" paths. > > The result is: They exhaust the complete infinite binary tree. There > is no node that remains uncovered. And there remains no node and no > set of nodes that could be used to construct a path that is not yet in > the tree. The tree is complete at the end. With the tip of your finger > you can follow every path, even 1/pi or 1/3.
While there is no node "uncovered" there are maximal totally ordered sets of nodes under the order induced by the the "parent of" relation (paths) which have not been accounted for. Namely all those which have infnitely many 1's in there binary representation. > > > > But then, that doesn't prove all too much, because it's obvious > > that there are many ways of not counting the reals. > > The point is that there is no way of counting the reals. > > As I said, I do not count the reals.
Because you can't.
> I cover all possible infinite > sequences of bits by means of a countable set of paths = a countable > set of binary sequences.
You have deliberately overlooked every sequence of bits which contains as many 1 bits as 0 bits. WHich is most of them. > > And your idea to start with some paths like 0.111... or 1/pi that do > not belong to the countable set (of "terminating" paths, i.e., such > with tails 000...) that I use for construction, shows us, that the > result is in no way different.
You have deliberately overlooked every sequence of bits which contains as many 1 bits as 0 bits. WHich is most of them
> But how can we explain that the result > is not influenced by those paths? In my opinion by recognizing that > such paths do not exist.
So a path of infinitely many 0's does exist but a path of infinitely many 1's does not?
Then your tree is not maximal, and omits some necessary nodes.