In article <firstname.lastname@example.org>, WM <email@example.com> wrote:
> On 9 Dez., 18:40, Zuhair <zaljo...@gmail.com> wrote: > > On Dec 9, 1:14 pm, WM <mueck...@rz.fh-augsburg.de> wrote: > > > > > > > > > On 9 Dez., 10:41, Zuhair <zaljo...@gmail.com> wrote: > > > > > > On Dec 9, 11:35 am, WM <mueck...@rz.fh-augsburg.de> wrote:> On 9 Dez., > > > > 05:55, Zuhair <zaljo...@gmail.com> wrote: > > > > > > > > You need to prove that the set of all paths is countable, and so > > > > > > far > > > > > > you didn't present a proof of that. > > > > > > > The set of all finite paths is countable. Therefore it is not > > > > > possible > > > > > to define an infinite path by adding nodes to any finite path. All > > > > > nodes to be added are already in finite paths. Therefore, by > > > > > following > > > > > the nodes of a path, you never define an infinite path. It is > > > > > interesting that practically everybody not yet brainwashed can > > > > > understand that. > > > > > > Every node is reachable by a finite path, that's correct. But that is > > > > irrelevant > > > > here, we are speaking here about the number of all "path"s in the > > > > Binary Tree > > > > and not about the number of all nodes. we know that the number of all > > > > nodes > > > > is countable, the question is: is the number of all paths (finite and > > > > infinite) > > > > is countable? > > > > > So it has become obvious now, that is not possible to define "all > > > paths" by nodes. Only the finite paths can be defined by their nodes. > > > How can you define all paths if not by nodes? > > > > Simply there are non finitely definable paths. > > No they are not anywhere. Your assertion is simpky false. I construct > one path through each node such that every node has its own path.
The only way this can be done is to assign to each node the unique path ending at that node, but there are paths that do not have any ending node, so that the node to path mapping is not a surjection onto the set of paths. Proving there are more paths than nodes!
> By > this construction every node is covered by its own path. And there is > no chance to define any further path by further nodes. There are no > further nodes available.
Then every node must be a terminal node of every path containing it, making for a one node tree. > > > > Anyhow what is the proof that ALL reals can be represented by paths of > > an infinite Binary Tree (actually two trees). It looks that only a > > countable subset of reals can be represented in that way. I'm not sure > > really. > > Only a countable subset can be represented by the Binary Tree. The > reason is that no path is really actually infinite.
That may be the case in Wolkenmuekenheim, but is definitely not the case in standard mathematics > > > > Anyway the diagonal argument of Cantor is provable in very weak > > systems of ZFC which are proved to be consistent. So uncountability of > > reals is a possibility. > > And it leads to a contradiction with the fact that all real numbers > that are paths in the Binary Tree form a countable set.
Actually WM's version of Complete Binary Trees do not exist in ZFC. only the Complete Binary Trees with all paths being infinite can exist in ZFC. WMs version can only exist in Wolkenmuekenheim, which is completely incompatible with ZFC. > >
> > > > This includes the main bulk of experts on foundations of mathematics. > > The main bulk of experts on foundations of astrology is by far more > trustworthy.
Perhaps to those living in Wolkenmuekenheim, but I do not find it so in this world. --