In article <firstname.lastname@example.org>, WM <email@example.com> wrote:
> On 10 Dez., 11:09, Zuhair <zaljo...@gmail.com> wrote: > > > > I have constructed all paths that can be defined by nodes in the > > > Binary Tree, because no node is missing in my construction. > > > > Which Binary Tree, Is it the binary tree on NxN grid? > > This Binary Tree contains all possible combinations of 0 and 1 that > exist in an actual infinite sequence. > > 0. > / \ > 0 1 > / \ / \ > 0 1 0 1 > / > 0 ... > > > which indeed has > > countably many nodes. If we hold that the number of paths cannot > > exceed the number of nodes, then clearly it follows that this Binary > > tree of yours is countable. > > Correct: At every mode a path comes in and there are 2 possibilities > to continue. 2 is also the sum of 1 (incoming path) and 1 (node). As > this holds for every node in the whole tree, there cannot be more than > countably many possibilities to continue. > > > But the question is: > > > > Why must we believe that ALL reals can be represented in that > > Countable binary tree of yours. > > Because it contains all infinite sequences that possibly can be built > from 0, 1 or, as Cantor originally did, from m, w. If there were reals > outside, they would be without interest - they could not appear as > anti-diagonals.
Except that no tree of only countably many such paths can be complete.
The only proper test for countability of the set of all such paths, or infinite binary sequences, is whether one can construct a complete list of all of them.
But, ever since Cantor, that is known to be impossible, so WM is, as usual, wrong! > > > Just diagonalize the set of all infinite paths of your binary tree, > > That is not possible.
If that means that no complete listing of all binary sequences is possible, the WM is, for once, correct.
Which condition proves the set of all binary sequences does not satisfy the definition of countability, and thus is not countable.
> Every path is already there by definition of the > tree. Every finite path can be continued by 0 or by 1. More is > impossible in binaries. Do you doubt that every real has a binary > representation?
That is not relevant to the question of whether the set of all binary sequences is countable. > > > and then you'll get a path that is not in your tree and that of course > > correspond to some real. So your tree provably cannot supply a > > representation for each real. > > > Look at the tree. Tell me what real number between 0 and 1 is missing. > You will never find any.
Since you do not give enough information to determine whether an arbitrary path is or is not in your tree, your tree is incompletely defined.
But your claim of its having only countably many paths, when the set of all possible paths is provably not countable, allows us to deduce that your tree is missing more paths than it contains. --