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.
> Just diagonalize the set of all infinite paths of your binary tree,
That is not possible. 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?
> 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.