In article <firstname.lastname@example.org>, WM <email@example.com> wrote:
> 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, ...? > > This is really a splendid idea! In this manner all real numbers of the > unit interval can be inserted into the infinite binary tree, 1/3 and > 3/137 and even 1/pi. In addidion the tree can be completed by all the > "terminating paths" (those with tails 000...) > > Alas, we have used for construction a countable set of paths (because > all real numbers you know or can construct by means of Cantor's > diagonal method form a countable set).
But every such countable set is of paths, like every countable set of other sorts of binary sequences, is provably incomplete, so there is no ay to count all of them. > > This leads to the result that there are not uncountably many paths in > the tree, unless they sneak in during / after construction.
They do! it is in forming the union of all the node sets of finite trees that the node set of the infinite tree gets its uncountably many paths.
> Refusing > such we magic, we find that there are no infinite paths at all.
That is only because WM is incapable of looking at anything outside his MathUnrealistic world and seeing what is there.
Once one has an infinite number of nodes making a complete infinite binary tree, the result that WM refuses to accept is inevitable.
> And > this implies that we cannot start off with 0.111....
In an attempt to list all such binary sequences we can start with any of them. The "cannot" is in trying to finish it off.