Virgil
Posts:
8,833
Registered:
1/6/11


Re: UNCOUNTABILITY
Posted:
Dec 23, 2012 4:13 PM


In article <b48ee3dd66ed4c00a97f27cbbab23f7c@f4g2000yqh.googlegroups.com>, WM <mueckenh@rz.fhaugsburg.de> wrote:
> On 22 Dez., 21:52, Virgil <vir...@ligriv.com> wrote: > > > Actually each finite initial segment will be contained in uncountably > > many complete paths. > > And none of them will contain more than finite initial segments. I.e., > every path consists only of finite initial segments!
Each path in an actual Complete Infinite Binary Tree is, effectively, the same as a mapping from N to {Leftchild, rightchild}.
That WM's tree does not contain any such paths proves that it is not a Complete Infinite Binary Tree at all. > > > But you tree is provably incomplete, since, if > > there are only countably many, > > There is no doubt among informed scholars that the set of finite > initial segments is countable and that no infinite path contains more > than countably many finite initial segments.
Which, while true, is totally irrelevant to the issue that WM is disputing.
Each complete path is the union of infinitely many "nested" finite initial segments, none of which is a path or even determinesany particular path on its own.
In fact each finite initial segment of any path is the finite initial segment of uncountably many paths, as many as in the whole tree, as is proved by a simple bijection between the sets of its tails and those paths. > > > one must by the very definition of > > countability be able to list its members and thus by Cantor's diagonal > > argument show that a least one is missing., > > Not by Cantor's diagonal argument, but by pointing to a path that is > not in the Binary Tree. You must be able to identify *by nodes* (or > edges, what is the same) a path that is not in the CIBT that I have > constructed from countably many paths.
Since your hidden construction does not reveal which paths occur in your tree, it also does not reveal the ones not in your tree, nor prove that your set of paths is countable.
To prove that your set of paths is countable you must present us with a complete listing of them, as that is the only legitimate way to justify your claim of countability.
And when, if ever, you do that, your will simultaneously have proved your listing is incomplete.
Thus WM loses again/ > > > > > Therefore Cantor's argument here fails in case of an obviously > > > countable set. > > > > You countable set of paths or binary sequences is provably incomplete by > > means of the very proof of its countability: > > To be coutnatble requires being listable, and for listable sets of binary > > sequences, Cantor proved inocmpleteness. > > And I proved by your very inability to identify a missing path that > Cantor's proof (precisely his notion of countability i.e. finished > infinity) is wrong.
WM has a set of paths whose individual identities he will not reveal, which he claims he can list and challenges others to find a nonmember path.
Well, if that set is listable, for any such listing Cantor has proved the existence of non members.
So apparently WM is claiming the existence of a set which he can prove to be countable but which he can also prove not to be listable.
But in the real world of real mathematics, rather than WM's WMytheology, a set being countable is DEFINED to mean that there is a surjection from N to that set.
So that whatever definition of "countable" WM is using, it is certainly not valid in the real world of real mathematics. 

