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!
> 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.
> 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. > > > 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 becoutnatble 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.