
Re: Matheology § 233
Posted:
Mar 31, 2013 4:31 AM


On 31 Mrz., 02:56, Virgil <vir...@ligriv.com> wrote: > In article > <2bc13fff5cbb43dda06a218c68c99...@m9g2000vbc.googlegroups.com>, > > > > > > WM <mueck...@rz.fhaugsburg.de> wrote: > > On 30 Mrz., 22:11, Virgil <vir...@ligriv.com> wrote: > > > > > > Thus there is always at least one bit of any listed entry disagreeing > > > > > with the antidiagonanl, just as the Cantor proof requires. > > > > > In a list containing every rational: Is there always, i.e., up to > > > > every digit, an infinite set of paths identical with the anti > > > > diagonal? Yes or no? > > > > The set of paths in any Complete Infinite Binary Tree which agree with > > > any particular path up to its nth node is equinumerous with the set of > > > all paths in the entire tree i.e., is uncountably infinite. > > > This was the question: In a list containing every rational: Is there > > always, i.e., up to every digit, an infinite set of paths identical > > with the antidiagonal? Yes or no? > > Lists and trees are different. And antidiagonals derive from lists, not > trees. > The entries in list are well ordered. > The entries in a Complete Infinite Binary Tree are densely ordered.
The entries in form of nodes are well ordered. Anything else is not added to the list.
> Those order types are incompatible.
This was the question: In a list containing every rational: Is there always, i.e., up to every digit, an infinite set of paths (rational numbers) identical with the antidiagonal? Yes or no?
Regards, WM

