
Re: Matheology § 233
Posted:
Mar 30, 2013 9:46 PM


On Mar 30, 5:56 pm, 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. > Those order types are incompatible. > So questions, like WM's, which confuse them, are nonsense. > At least outside Wolkenmuekenheim. > 
Zuhair simply brought forth an antidiagonal argument for the infinite balanced binary tree, and then the breadthfirst traversal or sweep was shown to iterate the paths that it didn't apply.
With f = lim_d>oo n/d, n > d, the elements of ran(f) are dense in [0,1] and wellordered.
Fuse the elements, or, unfuse them: don't confuse them.
Regards,
Ross Finlayson

