On Mar 30, 5:56 pm, Virgil <vir...@ligriv.com> wrote: > In article > <2bc13fff-5cbb-43dd-a06a-218c68c99...@m9g2000vbc.googlegroups.com>, > > > > > > > > > > WM <mueck...@rz.fh-augsburg.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 anti-diagonal? Yes or no? > > Lists and trees are different. And anti-diagonals 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 anti-diagonal argument for the infinite balanced binary tree, and then the breadth-first 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 well-ordered.
Fuse the elements, or, un-fuse them: don't con-fuse them.