Date: Mar 30, 2013 9:46 PM
Author: ross.finlayson@gmail.com
Subject: Re: Matheology § 233
On Mar 30, 5:56 pm, Virgil <vir...@ligriv.com> wrote:

> In article

> <2bc13fff-5cbb-43dd-a06a-218c68c99...@m9g2000vbc.googlegroups.com>,

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> WM <mueck...@rz.fh-augsburg.de> wrote:

> > On 30 Mrz., 22:11, Virgil <vir...@ligriv.com> wrote:

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> > > > > 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?

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> > > 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.

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> > 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.

Regards,

Ross Finlayson