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Topic: Matheology � 233
Replies: 36   Last Post: Apr 2, 2013 5:56 PM

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ross.finlayson@gmail.com

Posts: 915
Registered: 2/15/09
Re: Matheology § 233
Posted: Mar 30, 2013 9:46 PM
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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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>
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>  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.

Regards,

Ross Finlayson



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