Date: Mar 21, 2013 4:25 AM
Author: Virgil
Subject: Re: Matheology � 224

In article 
<35e0a797-93d4-4045-b394-c813757273f6@z3g2000vbg.googlegroups.com>,
WM <mueckenh@rz.fh-augsburg.de> wrote:

> On 21 Mrz., 03:41, Virgil <vir...@ligriv.com> wrote:
>

> > All that is required is a non-empty set of lines with no last line.
>
> But this set does not obey the basic rule that every non-empty set of
> natural numbers has a first element?


It is lines, not naturals, that are the members of such sets.

But it does satisfy the rule that every set of lines that covers all of
|N has a first line and also satifies that no such set of lines has a
last line..
>
> Or is this set a set without a first line?


Which one? There are infinitely many such sets, each having both a first
line and for each line a next line.

In fact every set of lines that covers all of |N has both a first line
and for each of its lines a next line.

And every set of lines that fails to include all of |N must either not
have a first line or must have some line for which there is no next line.


====================================================================

WM claims to know how to map bijectively the set of infinite binary
sequences, B, linearly to the set of reals and then map that image set
of reals linearly ONTO the set of all paths, P, of a Complete Infinite
Binary Tree.

But each binary rational in |R is necessarily the image of two sequences
in B but that one rational can then only produce one image in P, so the
mapping cannot be the bijection WM claims.

SO that WM is, as usual with things mathematical, wrong.
--