Date: Mar 21, 2013 1:26 PM
Author: Virgil
Subject: Re: Matheology � 224

In article 
<4782d0ea-065c-415f-b7a8-43ddbcc6ac95@gp5g2000vbb.googlegroups.com>,
WM <mueckenh@rz.fh-augsburg.de> wrote:

> On 21 Mrz., 11:36, William Hughes <wpihug...@gmail.com> wrote:
> > On Mar 21, 11:21 am, WM <mueck...@rz.fh-augsburg.de> wrote:> On 21 Mrz.,
> > 08:57, William Hughes <wpihug...@gmail.com> wrote:
> >
> > <snip>
> >

> > > > There is such a thing as a sufficient set of
> > > > lines  (all sufficient sets are composed
> > > > entirely of unnecessary lines, which means
> > > > that you can remove any finite set of lines

> >
> > > Why only finite sets?
> >
> > You can only use induction to prove
> > stuff about finite sets.

>
> For finite sets induction is not required.
> It is required however, to define the infinite set |N


Actually, a definition of |N, or some similar set, is required to define
induction.
>
> > Once we remove one line, we are left with
> > a new set of unnecessary lines.  We can
> > remove one of these lines.
> > From induction we get that
> > any finite set of lines can
> > be removed.

>
> Exactly. And we can prove by induction that the set of all removable
> lines is infinite, no?


EVERY infinite set of lines covers |N, so long as as one leaves
infinitely many lines unremoved from such a covering set, what remains
is still a covering set.

Why Wm cannot get this simple fact into his thick head is an insoluble
puzzle.


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

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.
--