Date: Feb 7, 2013 4:20 PM
Author: Virgil
Subject: Re: Matheology � 222 Back to the roots
In article

<8d610f94-e350-49b9-82c9-e2efc2a3b89e@fn10g2000vbb.googlegroups.com>,

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

> On 7 Feb., 20:12, William Hughes <wpihug...@gmail.com> wrote:

> > On Feb 7, 8:06 pm, WM <mueck...@rz.fh-augsburg.de> wrote:

> >

> > > On 7 Feb., 19:46, William Hughes <wpihug...@gmail.com> wrote:

> >

> > > > Gosh, you are really running away

> > > > from the fact that induction can

> > > > show d is not in the list.

> >

> > > Induction can show that *your* d does not exist.

> >

> > My d? You are the one who defined d to be

> > the antidiagonal of the list.

>

> The antidiagonal of a list is not always in the list, but the diagonal

> of the list

>

> 1

> 11

> 111

> ...

>

> is with certainty in this very list - since it is nothing else but a

> potentially infinite sequence of 1' and not longer than the lines.

If it is not longer than all lines, then there must be a first such line

that it is not longer than.

So which is the first line that the diagonal is as short as?

No first line implies no line at all.

> > You also

> > show by induction that the antidiagonal of

> > a list is not in the list.

Actually, one should not speak of THE antidiagonal as there are at least

as many antidiagonals as lines in the original list.

>

> No, that depends on the list.

>

> The antidiagonal of the list

Which one? There are lots of them

>

> 0.0

> 0.1

> 0.11

> 0.111

> ...

and none of them are in the list itself.

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