Date: Mar 13, 2013 1:33 PM
Author: mueckenh@rz.fh-augsburg.de
Subject: Re: Matheology § 222 Back to the roots

On 13 Mrz., 17:59, William Hughes <wpihug...@gmail.com> wrote:
> On Mar 13, 5:37 pm, WM <mueck...@rz.fh-augsburg.de> wrote:
>

> > On 13 Mrz., 13:19, William Hughes <wpihug...@gmail.com> wrote:
>
> <snip>
>

> > > If you wish to contest this, use my words not
> > > yours  (e.g.  I have never said "The list contains more
> > > numbers than fit into a single line",  I have said
> > > "There is no line in the list which contains every
> > > number in the list".)

>
> > Correct. The list has more numbers than a single line has. Since every
> > number that is in the list, must be in at least one line, this implies
> > that the numbers are in more than one line.

>
> To be precise, a set of lines, say K, that contains all the numbers
> contains at least two lines.


In actual infinity, this is not avoidable.
We note: At least two lines belong to the set that contains all
numbers. We call these lines necessary lines.
So the set of necessary lines is not empty.

>  However, this does *not* imply that
> there are two numbers that are not in a single line.


Why then should two lines be necessary?
One being the substitute in case the other falls ill?

> Nor does it imply that there is a necessary line in K.

If there is not one necessary line, then there are two or more
required.
Proof: If you remove all lines from the list, then there remains no
line and no number.

> Note that a sufficient set does not imply a necessary line
> even in potential infinity.  There is no line that is needed
> to make L have an unfindable last line.


So you believe that there can remain all numbers in the list after
removing all lines? That is a remarkable claim. I would not accept it
in mathematics.

Note in actual infinity it makes sense to talk about all lines and to
remove all lines.

Regards, WM