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