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.