In article <email@example.com>, WM <firstname.lastname@example.org> wrote:
> 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.
It is equally obviously not finite. > > > However, this does *not* imply that > > there are two numbers that are not in a single line. > > Why then should two lines be necessary?
Becasuse not all natural numbers are in any one line, or even in any finite set of lines.
> 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.
In order for a set of WM's "lines" to contain all naturals it is both necessary and sufficient for that set of lines to be infinite. So if WM denies the possibility of infinite sets, any of his sets of lines necessarily fail to include all naturals.
> Proof: If you remove all lines from the list, then there remains no > line and no number.
Which proves nothing but WM's incompetence at proofs. > > > 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?
He did not say that. He merely said any one line could be removed. In fact, as many can be removed as remain, but not more.
> That is a remarkable clatim. I would not accept it > in mathematics.
Wm has not been 'in' mathematics for so long, if he ever was, that he now has no idea of what goes on in it. > > Note in actual infinity it makes sense to talk about all lines and to > remove all lines.
In actual infinity a lot of things make sense that WM cannot make sense of.
In WM's version of only potential infiniteness, nothing makes sense.
WM has frequently claimed that a mapping from the set of all infinite binary sequences to the set of paths of a CIBT is a linear mapping. In order to show that such a mapping is a linear mapping, WM must first show that the set of all binary sequences is a vector space and that the set of paths of a CIBT is also a vector space, which he has not done and apparently cannot do, and then show that his mapping satisfies the linearity requirement that f(ax + by) = af(x) + bf(y), where a and b are arbitrary members of a field of scalars and x and y are f(x) and f(y) are vectors in suitable linear spaces.
By the way, WM, what are a, b, ax, by and ax+by when x and y are binary sequences?
If a = 1/3 and x is binary sequence, what is ax ? and if f(x) is a path in a CIBT, what is af(x)?
Until these and a few other issues are settled, WM will still have failed to justify his claim of a LINEAR mapping from the set (but not yet proved to be vector space) of binary sequences to the set (but not yet proved to be vector space) of paths ln a CIBT.
Just another of WM's many wild claims of what goes on in his WMytheology that he cannot back up. --