In article <firstname.lastname@example.org>, WM <email@example.com> wrote:
> On 17 Mrz., 12:07, William Hughes <wpihug...@gmail.com> wrote: > > > > > If the question is "Can a findable line be necessary" > > the question of whether a findable line is needed is > > certainly relevant.- > > Stop! > > I do not force anybody to accept my model. I cannot and I do not wish > to. > > The question in current mathematics was and is only this: Can a set of > more than one lines in the given list contain more than a single line? > In other words: Can more than one line be necessary? > > Regards, WM
If those lines are FISONs. any finite set of them is limited to the finitely many naturals contained in the largest of them, so one needs a non-empty set of FISONs that does not have a largest FISON, which can only be achieved but having an actually infinite set of FISONs.
Which means that in WM's world, it can never happen at all.
WM has frequently claimed that HIS 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 would first have to show that the set of all binary sequences is a linear space (which he has not done and apparently cannot do) and that the set of paths of a CIBT is also a vector space (which he also has not done and apparently cannot do) and then show that his mapping, say f, satisfies the linearity requirement that f(ax + by) = af(x) + bf(y), where a and b are arbitrary members of the field of scalars and x and y and f(x) and f(y) are arbitrary members of suitable linear spaces.
While this is possible, and fairly trivial for a competent mathematician to do, WM has not yet been able to do it.