In article <firstname.lastname@example.org>, WM <email@example.com> wrote:
> On 13 Mrz., 22:41, William Hughes <wpihug...@gmail.com> wrote: > > > > Let J be a set of the lines of L with no > > findable last line. At least two lines > > belong to J. Are any lines of J necessary? > > Remove all lines. > Can any numbers remain in the list? No. > Therefore at least one line must remain in the list. > > We do not know which it is, but it is more than no line. > In other words, it is necessary, that one line remains.
Given any finite set of lines, can one find a natural number not in any of them? Yes! In other words, it is necessary, that more than any finite set of lines remain. > > Now you say that not all numbers of the list can be in this one line. > This means that at least two lines are necessary. > We do not know which lines. > But if all numbers exist in the list, at least two lines must exist in > the list, containing these numbers. > In other words, two lines are necessary.
By the same argument expanded every finite number of lines is both necessary and not sufficient > > From the construction we know, that all numbers, that are in two > lines, are in one line. Therefore your claim, that more than two lines > must remain in the list, is contradicted.
Nope. While any finite set of lines contains no more that the last and largest line of that set, that clearly does not hold for any non-empty set of lines which does not have a last line.
So that it is both necessary and sufficient to have a non-empty set of lines ( finite initial segments of |N) which has no last line in order to include all naturals.
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. --