In article <f1e51e0a-48c1-4e0a-b031-77cb03dd41d0@he10g2000vbb.googlegroups.com>, WM <mueckenh@rz.fh-augsburg.de> wrote:
> On 11 Mrz., 11:31, William Hughes <wpihug...@gmail.com> wrote: > > On Mar 11, 10:47 am, WM <mueck...@rz.fh-augsburg.de> wrote: > > > > > On 10 Mrz., 20:52, William Hughes <wpihug...@gmail.com> wrote: > > > > > > Let l be a line of L > > > > > > Do you agree with the statement > > > > > > For every n, the nth FIS of d is > > > > contained in l iff > > > > l is coFIS to (d) > > > > > or, more precisely: to the sequence 1, 2, 3, ..., max defined by (d). > > > Yes, that is right. > > > > Do you agree with the statement > > > > If G is a subset of lines of L > > and G has a fixed last element > > then there is no line, l, in G > > for which it is true that > > For every n, the nth > > FIS of d is contained in l > > This holds if you fix a line l but do not fix the findable part of d.
But the "findable" part of d cannot be fixed because for any and every line l there is a FIS of d longer than that l. And WM cannot deny his.
> Otherwise for every findable part of d there is an identical line.
For every findable part of d there is a longer findable part. > > You can also say that for every findable part of d there is a line > twice as long, if you fix d_1, ..., d_n but do not fix the line. > > Why do you think it is more important or in any way preferable to fix > a line but to extend d than vice versa?
We don't think it important to "fix" either. For us , both For each FIS(d) there is an l longer than FIS(d) and For each l there is a FIS(d) longer than l. are equally true.
WHy do YOU think that there is any finite sequence of lines or finite FIS of d is incapable of being extended?
OUSIDE of Wolkenmuekenheim, n cannot be a natural number unless n+1 is also.
INSIDE of Wolkenmuekenheim , it seems there must always be a natural n for which there is no successor natural, n+1.
WM has 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 f(ax + by) = af(x) + bf(y), where a and b are arbitrary members of the field of scalars and x and y are binary sequences and f(x) and f(y) are paths in a CIBT.
By the way, WM, what are ax and 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. --
