In article <email@example.com>, WM <firstname.lastname@example.org> wrote:
> On 10 Mrz., 18:24, William Hughes <wpihug...@gmail.com> wrote: > > On Mar 10, 6:05 pm, WM <mueck...@rz.fh-augsburg.de> wrote: > > > > > On 10 Mrz., 17:40, William Hughes <wpihug...@gmail.com> wrote: > > > > > > There is no findable line that is > > > > coFIS to (d) > > > > > (d) is *not* an actual infinite sequence but only a description in > > > letters. > > > > > > g is a findable line. > > > > > > Do you agree with the statement > > > > > > g is not coFIS to (d) > > > > > Of course. The number m = max is not findable or fixable. > > > > So do you agree with the statement. > > > > If G is a set of lines of L with a findable > > last element, then there is no line s of > > G such that s is coFIS to (d) > > Yes. How often will you ask? > (d) is a prescription to find or to construct FIS d_1, ..., d_n. > > Would you expect that > "write 0. and then add the digit 1 with no end" is coFIS with a line > of > 0.1 > 0.11 > 0.111 > ... > > The *result* of this prescription is coFIS with a line of the above > list. Alas both are not findable. All we know is that for every line > there is an identical result of the prescription.
"It ain't what you don't know that hurts you most, its what you know for sure that jest ain't so." Mark Twain.
And WM is hurtng! > > Regards, WM
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. --