In article <firstname.lastname@example.org>, WM <email@example.com> 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. > > Note: since max is not findable, we can state that for every findable > part of d there is a line identical with that part. For the unfindable > 1, 2, 3, ..., max defined by (d) there is the unfindable last line 1, > 2, 3, ..., max defined by (l). > Nonsense! But if there are countably many unfindable naturals, and if there are any there must be countably many of them, there can be unfindable reals, and can be uncountably many of them, ever since Cantor.
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. --