In article <email@example.com>, WM <firstname.lastname@example.org> wrote:
> > 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, > > The field of real numbers (|R, +, *) should satisfy your wishes. > Written in the form of a tree with the decimal point common to all > paths that stretch from oo to -oo you get the same space as a decimal > tree. And if you translate that into binaries, you have the desired > fields.
WRONG! Two of your binary sequences both representing the same binary rational will map to a single real and then not be able to split again to give the two needed paths.
The following does what WM has shown himself incapable of doing, namely construction of a truly linear mapping and vector-space isomorphism from the set of all infinite binary sequences, B, as a subset of a vector space to the set of all paths, P, of a Complete Infinite Binary Tree, also as a subset of a vector space.
First: A general method for construction of linear spaces over a given field (F, +, *, 0, 1)
Given any field, (F, +, *, 0, 1) and any non-empty set S, one can form a linear space (vector space), out of the set of all functions from S to F, denoted here by F^S, with (F, +, *, 0, 1) as its field of scalars, as follows: Defining the VECTOR ADDITION of two vectors: For g and h being any two functions in F^S, from S to F, define their vector sum, k = f + g in F^S, by k(s) =(f+g)(s) = f(s) + g(s) for all s in S. Then k = g + h is the vector sum of g and h in F^S. Defining the SCALAR MULTIPLICATION of a scalar times a vector: For scalar f in F and vector g in F^S define h = f*g by h(s) =(f*g)(s)= f*g(s) for all s in S. Any f*g thus defined is then a scalar multiple of g and a member of vector space F^S.
It is straightforward and trivial to verify that, given the operations of addition of two vectors and scalar times vector multiplication as defined above, any such F^S is a true vector space over its field F.
Even WM should be able to understand and accept this. If WM cannot understand and accept this, then any attempts on our parts to upgrade his mathematical skills is domed to fail
Given the finite field, F_2, of characteristic two, thus having only the two members 0 and 1 required of every field, and the set |N of all natural numbers as S, F^S = F_2 ^ |N, which, as a set is the set of all infinite binary sequences, and becomes, with the above construction, automatically a linear space or vector space over the given field of two elements as its field of scalars, with scalar 0 times any vector giving the zero vector and scalar 1 times any vector giving back that vector again.
Thus the set of all binary sequences has become a linear or vector space over the unique field of chacteristic two, and the domain of our linear mapping is now a vector space.
One can also represent each path in a Complete Infinite Binary Tree by a binary sequence, e.g,, with a 0 for a left branch and a 1 for a right branch in succession from the root node onwards ad infinitum, and thus the set of such paths can similarly be formatted into a vector space with the same vectors over the same field of characteristic two.
Then the "identity" mapping on binary sequences, between these identical vector spaces of binary sequences, automatically becomes a vector space isomorphism and bijective linear mapping.
But without using the unique field of two elements field as the field of scalars for both the linear spaces involved, construction of an HONEST linear mapping from the given set of binary sequences as a linear space to the given set of paths as a linear space appears highly implausible, at least for someone of WM's demonstrated lack of mathematical creativity, or even common sense.
A pity that WM's mathematical skills are so miniscule, particularly when his ego is so gargantuan. --