Date: Apr 17, 2013 11:30 PM
Subject: Re: Matheology � 222 Back to the roots
WM <email@example.com> 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
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,
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
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.