Date: Apr 5, 2013 5:54 PM
Subject: Re: Matheology � 224
WM <firstname.lastname@example.org> wrote:
> On 5 Apr., 21:03, William Hughes <wpihug...@gmail.com> wrote:
> > On Apr 5, 6:04 pm, WM <mueck...@rz.fh-augsburg.de> wrote:
> > > On 5 Apr., 12:08, William Hughes <wpihug...@gmail.com> wrote:
> > <snip>
> > > > There is an infinite set of lines D
> > > > such that any finite subset of D can be removed.
> > > What has to remain?
> > This depends on the finite subset removed.
> > If the finite set removed is E then
> > D\E has to remain.
> Is E restricted to an upper threshold?
Is finiteness limited to an upper threshold in Wolkenmuekenhein?
It is not elsewhere!
> If not, how do you prove its finiteness?
By finding its largest member.
> > Note that whatever
> > subset E is chosen the number of lines
> > in D\E is infinite (but of course we
> > do not know which lines are in D\E).
> Can you prove for at least one fixed line that it cannot be removed?
> If not, why do you think that some (even infinitely many) lines must
Because if none remain, one has the empty set, which is clearly no
longer infinite, even in Wolkenmuekenheim.
WM seems to be getting daily stupider.
> Don't you feel a bit ridiculous, when you again and again claim
> infinitely many natural numbers none of which you can name?
Not as ridiculous as you askins whether a set after all its members have
been removed is still infinite.
> > However, D cannot be removed without
> > changing the union of the remaining lines.
> That is true if D is nothing but the union of the lines. But in that
> case we can prove
WM has yet to prove he can prove anything.
WM keeps claiming proofs that he never delivers.
Remember WM claiming a linear mapping from the set of all binary
sequences to the set of reals to the set of all paths in a Complete
Infinite Binary Tree?
He never delivered on that proof because it is impossible to do.
Though for eveenas incpmpetent a mathematician as WM claimed me to be,
I have constructed a linear mapping from the set of all binary sequences
to the set of reals to the set of all paths in a Complete Infinite
Binary Tree, though not through the reals.
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.