Date: Mar 20, 2013 11:06 PM Author: Virgil Subject: Re: Matheology � 224 In article

<d97b811f-45df-4158-a726-3d32a2251629@x15g2000vbj.googlegroups.com>,

WM <mueckenh@rz.fh-augsburg.de> wrote:

> On 20 Mrz., 22:11, William Hughes <wpihug...@gmail.com> wrote:

> > On Mar 20, 10:01 pm, WM <mueck...@rz.fh-augsburg.de> wrote:

> >

> > > It cannot contain a necessary line, since, as you said, there is no

> > > necessary line.

> > > But you think that it necessarily must contain a not necessary line.

> >

> > Correct. However, this not necessary line can be any line.

> >

> > > If so, then at least one line is necessary.

> >

> > No, knowing that you need one line does not mean you need

> > one particular line.

>

> You cannot know that you need one line, because every chice can be

> disproved.

We can know that we need at least one line, because no lines fails.

For any natural n in |N, we can know that n lines fail, but we can also

know that any infinite set of lines succeeds.

$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$

The following does what WM has shown himself incapable of doing, namely

constructon 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 avector 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 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 avector:

For scalar f in F and vector g in F^S

define h = f*g by

where 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 addtion 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 upgrede his mathematical skills is domed t 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 chasteristic two,

and the domain of our linear mapping is now a vector spacde.

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 successtion 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 cfreativity, or even common sense.

A pity that WM's mathematical skills are so miniscule,

particularly when his ego is so gargantuan.

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