```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 operationsof addtion of two vectors and scalar times vector multiplicationas 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 failGiven 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 vectorand 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 Treeby a binary sequence,  e.g.,  with a 0 for a left branch and a 1 for a right branchin successtion from the root node onwards ad infinitum, and thus the set of such paths can similarly be formatted into a vector spacewith 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,  constructionof an HONEST linear mapping from the given set of binary sequences as a linear space to the given set of pathsas 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.--
```