Date: Mar 18, 2013 3:10 AM Author: Virgil Subject: Re: Matheology � 224 In article <mOOdnU1k7dJ5ONvMnZ2dnUVZ_rednZ2d@giganews.com>,

fom <fomJUNK@nyms.net> wrote:

> On 3/17/2013 7:11 PM, Ross A. Finlayson wrote:

> >

> > A simple and trivial

> > continuous mapping was noted.

> >

> > Regards,

> >

> > Ross Finlayson

> >

>

>

> That is not enough Ross.

>

> By definition, a linear map must satisfy

>

> f(x+y) = f(x) + f(y)

> f(ax) = a*f(x)

>

> So, the domain must at least have the

> structure of a module since it needs

> to have an abelian addition of domain

> elements and a map from the domain

> into itself with a scalar multiplication.

>

> Furthermore, it is unlikely that one

> could take the scalar multiplication

> to be the Galois field over two

> elements since multiplication by

> zero would be the zero vector and

> multiplication by one would be

> the identity map.

>

> A morphism with that scalar field

> could not reasonably be expected

> to have a linear map with a

> system of real numbers.

>

> In order to build a scalar that

> could even possibly serve this

> purpose, given WM's claims related

> to various finite processes, one

> would have to invoke compactness

> arguments involving completed

> infinities.

>

> For example, for any non-zero

> sequence of zeroes and ones

> that becomes eventually constant

> with a trailing sequence of zeroes,

>

> 1001101000......

>

> we can replace that sequence with

> a trailing sequence of ones,

>

> 1001101111......

>

> We want to use these forms because

> of the products

>

> 1*1=1

> 1*0=0

> 0*1=0

> 0*0=0

>

> Then, coordinatewise multiplications

> along the trailing sequence of ones

> retains a trailing sequence of ones.

>

> In addition, on the interval

>

> 0<x<=1

>

> we can associate 1 with the constant

> sequence,

>

> 111...

>

> Given these facts, we can now say that

> a collection of infinite sequences is

> "compactly admissible" if for every

> finite collection of those sequences

> coordinatewise multiplication yields

> a sequence different from one

> consisting solely of an initial

> segment of zeroes followed by

> an initial segment of ones.

>

> In other words, even though

>

> 000000111...

>

> may be representationally

> equivalent to

>

> 000001000...

>

> for some purposes, compact

> admissibility has to ignore

> what happens in this conversion.

> The situation above is

> interpreted as corresponding

> with a non-compact set of

> sequences.

>

> Given this, sequences like

>

> 1000...

> 11000...

> 110000...

> 1101000...

>

> yield

>

> 1111..

> 11111...

> 110111...

> 1101111...

>

> whose coordinatewise product

> is

>

> 1101111...

>

> So that the original sequence

> is compactly admissible.

>

> Given a construction along these

> lines, one could then think of

> compactly admissible collections

> as possibly forming a sequence space

> as described here

>

> http://en.wikipedia.org/wiki/Hilbert_space#Second_example:_sequence_spaces

>

> Obviously, the compactly admissible

> collections are not defined as

> converging in the sense of a sequence

> of partial sums.

>

> Equally obviously, I have not done

> all the work necessary to decide

> whether or not this would work.

>

> My purpose here is to explain that

> the scalar multiplication would

> require a construction along these

> lines just to even begin to talk

> about whether or not WM could

> do what Virgil is asking.

MY points are

(1) The bijective mapping from the set of binary sequences to the set of

paths of a Complete Infinite Binary Tree, was NOT a linear mapping as it

was originally formulated by WM.

(2) WM is not competent enough to be able to reformat it correctly .

i.e., to make it an actual and obvious linear mapping.

(3) It is not all that difficult to create an actual and obvious linear

mapping there for someone who knows something more about linear spaces

that WM does.

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