Date: Mar 18, 2013 3:32 AM Author: fom Subject: Re: Matheology § 224 On 3/18/2013 2:10 AM, Virgil wrote:

> 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.

>

Since you use the word "obvious" I am assuming

that I did much too much work attempting to

establish a scalar multiplication.

But, I would assume that you would be

basing it on the representations as real

numbers directly. I tried to avoid that

because of WM's finitist claims and the

abstract definition of tree paths.