Date: Mar 18, 2013 3:32 AM
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.
>>> 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
>> For example, for any non-zero
>> sequence of zeroes and ones
>> that becomes eventually constant
>> with a trailing sequence of zeroes,
>> we can replace that sequence with
>> a trailing sequence of ones,
>> We want to use these forms because
>> of the products
>> Then, coordinatewise multiplications
>> along the trailing sequence of ones
>> retains a trailing sequence of ones.
>> In addition, on the interval
>> we can associate 1 with the constant
>> 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
>> may be representationally
>> equivalent to
>> 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
>> Given this, sequences like
>> whose coordinatewise product
>> 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
>> 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.