```Date: Mar 18, 2013 1:28 AM
Author: fom
Subject: Re: Matheology § 224

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 satisfyf(x+y) = f(x) + f(y)f(ax) = a*f(x)So, the domain must at least have thestructure of a module since it needsto have an abelian addition of domainelements and a map from the domaininto itself with a scalar multiplication.Furthermore, it is unlikely that onecould take the scalar multiplicationto be the Galois field over twoelements since multiplication byzero would be the zero vector andmultiplication by one would bethe identity map.A morphism with that scalar fieldcould not reasonably be expectedto have a linear map with asystem of real numbers.In order to build a scalar thatcould even possibly serve thispurpose, given WM's claims relatedto various finite processes, onewould have to invoke compactnessarguments involving completedinfinities.For example, for any non-zerosequence of zeroes and onesthat becomes eventually constantwith a trailing sequence of zeroes,1001101000......we can replace that sequence witha trailing sequence of ones,1001101111......We want to use these forms becauseof the products1*1=11*0=00*1=00*0=0Then, coordinatewise multiplicationsalong the trailing sequence of onesretains a trailing sequence of ones.In addition, on the interval0<x<=1we can associate 1 with the constantsequence,111...Given these facts, we can now say thata collection of infinite sequences is"compactly admissible" if for everyfinite collection of those sequencescoordinatewise multiplication yieldsa sequence different from oneconsisting solely of an initialsegment of zeroes followed byan initial segment of ones.In other words, even though000000111...may be representationallyequivalent to000001000...for some purposes, compactadmissibility has to ignorewhat happens in this conversion.The situation above isinterpreted as correspondingwith a non-compact set ofsequences.Given this, sequences like1000...11000...110000...1101000...yield1111..11111...110111...1101111...whose coordinatewise productis1101111...So that the original sequenceis compactly admissible.Given a construction along theselines, one could then think ofcompactly admissible collectionsas possibly forming a sequence spaceas described herehttp://en.wikipedia.org/wiki/Hilbert_space#Second_example:_sequence_spacesObviously, the compactly admissiblecollections are not defined asconverging in the sense of a sequenceof partial sums.Equally obviously, I have not doneall the work necessary to decidewhether or not this would work.My purpose here is to explain thatthe scalar multiplication wouldrequire a construction along theselines just to even begin to talkabout whether or not WM coulddo what Virgil is asking.
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