```Date: Mar 17, 2013 7:42 PM
Author: Virgil
Subject: Re: Matheology � 224

In article <0ff7b4e3-a20a-44a5-b93d-dfe1b8e94795@a14g2000vbm.googlegroups.com>, WM <mueckenh@rz.fh-augsburg.de> wrote:> On 17 Mrz., 22:39, William Hughes <wpihug...@gmail.com> wrote:> > You are contradicting yourself.> >> > You say there are no necessary findable lines> > because of the last line (an unfindable line)> > In pot. inf. there is always a last line.But as every line implies the actual existence of a successor line, no such last line can exist ong enough to be seen.> It is unfindable or> unfixable. An non-existable!> But I do not wish to discuss potential infinity but actual infinity> here.> >> > You say that if a set of lines contains an unfindable> > line it is necessary that there are> > two findable lines.> > No. I say that in actual infinity a list contains all natural numbers.> But they cannot be in one line, because there is no actually infinite> line. This is a contradiction.Only if one claims that a FISON need not be a FISON.> > Please kindly note: Even if my personal theory was self-contradictoryWhich it is!> that would not improve the situation presently adopted in mathematics.> So please concentrate on defending your position.We define the set of naturals so that it has a unique first member and for each member there is a successor member larger that that predecessor. According to that definition, all WMytheology is nonsense.######################################################################WM has frequently claimed that HIS mapping from the set of all infinite binary sequences to the set of paths of a CIBT is a linear mapping.In order to show that such a mapping is a linear mapping, WM would first have to show that the set of all binary sequences is a linear space (which he has not done and apparently cannot do) and that the set of paths of a CIBT is also a vector space (which he also has not done and apparently cannot do) and then show that his mapping, say f,  satisfies the linearity requirement that   f(ax + by) = af(x) + bf(y),where a and b are arbitrary members of the field of scalars and x and y and f(x) and f(y) are arbitrary members of suitable linear spaces.  While this is possible, and fairly trivial for a competent mathematician to do, WM has not yet been able to do it.But frequently claims already to have done it.--
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