Date: Mar 10, 2013 5:10 PM
Subject: Re: Matheology � 222 Back to the roots
WM <firstname.lastname@example.org> wrote:
> On 10 Mrz., 20:20, William Hughes <wpihug...@gmail.com> wrote:
> > > > So do you agree with the statement.
> > > > If G is a set of lines of L with a findable
> > > > last element, then there is no line s of
> > > > G such that s is coFIS to (d)
> > > Yes. How often will you ask?
> > > (d) is a prescription to find or to construct FIS d_1, ..., d_n.
> > > Would you expect that
> > > "write 0. and then add the digit 1 with no end" is coFIS with a line
> > > of
> > > 0.1
> > > 0.11
> > > 0.111
> > > ...
> > No, the other way round.
> There is no way. This is a sequence of less than 10 words: "write 0.
> and then add the digit 1 with no end". It is not coFIS with any line
> of the list. But it defines the lines of the list.
> > Recall
> > We will say x is coFIS to (y) iff
> > i. We have (x) associated to x and
> > (y) associated to y
> > ii. For every n, (x) and (y) produce the same
> > finite string.
> (x) and (y), if describing infinite sequences, are phrases of few
> words. They are probably not coFIS.
On the other hand, if they are references to the infinite sequences
themselves, then you must answer for the things being referred to.
If one cannot refer to things without actually having them present, the
language is of little use.
> > The statement x is coFIS to (y) means approximately
> > that x and the potentially infinite sequence described
> > by (y) are COFIS.
> > Do you agree with the statement
> > For every n, the nth FIS of x is
> > contained in g iff
> > g is coFIS to (x)
> Let us stay in the concrete example:
> L is the list
> and d is the diagonal 1,2,3,...,max.
Lets not. Unless WM allows that every natual number has successor
natural number, he is not working with any standard sort of natural
numbers but some oddball creation of his own invention and of no
> For every n, the nth FIS of d is contained in the list L and,
> therefore, in the last, unfixable and unfindable, line 1,2,3,...,max.
Except there is no "max", since it would have to be a natural number
with no successor, which cannot exist, at least not outside the
corruptions of Wolkenmuekenheim.
> Regards, WM
WM has claimed that a 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 must first
show that the set of all binary sequences is a vector space and that the
set of paths of a CIBT is also a vector space, which he has not done and
apparently cannot do, and then show that his mapping satisfies
f(ax + by) = af(x) + bf(y), where a and b are arbitrary members of the
field of scalars and x and y are binary sequences and f(x) and f(y) are
paths in a CIBT.
By the way, WM, what are ax and by and ax+by when x and y are binary
If a = 1/3 and x is binary sequence, what is ax ?
and if f(x) is a path in a CIBT, what is af(x)?
Until these and a few other issues are settled, WM will still have
failed to justify his claim of a LINEAR mapping from the set (but not
yet proved to be vector space) of binary sequences to the set (but not
yet proved to be vector space) of paths ln a CIBT.