Date: Mar 10, 2013 4:58 PM
Subject: Re: Matheology � 222 Back to the roots
WM <firstname.lastname@example.org> wrote:
> On 10 Mrz., 11:12, William Hughes <wpihug...@gmail.com> wrote:
> > On Mar 10, 10:40 am, WM <mueck...@rz.fh-augsburg.de> wrote:
> > > On 10 Mrz., 10:28, William Hughes <wpihug...@gmail.com> wrote:
> > > > On Mar 10, 10:02 am, WM <mueck...@rz.fh-augsburg.de> wrote:
> > > > > On 9 Mrz., 23:53, William Hughes <wpihug...@gmail.com> wrote:
> > > > > > 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.
> > > > > "Every given n" is tantamount to "there is a last given n".
> > > > I do not talk about "every given n" but about "every n"
> > > > (this means from 1 to n for every n). Note that
> > > > "there is a last n" but it is not a findable natural
> > > > number.
> > > Better say "a not fixable natural number".
> > > > Note that you do not need the x_n to exist to say
> > > > something about them.
> > > That is true. For instance we can say that a natural number is either
> > > even or odd.
> > > > For example, you can say no
> > > > x_n that will ever exist will be equal to 0.
> > > Yes, that is another example.
> > > > If you say x is coFIS to y you are saying something
> > > > about x_n and y_n that may not exist at this time.
> > > That is the hardest problem. There are some properties which can be
> > > determined (like the examples above). There are other which cannot.
> > > But in every case we know that there is a line of the list that is
> > > identical with the FIS of d, both existing or not existing yet.
> > However this is not a findable line.-
> It is not a fixable line, say.
> But here we have the case where we can know properties of not (yet)
> existing entities. We know that every natural number (without 0) is
> positive. In a similar way we know that for every line of the list
> there is a FIS of the diagonal and vice versa. That's why I refuse to
> accept that the diagonal is longer than every line.
Is there any line which is not a PROPER subset of the diagonal?
Outide of Wolkenmuekenheim, the answer is clearly "NO!"
> We can also say, the diagonal is coFIS with the list. But that would
> not change anything, because the list has always a finite maximal line
> (which has *never* more elements than can be counted by a natural
> number). You would like to state that the diagonal has more FISs than
> any natural number can count. That is not possible.
It is possible, and even necessary, once one frees ones self from the
unnatural restrictions of Wolkenmuekenheim.
Just as every natural has a successor outside of Wolkenmuekenheim, so
every FIS of d has a successor there, and no one is constrainted to stay
incarcerated in Wolkenmuekenheim against their will.
> The resaon is that
> the list has no actually infinite line which is correct even when
> accepting actual infinity. And just this fact contradictis actual
WRONG! The set of naturals manages to be actually infinite outside
Wolkenmuekenheim without ever needing an infinite member.
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.