```Date: Mar 10, 2013 4:58 PM
Author: Virgil
Subject: Re: Matheology � 222 Back to the roots

In article <0395d841-2ee7-4e25-84b3-f9ec767f996c@x15g2000vbj.googlegroups.com>, WM <mueckenh@rz.fh-augsburg.de> 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> infinity.  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 satisfiesf(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 sequences? 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.--
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