In article <firstname.lastname@example.org>, WM <email@example.com> wrote:
> On 11 Mrz., 21:22, William Hughes <wpihug...@gmail.com> wrote: > > > l is a line of G and hence findable. > > d_max is not findable and used ("for every n") > > l_max is used too ("for every line"). > > > > Do you agree with the statement > > > > If G is a subset of lines of L > > and G has a findable last element > > then there is no line, l, in G > > for which it is true that > > For every n, the nth > > FIS of d is contained in l > > I agree with this statement: > > For every findable line of L there is an identical findable FIS of the > diagonal up to that line. And for every findable FIS of the diagonal > there is an identical line. Same holds for the diagonal 1, ..., max of > L and the last line 1, ..., max of L.
And, equally, for every findable line there is a findable FIS of d longer than that line, and for every findable FIS do d the is a findable line longer than that FIS. > > I do not see any use in answering your questions which try to make a > difference between changing the FIS of the diagonal and changing the > due line. When changing the FIS of the diagonal you speak of the same > diagonal, but when changing the line, you speak of different lines. > > This is unjustified. In order to see it, write the list in the form > > 1,2,3,...,max
That is not the whole list, as in a whole list every member must have a successor. > > where every line and the diagonal are written in one and the same > line. Does this answer your problems? If you have pleasure in > continuing to "prove" that there is a difference between line(s) and > diagonal, please go on, but leave me out of the play - since I do not > see a difference and will not change my mind in this respect.
Those, like WM, who choose not to see, are worse off that the blind who cannot choose to see. > > And a last remark: You will never succeed in proving that pot. inf. is > the same as act. inf, since your unsurmountable obstacle is the > requirement that all natural numbers have to be in the list, but > cannot be in one line but must be in one line.
We do not claim that they are the same, but only note that potential infiniteness requires the existence of natural numbers which have no successors and the nonexistence of inductive arguments. > > Meanwhile I am tired to answer your questions. 600 postings are > enough, and there are many further §§ of matheology waiting to be > published as soon as the current discussions will have ceased. > > Regards, WM
Wm is certainly a prolific enough author of his matheology, though we call it WMytheology.
But he is not much good at publishing mathematics, which is the supposed point of posting to sci.math.
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 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. --