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.
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
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.
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.
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.
