In article <email@example.com>, WM <firstname.lastname@example.org> wrote:
> On 24 Jan., 13:36, "Jesse F. Hughes" <je...@phiwumbda.org> wrote: > > WM <mueck...@rz.fh-augsburg.de> writes: > > > Well, what you present below is *not* a proof of (*). > > That is wrong. You have no reason to believe that your definition of > proof is correct or the only one.
We have lots of reason to beeive that anything WM presents as a proof that is not copied from soemone more competent, is incorrect.
Some of those reasons are the obvious flaws in logic that WM is know for. > > > > > Clearly, for all j, d(j) != t_j(j) and hence d != t_j for any j in > > N. > > > > Is this what you mean up 'til now? > > Yes. > > > > > > 4) Certainly you agree that, since all t_i = (t_i1, t_i2, ..., t_in) > > > have only a finite, though not limnited, number n of digits, the > > > diagonalization for every t_i yields a finite d_i =/= t_ii. > > > (The i on the left hand side cannot be larger than the i on the right > > > hand side. In other words, "the list" is a square. Up to every i it > > > has same number of lines and columns. ) > > > > No idea what you mean by the parenthetical remark. > > You will have have recognized that here the diagonal argument is > applied. It is obvious that up to every line = column the list is a > square.
Not at all. there is no reason why line n, for any n > 1, must be of length >= n.
In decimal notation, one could start with 10 lines of length 1 followed by 90 lines of length 2, followed by 900 lines of length 3, etc., and never repeat a number.
> > > > I do agree that d_i is defined for every i in N. In particular, (d_i) > > is an infinite sequence of digits. Is this what you're claiming, too? > > You've lost me. I don't know what you mean when you say, "everything > > here happens among FISs." And I'm also puzzled by the meaning of the > > next sentence. > > Every t_i is finite. Hence, in a square, if the width is finite, also > the length must be finite.
But a "diagonal" need not be, and will not be finite. > > > > Here are some obvious things. > > > > d(j) is defined for every j in N. > > d(j) != 0 and d(j) != 9 for any j in N. > > > > Hence the number d does not have a terminating decimal > > representation. > > Neither the set of t_i does have a largest element. Nevertheless there > is no t_i of actually infinite length. > > > > This looks like I do *not* agree with your claim that "d cannot be > > longer than every t_i". > > A sequence of squares will never result in a square such that all > sides are finite but the diagonal d is infinite. The overlap of d and > t_i cannot be larger than t_i. > > In particular, what would be changed in the length of d if we admitted > also non-terminating t_i (of infinite length)?
The diagonal is already required to be infinitely long, so its length need not change. --