In article <firstname.lastname@example.org>, WM <email@example.com> wrote:
> On 11 Apr., 23:52, William Hughes <wpihug...@gmail.com> wrote: > > On Apr 11, 9:51 pm, WM <mueck...@rz.fh-augsburg.de> wrote: > > > > > On 11 Apr., 20:52, William Hughes <wpihug...@gmail.com> wrote: > > > > > > On Apr 11, 8:17 pm, WM <mueck...@rz.fh-augsburg.de> wrote: > > > > > > > On 11 Apr., 19:19, William Hughes <wpihug...@gmail.com> wrote: > > > > > > > > > So you agree that B implies A. > > > > > > > > yep, but A does not imply C: P(d_1,d_2,d_3..) > > > > > > > That is your interpretation. > > > > > > One you agree with, > > > > > Not at all! > > > > Yep, you have agreed that saying something is true for > > every element of a collection does not imply that it > > is true for the collection. > > It is true for all elements. No mentioning of collection.
One cannot have 'all elements' without having a way of distinguishing those elements from non-elements, which is all that is needed to create/justify a set of all those elements.
> all interested whether these elements form something like a > collection. Of course you cannot say that the collection of real > numbers is a real number. But when all real numbers have been removed, > the collection has gone too. > > > > Now let P be (can remove the collection without changing > > the union of the remaining lines). We have there is no > > contradiction in saying A: for all n, the nth line can be removed > > because A does not imply C and only C is a contraction. > > Then Cantor's argument fails from the scratch. Cantor uses the fact > that removing every line from the set L of possible duplicates of the > anti-diagonal d does not leave any line. So d is not in the set L. Of > course, we have also that d is not the set L, but that is not in > question. > > Regards, WM --