On Nov 25, 8:03 pm, WM <mueck...@rz.fh-augsburg.de> wrote: > On 25 Nov., 17:34, Tonico <Tonic...@yahoo.com> wrote: > > > is possible to say: ALL the elements of S_k, for any k, are contained > > in any S_(k+j), > > > for all j > 0. > > So it is. > > > > > From where "the theorem" follows: no single S_K can contain ALL the > > natural numbers, > > > so....how eaxctly have you proved that all the naturals are contained > > in one single > > > S_k, again? > > I proved that in order to contain the actual infinite set |N no S_k is > necessary. That includes that not more than one S_k are necessary.
**** I am going to show this to other people, without remarking your name, of course, lest they won't believe me...or are you joking, dude?
I've no desire to continue with this nonsense. I oficially give up this thread and shal continue with others, in the meantime.
> That is what I need in order to show those wrong, who claim infinitel > many S_k would be necessary. > > > More important, perhaps: how the above gets even slightly close to > > proving that the > > > union of all S_k is NOT the whole set IN?? > > Obviously the union is not necessary (and not sufficient). > > > How can anyone after high school mathematics (well taught and well > > learnt, of course) > > > can believe, before or after the above, what you wrote at the > > beginning of the post > > > "Same holds for my proof. > > The set of natural numbers that is not covered by one finite initial > > segment *alone* is empty." ??? > > The actually infinite set |N does not exist. If it would exist it > would have to have infinite numbers. Those are not available in any > S_k. That is the essence of my proof. > > Regards, WM