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Re: Effect of gravitation in set theory
Posted:
Nov 25, 2011 7:11 PM
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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.
Tonio ****
> 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
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