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Topic: ? 533 Proof
Replies: 8   Last Post: Aug 4, 2014 2:04 PM

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Virgil

Posts: 3,203
Registered: 6/8/11
Re: ? 533 Proof
Posted: Aug 4, 2014 1:27 PM
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In article <db0da8bf-3e3b-4792-adc6-e8284c6d8a1a@googlegroups.com>,
mueckenh@rz.fh-augsburg.de wrote:

> On Monday, 4 August 2014 18:37:58 UTC+2, Zeit Geist wrote:
>

> > > > You have Shown this For Sets of All Finite Cardinalities.
> > > > You have Not Shown that this is True of N itself.

> >
> >

> > > Nobody can show anything for N itself. All proofs in set theory apply all
> > > natural numbers only.

> >
> > Now, what on Earth are you Talking about?
> >
> > We can Prove lots of Theorems about N.

>
> Always you need the elements, at least for proving bijections.


One can prove a bijection between N and Q by proving Q can be
well-ordered which it can be.
> >
>
> > > > Only a Fool would Think Otherwise!
> >
> > >
> >
> > > You seem to think otherwise. But you cannot show any proof that proves
> > > the bijection for N itself, can you? (Further you cannot show an
> > > uncountability proof that does not construct a defined element, although
> > > you boasted you knew such.)

>
> >
> > First, there are Many Enumations of Q, or at least Q+. For instance Take
> > the Stern-Brocot Tree

>
> A tree? Do you use a mollusc called tree? Or do you use its nodes and path?
> It is simply foolish to claim any bijection without considering the elements.


Any two sets which are well-orderable with unique non-successor
and every element a predecessor biject with each other.
N is natually such set.
Q becomes such a set when properly reordered:
Each member of Q is of form m/n, with m an integer, n a natural, and
with no common factor greater than 1.

Order them by increasing values of |m|+n and for equal values of |m|+n
by increasing values of m.

This is a well-ordering of Q with a first rational, 0/1, and for each
rational a unique successor rational, and none left out.

Thus each rational is now enumerated by the natural number marking its
position in that well-ordering, and Mueckenheim is proved wrong AGAIN!.

> You need all natural numbers to prove a bijection N-Q.

The normal well-ordeing of N provides them.

>
> > > Show a proof of bijection N and Q according to your delusions.
>
> > Done

And done!.
>
> You are either very stupid or very, very stupid. In both cases it is
> appropriate to stop this nonsense.


Since WM is the only one producing any nonsense, he is the only one able
to stop it.
>
> Regards, WM

--
Virgil
"Mit der Dummheit kampfen Gotter selbst vergebens." (Schiller)



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