Virgil
Posts:
10,821
Registered:
6/8/11


Re: ? 533 Proof
Posted:
Aug 1, 2014 2:39 PM


In article <c3cd939e5f6d4cd0937897be55e3f7e6@googlegroups.com>, mueckenh@rz.fhaugsburg.de wrote:
> On Thursday, 31 July 2014 13:30:35 UTC+2, Michael Klemm wrote: > > WM wrote; > > > > > > > > > For every k in N there is n_0 in N such that for n >= n_0: (nk, n] c > > > s_n, i.e., an interval that does not contain rational numbers indexed by > > > n > > > or smaller naturals. > > > > Why so complicated? For any finite number of reals there are always gaps > > between any two of them. > > There are always gaps between real numbers. Impossible to close them. > > But here we have the overwhelming evidence that the rationals cannot be > enumerated by the naturals.
WM's claim above proved false below by wellordering both Q+ and Q.
> The number of unit intervals, each one containing > infinitely many rationals without index =< n, increases infinitely, i.e., > beyond any upper bound. Of course a matheologian will brush this aside by the > standard blether "cardinals are not continuous". But everybody with a > critical intellect will ask *why* he should believe this. The honest answer > is: it is forced by the claim that infinity can be finished.
The honest answer outside of WM's worthless world of WMytheology is that the set of rationals can easily be wellordered with unique nonsuccessor, which makes it orderisomporphic to , thus bijected with the set of naturals.
Here is a straightforward way to construct a wellordering of the positive rationals. Write each one as p/q where naturals p and q have no common factor (other than 1). Order them by ascending value of (p+q), then within each set of p+q values, order by ascending p.
So you get: 1/1, 1/2, 2/1, 1/3, 3/1, 1/4, 2/3, 3/2, 4/1...
You can also wellorder the whole set of rationals by putting zero at the start and interleaving each negative after its corresponding positive.
And here we have the overwhelming evidence that the rationals CAN be, and have been, enumerated by the naturals.
Every and any wellordered set having no more than finitely many nonsuccessor elements can be ennumerated by the naturals, and any wellordered set with only one such nonsuccessor has been thus ennumeated by that wellordering.
In spite of All WM's waiings to the contrary!  Virgil "Mit der Dummheit kampfen Gotter selbst vergebens." (Schiller)

