On Thursday, June 13, 2013 3:01:00 AM UTC-7, muec...@rz.fh-augsburg.de wrote: > On Wednesday, 12 June 2013 23:51:52 UTC+2, Zeit Geist wrote: > > > On Wednesday, June 12, 2013 12:43:37 PM UTC-7, muec...@rz.fh-augsburg.de wrote: > > > > > > Choose any q e Q. > > > > > > For you algorithm to work, we have to be able > > > > > > find a step n_1 where that 1/2 of the rationals > > > > > > less than q are well-ordered, a step n_2 where > > > > > > 2/3 of the rations less than q are well-ordered, > > > > > > and so on for all numbers in form of n/n+1. > > > > Same requirement could be required and not be satisfied in the process of enumerating the rationals. Never 2/3 of all rationals less than q are enumerated. That's just what I wanted to show. >
You are so lost. I didn't say 1/2 then 2/3 then 3/4 then ... of all Q.
I said for each particular q e Q, you must be able to produce a natural number m_(1/2) such that at step m(1/2) we have for at least one-half of all rationals p, p < q, p is in the natural order.
Indeed for any real, r, you must be able to find a natural number m_(r); where if E(q) is the natural number index of q in the enumeration, then at step m_(r) you have ceiling( r * E(q) ) of the rationals less than q in their proper position in the enumeration. Actually, you just need to show they are in a position less than E(q) at step m_(r). Since the property must be true for all q e Q, the fact that must be eventually be in proper order eventually does follow.
> > > This will never happen in you algorithm, > > > > > > so you never well-order all rationals in their > > > > > > order of magnitude. > > > > And this will never happen in countaing the rationals. Thanks. >
For a general well-ordering of the rationals, not in order of magnitude, we don't have to worry about, say, 1/2 of the rations less than q. We just have to worry about, say, 1/2 the rationals, p, such that E(p) < E(q).
It is the "in order of magnitude" part that make it a different situation.