On Wednesday, June 12, 2013 12:39:58 PM UTC-7, muec...@rz.fh-augsburg.de wrote: > On Wednesday, 12 June 2013 17:59:40 UTC+2, Zeit Geist wrote: > > > > Every configuration can be well-ordered by size. This implies there is no q that stays outside the well order. > > > > > Every finite one can, but not the entire set of Q. > > > > I prove that all elements of Q can be well-ordered by size. You cannot disprove it, because you cannot find any element that stays outside, can you? >
You proved nothing! You proposed an algorithm, But never showed that it conclusively does what you say.
> > > Any finite set of real numbers can be well- according to magnitude. Infinite ones, not necessarily. > > > > Either find a q that stays outside of the well-order by size. If such a q exists and is enumerated as q_n, then there is a first natural number n that cannot be put in a permutation such that all q's are in order by size. If such a q exists and cannot be enumerated, then countability is nonsense. > > > > I claim: Every set of natural numbers can be put in every desired permutation. Among them there is the permutation that orders Q by size. >
Sure thing. If the enumeration has f(j) = 1/2, what is f(j+1), f(j+2) and f(j+3)?
If f keeps the order we must have
| f(m) | <= | f(m+1) |
And no rational r, such that the magnitude Of r is in between that of f(m) and f(m+1).