On Tuesday, June 11, 2013 11:49:59 AM UTC-7, muec...@rz.fh-augsburg.de wrote: > On Tuesday, 11 June 2013 20:23:17 UTC+2, Julio Di Egidio wrote: > > >> After each one has been > attached to a natural, we have the set of naturals that obviously can > be re-ordered in any desired way. > > > > > Obviously and crucially not in any finite number of steps (you will get the rationals well-ordered by magnitude from, say, the rationals lexicographically ordered). Anyway, I'm already out of my depths here, so I'll leave this to more competent mathematicians. > > > > It is obvious: If we show in set theory a proposition P(n) for the first n elements of a well-ordered set (where n is an arbitrarily large natural number), then we do show it for all elements of the set. >
This is not obvious. Indeed, it is false.
If we show in set theory a proposition P(n) for the first n elements of a well-ordered set (where n is an arbitrarily large natural number), then we do show it for EACH element of the set.
The validity P may not carry over to the limit point of Omega.
For all n e |N, FIS_n(|N) is finite. But, |N is not finite.
For all n e |N, the first n digits of 2^( 1/2 ) forms a ration number. But, 2^( 1/2 ) is not a rational number.
> > If we enumerate the rationals, we do it up to n (since there is no infinite natural number). If we apply the diagonal argument, we do it up to n (since there is no infinite natural number). If we well-order the rational numbers by magnitude, we do it up to n. > > > > There is absolutely no difference. >
You right, no difference. Both fail at the limit stage.
> > > (The power set of |N should even include the singleton of the last natural > as an element.) There is no such thing as the last natural number, and not even a next to last, and not even a next to next to last, and so on. > > > > I know. But there are more than any natural number of numbers. So if we take the numbers 1 to n for any natural number, then we have less than aleph_0 naturals. What remains to take all? >
Omega is Limit Ordinal.
So, for all a < Omega, there exist b such that a < b < Omega.
When we take it up to each every natural, we have it all.
> > This is but *one* simple aspect which shows that the jerks of matheology should really be imprisonde in a mad-house. > > > > It's a bit of a pain to have to reformat your posts and quotes every time: it would help a lot if you could at least split your paragraphs in short lines using carriage returns. > > > > That's due to the new Google which is an outspoken shit. Unfortunately it seems that I cannot return to the old version. The developer of that mess be cursed. > > > > Regards, WM