<email@example.com> wrote in message news:firstname.lastname@example.org... > 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. > > 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.
The obvious and essential difference is that in the second paragraph you drop the "where n is an arbitrarily large natural number" to equivocate on the "up to n": your usual and hardly candid word salads.
(Plus, I had something quite different in mind with the "not in any finite number of steps"/"not effectively", but that's incidental and I cannot make it more precise, anyway.)
> > (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?
Same word salad. Argue with/against the math if you can: