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Well, there's a transfinite implicit in the series of exponentiations -- we have aleph-0 exponentiations, don't we? And if we knock off that first one, we've go
by cstross 16y ago
Well, there's a transfinite implicit in the series of exponentiations -- we have aleph-0 exponentiations, don't we? And if we knock off that first one, we've got (aleph-0 minus 1) exponentiations.
(NB: I don't understand your use of the term "convergence" -- must be either something I've forgotten in the third of a century since I studied maths, or something above the level I reached. (Sub A-level.) (See also: Dunning-Kruger effect.))
- JoachimSchipper 16y agoTechnically, f(x) converges to y as x -> infinity if for every epsilon there is an X such that |f(x) - y| < epsilon for all x > X ("the error becomes arbitrarily small"). For instance, 1/x converges to 0 as x -> infinity (pick X = 1/epsilon), but the sequence 0, 1, 0, 1, 0, ... does not converge to any y (pick epsilon < 1/2 to see this). In particular, the sequence r, r^r, r^r^r, ..., where r^2 = 2, does converge (to 2; the first terms are approximately 1.41, 1.63, 1.76, ...); the sequence 2, 2^2, 2^2^2, ..., however, does not (the first terms are 2, 4, 16, ... - and it only grows faster from there.)
- gjm11 16y agoConvergence is what happens when you have a sequence of numbers, and they get closer and closer to some fixed number. Some sequences converge and some don't, and part of what's going on with this "paradox" is that one of the sequences it implicitly asks you to consider doesn't converge. Yes, there's a transfinite number implicit in the sequence of exponentiations, namely aleph-0. (It might be better to say omega-0 -- an infinite ordinal rather than an infinite cardinal -- but that's a technicality that doesn't really matter here.) But the problem with the "proof" doesn't have anything to do with the existence of infinite cardinals larger than aleph-0.