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Jan 18, 2023 at 10:27 history edited Martin Sleziak CC BY-SA 4.0
http -> https (the question was bumped anyway)
Apr 11, 2013 at 8:50 comment added Mikhail Katz ... I concur with Joel David's opinion that this is in fact "a primordial Transfer Principle". The connection between Leibniz's Law of continuity and the Los-Robinson Transfer Principle was explored in a paper by David Sherry and myself in the Notices of the AMS, see ams.org/notices/201211
Apr 11, 2013 at 8:48 comment added Mikhail Katz @Tom Leinster: you commented above that "As I understand it, a crucial difference is that the infinitesimals of SDG can, for instance, have square equal to 0, but the infinitesimals of NSA can't. I'd guess that to be an important part of providing that 'missing legitimacy'." From the point of view of nilpotency, Leibnizian calculus is closer to NSA than SDG. This is because Leibnizian infinitesimals aren't nilpotent but on the contrary are assumed to have all the properties of ordinary numbers, a principle Leibniz referred to as the "law of continuity" ...
Apr 7, 2013 at 15:51 history edited Todd Trimble CC BY-SA 3.0
corrected spelling
Feb 27, 2010 at 12:56 comment added Joel David Hamkins Tom, I didn't mean to suggest that NSA is the only possible justification of infinitesimals. It is true, however, that many early calculus writers described infinitesimals as having all the properties of real numbers, except infinitely smaller (smaller than 1/n for every natural number n). That idea sounds like a primordial Trasfer Principle, which holds for NSA but not SDG.
Feb 27, 2010 at 2:21 comment added Tom Leinster Joel, I enjoyed your answer and learned from it, but I wondered whether "nonstandard analysis exactly provides the missing legitimacy" [of early calculus] was overstating it. I'm more familiar with the "other" way of putting infinitesimals on a firm footing, that of Synthetic Diff Geom (as e.g. in Bell's text A Primer of Infinitesimal Analysis). As I understand it, a crucial difference is that the infinitesimals of SDG can, for instance, have square equal to 0, but the infinitesimals of NSA can't. I'd guess that to be an important part of providing that "missing legitimacy". Any thoughts?
Feb 25, 2010 at 21:32 comment added Joel David Hamkins I should say that the reverse implication in that equivalence uses countable choice, because if you have an ultrafilter, you still need countable choice to verify that the ultrapower satisfies the Los theorem, which is what gives you the Transfer principle. But the transfer principle in any case gives you ultrafilters, which is an interesting little argument.
Feb 25, 2010 at 21:26 comment added Joel David Hamkins One needs no choice at all to construct nonstandard models of arithemtic. For the reals, however, the existence of a nonstandard model of the reals with the transfer principle is equivalent to the existence of a nonprincipal ultrafilter on omega, which would be a weak choice principle. Nevertheless, one needs at least DC to have a decent theory of Lebegue measure, so there seems to be choice all around here.
Feb 25, 2010 at 21:19 comment added Joel David Hamkins Well, many ordinary uses of epsilon-delta also use choice. For example, to know that the epsilon-delta definition of continuity for a function on the reals is equivalent to the convergent sequence characterization relies on AC, since you need to pick the points inside those delta-balls.
Feb 25, 2010 at 21:01 comment added HJRW Regarding NSA vs epsilons and deltas, isn't 'not using the Axiom of Choice unnecessarily' a good reason to use the latter?
Feb 25, 2010 at 2:37 history answered Joel David Hamkins CC BY-SA 2.5