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I wonder whether non-standard analysis, non-archimedean extensions of reals such as surreal or hyperreal numbers can help us in obtaining standard-analytic results which are not possible to obtain by means of standard analysis?

  • Finding limits in closed form which are impossible to find in standard analysis

  • Generalizing functions to the values where standard analysis does not provide any satisfactory method

  • Eliminating undetermined forms where it is impossible by standard-analysis means

  • Finding identities and relations between standard numbers and functions that are inpossible by standard means

Or non-standard analysis is just as powerful as epsilon-delta method in regard of standard numbers?

I am asking this because evidently complex numbers gave a lot of new powerful methods and results - from solving algebraic equations towards Fourier transforms, integral forms for finite differences, number theory etc.

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    $\begingroup$ IST, for example, is a conservative extension of ZFC, so that any standard theorem with a nonstandard proof also has a standard proof. $\endgroup$ – M T Apr 5 '15 at 18:27
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    $\begingroup$ The new results about real analysis that are provided by complex analysis are not impossible to obtain in real analysis. You could take the complex-analysis proof and rewrite everything in terms of real numbers (real and imaginary parts) without ever mentioning complex numbers. Of course those proofs would be ugly, hard to find and hard to understand, but they would not be impossible. Similar comments apply to nonstandard analysis. $\endgroup$ – Andreas Blass Apr 5 '15 at 19:41
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    $\begingroup$ Sure, NSA gives you new methods and new ways to think, but the answer to whether we can "obtain standard-analytic results which are not possible to obtain by means of standard analysis" is no, at least in the IST formulation, because every standard theorem has a standard proof. NSA might make those theorems easier to discover however: the classic example is Bernstein and Robinson's work on the invariant subspace problem en.m.wikipedia.org/wiki/Invariant_subspace_problem $\endgroup$ – M T Apr 5 '15 at 19:45
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    $\begingroup$ @LeeMosher I'm slightly amused to hear that ultralimits are "becoming standard" since in some corners of algebraic analysis they have been merrily using variants of the ultrapower construction for a long time... $\endgroup$ – Yemon Choi Apr 5 '15 at 19:56
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    $\begingroup$ @Anixx Your comment is correct as long as one doesn't distinguish between provability and truth. If one does make such a distinction, though, then arithmetic is weaker than set theory, but real analysis is not weaker than complex analysis. $\endgroup$ – Andreas Blass Apr 22 '15 at 19:42
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To respond to the title question "Are there any standard analysis facts that can be proven or arrived only by means of non-archimedean extensions of reals and non-standard analysis?", while it is true that in principle every NSA proof has a standard translation, in practice such translation may exponentially increase the complexity of the proof, so a real resercher, even equipped with a supercomputer, may well come across a result that can be proved using NSA but cannot in practice be proved in a traditional framework. There are certainly results in the literature that currently only have NSA proofs. User Gerry Myerson asked for examples. Such examples are provided by the work of Renling Jin and coauthors in additive combinatorics, and that of Terence Tao and coauthors on approximate groups. Whoever can add a single layer of quantifiers and thereby eliminate the infinitesimals and/or ultraproducts from the proofs given by Jin and Tao will earn my admiration.

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    $\begingroup$ I think this would be a better answer if some examples (of what's alleged in the last sentence) were provided. $\endgroup$ – Gerry Myerson Oct 7 '15 at 4:57
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    $\begingroup$ This sounds to me like it might be overstating the reduction in complexity achieved with NSA. I think in most cases what is achieved is something like eliminating one layer of quantifiers. $\endgroup$ – Ben Crowell Oct 7 '15 at 14:59
  • $\begingroup$ @BenCrowell, I gave a response in the body of my answer. $\endgroup$ – user81136 Oct 11 '15 at 7:23

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