# The value of “hard” nonstandard analysis

First of all, I want to make clear what I'm NOT asking. I'm not hoping to do a rehash of the implications of nonstandard analysis on calculus. Rather, I'm interested in its use in "harder" math. I'm currently reading through Goldblatt's Lectures on the Hyperreals and working on the later sections, wherein he discusses ways of rephrasing other areas of math in nonstandard language (e.g. Loeb measures). I'm trying to understand what the purpose of this is.

I understand that nonstandard doesn't get us new results, that is there's nothing we can prove in a nonstandard framework that we can't prove over old-fashioned ZFC. I also understand that generally nonstandard allows us to see the spaces we work in "more intuitively", e.g. Loeb measures allow us to see Lebesgue measure in a more finitary light, but I don't have much of a sense for what this more intuition looks like when we're actually trying to prove statements.

So what is the use of nonstandard analysis in its broadest sense? To those of y'all who study/use/teach it, what do you see it as buying you over "standard" analysis?

• When I saw your subject line I thought you meant "hard" as opposed to "soft", and that I might find out what you mean by that by reading your question. Now I suspect you mean "hard" as opposed to "easy". – Michael Hardy Dec 9 '16 at 0:41
• "over old-fashioned ZFC" is not what NSA is opposed to. NSA is "over old-fashioned ZFC". Rather, the thing to contrast with NSA is not "old fashioned ZFC", but rather epsilon-delta methods and the like, i.e. the old-fashioned way of making calculus logically rigorous. $\qquad$ – Michael Hardy Dec 9 '16 at 0:43
• @MichaelHardy I meant hard to be closer to "not soft" than "not easy", but I was looking for a way to basically distinguish from your basic analysis of maps from $\mathbb{R}$ to $\mathbb{R}$ and whether they're continuous or differentiable, which seemed to be the topic of many posts I found, and more about how the general concept of taking a universe and enlarging it works. As for the "old-fashioned ZFC" comment, that point came up frequently in other threads so it seemed it mightn't be a bad idea to repeat one more time. – AJY Dec 9 '16 at 2:10

Robinson's framework does give you new results. For example see this recent result by Terry Tao and Van Vu in Discrete analysis, where the authors use the language of infinitesimals because a paraphrase in an epsilontic framework would have been nearly unmanageable.

Robinson did prove a theoretical result that a theorem proved in his framework can be proved without infinitesimals, as well. However, practically speaking such a translation may be unreadable because of an explosion of complexity. A related point was discussed in detail in an article by Keisler and Henson in 1986:

Henson, C. Ward; Keisler, H. Jerome. On the strength of nonstandard analysis. J. Symbolic Logic 51 (1986), no. 2, 377–386.

• So although technically a result of nonstandard will have an analogue in standard with a standard proof, in practice it may be the case that this translation would require so much "epsilon management" (which nonstandard techniques tend to conceal) that we might as well just let it stay nonstandard? Also, what was the Keisler & Henson article? – AJY Jan 5 '17 at 16:13
• Exactly. I will add the Keisler-Henson reference. – Mikhail Katz Jan 5 '17 at 16:16

I'm just going to give a fragment of an answer. "what do you see it as buying you over 'standard' analysis?" Here's one small insight that I think might never have come about without Robinson's NSA:

• (This part was there before Robinson's NSA, in the form of an intuitive statement.) Suppose $f:\mathbb R \to \mathbb R.$ Then continuity of $f$ at $a$ means if $\varepsilon$ is infinitely small, then so is $f(a+\varepsilon) - f(a)$. Thus $f$ is everywhere continuous if that holds for every real number $a$.
• But $f$ is uniformly continuous if the same is true not just at every real number $a$, but also every nonstandard real number $a$, including those infinitely close to some real number, and also including those that are infinitely large. For example, suppose $f(x)=e^x$. Then if $a>0$ is infinite, then you can have $f(a+\varepsilon) -f(a)=1$ even though $\varepsilon$ is infinitely small, since the growth rate of $a\mapsto e^a$ is infinitely large when $a$ is infinitely large. Thus $a\mapsto e^a$ is not uniformly continuous. Likewise, suppose $f(x) = \sin(1/x).$ Then when $a>0$ is infinitely close to $0$, you can have $f(a+\varepsilon)-f(a) = 2$ even though $\varepsilon$ is infinitely small. Thus $f$ is not uniformly continuous.
• I'm familiar with this. I'm trying to move past basic calculus and undergrad analysis to the "wider world" of nonstandard. – AJY Dec 9 '16 at 2:12

In the 1960's Bernstein and Robinson used NSA to prove for the first time that a polynomially compact operator on Hilbert space has an invariant subspace.

Halmos subsequently rewrote their proof in standard terms. In the abstract he says

The purpose of this paper is to show that by appropriate small modifications the Bernstein-Robinson proof can be converted (and shortened) into one that is expressible in the standard framework of classical analysis.

PACIFIC   JOURNAL  OF  MATHEMATICS
Vol.   16, No.  3, 1966
INVARIANT SUBSPACES OF POLYNOMIALLY COMPACT OPERATORS


http://msp.org/pjm/1966/16-3/pjm-v16-n3-p05-p.pdf

... but the nonstandard proof to be modified came first.

• I wasn't looking for "historical" reasons, I was looking for why you use it here and now. – AJY Dec 10 '16 at 2:08
• @AJY I knew that but thought the story worth telling. – Ethan Bolker Dec 10 '16 at 2:46
• @EthanBolker, you might want to take a look at this article on Halmos and his antics. – Mikhail Katz Jan 5 '17 at 16:19