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Some of the more organic theories considered in model theory (other than set theory, which, from what I've seen, seems to be quite distinct from "mainstream" model theory) are those which arise from algebraic structures (theories of abstract groups, rings, fields) and real and complex analysis (theories of expansions of real and complex fields, and sometimes both).

While relationships with algebra seem quite apparent, I wonder what are some interesting results in real and complex analysis that have nice model-theoretical proofs (or better yet, only model-theoretical proofs are known!)?

Of course, there's nonstandard analysis, but I hope to see some different examples. That said, I wouldn't mind seeing a particularly interesting application of nonstandard analysis. :)

I hope the question is at least a little interesting. I have only the very basic knowledge of model theory of that type (and the same applies to nonstandard analysis), so it may seem a little naive, but I got curious, hence the question.

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Whereas algebra deals with general structures, analysis often deals with concrete objects. It follows that the tools of model theory are immediately better for algebra. That being said there is a very interesting connection between Banach spaces and model theory, but I know far too little about it to write a proper answer. –  Asaf Karagila Dec 23 '12 at 6:34
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There's the article Applications of Model Theory to Functional Analysis by José Iovino that seems to contain an introduction to the results that Asaf Karagila mentions. –  Martin Dec 23 '12 at 12:02

2 Answers 2

up vote 8 down vote accepted

There is a result in functional analysis whose first known proof uses non-standard techniques:

Theorem If a bounded linear operator $ T $ on a Hilbert space $ \mathcal{H} $ is polynomially compact, i.e., $ P(T) $ is compact for some non-zero polynomial $ P $, then $ T $ has an invariant subspace. This means that there is a non-trivial proper subspace $ W $ of $ \mathcal{H} $ such that $ p(T)[W] \subseteq W $.

The proof was given by Allen Bernstein and Abraham Robinson. Their result is significant because it is related to the so-called Invariant-Subspace Conjecture, an important unsolved problem in functional analysis. Paul Halmos, a staunch critic of non-standard analysis, supplied a standard proof of the result almost immediately after reading the pre-print of the Bernstein-Robinson paper. In fact, both proofs were published in the same issue of the Pacific Journal of Mathematics!

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Nitpick: "the so-called Invariant-Subspace Conjecture" is usually phrased as a question since it is still wide open in the Hilbert space case. It's therefore more common to refer to it as the Invariant Subspace Problem (link goes to a discussion on mathoverflow.net). It is known that for Banach spaces both positive and negative answers are possible. More details on mathoverflow. –  Martin Dec 23 '12 at 10:30
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Thanks. I've heard about the theorem, but not about it being proven by nonstandard means. The paper in question can be found on project Euclid, and is openly accessible. –  tomasz Dec 23 '12 at 13:36

Ax found the following application in complex analysis:

Theorem: If $f : \mathbb{C}^n \to \mathbb{C}^n$ is an injective polynomial function, that is there exist $f_1,...,f_n \in \mathbb{C}[X_1,...,X_n]$ such that $f=(f_1,...,f_n)$, then $f$ is surjective.

You can show that the theorem holds for $f : k^n \to k^n$ where $k$ is a locally finite field, therefore it holds for the algebraic closure $\overline{\mathbb{F}_p}= \bigcup\limits_{n \geq 1} \mathbb{F}_{p^n}$ of $\mathbb{F}_p$. Then, it holds for the (non trivial) ultraproduct $K=\prod\limits_{p \in \mathbb{P}} \overline{\mathbb{F}_p} / \omega$ where $\omega$ is a non principal ultrafilter over the set of primes $\mathbb{P}$, because the theorem can be expressed as a set of sentences of the first order. But $K$ is an algebraic closed field of caracteristic $0$ and the theory of algebraic closed fields of caracteristic $0$ is complete, so $K$ and $\mathbb{C}$ are elementary equivalent. Finally, Ax theorem is proved.

Ax theorem was generalized by Grothendieck.

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I have heard that the analytic proof of this theorem is very long and difficult. I cannot attest this myself, but it seems like a relevant point to this answer. –  Asaf Karagila Dec 23 '12 at 11:09
    
I've actually seen this result proved during a mathematical logic course, but have since forgotten it. Thanks. :) –  tomasz Dec 23 '12 at 13:32
    
A minor remark (the proof is short enough that it can actually be spelled out): the theorem holds for $\overline F_p$ because for each $k$, we have that $F_{p^k}^n$ is preserved by any polynomial function, but for finite sets an injection into itself is automatically a bijection. –  tomasz Dec 23 '12 at 13:45

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