11

Consider the smallest ordered field that contains R and does not satisfy the Archimedean property. I assume this is a much simpler construction than ultrafilters and other big caliber artillery used in non-standard analysis. Why does this approach fail?

Mikhail Katz
  • 42,112
  • 3
  • 66
  • 131
Tegiri Nenashi
  • 1,030
  • "Smallest" in what sense? Inclusion? How do you know there is a "smallest" such field? Maybe I can have two ordered fields $F$ and $F'$ that contain $\mathbb{R}$ and don't satisfy the archimedean property, but with $F\cap F'=\mathbb{R}$; which one is smallest, then? (And it's easy to see that if there is one field $F$ that is ordered, contains $\mathbb{R}$, and does not satisfy the Archimedean property, then one can find an $F'$ that also has those properties and intersects $F$ at exactly $\mathbb{R}$ by using transport of structure). – Arturo Magidin Sep 12 '11 at 20:19
  • How about field extension of R adjoining just a single non-Archimedean element? – Tegiri Nenashi Sep 12 '11 at 20:21
  • @Arturo: You might want to specify that $F$ and $F'$ are not isomorphic as well. – Asaf Karagila Sep 12 '11 at 20:24
  • @Arturo, a quick glance at the linked article suggests to me that the field in question is asserted to be "smallest" with respect to the existence of injective field homomorphisms, not by set inclusion. – hmakholm left over Monica Sep 12 '11 at 20:24
  • 1
    @Henning: You would still need to establish somehow that such a thing exists in the first place. This does not seem to be something you can simply "define into existence". – Arturo Magidin Sep 12 '11 at 20:26
  • 2
    @Arturo, sure, and there's a concrete construction in the article. Without checking all the details, it looks like the fraction field of $\mathbb Q[X]$, equipped with an appropriate ordering. – hmakholm left over Monica Sep 12 '11 at 20:28
  • @Tegiri: Field extensions obtained by adjunction are defined as subfields of a specific extension; what extension are you considering so that you can talk about "adjunction by a single element"? Or else they are defined as specific quotients of explicit objects. Which object/quotient will give me your construction? I can define "the opensure" of a subset $X$ of a topological space to be the smallest open subset that contains $X$. Doesn't mean that $[0,1]$ has an opensure in the usual topology of the real line, though. – Arturo Magidin Sep 12 '11 at 20:29
  • @Henning: At best, the article is attempting to define a minimal structure, without asserting it is a minimum structure. – Arturo Magidin Sep 12 '11 at 20:30
  • 1
    @Arturo, it contains the claim: "if $\mathbb{F}$ is any ordered field that fails the Archimedean property and $y$ is any positive element of $\mathbb{F}$ that is smaller than every $1/n$, then there is a unique field embedding from $\mathbb{Q}[x]$ to $\mathbb{F}$ that takes $x$ to $y$. This gives us a precise sense in which our example is "smallest"." – hmakholm left over Monica Sep 12 '11 at 20:32
  • @Tegiri: You would be defining a transcendental extension of $\mathbf{R}$. That would not be real closed field (=all the odd degree polynomials must have a root). So e.g. Bolzano's theorem would fail. Consequently we would also be in the situation of Henning's answer: you would not get an algebraically closed field by adjoinins $\sqrt{-1}$. – Jyrki Lahtonen Sep 12 '11 at 20:34
  • 1
    @Henning: If there is a construction, and the construction pans out, fine; but it needs to go beyond merely stating "the smallest a such that b..." – Arturo Magidin Sep 12 '11 at 20:36
  • 2
    @All: the context of the article seems to be the construction of a field which is an initial object in the category of ordered fields equipped with a distinguished non-Archimedean element. It is of course not true that there is a smallest non-Archmidean ordered field, because if $K$ is such a field and $x$ is a non-Archimedean element of $K$ then $\mathbb{Q}(x^2)$ is a strictly smaller non-Archimedean field. – Pete L. Clark Sep 12 '11 at 20:43
  • @Pete: it's only smaller as measured by $\subseteq$, which is probably not the measure we want. There's an ordered field isomorphism $\mathbb{Q}(x^2) \to \mathbb{Q}(x)$, for example. –  Sep 13 '11 at 07:45
  • @Hurkyl: I agree. – Pete L. Clark Sep 13 '11 at 14:38

4 Answers4

13

To perform nontrivial analysis in a nonstandard extension of $\:\mathbb R\:$ requires much more than simply the existence of infinitesimals. One needs some efficient general way to transfer (first-order) properties from $\:\mathbb R\:$ to the nonstandard extension. This is achieved by a powerful transfer principle in NSA. Lacking such, and other essential properties such as saturation, one faces huge obstacles.

One need only examine earlier approaches to see examples of such problems. For example, see the discussion of the pitfalls of Schmieden and Laugwitz's calculus of infinitesimals in Dauben's biography of Abe Robinson (lookup "Laugwitz" in the index) and, for much further detail, see D. Spalt: Curt Schmieden’s Non-standard Analysis, 2001. In short, viewed in terms of ultrapowers $\mathbb R^\mathbb N,\:$ the S&L approach mods out only by a Frechet filter on $\mathbb N\:$ instead of a free ultrafilter. Thus one loses full transfer of first-order properties, e.g. one obtains only a partially ordered ring, with zero-divisors (i.e. there are $r,s\neq 0\,$ with $rs = 0)$. Without all of the essential properties of $\mathbb R,\:$ and without a general transfer principle, one obtains a much weaker and much more cumbersome theory as compared to Robinson's NSA.

Another point which deserves emphasis is the role played by logic, in particular the concept of formal languages. One of the major problems with early approaches to infinitesimals is that they lacked rigorous model theoretic techniques. For example, without the notion of a (first-order) formal language it is impossible to rigorously state what properties of reals transfer to hyperreals. This logical inadequacy is one of the primary sources of contradictions in the earlier approaches.

Abraham Robinson wrote much on these topics. It is said that he knew more about Leibniz than anyone. His lofty goal was to vindicate Leibniz' intuition, and to reverse the historical injustices done to him by many Whiggish historians. See Abby's collected papers for much more, and see also Dauben's superb biography of Robinson.

For a brief introduction to the ultraproduct approach to NSA see Wm. Hatcher: Calculus is Algebra, AMM 1982, and Van Osdol: Truth with Respect to an Ultrafilter or How to make Intuition Rigorous. For a much more comprehensive introduction to ultraproducts see Paul Eklof. Ultraproducts for Algebraists, 1977.

Bill Dubuque
  • 272,048
10

The heavy artillery of the model-theoretic approach to non-standard analysis is doing much more than just producing any one non-Archimedean ordered field or even a non-Archimedean ordered field which is elementarily equivalent to $\mathbb{R}$ (an ordered field is elementarily equivalent to $\mathbb{R}$ iff it is real-closed, so starting with any non-Archimedean ordered field, e.g. $\mathbb{R}(t)$, and taking the real-closure produces such a field). There is an elaborate machinery of transfer, standard parts, etc. to overcome the fact that calculus does not work as we want it to in any non-Archimedean ordered field.

Note that the unique (up to unique isomorphism!) Dedekind-complete ordered field is $\mathbb{R}$, so any other ordered field is not Dedekind-complete. More directly, if $K$ is a non-Archimedean ordered field, then the set of positive integers is bounded above (by any infinite element of $K$) but has no least upper bound, so $K$ is not Dedekind complete. This spells doom for many of the basic results of real analysis.

For instance, for an ordered field $(K,\leq)$, the following are equivalent:
(i) $K$ is isomorphic to the real numbers.
(ii) $K$ satisfies the least upper bound axiom (i.e., $K$ is Dedekind complete).
(iii) Any bounded monotone sequence in $K$ is convergent.
(iv) (Bolzano-Weierstrass) Every bounded sequence in $K$ has a convergent subsequence.
(v) Any interval in $K$ is connected in the order topology.
(vi) Any closed bounded interval $[a,b]$ is compact in the order topology.

So working in a non-Archimedean field -- and not using model-theoretic cleverness to carefully restrict to certain kinds of statements -- calculus as we know it fails dramatically.

Pete L. Clark
  • 97,892
9

The construction you link to will work as far as it goes, if your only goal is to produce some non-Archimedian ordered field. However, the model-theoretic construction is much stronger; it will guarantee that every formal statement that is true about the standard reals is also true for the non-standard reals, as long as you replace all constants in the statement with their non-standard counterparts. Without this, it could be difficult to transfer results between the standard and the non-standard worlds.

For example, if one takes your simpler construction and adjoins a square root of $-1$, will the resulting field satisfy the Fundamental Theorem of Algebra? You would have to repeat the entire proof from the real case, if it goes through at all. With the heavy-caliber nonstandard analysis, you can just says that every nonconstant degree-$n$ polynomial over $\mathbb C$ has a root in $\mathbb C$, thus every nonconstant degree-$n$ polynomial over $\mathbb C^*$ has a root in $\mathbb C^*$. Case closed.

hmakholm left over Monica
  • 286,031
  • I don't think it is a sensible definition, myself... if you use inclusion as your order relation, then there is no such thing in general and you would first need to show you can make it work (say, by working inside some specific field, much like we do with "Galois closures"); if you use "there is an injective homomorphism" as your order relation, you need to show that there is a minimum element of the collection... – Arturo Magidin Sep 12 '11 at 20:24
  • Um, what is not a sensible definition? I'm not defining anything here -- I'm assuming that the reader knows what the heavy-caliber ultrafilter stuff does, and just explaining how that makes it superior. – hmakholm left over Monica Sep 12 '11 at 20:30
  • Sorry for being unclear; I don't think "The smallest ordered field that contains $\mathbb{R}$ and does not satisfy the Archimedean property" is a sensible definition. By saying "it will work as far as it goes", you seem to be saying that this definition is sensible. – Arturo Magidin Sep 12 '11 at 20:32
  • That's what I get for not reading through pages of stuff; the original post here reads like it attempts to define this thing into existence by simply saying the magic words. Sorry about that. – Arturo Magidin Sep 12 '11 at 20:38
  • When you say "every formal statement" I guess you are referring to Łoś' theorem -- so you need to restrict to first order statements only. The hyperreals $^* \mathbb{R}$ don't satisfy the (second order) completeness axiom for example (a good thing too, otherwise they'd be isomorphic to the reals again). – Matthew Towers Sep 12 '11 at 20:42
  • @mt, yeah, I'm doing first-order logic by default, and not in the habit of pointing that out explicitly. The development I have the clearest recollection of created an entire non-standard set theory anyway... – hmakholm left over Monica Sep 12 '11 at 20:46
  • Is this really an answer to the OP's question? Shouldn't one say something about why calculus breaks down for non-Archimedean fields like $\mathbb{Q}(t)$? Also, of course the field $\mathbb{Q}(t)[\sqrt{-1}]$ does not satisfy the fundamental theorem of algebra: the algebraic closure of $\mathbb{Q}(t)$ has infinite degree over $\mathbb{Q}(t)$, as one can see already by noting that the polynomials $x^n - t$ are all irreducible. – Pete L. Clark Sep 12 '11 at 20:51
  • @Pete, if you can write a more complete answer, by all means do so. I'm not claiming a monopoly on the question. :-) – hmakholm left over Monica Sep 12 '11 at 20:54
  • For more or less the simplest use of such an extension field $\mathbb{R}^\ast$, we need, for every differentiable function $f$ on $\mathbb{R}$ an extension $f^\ast$ to $\mathbb{R}^\ast$ with the property that for any infinitesimal $\epsilon$, the standard part of $(f^{\ast}(x+\epsilon)-f(x))/\epsilon$ exists and is equal to $f'(x)$. This is a very strong requirement. Sure, we need failure of the Archimedean property, but we need enormously more. – André Nicolas Sep 12 '11 at 20:55
  • Why Fundamental theorem of algebra is needed in calculus, i.e. where irreducibility of x^^n-t matters? – Tegiri Nenashi Sep 12 '11 at 21:04
  • @Tegiri: well, for instance, the Intermediate Value Theorem holds for all polynomials with coefficients in an ordered field $K$ iff $K(\sqrt{-1})$ is algebraically closed. – Pete L. Clark Sep 12 '11 at 21:09
  • @Tegiri, FTA was a more or less arbitrary example of a property you might want to use in the non-standard world. It tends to crops up everywhere if you dig a bit for it. As a single just-as-arbitrary example, if you're doing vector analysis, sooner or later you'll probably want to look for roots of the characteristic polynomial of a Jacobian matrix. – hmakholm left over Monica Sep 12 '11 at 21:11
  • Sorry to be obtuse but isn't Intermediate Value Theorem scope is any function, not just polynomial? Is there a more elementary counterexample (copmpared to Jacobian matrix). If you demonstrate, say, that there is a problem finding derivative of a square function (or any other derivatives rule for that matter), than it is much more convincing:-) – Tegiri Nenashi Sep 12 '11 at 21:26
  • @Tegiri: right, so whether IVT holds for polynomials only is a weaker property of the field. (But there are branches of mathematics in which one is interested primarily in polynomial functions, e.g. real algebraic geometry.) Related to my answer, IVT holds for all continuous functions over an ordered field $K$ iff $K \cong \mathbb{R}$. – Pete L. Clark Sep 12 '11 at 21:43
2

If the assumption of the question is that using a simple-minded non-Archimedean field cannot be used to get interesting mathematical results, that assumption can be challenged. For example, Dehn used a simple-minded non-Archimedean field to resolve one of Hilbert's famous problems; see here.

Mikhail Katz
  • 42,112
  • 3
  • 66
  • 131
  • Note that the title refers to it working for "calculus foundations*, not "interesting mathematical results". But no doubt other applications may also be of interest to readers. – Bill Dubuque Jun 09 '17 at 14:32
  • @BillDubuque, the OP's concluding question is "Why does this approach fail?" This question is what my answer is a response to. – Mikhail Katz Jun 11 '17 at 06:55
  • 1
    Yes, it is clear now that you are interpreting the question "why does this approach fail" more generally than the scope given in the title ("doesn't work for calculus foundations"). – Bill Dubuque Jun 11 '17 at 15:16