# Non-Standard analysis and infinitesimal

Can someone please explain how Non Standard Analysis is used to justify infinitesimals?

The "infinitesimal", or a quantity which is smaller than any non-zero real number but still non-zero, is a non-rigorous idea that had been a weak link in early calculus, which extensively relied upon it. The Leibniz notation $dy/dx$, for example, is originally supposed to mean "an infinitesimal change in $y$ divided by an infinitesimal change in $x$". Well, it worked, but exactly why eluded the early mathematicians who developed calculus.

The infinitesimal problem was first resolved by looking at limits not as precise evaluations of a function at an infinitesimal value, but a somewhat more complicated definition (the $\epsilon$-$\delta$ definition) that made the concept rigorous and logically sound. However, people have always wondered if it is at all possible to actually extend the real numbers in a way (much like the extension to complex numbers) which makes it satisfy two properties:

1. It has to preserve certain properties of the real numbers familiar to us, and
2. It has to contain elements such that any "infinitesimal number" that arises in early calculus corresponds to a precisely-defined element.

Such an extension is called the hyperreals. It justifies the infinitesimal numbers because they are now well-defined: say one wishes to take the derivative of $x^2$ at $x=3$. The definition gives us

$$\lim_{\Delta x\to 0}\frac{(3+\Delta x)^2-3^2}{\Delta x}$$

Doing some quick algebra works this out to $6+\Delta x$ where $\Delta x$ is infinitesimal. In the standard real numbers this word has no precise meaning: it's simply a figure of saying that we happen to be able to manipulate as if it is zero, but without the restrictions (in the simplification above, for instance, we have to divide the numerator and denominator by $\Delta x$, which we cannot do if it is zero). In the hyperreals, however, $6+\Delta x$ is a specific element which is not equal to $6$. However, an equivalence relation called the standard part on a subset of the hyperreals (it is only defined on a subset because the hyperreals actually contain infinite elements as well) gives a homomorphism from the hyperreals to the reals, which can be considered as the hyperreals modulo this equivalence relation. So taking the standard part of $6+\Delta x$ easily shows that the real derivative is $6$. (In fact, any hyperreal construction necessarily admits infinitely many infinitesimals in the equivalence class of any real number.) This would give us property 2 as we discussed above.

What, then, about property 1? It turns out that the hyperreals can preserve many properties of the reals, but not all of them. The ones that they do preserve are the first-order properties. These are properties that can be expressed as a finite formula using quantifiers on the elements, but not on subsets. For instance, "for any two real numbers there is a real number strictly in between" is first-order, hence it's preserved in the hyperreals. On the other hand, the infinitary statement "for any real number $x$ there exists a natural number $n$ such that $x<\underbrace{1+1+\cdots+1}_n$" is not preserved in the hyperreals. (In particular, the field axioms are first-order, so the hyperreals form a field.) This preservation is called the "transfer theorem", a straightforward consequence of the standard construction of the hyperreals as an ultrapower of $\omega$ copies of $\mathbb{R}$ modulo some suitable ultrafilter; Łoś's theorem shows that the first-order properties are all preserved.

• The comment On the other hand, the infinitary statement "for any real number $x$ there exists a natural number $n$ such that $x<\underbrace{1+1+\cdots+1}_n$" is not preserved in the hyperreals is incorrect. The formula $(\forall x\in\mathbb{R})(\exists n\in\mathbb{N}) (x<n)$ is a first-order statement and therefore transfers to the formula $(\forall x\in{}^\ast\mathbb{R})(\exists n\in{}^\ast\mathbb{N}) (x<n)$ which is true in the hyperreals by Łoś's theorem. – Mikhail Katz Dec 4 '14 at 13:49
• @user72694 First of all, your statement is not a first-order statement (or really a statement in any formal language). Secondly, your statement is not the same as my statement. In my statement I don't quantify over the naturals as a subset of the reals, but there is an infinitary disjunction over $\omega$-many formulae which I abbreviated as "there exists a natural number $n$". Finally, the crucial problem with your statement is that $\mathbb{N}$ is not definable in $\mathbb{R}$, for otherwise $\mathbb{Q}$ is definable. So you can't even translate your statement into a first-order sentence. – Marc Dec 4 '14 at 19:46
• @user72694 On the other hand, your statement is correct if we take a new unary predicate symbol, interpret it as the natural numbers in the reals, then formalise your statement in the expanded language. This is, however, not what I meant in my answer, as explained in my first comment above. – Marc Dec 4 '14 at 20:04
• I was obviously referring to the version of first-order as it is used by Robinson since we are talking about nonstandard analysis. In Robinson's version, we have symbols in the language for $\mathbb{N}$ as well as every standard object. With your restrictive notion of first-order, you don't get a correct statement of the transfer principle but rather get a much weaker theorem not useful in analysis. – Mikhail Katz Dec 5 '14 at 9:44

A much simpler, if less complete, way of explaining how non-staandard analysis implies infintesimals would be just to note that non-standard models include transfinite integers, but as soon as one has those, one merely needs to take their reciprocals to produce the infintesimals. Strangely enough, Cantor himself stubbornly refused to grant such infintesimals even though they are an immediate consequence of his tranfinite numbers!!

• Cantor's transfinite numbers are basically totally unrelated to the transfinite integers of nonstandard analysis. – Eric Wofsey 2 days ago
• "Cantor's transfinite numbers are basically totally unrelated to the transfinite integers of nonstandard analysis." Huh, both Cantor and Robinson add transfinite integers to their concpetualizations. And Robinson uses them to define the infintessimals as do most who follow after him in the field of non-standard analysis. "totally unrelated"?? It is true that Cantor specifies many other types of transfinite cardinals, but his specification of the ordinal omega explicitly introduces the same transfinite integers as operate in non-standard analysis. – Wd Fusroy 2 days ago
• Yes, they are both numbers that are "infinitely large" in some sense. Other than that, they are totally unrelated. In particular, Cantor's transfinite numbers are completely useless for talking about infinitesimals in analysis, which is what this question is about. – Eric Wofsey 2 days ago
• Your claim that "his specification of the ordinal omega explicitly introduces the same transfinite integers as operate in non-standard analysis" is outright wrong and shows you have no idea what you are talking about. – Eric Wofsey 2 days ago