Does evaluating hyperreal $f(H)$ boil down to $f(±∞)$ in the standard theory of limits? [closed]

Does evaluating $f(H)$ at an infinite hyperreal $H$ when doing calculus in Robinson's (Keisler's) framework amount merely to assigning $f(\infty)$ in the standard theory of limits?

This question has its source in a comment exchange following this answer: Concept behind the limit to infinity?

• A suggestion: please read the linked comment exchange if you intend to post an answer here. The link leads to an extensive discussion of questions like whether $f(H)$ exists independent of its standard part (which is the conventional limit) and whether this makes the nonstandard constructs superfluous to the computations as done in Keisler's book. – zyx Jan 1 '16 at 20:25
• I don't think that Note 2 improves the question, and I fear that it makes the question needlessly argumentative. – Carl Mummert Jan 8 '16 at 0:43
• @CarlMummert, I am happy with the part of Note 2 that the OP quoted but didn't write himself, and don't mind it being in the question. The severe panic reaction it is receiving indicates that it is outside the usual well rehearsed sort of commentary on NSA, and as such would be a good thing to sort out here. The problem is it's not part of any question; the OP isn't asking anything about it, and presents it only by way of declaring it wrong (without evidence) and soliciting criticism. The snarky editorial after the quotation should be removed. If OP has a real question, post that. – zyx Jan 8 '16 at 3:09
• Am I the only one who feels that this question is a bit of soapboxing to why NSA and hyperreals are "good" and "useful"? – Asaf Karagila Jan 10 '16 at 21:19
• This, and numerous other questions here and at MO, the link farming connecting the graph of those, plus a larger number of soapbox answers at both places. Taking over all Wikipedia articles on infinitesimals and NSA. Articles in the AMS Notices. Astroturfing (3 coauthors post answer within an hour) on recent MSE thread. Similar activity in other online forums. Flood of papers in mathematical philosophy/history/education journals. @AsafKaragila – zyx Jan 10 '16 at 23:12

The answer is no. First of all, in the system of hyperreals, there isn't just one infinite hyperreal $H$, but in fact infinitely many (uncountably many in fact, but that's not important now), and applying $f$ to them can lead to many results.

Secondly, more importantly, $f(\infty)$ and $f(H)$ will often disagree. Consider the simplest example: $f(x)=\frac{1}{x}$. In what you call "standard theory of limits" we will define $f(\infty)=\lim\limits_{x\rightarrow\infty}f(x)=0$, while in system of hyperreals we have $f(H)=\frac{1}{H}$, which is not equal to zero. Instead, it is an infinitesimal hyperreal number.

• But when the limit exists in the classical sense, it is the standard part of $f(H)$ regardless of what $H$ is. – Ian Jan 1 '16 at 19:31
• No worries. I upvoted and agree with the answer, though my interpretation of the implications might be slightly different. – zyx Jan 1 '16 at 20:47
• @zyx What? Hyperreals aren't "metaphors." They have very precise definitions. Have you looked up how they are constructed? (P.S. You need the axiom of choice.) – Akiva Weinberger Jan 5 '16 at 2:45
• @AkivaWeinberger, I have the impression that for zyx the terms "metaphor", "fiction", and "ZFC" are all synonymous. What could be pointed out is that, as proved by Kanovei and Shelah, there does exist a definable hyperreal line. This is an article from 2005 that is frequently overlooked in this type of discussion. – Mikhail Katz Jan 5 '16 at 9:02
• @zyx So I'm guessing you're having trouble with the nonstandard algorithms to compute limits (which are essentially the standard algorithms in a very thin disguise, and generally do not require an understanding of the ultrafilter construction), rather than the actual theorems (i.e. the proof of the mean value theorem or of the De Bruijn–Erdős theorem, where it is not immediately obvious how to turn them into standard proofs). – Akiva Weinberger Jan 5 '16 at 11:19
$\DeclareMathOperator{\st}{st}$

This is a bit too long for a comment, but I'm familiar with the Nelson approach to Nonstandard Analysis, Internal Set Theory (IST), and I believe an answer in this framework easily translates. In IST, instead of creating new sets such as hyperreals, IST adds axioms to set theory that reveal new numbers in existing sets, such as $\mathbb{R}$. In addition, IST introduces the terms "standard" and "nonstandard" as new terms that can be used to describe sets. I hope that by explaining the situation in IST, it is translatable to the other systems or useful to someone who finds this question.

I will assume we have some $f:\mathbb{R}\rightarrow\mathbb{R}$ for simplicity.

First, if $H$ is an unlimited real (guaranteed to exist for us by the Idealization axiom), it's important to note that $f(H)$ still exists even if $\lim_{x\rightarrow\infty}f(x)$ does not exist.

For example, let $f(x)=\sin x$. $\lim_{x\rightarrow\infty}f(x)$ does not exist, but $f(H)$ is still well defined for all unlimited $H$. For example, let $H=\pi n$ for $n$ any unlimited $n\in\mathbb{N}$, then $f(H)=0$.

However, suppose the limit does exist. Let us suppose $\lim_{x\rightarrow\infty}f(x)=a$ for some standard $a$. If $H$ is an unlimited (positive unlimited, I will let this be implied now) real, it is not true in general that $f(H)=a$. As others have mentioned, for a function like $f(x)=\frac{1}{x}$, $f(H)$ will be an infinitesimal ($f(H)=\frac{1}{H}$). However, some additional facts will be true since the limit exists.

If $H$ is unlimited, then we have $f(H)\simeq a$ where we read $\simeq$ as "infinitely close to". We say $a\simeq b\iff\forall ^\mathsf{s}c>0,\lvert a-b\rvert <c$. Here, $\forall ^\mathsf{s}$ is "for all standard".

Nonstandard analysis can be made to rephrase standard analysis limits. Certainly this relates the idea of evaluating functions "at infinity" to limits. We prove the following statements are equivalent for a standard $f(x)$.

1. $\lim\limits_{x\rightarrow\infty}f(x)=a$
2. $a$ is standard and $f(x)\simeq a$ for all unlimited $x$

Note that if $a$ is standard, then (2) implies $\st (f(x))=a$ by definition of the standard part. That is, (2) also means the standard part of $f(x)$ is equal to $a$ for all unlimited $x$.

First, we prove $(1)\implies(2)$.

I first note why $a$ must be standard: Since $f(x)$ is standard and converges by a classical formula, its limit must be standard by the transfer axiom (If there is a unique $a$ that satisfies the classical convergence formula for our standard parameters, $a$ must be standard).

$(1)$ tells us $\forall\varepsilon >0, \exists M\in\mathbb{R},\forall x\geq M,\lvert f(x)-a\rvert <\varepsilon$. Let $\varepsilon$ have a standard fixed value, so this is a classical formula of $M$ taking fixed standard values: $$\forall ^\mathsf{s}\varepsilon >0, \exists M\in\mathbb{R},\forall x\geq M,\lvert f(x)-a\rvert <\varepsilon$$ The transfer axiom of IST tell us that if there exists an $M$ satisfying this formula, there must exist a standard $M$ satisfying it. In this case, $x\geq M$ is automatically satisfied for any unlimited $x$. So, we may write: $$\forall ^\mathsf{s}\varepsilon >0,\lvert f(x)-a\rvert <\varepsilon\text{ for all unlimited x}$$ By our definition of $\simeq$, this means $f(x)\simeq a$ for all unlimited $x$. $$\tag*{\square}$$

Now, we show $(2)\implies(1)$.

We assume that for any unlimited $x$, $f(x)\simeq a$ and $a$ is standard. Suppose we have a standard $\varepsilon >0$, then the following is true $$\forall ^\mathsf{s}\varepsilon >0, \exists M\in\mathbb{R},\forall x\geq M,\lvert f(x)-a\rvert <\varepsilon$$ This is true because we can pick any unlimited $M$, and the statement is satisfied by our assumption. Now, we have a completely classical formula except for $\varepsilon$. Since this is a classical formula with standard parameters, the transfer axiom tells us this formula is true $\forall ^\mathsf{s}\varepsilon >0$ as soon as it is true for $\forall \varepsilon >0$, so we have $\forall\varepsilon >0, \exists M\in\mathbb{R},\forall x\geq M,\lvert f(x)-a\rvert <\varepsilon$. In other words, $$\lim_{x\rightarrow\infty}f(x)=a$$ $$\tag*{\square}$$

Please comment if I can clarify any IST-specific issues for you that aren't clear from a different NSA framework.

• Nice answer. There is a terminological issue that different authors propose different ways of resolving. As you point out, Nelson refers to such $H$ as "unlimited" or perhaps "illimited" rather than infinite, because being regular real numbers (albeit nonstandard) it does not seem logical to call them infinite. There is a similar problem with infinitesimals and the relation of infinite closeness that you mentioned... – Mikhail Katz Jan 5 '16 at 7:51
• @user72694 In fact, the terminology in IST is that unlimited numbers are finite. I struggled with that terminology when learning, but I think it's necessary with the way IST works. You can prove they are finite in IST by any reasonable classic definition of the word, but you can also understand why it is necessary (as you allude) if you are to maintain statements such as "all natural numbers are finite." I think I struggled a bit more though with the axiomatic gymnastics of giving IST rigor. I am not well educated enough on other NSA approaches to compare their difficulty, though. – GPhys Jan 5 '16 at 8:49
• What are we supposed to say instead of "infinitely close" in IST? – Mikhail Katz Jan 5 '16 at 8:57
• @user72694 "Infinitely close" is what the authors I learned from used. – GPhys Jan 5 '16 at 9:05
• In later works, Nelson also used "i-small" and "i-large" for infinitesimals and unlimited numbers. – Dr. Lutz Lehmann Jan 5 '16 at 10:09

I agree with @Wojowu that evaluating $f$ at a hyperreal $H$ carries more information than could be contained by $f(\infty)$ in an Archimedean framework, and would like to provide an additional reason.

Uniform continuity is known to be a global property of $f$ in the sense that there is no pointwise notion of uniform continuity in an Archimedean framework. On the other hand, in a hyperreal framework, one can come pretty close, as follows.

It turns out that a real function $f$ is uniformly continuous on $\mathbb{R}$ if and only if the natural extension ${}^{\ast}\!f$ is S-continuous at every point of ${}^{\ast}\mathbb{R}$. Here S-continuity means that every infinitesimal change of the input leads to an infinitesimal change of the output.

Thus, the squaring function on $\mathbb{R}$ is not uniformly continuous because at an infinite point $H$, and infinitesimal change $\alpha=\frac{1}{H}$ of the input leads to a change in the output which is not infinitesimal but rather appreciable: $$(H+\alpha)^2-H^2=H^2+2H\alpha+\alpha^2-H^2=2+\alpha^2.$$ Thus the failure of the squaring function to be uniformly continuous is captured by its failure to be S-continuous at a single infinite point.

The syntax of the Archimedean framework is too poor to express this distinction; certainly one can't get anywhere here with merely $f(\infty)$.

Cauchy and Robinson defined continuity in similar way procedurally, namely infinitesimal increment always leading to infinitesimal change in the function.

Depending on how broad a class of functions one wishes to encompass, one can develop infinitesimals using stronger or weaker tools (always in the context of the standard ZFC). Thus, Skolem's arithmetic with infinite numbers is completely constructive, as analyzed in Stillwell's

Stillwell, John. Concise survey of mathematical logic. J. Austral. Math. Soc. Ser. A 24 (1977), no. 2, 139–161 (see http://www.ams.org/mathscinet-getitem?mr=465766).

Of course, to handle the squaring function we don't need more than Skolem's infinities.

• There isn't anything nonstandard about this. The calculation is true for all nonzero $H$. In general, arguments and definitions that are independent of $H$ go through with the (simpler, older) standard Cauchy definitions of infinitesimal/infinite using sequences, or the yet simpler notion of "large enough finite $H$". At the level of a book like Keisler, and considerably further than that, there is no way to tell the difference between the Robinson and Cauchy style of argument because nothing depends on the equivalence relation between sequences being an ultrafilter. – zyx Jan 3 '16 at 7:12
• Also, is there an example in which $\lim f = f(\infty)$ exists and is finite? In the source of this question at the cited comment thread, conversation was about the more general matter of whether things boil down to "standard analysis". This question uses the narrower words "boil down to $f(\infty)$ in the standard theory of limits" which is fine as an abbreviation, but not as an assertion that $\lim f$ is the extent of the standard theory that NSA should be compared to, when the limit does not exist. – zyx Jan 3 '16 at 7:24
• @zyx, I replied as part of my answer. – Mikhail Katz Jan 3 '16 at 15:09
• The addition doesn't disagree with the statement that Robinson nonstandard analysis = Cauchy standard analysis in this context. It is also unclear what Skolem arithmetic has to do with the question of "evaluating f(H) ... when doing calculus in Robinson (or Keisler's) framework". Maybe I am missing something since I know NSA primarily from Robinson and Nelson's books. For Skolem arithmetic, which is probably best discussed on a different page, the issues are more that it is not a framework for analysis and that the operations in any form of nonstandard arithmetic are uncomputable. – zyx Jan 3 '16 at 22:07
• @zyx, discussing Skolem at a different page sounds like a good idea. Would you like to post such a question? Note also that Cauchy's infinitesimal procedures are very similar to Robinson's. – Mikhail Katz Jan 4 '16 at 7:52