5
$\begingroup$

Question 1. Let $g : \mathbb{R} \to \mathbb{C}$ with $g(y) = \lim_{x \to \infty} f(x,y)$, where $f : \mathbb{R}^2 \to \mathbb{C}$. Is it correct that the nonstandard extension $^*g$ will have $x \in \mathbb{R}$ rather than $x \in \mathbb{^*R}$? So $$^*g(y) = [\langle \lim_{x \to \infty} f(x,y_1), \lim_{x \to \infty} f(x,y_2), \ldots \rangle] ?$$

Question 2. Suppose that I want to let $x$ be nonstandard and take the "nonstandard" limit, say $\lim^*$, as $x \to \infty$. That is, I want something like $h : \mathbb{^*R} \to \mathbb{^*C}$ with $$h(y) = \underset{x \to \infty}{\mathrm{lim^*}}\,^*f(x,y),$$ where $^*f : \mathbb{(^*R)}^2 \to \mathbb{^*C}$. Do people do this? How does $\lim^*$ work? Is the incompleteness of the hyperreals problematic? If this has a long answer, a reference is fine. I've found it impossible to search precisely for info on this.

Question 3. Is $h$ different from $^*g'(x,y)$ where $$g'(x,y) = \lim_{x \to \infty} f(x,y)?$$ Intuitively, it seems $h \neq \,^*g = \,^*g'$ because $^*g'$ is not really a function of $x$, but my concentration keeps dying before I can see the structure clearly.

  • 0
    I think it would help if you could clean up your question a little. It has some features that appear to be inessential, such as making $g : \mathbb{R} \to \mathbb{C}$ rather than just $\mathbb{R} \to \mathbb{R}$. I also don't understand your notation in the final line of Question 1.2012-08-22
  • 0
    @Ben: I think that Rachel is taking $\mathbb{^*R}$ to be an ultrapower; the square brackets indicate the equivalence class mod $\mathscr{U}$ of the sequence inside, where $\mathscr{U}$ is the ultrafilter. And it is essential for her question that $g$ map $\Bbb R$ either to $\Bbb C$ or $\Bbb R^2$ rather than to $\Bbb R$, though I’m not sure why she chose $\Bbb C$ rather than $\Bbb R^2$.2012-08-22
  • 0
    @BenCrowell: Yes, I just didn't think to simplify that bit when I generalized the question. It was originally related to a particular complex-valued function, but I suppose it is not essential. Brian is correct that $[\langle x_1, x_2,\ldots\rangle]$ denotes the equivalence class of a sequence mod an ultrafilter.2012-08-22

2 Answers 2

2

EDIT: Due to the apparent confusion, I've rewritten my answer.


This is a job for the transfer principle! The basic idea is that every standard object and notion of standard analysis can be translated perfectly to non-standard model.

In particular, the standard notion of limit applies equally to the non-standard model: simply by applying the *-transfer to the ordinary notion of limit, we get an operator

$$ \lim_{x \to a} f(x) $$

where $f$ is an (internal) function and $a$ is a hyperreal, and the dummy variable $x$ ranges over hyperreals near $a$.

Again by the transfer principle, this operator even agrees with the usual $\epsilon$-$\delta$ definition (with everything varying over hyperreals rather than reals). Although this fact is less useful: one usually doesn't use $\epsilon$-$\delta$ arguments to prove things about non-standard limits of non-standard functions: one instead proves things about standard limits of standard functions, and then transfers those results to the non-standard model.

Of course, this only applies to internal statements. External ones don't transfer: e.g. the property that

$$ \lim_{x \to 0} f(x) = \text{st}(f(\epsilon)) $$

for nonzero infinitesimal $\epsilon$ can only be expected to hold when $f(x)$ is a standard function.

The incompleteness of the hyperreals is not a problem here, because that is an external fact. The hyperreals, in fact, satisfy the internal version of the completeness axiom -- e.g. every bounded (internal) subset of the hyperreal numbers has a least upper bound. You just have to be careful to work with internal objects. e.g. $\mathbb{N}$ and $\mathbb{R}$ are external subsets of ${}^\star \mathbb{R}$, so it's not surprising that they don't have least upper bounds. In fact, this relates to the overspill principle -- e.g. any subset of ${}^\star \mathbb{R}$ that contains all of $\mathbb{R}$ must also contain an infinite hyperreal.

Limits at infinity or converging to infinity, such as

$$ \lim_{x \to +\infty} x^2 = +\infty$$

also transfer. As usual, one can work either with the ad-hoc definitions of such limits given by introductory classes, or one can transfer the extended real numbers. The latter yields an interesting conceptual fact: the infinite hyperreals are not "beyond $+\infty$": the standard extended real number $+\infty$ is still larger than every hyperreal number. They infintie hyperreals merely closer to $+\infty$ than any standard number, in a sense very similar to the fact that infintiesimals are closer to $0$ than any nonzero standard number.

As for your first question, you are correct. In the representation of the value of ${}^\star g(y)$ as a sequence of real numbers, each component is indeed a standard limit: in particular, the standard limit that defines $g(y_i)$.

  • 0
    What do you mean by the bottom * in a notation like $\lim_{x \to {}^\star\infty}^\star$? Was that just a typo? Since $\infty$ isn't a member of the reals, or a relation on the reals, or part of the superstructure of the reals, the mapping * can't be applied to it.2012-08-22
  • 0
    @Ben: $+\infty$ is a member of the extended real numbers.2012-08-22
  • 0
    I don't think that addresses my question.2012-08-22
  • 0
    I don't think your approach is the right way to go at this. Just because the transfer principle can be applied to something like the limit operator, that doesn't mean that the resulting object $\star\lim$ is useful or has nice properties, only that it has the same first-order properties. Part of Rachel's question is "Do people do this?" She's not asking whether it *can* be done but whether it *should* be done. In the sample of 4 NSA books on my shelf, I find none that do this. They define limits as a way of making contact with standard analysis. They don't define $\star\lim$. Counterexamples?2012-08-22
  • 0
    Apologies, I think my question was unclear. I tried to clarify just now in a comment to Ben's answer (which I know not how to link to). Like Ben, I also am curious about your starring of infinity. Is this because you are thinking of it as an argument of $\lim$? I had overlooked this, but I think that makes sense. And $^*\infty$ is then greater than every infinite hyperreal?2012-08-22
  • 0
    @Ben: And another part of her question is whether $\lim^\star$ makes sense, and it does. If one never takes the limit of a non-standard function, it should be because one simply never feels the need to, rather than because one keeps the standard and nonstandard models so separate in their mind that one can't use the transfer of standard methods. A major part of why NSA has been effective is precisely because all of the standard notions *do* transfer: e.g. we can do things like write down $\sin \sqrt{\epsilon}$ for non-standard $\epsilon$ without any reason for hesitation.2012-08-22
  • 0
    @Rachel: IMO, limits involving $\pm \infty$ are best understood using the extended real numbers: the ad-hoc definitions and methods of elementary calculus are just ways to state things without explicitly introducing the extended number system. And yes, $+\infty$ is greater than every infinite hyperreal. This, IMO, is an important conceptual point: the infinite hyperreals are not "beyond" the $+\infty$ of real analysis: they are merely *closer* to $+\infty$ than any standard real. In a very similar sense, really, to the fact that infinitesimals are closer to 0 than any nonzero real.2012-08-22
  • 0
    Okay, but then $x$ (the variable approaching whatever) should be ranging over the hyperreals, yes? When extending $\lim_{x \to a}f(x)$, you are letting $a$ be nonstandard, yes? It seems that $x$ then should range over all hyperreals, or else the starring of $a$ is rather strange. I guess I should write out what the $\lim$ notation really says formally. (Also, hey, stranger! :^))2012-08-22
  • 0
    @Hurkyl: "And yes, +∞ is greater than every infinite hyperreal...the infinite hyperreals are not "beyond" the +∞ of real analysis: they are merely closer to +∞ than any standard real." This is totally wrong. The extended reals and the hyperreals are separate number systems. They are not compatible, since, e.g., the inability to compute $\infty-\infty$ violates the transfer principle. Since they're separate systems, there is no meaningful sense in which you can compare $\infty$ to a hyperreal number. It's as nonsensical as comparing $\aleph_0$ to a point at infinity in projective geometry.2012-08-22
  • 0
    @Rachel: "And $\star\infty$ is then greater than every infinite hyperreal?" No, there is nothing called $\star\infty$.2012-08-22
  • 0
    @BenCrowell: I presumed that Hurkyl was talking about the extended hyperreals, i.e., $^*\infty$ is being considered as a member of an ultrapower of the extended reals. Are you saying that this makes no sense?2012-08-22
  • 0
    @Rachel: Ah, I was wondering if that was you :) Yes, when transferring the notion of limit to the nonstandard model, the dummy variable ranges over nonstandard numbers. Also, $f$ can be any internal function, not merely a standard one. External functions are forbidden, though! Also, formulas of the sort $\lim_{x \to 0} f(x) = \text{st}(f(\epsilon))$ are only expected to hold for standard $f$.2012-08-22
  • 0
    @Ben: Rachel is exactly right. The extended hyperreals are a perfectly good object of study in real analysis, and they transfer to the nonstandard model just like everything else does. The expression $x < +\infty$ for a non-standard hyperreal is meant to be interpreted as involving *-transfer of $+\infty$ (as well as the *-transfer of <).2012-08-22
  • 0
    Thanks for the clarification. I'm not sure why I didn't just write down the $\epsilon\mathrm{-}\delta$ definition and star things. That was very clear. I still am curious about general nonstandard convergence and limits (i.e., the obvious notion of the outputs being in the same halo for all inputs in a certain set but with laxer constraints on the input set than is needed for transfer (e.g., saying a hypersequence converges if there is *any* hypernatural index after which the values remain inf. close)). But that can wait for another time.2012-08-23
  • 0
    @Rachel: You can always set up a hierarchy of non-standard models. Each successive one will have elements that are infinitesimal with respect to the previous models, and so limits in each model can be defined in terms of halos in the next. I've actually had a use for this sort of hierarchy before, but it was in a sufficiently algebraic context so I could use the simpler theory of real closed fields instead.2012-08-23
  • 0
    This new version fixes some of the most incorrect and misleading features of your original one, but IMO is still quite poor. "The extended hyperreals are a perfectly good object of study in real analysis[...]" I would welcome some evidence of this. A google search on "extended hyperreals" yields exacty one hit, which is talking about something different. "[...]interpreted as involving *-transfer of $+\infty$" This is wrong for the reasons I explained above. $+\infty$ is not a member of the superstructure of the reals. Therefore it's not in the domain of the $\star$ operator.2012-08-23
  • 0
    @Ben: If wikipedia isn't a good enough source that the extended real line is used in real analysis, then try Royden. It's such a fundamental idea of real analysis that he introduces them before he even talks about limits. A formal set theoretic construction of the extended reals is a trivial matter: their underlying set is the disjoint union of $\mathbb{R}$ and two one-point sets. Alternatively, the set is bijective with the interval $[0,1]$. In both models, defining the related objects (e.g. $<$) is a trivial matter. As you're practically trolling me, this is the last I have to say to you.2012-08-23
  • 0
    "If wikipedia isn't a good enough source that the extended real line is used in real analysis[...]" The issue is the "extended hyperreals" you're talking about, not the extended reals.2012-08-26
  • 0
    @Ben: I'll try again. The extended real line is the underlying idea behind all of the ad-hoc variations of limits involving infinity, unifying all of them into the usual definition of limits in terms of open sets. Their construction is easy to carry out, and transfers without problem. The transfer *still* unifies limits with the standard definition, and has the bonus feature of unifying limits of standard functions them with the external definition in terms of monads as well. It's one thing to argue your opinion that they aren't useful to NSA, but another thing to make absurd technical claims.2012-08-27
  • 0
    "Their construction is easy to carry out, and transfers without problem." See my post from Aug 23.2012-08-28
  • 0
    I can see how the fact I made a typo in a comment and said "extended hyperreals" in a case where I meant "extended reals" is a source of a small amount of confusion, but this typo is not present in the main answer. Also, it doesn't explain your mystifying objection to the notion of transferring the extended reals. As an aside, if I'm not in error, the usual model of the hyperreals, viewed externally, really is a member of the superstructure anyways. (and correspondingly so is the extended hyperreals)2012-08-28
  • 0
    @Ben: (whoops, forgot to direct my comment)2012-08-28
0

As explained in the comment above, I'm not clear on your notation in Q1, but I think the general answer is that $x$ is a feature of some particular way of writing down a definition of g, but it's not a property of g. The properties of g* depend only on g itself, not on the way in which g's definition was written down.

As a general philosophical thing, NSA is meant as a "back to the future" approach, in which we pretend the epsilons and deltas were never invented, and go back to using infinitesimals in a manner similar to the way they were used in the 18th and 19th centuries. Since the whole point is to get rid of limits, there is little motivation for introducing something like lim*. When you get to something that in standard analysis would be described as a limit, the style in NSA is to think of it as a process where you throw away information, often by taking the standard part of something. For example, $\lim_{x\rightarrow\infty}(1/x)=0$ is expressed as $st(1/x)=0$ for $x$ infinite. By taking the standard part, we throw away the information about the infinitesimal value of $1/x$. The fact that $\lim_{x\rightarrow\infty}\sin x$ is undefined is expressed by saying that $\sin x$ can't be determined simply by knowing that $x$ is infinite. In other words, throwing away the information about which infinite value $x$ has entails the inability to find the value of $\sin x$.

I think your question boils down to one about how to express the noninterchangeability of limits in NSA. Here's an example from WP's article on interchange of limits: $a_{mn}=2^{m-n}$, where $m$ and $n$ are integers, and it makes a difference which one you let approach infinity first. In NSA, the function $a$ has a counterpart $*a$ defined on the hyperintegers. To express the idea that the order of limits matters, we would just say the following. Let $m$ and $n$ be positive, infinite hyperintegers, but let no other information be given. Specifically, we don't have the information about whether $m>n$, $m, or $m=n$. Then we can't determine whether $*a$ is finite or infinite.

  • 0
    Apologies, I think I was not clear enough in my question that I don't mean $\mathrm{lim^*}$ to necessarily be the extension of a standard function. Perhaps putting the star on the opposite side was a bad idea. I want to think about "taking the limit" of a nonstandard function. I only mention the standard $\lim$ function because I am curious how its extension $^*\lim$ relates to this other thing $\mathrm{lim^*}$. The difference that I am insisting on is that, in $\mathrm{lim^*}_{x \to a}$, $x,a \in\,^*\mathbb{R}$, which apparently doesn't happen for $^*\lim$.2012-08-22
  • 0
    The thing to remember is that NSA is only interested in properties of standard functions at standard points. This is one criticism that has been made of NSA. For example, the standard treatments have very little to say about the derivative of a function at a nonstandard point. The only *use* nonstandard points to get information about standard points; the nonstandard points never have equal status. Thus, for example, even if the "physical real line" has infinitesimals, NSA has little to say about them.2012-08-23
  • 0
    @CarlMummert: Yes, I have noticed this and find it disappointing. I find the hyperreal space interesting in itself, but I am never able to find any literature investigating it. (If anyone knows of some examples, please share!)2012-08-23