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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.

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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. –  Ben Crowell Aug 22 '12 at 15:36
    
@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 $\Bbb R^2$. –  Brian M. Scott Aug 22 '12 at 17:21
    
@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. –  Rachel Aug 22 '12 at 19:28
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2 Answers 2

up vote 1 down vote accepted

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 $g(y_i)$.

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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. –  Ben Crowell Aug 22 '12 at 18:29
    
@Ben: $+\infty$ is a member of the extended real numbers. –  Hurkyl Aug 22 '12 at 18:44
    
I don't think that addresses my question. –  Ben Crowell Aug 22 '12 at 19:30
    
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? –  Ben Crowell Aug 22 '12 at 19:35
    
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? –  Rachel Aug 22 '12 at 20:16
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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 $m>n$, $m<n$, or $m=n$. Then we can't determine whether $*a$ is finite or infinite.

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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 $^*\lim$. –  Rachel Aug 22 '12 at 20:04
    
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. –  Carl Mummert Aug 23 '12 at 2:53
    
@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!) –  Rachel Aug 23 '12 at 3:26
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