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Periodically I've tried to wrap my head around nonstandard calculus and hyperreals, but I always thought I needed a lot more of a background in formal logic and/or set theory to understand what's going on with them. Last night I had a sudden flash of insight about how they work, based on the ultrapower construction, but it was so anticlimactic that I wonder if I'm missing something. I want to outline my understanding to see if I actually have the basic idea correct. The construction I'm following is per Wikipedia, and I'm heavily paraphrasing:

One way to construct the hyperreals is as equivalence classes of real-valued sequences $\mathbf{x} = \{ x_i \}_{i \in \Bbb{N}} \in \Bbb{R}^\Bbb{N}.$ The construction is not dissimilar to using Cauchy sequences of rationals to define the reals, except that our equivalence relation doesn't necessarily treat Cauchy sequences that converge to the same number as being "equal"; rather, they are considered to be "infinitesimally different".

Any operation or function that we can define on the real numbers - addition, multiplication, comparison, absolute value, floor function - is extended to these sequences componentwise. This gives us a ring with a lot of the same nice structure and operations as we had on $\Bbb{R}$ originally; we can identify the elements $r$ of $\Bbb{R}$ with the constant sequences $(r, r, r, r, ...)$; and we can put a partial order on these sequences via componentwise comparison $\leq$: $\mathbf{a} \leq \mathbf{b}$ iff $a_i \leq b_i$ for all $i \in \Bbb{N}$. This partial order also lets us talk about sequences like $\mathbf{ε} = (1, 0.5, 0.25, 0.125, ..., 2^{-i}, ...)$, which are "greater than" the sequence $\mathbf{0} = (0, 0, 0, 0, ...)$ but "less than" any sequence $\mathbf{r} = (r, r, r, r, ...)$ for $r > 0$, as being infinitesimal in some sense, since $\mathbf{0} < \mathbf{ε} < \mathbf{r}$ for any positive $r$.

However, most sequences in $\Bbb{R}^\Bbb{N}$ are not comparable in this partial order, and the ring of such sequences is also rife with zero divisors, both of which are serious deficiencies if we hope to do any calculus. If possible, we would also like two sequences $\mathbf{a}, \mathbf{b} \in \Bbb{R}^\Bbb{N}$ to be considered "equivalent" (i.e. different names for the same sequence) if $a_i = b_i$ for all but finitely many $i$. We use an ultrafilter to accomplish this, and also to extend the partial order $\leq$ on sequences to a total order on equivalence classes of sequences (where the equivalence classes are given by "$\mathbf{a} \sim \mathbf{b}$" iff "$\mathbf{a} \leq \mathbf{b}$ and $\mathbf{b} \leq \mathbf{a}$"). The set of all equivalence classes, $\Bbb{R}^\Bbb{N} / \sim$, is the hyperreal numbers $^*\Bbb{R}$. Every function $f: \Bbb{R} \to \Bbb{R}$ has a nonstandard counterpart $^*f: ^*\Bbb{R} \to ^*\Bbb{R}$ given by $$^*f([x_1, x_2, x_3, ...]) := [f(x_1), f(x_2), f(x_3), ...]$$ where $[x_1, x_2, x_3, ...]$ means the equivalence class of the sequence $(x_1, x_2, x_3, ...) \in \Bbb{R}^\Bbb{N}$.

Under this interpretation, every infinitesimal hyperreal is the equivalence class of a Cauchy sequence with limit zero; every finite hyperreal is the equivalence class of a Cauchy sequence; and the "infinite hyperreals" are equivalence classes of unbounded sequences that either approach $+\infty$ or $-\infty$ (which gives an interesting way to rigorously interpret Big O notation). The standard part of a finite hyperreal is just the limit of a Cauchy sequence in its equivalence class.

Now that I understand how this works, it seems pretty clever to treat "essentially different" Cauchy sequences as different numbers "infinitesimally close" to a given one, since these sequences are the foundation of the limiting processes in calculus. But it also seems surprising to me that such heavy machinery was applied to proving that nonstandard calculus is logically equivalent to regular calculus, and such pedagogical emphasis (in the sources I've perused, like Robinson's original Nonstandard Analysis) is placed on the transfer principle. Because now that I have this understanding, nonstandard analysis almost seems like a trivial rewording of regular calculus; and I have such confidence that it works just like regular calculus, because the rewording is so mechanical once you understand that "standard part of a finite hyperreal" means "limit of a Cauchy sequence in its equivalence class":

Continuity

  • Standard: $f: \Bbb{R} \to \Bbb{R}$ is continuous iff $\{ f(x_i) \}_{i \in \Bbb{N}}$ is Cauchy whenever $\{ (x_i) \}_{i \in \Bbb{N}}$ is Cauchy (using the standard definition of a Cauchy sequence.)

  • Nonstandard: $^*f: ^*\Bbb{R} \to ^*\Bbb{R}$ is continuous iff $\operatorname{st}(^*f(\mathbf{x})) = f(\operatorname{st}(\mathbf{x}))$.

Differentiability

  • Standard: $f: \Bbb{R} \to \Bbb{R}$ is differentiable iff $\{ \frac{f(x + h_i) - f(x)}{h_i} \}_{i \in \Bbb{N}}$ is a Cauchy sequence converging to a specific number $f'(x)$ independent of the sequence $\{ h_i \}_{i \in \Bbb{N}}$, whenever $\{ h_i \}$ is Cauchy and converges to $0$.

  • Nonstandard: $^*f: ^*\Bbb{R} \to ^*\Bbb{R}$ is differentiable iff for any infinitesimal $\mathbf{h}$, $\operatorname{st}\left(\frac{^*f(\mathbf{x}+\mathbf{h}) - ^*f(\mathbf{x})}{\mathbf{h}}\right) = f'(\operatorname{st}(\mathbf{x}))$ independent of the value of $\mathbf{h}$.

So I guess my questions are:

  1. Am I understanding this right and, on some level, the "infinitesimals" are just a thinly disguised version of Cauchy sequences with limit zero? That seems significantly more mundane than the name's suggestion of "infinitely small numbers" would imply.

  2. What is the gain in clarity, elegance, speed, etc. one gets by working with "finite hyperreals" rather than the representative Cauchy sequences, per se? That is, what are some specific instances when one gains meaningful new insight into a result from standard analysis (as opposed to a spiffy new phrasing of the result) by working in the nonstandard/hyperreal setting?

Rivers McForge
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  • Well, since the ring of finite hyperreals isn't literally isomorphic to the ring of reals, they can't genuinely be the same things as Cauchy sequences. So I don't see how it could possibly be a "trivial rewording" of classical calculus. How would you describe it? (Fine, technically we should speak of "a hyperreal field"/"a ring of finite hyperreals" rather than "the hyperreals"/etc., but that's a side issue.) – Noah Schweber Nov 06 '20 at 23:11
  • @NoahSchweber As I understand it: 1) The hyperreals $^\Bbb{R}$ are the field given by starting from the sequence ring $\Bbb{R}^\Bbb{N}$ under componentwise multiplication and addition, and modding out a maximal ideal which comes via an ultrafilter. 2) Any "finite" elements of this field (elements $\mathbf{x}$ so that there exists $r \in \Bbb{R}$ with $|\mathbf{x}| \leq ^\mathbf{r}$) are in the equivalence class of a Cauchy sequence, i.e. for finite $\mathbf{x} \in ^* \Bbb{R}$ there exists $(x_1, x_2, x_3, ...)$ Cauchy so that $[x_1, x_2, x_3, ...] = \mathbf{x}$, unless I'm greatly mistaken. – Rivers McForge Nov 06 '20 at 23:17
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    (1) isn't really correct. First of all, even if we restrict attention to nonprincipal $\mathbb{N}$-ultrapowers of $\mathbb{R}$, that still need not pin down a unique field up to isomorphism; it's consistent with $\mathsf{ZFC}$ that there are nonprincipal ultrafilters $\mathcal{U}_0,\mathcal{U}_1$ on $\mathbb{N}$ such that $\prod\mathbb{R}/\mathcal{U}_0\not\cong\prod\mathbb{R}/\mathcal{U}_1$. And more generally those aren't the only hyperreal fields out there. A hyperreal field is any proper field extension of $\mathbb{R}$ equipped with an appropriate transfer map. – Noah Schweber Nov 06 '20 at 23:37
  • But for simplicity, suppose we pick a specific nonprincipal ultrafilter $\mathcal{U}$ on $\mathbb{N}$ and let "the hyperreals" refer to $\prod\mathbb{R}/\mathcal{U}$. Your (2) is now true, but misleading since Cauchy sequences no longer mean what they used to: two sequences (Cauchy or otherwise) only name the same hyperreal if they coincide $\mathcal{U}$-frequently. So saying "every finite hyperreal corresponds to a set of Cauchy sequences," while true, is misleading. – Noah Schweber Nov 06 '20 at 23:40
  • Separately, it might help if you can explain more precisely what you mean by "a trivial rewording of classical analysis." – Noah Schweber Nov 06 '20 at 23:41
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    @RiversMcForge: To emphasize Noah’s point in his last comment but one, let $\sigma=\langle x_n:n\in\omega\rangle$ be any sequence of positive reals converging strictly monotonically to $0$. For $n\in\omega$ let $y_n=x_{n+1}$; $\tau=\langle y_n:n\in\omega\rangle$ is just $\sigma$ without its first term, yet $[\sigma]\ne[\tau]$ in $\prod\Bbb R/\mathscr{U}$, since ${n\in\omega:x_n=y_n}=\varnothing\notin\mathscr{U}$. – Brian M. Scott Nov 07 '20 at 00:22
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    I intend to write an answer to your question. But before I put in the effort, I'd like to understand/fix what you mean by rewording. Can you write down, as precisely as you can, the exact procedure you use to "reword" any nonstandard statement/definition into the corresponding standard statement? – Z. A. K. Nov 07 '20 at 00:25
  • @Noah Right, different Cauchy sequences can give different hyperreals. I’m actually going the other way and saying that for any finite hyperreal (= equivalence class of real sequences) there is a representative of its equivalence class that is Cauchy as an ordinary sequence of reals. I understand that there may be multiple non-isomorphic constructions of a hyperreal field; my question regards a particular construction with a particular ultrafilter. But I understand these niceties don’t substantially affect the way the symbols are pushed around to do nonstandard calculus. – Rivers McForge Nov 07 '20 at 00:51
  • @brian I think you also misunderstood that I was asking if Cauchy sequences with the same limit correspond to the same finite hyperreal. I’m actually asking if every finite hyperreal has an ordinary Cauchy sequence as one of its representatives in the “equivalence class of real sequences” construction. – Rivers McForge Nov 07 '20 at 00:55
  • @RiversMcForge I understood which way you were going: as I said, you are correct that every finite hyperreal does indeed get named by a Cauchy sequence. But that's misleading since Cauchy sequences are no longer "doing" what they're normally intended to. "I understand these niceties don’t substantially affect the way the symbols are pushed around to do nonstandard calculus." That is true. I'm still confused about your broader question: what do you mean by "trivial rewording?" (See also Z.A.K.'s comment.) – Noah Schweber Nov 07 '20 at 00:55
  • @zak Anywhere I need to discuss the convergence of some limit, say $$\lim_{x \to a} f(x) = L,$$ I instead make a statement that for any hyperreal $\mathbf{x}$ with standard part $a$, its image $^*f(\mathbf{x})$ has standard part $L$. – Rivers McForge Nov 07 '20 at 01:00
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    @RiversMcForge: No, I did not misunderstand, and neither did Noah. I was pointing out that Cauchy sequences that we would normally consider trivially equivalent can correspond to distinct infinitesimals, so it is not at all clear in what sense we’re dealing with a ‘trivial rewording’. – Brian M. Scott Nov 07 '20 at 01:01
  • @BrianM.Scott Take continuity. $f(x)$ is continuous at $x = a$ iff the image of every Cauchy sequence whose limit is $a$ [every finite hyperreal with standard part $a$] is a Cauchy sequence whose limit is $f(a)$ [a finite hyperreal with standard part $f(a)$]. That's what I mean by a "trivial rewording". I'm not saying that hyperreals with the same standard part are the same hyperreal, or anything like that. – Rivers McForge Nov 07 '20 at 01:21
  • @RiversMcForge But that "rewording" is going the wrong way: you're turning standard concepts into nonstandard ones. If you want to argue that nonstandard analysis is just a rewording of standard analysis, you need to explain how to view an arbitrary nonstandard concept as a reworded standard one. On the face of things, nonstandard analysis is more expressive than standard analysis. – Noah Schweber Nov 07 '20 at 01:24
  • @NoahSchweber Right, the question is about rephrasing classical analysis in the language of nonstandard analysis. I edited to clarify. – Rivers McForge Nov 07 '20 at 01:25
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    @RiversMcForge Ah! Now I see. Yes, part of the value of nonstandard analysis is exactly that it lets us repackage clunky limit-based definitions as snappier infinitesimal-based definitions, with the transfer principle guaranteeing that this works properly. – Noah Schweber Nov 07 '20 at 01:26
  • @NoahSchweber: "you are correct that every finite hyperreal does indeed get named by a Cauchy sequence" Your claim is actually incorrect; see my answer. – Mikhail Katz Sep 20 '23 at 13:18
  • You might find of interest the literature I cite here. – Bill Dubuque Sep 21 '23 at 18:57

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The OP wrote: "Under this interpretation, every infinitesimal hyperreal is the equivalence class of a Cauchy sequence with limit zero." This is actually incorrect in general. To make this correct, one needs to impose further conditions on the nonprincipal ultrafilter, such as being a P-point. The existence of P-point ultrafilters is independent of ZFC, so it would not be entirely satisfactory to require this to do infinitesimal calculus or analysis.

Mikhail Katz
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