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As is well known, fields of hyperreals $\mathbb R^*$ can be formed by an ultrapower construction, as quotients of the space of sequences of real numbers by a nonprincipal ultrafilter. In fact, some arguments using $\mathbb R^*$ can be reformulated using sequences. For example, the nonstandard definition of compactness of a set $S$ is that every point of $S^*$ is nearstandard in $S$, i.e., infinitely close to a point of $S$. This is parallel to the fact that every sequence in $S$ has a subsequence converging to a point of $S$ (an equivalent characterisation of compactness).

Other arguments cannot be reformulated as easily. For example, one can prove that a function is continuous at a point $c$ if and only if it is both right-continuous and left-continuous at $c$. Namely, an arbitrary infinitesimal $\epsilon\not=0$ is either positive or negative. If $\epsilon>0$ then right-continuity at $c$ implies that $f(c+\epsilon)\approx0$, and if $\epsilon<0$ then left-continuity at $c$ implies that $f(c+\epsilon)\approx0$, proving continuity (according to Robinson's criterion of continuity). If one tries to find a parallel argument in terms of sequences, one can immediately deal with, say, increasing or decreasing sequences in a similar way, but continuity via sequences would require testing all sequences converging to $c$, and therefore the nonstandard argument does not translate as cleanly.

Question. Are there other accessible arguments using hyperreals that would illustrate the advantage of using nonstandard definitions as compared to sequential ones?

Mikhail Katz
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    Can't you just decompose a sequence converging to $c$ into (if they exist) subsequences consisting of values that are smaller or larger, respectively, than $c$? – Michael Greinecker Dec 06 '22 at 12:03
  • @MichaelGreinecker, Certainly, one can choose the two subsequences, show that f maps each to a sequence converging to f(c) based on the hypothesis of one-sided continuity, and conclude that the original sequence has the same property. I imagine this is how it would be done in a textbook using Heine-type definitions... – Mikhail Katz Dec 06 '22 at 12:17
  • ...Such an argument may or may not be as "clean" as the hyperreal one, but in any case my main point was to ask for examples illustrating the advantage of the hyperreal approach over the sequential one. Of course, there are advanced examples, such as models of hyperreals where R is contained in a hyperfinite set, Loeb measures, etc., which cannot be done in the simple-minded ultrapower by an ultrafilter over a countable index set (and in particular not translatable into the sequential framework); but I was hoping to get some more elementary examples. @MichaelGreinecker – Mikhail Katz Dec 06 '22 at 12:17
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    P.S. I think my proof of compactness of [0,1] is shorter than yours https://math.stackexchange.com/a/189053/72694 :-) @MichaelGreinecker – Mikhail Katz Dec 06 '22 at 12:29

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$ \newcommand\R{\mathbb R} \newcommand\ext{{}^*} $

I really like Heine-Borel.

Let $S \subset \mathbb R$. I will use near to mean infinitesimally close, call elements of $\R$ standard, and use nearstandard to mean that an element of $\ext\R$ is near some element of $\R$.

  • $S$ is closed if every element of $\R$ near some element of $\ext S$ is actually in $x \in S$.
  • $S$ is bounded of $\ext S$ has no infinite elements.
  • $S$ is compact if every element of $\ext S$ is nearstandard to something in $S$. This is a direct generalization of finiteness since $S$ is finite iff every element of $\ext S$ is standard.
  • An element of $\ext\R$ is finite iff it is nearstandard.
  • If standard $x$ is nearstandard to $y$ then $x = y$.

Compact $\implies$ Closed+Bounded: Every element of $\ext S$ is nearstandard by compactness, so there are no infinite elements, so $S$ is bounded. If $x \in \R$ has some $y \in \ext S$ such that $x\approx y$, then by compactness there is a standard $y' \in S$ such that $x\approx y\approx y'$; but that means $x = y' \in S$, so $S$ is closed.

Closed+Bounded $\implies$ Compact: Let $x \in \ext S$. Since $S$ is bounded, $x$ is finite and so nearstandard to some $y$. By closure, $y \approx x$ means $y \in S$. Thus $S$ is compact.

Nicholas Todoroff
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