4
$\begingroup$

The hypperreal numbers are an extension of the reals that allow for a rigorous treatment of infinitely small and infinitely large values. Specifically it includes the number $\varepsilon$ where $$ 0 < \varepsilon < x \tag{1} $$ for all $x \in \Bbb R$. As such, the hypperreal field $^*\Bbb R$ is both non-Archimedean and non-complete, but we can still do analysis in it using the machinery of non-standard analysis. It also includes the reciprocal of $\varepsilon$, called $\omega$, which is larger than every real. Likewise it include $\varepsilon^2$, $\varepsilon^{-2}$, and so on.

The field $^*\Bbb R$ is typically constructed using an ultrapower construction. This means that we look at sequences $\Bbb R^{\Bbb N}$ and our numbers will be certain equivalence classes of those sequences. The next part is where my understanding gets shaky, so maybe that's why I'm struggling to see the motivation.

The equivalence classes are chosen using the usual ordering on $\Bbb R$ and a free ultrafilter, where an ultrafilter is a set of "sufficiently large" subsets of the natural numbers and a free one does not include any finite sets. So we say that if our ultrafilter is $\cal U$, then we will define an ordering on $\Bbb R^{\Bbb N}$ by saying that $(a_n) \leq (b_n)$ if and only if the set $\{n \mid a_n \leq b_n\}$ is in $\cal U$, i.e., it is "sufficiently large." We then define the equivalence relation by $(a_n) \sim (b_n) \iff (a_n) \leq (b_n) \land (b_n)\leq(a_n)$.

As for what "sufficiently large" means, this means that $\cal U$ is upwards closed (if $A$ and $B$ are subsets of $\Bbb R^{\Bbb N}$ where $A \subset B$ and $A \in \cal U$, then $B \in \cal U$), that it is closed under intersection, that $\varnothing \notin \cal U$ and that $\Bbb R^{\Bbb N} \in \cal U$. These are the "filter" part. The "ultra" part basically means maximal: there are no other filters on $\Bbb R^{\Bbb N}$ that strictly contain $\cal U$.

Using this construction, we define addition and multiplication component-wise. We then define $$ \varepsilon = \left[1,\; {1 \over 2},\; {1 \over 3},\; \cdots\right], $$ and the whole thing is provably a field.

So this is all fine and dandy, but I'm wondering why it's necessary. Wouldn't it be sufficient to define a transcendental simple field extension $^*\Bbb R = \Bbb R(\varepsilon)$ and extend the ordering by hand as in (1)? Won't this recover all of the algebraic and analytic properties that we're looking for in a much simpler way? The result is guaranteed to be a field containing $\Bbb R$ as a subfield, and doing this a a field extension intuitively captures that fact that we want "the reals, but with infinitesimals." So why do we need the ultrafilter and ultraproduct? Are these two constructions not isomorphic?

$\endgroup$
7
  • 4
    $\begingroup$ How exactly do you order a transcendental extension $\Bbb R(\epsilon)$? How do you prove all the other model theoretic stuff like, something something something first order statements about $\Bbb R$ remain true in $^\ast\Bbb R$? $\endgroup$
    – FShrike
    Mar 1 at 22:29
  • 3
    $\begingroup$ math.stackexchange.com/questions/1838272/… $\endgroup$
    – FShrike
    Mar 1 at 22:29
  • 1
    $\begingroup$ If $f$ is a nice real function, how would you define $f(\varepsilon)$ in your "simple field extension" setup? $\endgroup$
    – Mark S.
    Mar 1 at 22:32
  • 1
    $\begingroup$ @FShrike the field orderings on a simple transcendental extension $\mathbf R(x)$ arise by picking a real number $a$ and declaring $f(x) >_{a^+} g(x)$ if $f(t) > g(t)$ for all real $t$ in some interval to the right of $a$. Two different rational functions have a common ordering in their numerical values at all $t$ close to $a$ on the right, and that's what this ordering means. We can define an ordering $f(x) >_{a^-} g(x)$ in a similar way using numerical values at real $t$ near $a$ on the left. And you can do something similar as $t \to\infty$ or $t\to-\infty$. Those are all the options. $\endgroup$
    – KCd
    Mar 1 at 23:08
  • 1
    $\begingroup$ "Are these two constructions not isomorphic?" No. For example, where in $R(\epsilon)$ can you find the element $[1, 1/2, 1/2^{2}, 1/3^{3}, 1/4^{4}, \dots]$? This element is smaller than all of $\epsilon, \epsilon^{2}, \dots$ using your definition of $\epsilon$. (But you are asking a very natural question here.) $\endgroup$
    – Pilcrow
    Mar 2 at 12:02

1 Answer 1

7
$\begingroup$

You can build the hyperreals without directly mentioning (nonprincipal) ultrafilters by using something equally nonconstructive: maximal ideals. In the product ring $R = \prod_{n \geq 1} {\mathbf R}$, let $I = \bigoplus_{n \geq 1} \mathbf R$ be the ideal of the elements $(a_n)$ in $R$ where $a_n = 0$ for all large $n$. This is a proper ideal since it doesn't contain $(1,1,1,\ldots)$, so it is contained in some (in fact, many) maximal ideal $M$. The quotient $R/M$ is a field and this can be used as the hyperreals. The link between this viewpoint of working mod $M$ and using ultrafilters is that the sets of indices where some element in $M$ vanishes exactly at the positions in that index set form a filter on the positive integers, and maximality of $M$ as an ideal is precisely what makes this filter an ultrafilter. And having $M$ contain $I$ prevents $M$ from being a principal ultrafilter (corresponding to a principal maximal ideal in $R$). You can think about the index sets where elements in $M$ are $0$ as a collection of choices of "big" subsets of the positive integers.

I learned this algebraic way of thinking about the hyperreals, using maximal ideals in $R$, from the chapter on hyperreals in the book "Numbers" by Ebbinghaus et al. See also the last few paragraphs in Section $7$ here.

$\endgroup$
4
  • 1
    $\begingroup$ Readers who like this approach might also enjoy my video on the hyperreals, which was inspired by Ebbinghaus' book. $\endgroup$
    – blargoner
    Mar 2 at 4:18
  • $\begingroup$ @blargoner Nice video, but you didn't list that book in your references at the end. $\endgroup$
    – KCd
    Mar 2 at 12:31
  • $\begingroup$ I did (the chapter authored by Prestel). $\endgroup$
    – blargoner
    Mar 2 at 15:11
  • $\begingroup$ Ah, okay, I see now. $\endgroup$
    – KCd
    Mar 2 at 21:27

Not the answer you're looking for? Browse other questions tagged .