Currently reading Infinitesimal Calculus by Henle and Kleinberg. In this text, page 25, they note that they define a hyperreal number system, not the hyperreal number system. This is because "there are many different hyperreal number systems" (25), where by 'different' I suppose they mean superficially different systems (i.e. different sets of axioms) whose models are elementarily equivalent.

So, just how many different systems of hyperreals are there? A little more precisely (maybe not much), given a real closed field $S$ = $\langle$$R$, +, $\cdot$, 0, 1, <,$\rangle$, how many elementary extensions $S$* of $S$ are there? And more specifically how many hyperreal elementary extensions of $S$ are there?


P.S. I'm also assuming there's a substantial difference between an elementary extension of $S$ and an elementary extension by way of hyperreals. This might be contentious?

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    $\begingroup$ Unless you aim to bound their cardinality, class many extensions. $\endgroup$
    – Asaf Karagila
    May 30, 2015 at 11:42
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    $\begingroup$ @Asaf: So let's just say of cardinality $2^{\aleph_0}$. $\endgroup$ May 30, 2015 at 12:01
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    $\begingroup$ @Henning: So a hyperreal field is a model of $\rm RCF$ which is (1) not a subfield of $\Bbb R$; and (2) has size $2^{\aleph_0}$? $\endgroup$
    – Asaf Karagila
    May 30, 2015 at 12:03
  • $\begingroup$ @Asaf: I'm not sure, actually. The Wikipedia article seems to skirt around actually giving a definition. (And I've always found non-standard analysis confusing, zig-zagging between the standard and hyper worlds all the time with no visible principle for when to work where, except for incessant loud and bombastic proclamations that it is all so intuitive that you don't need any principle to guide you ...) $\endgroup$ May 30, 2015 at 12:36
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    $\begingroup$ I think hyperreal just means non-archimedean. Starting from any RCF, you can iterate the Puiseux series construction indefinitely to get $\omega$ non-isomorphic elementary extensions. I haven't checked the details, but I expect you can continue iterating the construction transfinitely to get a family of non-isomorphic elementary extensions indexed by ordinals. $\endgroup$
    – Rob Arthan
    May 30, 2015 at 16:40

1 Answer 1


First of all, I want to clarify the terms. The natural meaning of "different" here is non-isomorphic - it has nothing to do with sets of axioms. And I'll take Henle and Kleinberg's definition of "hyperreal number system": a structure elementarily equivalent to $\langle \mathbb{R}, +,\cdot,0,1,<\rangle$ and containing both $\mathbb{R}$ and an infinitesimal element (an element $\epsilon$ such that $0<|\epsilon|<1/n$ for all $n$). Other authors assume some saturation in the definition of hyperreal number system, see the end of this answer for a note on that.

Now a field $R^*$ is a hyperreal number system under the Henle and Kleinberg definition if and only if it is a proper elementary extension of $\mathbb{R}$. Indeed, if $R^*$ is a proper elementary extension, then there is a "standard part" map $\text{st}\colon R^*\to \mathbb{R}\cup\{\pm\infty\}$. Since $\mathbb{R}$ is complete, every element of $R^*$ which is not greater than all elements of $\mathbb{R}$ or less than all elements of $\mathbb{R}$ is infinitesimally close to a unique element of $\mathbb{R}$. That is, if $a\in R^*\setminus \mathbb{R}$, either $a$ is "infinite" ($\text{st}(a) = \pm\infty$), or $a - \text{st}(a)$ is infinitesimal. And the multiplicative inverse of an infinite element is infinitesimal and vice versa. The point is that any proper elementary extension of $\mathbb{R}$ is non-archimedean, with infinite and infinitesimal elements, and hence satisfies the Henle and Kleinberg definition.

Conversely, if $R^*$ contains $\mathbb{R}$ and is elementarily equivalent to it, it is an elementary extension, by model completeness of the theory of real closed fields. And if it contains an infinitesimal, it is a proper elementary extension.

Now we want to characterize the proper elementary extensions of $\mathbb{R}$ as the models of a first-order theory. We add to the language a constant symbol $c_r$ for all $r\in \mathbb{R}$, and let $T = \text{Th}(\langle\mathbb{R},+,\cdot,<,\{c_r\}_{r\in\mathbb{R}}\rangle)$. An elementary extension of $\mathbb{R}$ is the same thing (up to isomorphism) as a model of $T$, so if we want to count hyperreal number systems, we just have to count models of $T$ (remembering to throw away the standard model $\mathbb{R}$).

Of course, just like any first-order theory with infinite models, $T$ has models of all infinite cardinalities greater than or equal to the size of the language, so $T$ has a proper class of models. But we can ask the more refined question "given a cardinal $\kappa$, how many models does $T$ have of size $\kappa$?"

Now it's time to bring out the sledgehammer: $T$ is an unstable complete first-order theory (this means that $T$ defines a linear order on an infinite set of elements in some model - for our $T$, this is obvious, since for any model of $T$, the whole model is linearly ordered by $<$). It's a cornerstone result of Shelah's Classification Theory that if $T$ is unstable and $\kappa$ is an uncountable cardinal greater than or equal to the size of the language, $T$ has $2^\kappa$ many models of size $\kappa$ up to isomorphism (some counting shows that this is the maximal number possible). If you want to get an idea of how things like this are proven, Section 5.3 of Marker's Model Theory contains a version of this argument.

So, for example, there are $2^{2^{\aleph_0}}$-many hyperreal number systems of size $2^{\aleph_0}$.

I want to end by pointing out that it would be reasonable to define a "hyperreal number system" to be not just any proper elementary extension of $\mathbb{R}$, but only one of size $2^{\aleph_0}$ which is $\aleph_1$-saturated, meaning that for any countable set of formulas (with parameters) such that any finite subset is consistent, there is an element of $R^*$ satisfying all of them. You get a structure with these properties when you take an ultrapower of $\mathbb{R}$ by a nonprincipal ultrafilter on a countable set.

See, for example, this MO answer, where it is asserted that the field of real Puiseux series is not a hyperreal field, because it is not $\aleph_1$-saturated.

If you want to count hyperreal number systems in this narrower sense, the answer depends on set theory. If you assume the continuum hypothesis, then any such field is saturated in its own cardinality (since $2^{\aleph_0} = \aleph_1$), and hence there is a unique hyperreal field up to isomorphism! See here for discussion. On the other hand, if CH is false, I believe you still get the maximum number ($2^{2^{\aleph_0}}$) $\aleph_1$-saturated real closed fields of cardinality $2^{\aleph_0}$.

  • $\begingroup$ Thanks, Alex, you more than adequately answered all my questions. I wasn't sure how to read Henle and Kleinberg's statement. I considered "non-isomorphic", but they had just given a set of conditions which I read as the axioms for the extension of the theory for the reals to hyperreals, and wasn't sure if I could assume that was the appropriate meaning. But I was definitely asking how many non-isomorphic elementary extensions there were, and you definitely gave me that much. And thanks for the reference. Have yet to pick up Marker, will do soon at your suggestion. $\endgroup$ Jun 1, 2015 at 1:41
  • $\begingroup$ You say "any proper elementary extension of $\mathbf R$ is non-archimedean, with infinite and infinitesimal elements, and deserves the name hyperreal number system." Being an elem. extension is not relevant to have infinitesimals: every ordered field $R$ strictly containing $\mathbf R$ has infinite and infinitesimal elements, every $a$ in $R$ that is not infinite is infinitely close to a unique real number ${\rm st}(a)$ (say) and ${\rm st}$ is a ring homomorphism from the finite elements of $R$ onto $\mathbf R$ with kernel the infinitesimals in $R$. You want an elem. extn. just for NSA, right? $\endgroup$
    – KCd
    Apr 9, 2017 at 15:56
  • $\begingroup$ @KCd Sure. Personally, I wouldn't call just any ordered field containing $R$ "hyperreal" - I would require it to be an elementary extension. But as noted in the comments to the question, there doesn't seem to be a standard definition of "hyperreal field". $\endgroup$ Apr 9, 2017 at 17:33
  • $\begingroup$ I agree being elementarily equivalent to $\mathbf R$ should be part of calling any candidate field a hyperreal field. It seems that and having infinitesimals are the only conditions mattering. A concrete example of such a field w/o using ultrafilters is the real Puiseux series $\bigcup_{n \geq 1} \mathbf R((x^{1/n}))$, ordered by $\sum_{i \geq i_0} c_ix^{i/n} > 0$ if the first nonzero coeff. $c_{i_0}$ is positive. Then $x$ is positive and infinitesimal (infinitesimal means $i_0 \geq 0$) and this field is elem. equiv. to $\mathbf R$ since it's real closed and all RCF are elem. equiv. [cont.] $\endgroup$
    – KCd
    Apr 9, 2017 at 19:56
  • $\begingroup$ In that case why not just use that as a concrete model for the hyperreals? For students (or anyone!) this would be far less opaque than a construction via ultrapowers or a nonprincipal maximal ideal, neither of which can be described concretely. The real Puiseux series are explicit, every series being analogous to a decimal-type expansion (on top of the decimal expansions of the real coefficients). In the ultrapower construction you can't write down a single example of an infinitesimal. The cardinality is the same as $\mathbf R$ too (unless I messed up), so it's not too big either. $\endgroup$
    – KCd
    Apr 9, 2017 at 19:59

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