Let $F$ be a nonarchimedean ordered field. Is there always a hyperreal field $^*\mathbb{R}$ such that there is an embedding of $F$ in $^*\mathbb{R}$?

As far as I understand it, the answer here implies that the answer is positive in the special case where $F$ is the Dehn field or the Levi-Civita field.

  • 3
    $\begingroup$ What definition of "hyperreal field" are you using? There are a couple different versions floating around. (That said, as long as it's something like "an elementary extension of $(\mathbb{R};+,\times,...)$" the answer to your question will be yes: embed $F$ in its real closure $F^+$ and apply compactness to the atomic diagram of $F^+$ plus $Th(\mathbb{R};+,\times,...)$.) $\endgroup$ Aug 16, 2021 at 7:10
  • $\begingroup$ I don't know. The literature that prompted my question uses "the hyperreals" and "nonstandard reals" fairly casually. So the notion of "hyperreal field" I'm using is whatever notion they are using, or are at least committed to. I don't know enough to be more precise. $\endgroup$
    – David M
    Aug 16, 2021 at 7:22
  • $\begingroup$ Objects defined only "fairly casually" cannot have theorems proved about them. $\endgroup$
    – GEdgar
    Aug 16, 2021 at 19:57

1 Answer 1


A definition:* A hyperreal field is an extension $^*\mathbb{R}$ of $\mathbb{R}$ equipped with an operator $*$ defined on the structure $\mathcal{S}(\mathbb{R}):=\bigcup \limits_{n \in \mathbb{N}} \mathcal{S}_n(\mathbb{R})$ where for any set $E$, we have $\mathcal{S}_0(E)=E$ and $\mathcal{S}_{n+1}(E)= \mathcal{S}_n(E)\cup\mathcal{P}(\mathcal{S}_n(E))$ for all $n \in \mathbb{N}$. This operator fixes $\mathbb{R}$ pointwise and assigns a an element $^*X \in \mathcal{S}(^*\mathbb{R})$ to each element $X$ of $\mathcal{S}(\mathbb{R})$ (with $^*\mathbb{R}$ being the value of this operator at $\mathbb{R}$), such that the following transfer principle is satisfied:

For all (set-theoretic, first order) formulas $\varphi[x_1,...,x_n]$ where each quantifier appears in a bounded form $\forall x \in y,[...]$ and $\exists x \in y,[...]$, we have that for all $X_1,...,X_n \in \mathcal{S}(\mathbb{R})$, $\varphi[X_1,...,X_n]$ holds if and only if $\varphi[^*X_1,...,^*X_n]$ holds.

(in particular, $*$ preserves the "type" of sets, e.g. sends $n$-uples of reals to $n$-uples of hyperreals and so on)

One way (in fact, the way) to obtain this type of field is through ultrapowers of $\mathbb{R}$. One can also characterize them as quotients of rings $\mathbb{R}^I$ of functions $I \rightarrow \mathbb{R}$ by maximal ideals, where $I$ is a non-empty set. By a result of Keisler-Kunen, for all cardinals $\kappa$, there exists a ultrafilter $\mathcal{U}$ on $\kappa$ such that the corresponding ultrapower $\mathbb{R}^{\mathcal{U}}$ is $\kappa^+$-saturated. This implies that any ordered field of cardinality $\leq \kappa$ embeds into $\mathbb{R}^{\mathcal{U}}$. So the answer is yes in this sense.

See A ring of homomorphisms is enough to get non-standard analysis for more details. See also A purely algebraic characterization of fields of the hyperreal numbers for a somewhat more explicit description of those fields.

*this is different from Dale and Woodin's definition of hyper-real fields in their book Super-real fields. However the result is still true with this definition, since hyperreal fields in the previous sense are hyper-real fields in their sense.


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