# First order axiomatization of $\mathbb{R}$

Recently I've been trying to come to terms with the seemingly contradictory facts that (1) $\mathbb{R}$ is the only Dedekind complete ordered field up to isomorphism, and (2) $^*\mathbb{R}$, the ultrapower of $\mathbb{R}$, is a nonstandard model of the reals, apparently contradicting the uniqueness of $\mathbb{R}$. As I understand it, Dedekind completeness is a second-order axiom as it quantifies over sets of reals, and $\mathbb{R}$ is unique in the second-order theory, but not in a first-order axiomatization (similar to how $\mathbb{N}$ is unique in second-order PA, but there exist nonstandard models of first-order PA). In particular, Dedekind completeness holds for every member of $^*\mathcal{P}(\mathbb{R})$, which is a proper subset of $\mathcal{P}(^*\mathbb{R})$.

So my question is, what is the first-order axiomatization of the real numbers that admits both $\mathbb{R}$ and $^*\mathbb{R}$ as models?

• Carl's answer is correct, and answers your question; I just want to briefly confirm what you wrote, in case you aren't certain. Your comment about second- vs. first-order is exactly correct. Moreover, you are correct that Dedekind-completeness holds for every member of $^*\mathcal{P}(\mathbb{R})=\{^*X: X\in\mathcal{P}(\mathbb{R})\}$; this is because $X$ satisfies (in $\mathbb{R}$) "If I am nonempty and bounded above, then I have a least upper bound", and this is a first-order statement in $X$, so true in $^*\mathbb{R}$ about $^*X$ by transfer. – Noah Schweber Jan 4 '17 at 18:52

Because $\mathbb{R}$ and $\mathbb{R}^*$ satisfy exactly the same set of sentences in the language of ordered fields, you could take the axioms to be any set of sentences that are true in $\mathbb{R}$, such as:
1. The axioms of a real-closed ordered field. By a classical theorem of Tarski, these axioms prove every sentence in the language of ordered fields that is true in $\mathbb{R}$. This is probably the most natural choice. The axioms use formulation (2) on Wikipedia. They say that $<$ is a total order on the field, making it an ordered field, such that, in this ordering, every positive element has a square root and each polynomial of odd degree has a root.
2. You could use any weaker set of axioms, such as just the axioms for an ordered field of characteristic zero, or just the axioms for an Abelian group. But these will no longer prove every statement true in $\mathbb{R}$.
An ultrapower of a set is elementarily equivalent to it for any first-order structure you put on it. So you could take any collection of finitary relations and operations on the set $\mathbb{R}$, and any set of axioms that those relations and operations satisfy, and ${}^*\mathbb{R}$ will satisfy the same axioms for the induced relations and operations on the ultrapower.
As Carl Mummert said in his answer, the "standard" first-order axioms usually taken on $\mathbb{R}$ are those of a real-closed ordered field (with respect to the operations of addition and multiplication and the order relation). But the fact that an ultrapower ${}^*\mathbb{R}$ of $\mathbb{R}$ satisfies the same first-order axioms as $\mathbb{R}$ is true for any set of axioms at all, with respect to any collection of finitary relations and operations. So, just as an example, if you wanted to extend your axiomatization of $\mathbb{R}$ to also include axioms about the exponential function $\exp:\mathbb{R}\to\mathbb{R}$, ${}^*\mathbb{R}$ would still satisfy those additional axioms.
• Indeed, one (standard?) way to do the ultrapower $\mathbb{R}^*$ is to begin by extending the language of the ordered field $(\mathbb{R}, +, \cdot, <, 0, 1)$ to have a symbol for every function from $\mathbb{R}^n$ to $\mathbb{R}$, and a symbol for every subset of $\mathbb{R}^n$, for all $n \in \mathbb{N}$, and then take $\mathbb{R}^*$ to be the ultrapower of that structure, which will still satisfy every sentence in the extended language that was true in $\mathbb{R}$. – Carl Mummert Jan 4 '17 at 18:57