An ordered field with a infinitesimal element elementarily equivalent to the reals

Question

Let $\mathcal{L} := \{0, 1, +, -, \cdot, \leq \}$ be a language, where $0, 1$ are constants and $+, -, \cdot$ are binary functions and $\leq$ is a binary relation.

We consider the ordered fields in this language. An element $\varepsilon$ is said to be infinitesimal if $0 < \varepsilon < \frac{1}{n}$ for all integers $n \geq 1$.

The exercise is now the following: Show the existence of an ordered field $K$ such that $K$ admits an infinitesimal element and is elementarily equivalent to $\mathbf{R}$.

First I thought it might be good to mimic the construction of the hyperreals, but I do not know whether they are elementarily equivalent to $\mathbf{R}$ and the construction looks quite lengthy.

Can anyone give me a hint on how to start the proof?

Thanks!

Answer 1

After the very helpful discussion in the comments, I will try and put a whole answer here (if any of the commenters want to post an answer, please do so and I'll accept it):

Let $\mathcal{L}' := \mathcal{L} \cup \{ \varepsilon \}$ with $\varepsilon$ a constant. Let $T' := Th(\mathbf{R}) \cup \{ 0 < \varepsilon < \frac{1}{n} \}$.

We use the compactness theorem: A theory $T$ is consistent if and only if every finite subset of $T$ is consistent.

Let $T_0$ be a finite subset of $T'$. There is a $N \geq 1$ such that for any axiom $0 < \varepsilon < \frac{1}{n} \in T_0$ we must have $n \leq N$. The reals form a model of $T_0$ (they clearly satisfy the axioms of $T$ and for $\varepsilon$ take $\frac{1}{N +1}$).

By the compactness theorem $T'$ is consistent. Thus there exists a model $K$ for $T'$.

A fortiori $K \models T$ for $K$ viewed as an $\mathcal{L}$-structure and $Th(\mathbf{R}) \subset Th(K)$. Let $F \in Th(K)$. We have $\mathbf{R} \models F$ or $\mathbf{R} \models \neg F$. But if $\mathbf{R} \models \neg F$, then $\neg F \in Th(K)$. This is of course a contradiction, hence $Th(\mathbf{R}) = Th(K)$.

Added: I found this math overflow link: https://mathoverflow.net/questions/39504/what-are-examples-of-ordered-fields-that-do-not-have-the-archimedean-property, which gives the explicit example $\mathbf{R}(X)$ for an ordered field with infinitesimal element $\frac{1}{X}$.

Answer 2

Ordered fields elementarily equivalent to the reals are called real closed fields.

Any ordered field $F$ has a real closure $F^r$ which is an algebraic extension that is real closed.

So, if you let $F$ be any non-archimedean ordered field, then $F^r$ is an example.

A standard, explicit construction of examples is that if $F$ is any real closed field, then so is the field of Puiseux series over $F$.

The field of Puiseux series over $F$ is the real closure of the field $F((t))$ of Laurent series over $F$, ordered so that $t$ is a positive infinitesimal.