[proofplan] Assume that such a theory $T$ exists. We enlarge the language by a new constant symbol $c$ and add the infinite family of sentences saying that $c$ is positive and smaller than every positive integer reciprocal. Every finite part of this enlarged theory is satisfiable in the standard ordered real field by interpreting $c$ as a sufficiently small positive real number. Compactness then gives a model of the whole enlarged theory, whose reduct satisfies $T$ but contains a positive infinitesimal, contradicting the assumed Archimedeanness of all models of $T$. [/proofplan] [step:Assume an elementary axiomatization and add a constant for an infinitesimal] Let \begin{align*} \mathcal{L}_{\mathrm{or}} = \{0,1,+,-,\cdot,<\} \end{align*} be the language of ordered rings. Throughout the proof, let $\mathbb{N}$ denote the set of positive integers. Suppose, toward a contradiction, that there exists a first-order $\mathcal{L}_{\mathrm{or}}$-theory $T$ such that an $\mathcal{L}_{\mathrm{or}}$-structure $M$ satisfies $T$ exactly when $M$ is an Archimedean real closed field. Let $\mathcal{L}_c := \mathcal{L}_{\mathrm{or}} \cup \{c\}$, where $c$ is a new constant symbol. For each $n \in \mathbb{N}$, define the $\mathcal{L}_c$-sentence $\varphi_n$ by \begin{align*} \varphi_n \;:\;\; 0 < c \;\wedge\; n \cdot c < 1, \end{align*} where $n \cdot c$ denotes the term \begin{align*} \underbrace{c + \cdots + c}_{n \text{ summands}}. \end{align*} Define the $\mathcal{L}_c$-theory \begin{align*} \Sigma := T \cup \{\varphi_n : n \in \mathbb{N}\}. \end{align*} [guided] We argue by contradiction. The assumed theory $T$ is a set of first-order sentences in the ordered-ring language \begin{align*} \mathcal{L}_{\mathrm{or}} = \{0,1,+,-,\cdot,<\}. \end{align*} Throughout the proof, $\mathbb{N}$ denotes the set of positive integers. The assumption says that $T$ recognizes exactly the Archimedean real closed fields: for every $\mathcal{L}_{\mathrm{or}}$-structure $M$, the relation $M \models T$ holds precisely when $M$ is an Archimedean real closed field. To force a contradiction, we try to build a model of $T$ that contains a positive infinitesimal. First-order logic cannot quantify over all natural numbers inside a single external scheme, so we introduce one sentence for each positive integer. Let $\mathcal{L}_c := \mathcal{L}_{\mathrm{or}} \cup \{c\}$, where $c$ is a new constant symbol. For each $n \in \mathbb{N}$, define the sentence \begin{align*} \varphi_n \;:\;\; 0 < c \;\wedge\; n \cdot c < 1, \end{align*} where $n \cdot c$ is the ordered-ring term obtained by adding $c$ to itself $n$ times: \begin{align*} n \cdot c = \underbrace{c + \cdots + c}_{n \text{ summands}}. \end{align*} Thus $\varphi_n$ says that the element named by $c$ is positive and smaller than $1/n$ in the ordered-field sense. We then set \begin{align*} \Sigma := T \cup \{\varphi_n : n \in \mathbb{N}\}. \end{align*} A model of $\Sigma$ would be a model of $T$ together with an element smaller than every positive rational reciprocal. [/guided] [/step] [step:Show every finite subset of the infinitesimal theory is satisfiable in the real field] Let $\Sigma_0 \subset \Sigma$ be a finite subset. Since $\Sigma_0$ contains only finitely many sentences of the form $\varphi_n$, there exists $N \in \mathbb{N}$ such that every $\varphi_n \in \Sigma_0$ has $n \leq N$. Let $\mathbb{R}_{\mathrm{or}}$ denote the standard ordered real field as an $\mathcal{L}_{\mathrm{or}}$-structure. Since $\mathbb{R}_{\mathrm{or}}$ is an Archimedean real closed field, the defining property of $T$ gives \begin{align*} \mathbb{R}_{\mathrm{or}} \models T. \end{align*} Expand $\mathbb{R}_{\mathrm{or}}$ to an $\mathcal{L}_c$-structure $\mathbb{R}_{c}$ by interpreting $c$ as the real number \begin{align*} c^{\mathbb{R}_c} := \frac{1}{N+1}. \end{align*} For each $n \leq N$, \begin{align*} 0 < \frac{1}{N+1} \end{align*} and \begin{align*} n \cdot \frac{1}{N+1} \leq N \cdot \frac{1}{N+1} < 1. \end{align*} Hence $\mathbb{R}_c \models \varphi_n$ for every $\varphi_n \in \Sigma_0$. The sentences in $T \cap \Sigma_0$ do not mention $c$, so they remain true in the expansion $\mathbb{R}_c$. Therefore \begin{align*} \mathbb{R}_c \models \Sigma_0. \end{align*} Thus every finite subset of $\Sigma$ is satisfiable. [guided] Fix a finite subset $\Sigma_0 \subset \Sigma$. The set $\Sigma_0$ may contain finitely many sentences from $T$ and finitely many of the special sentences $\varphi_n$. Because only finitely many $\varphi_n$ occur, there is an integer $N \in \mathbb{N}$ such that whenever $\varphi_n \in \Sigma_0$, we have $n \leq N$. Let $\mathbb{R}_{\mathrm{or}}$ be the standard ordered real field, viewed as an $\mathcal{L}_{\mathrm{or}}$-structure. The real field is Archimedean and real closed, so the assumed defining property of $T$ gives \begin{align*} \mathbb{R}_{\mathrm{or}} \models T. \end{align*} We now expand this structure to the larger language $\mathcal{L}_c$ by choosing an interpretation for the new constant symbol. Define $\mathbb{R}_c$ to be the $\mathcal{L}_c$-structure whose $\mathcal{L}_{\mathrm{or}}$-reduct is $\mathbb{R}_{\mathrm{or}}$ and in which \begin{align*} c^{\mathbb{R}_c} := \frac{1}{N+1}. \end{align*} This choice is positive. Also, for every $n \leq N$, \begin{align*} n \cdot c^{\mathbb{R}_c} = n \cdot \frac{1}{N+1} \leq N \cdot \frac{1}{N+1} < 1. \end{align*} Therefore $\mathbb{R}_c \models \varphi_n$ for each infinitesimal sentence $\varphi_n$ that appears in $\Sigma_0$. The other sentences in $\Sigma_0$ belong to $T$. They are written in the original language $\mathcal{L}_{\mathrm{or}}$ and do not mention the new constant symbol $c$. Expanding a structure by interpreting an additional symbol does not change the truth of sentences in the old language. Hence all sentences in $T \cap \Sigma_0$ remain true in $\mathbb{R}_c$. Combining the two parts gives \begin{align*} \mathbb{R}_c \models \Sigma_0. \end{align*} So every finite subset of $\Sigma$ has a model. [/guided] [/step] [step:Apply compactness to obtain a model with a positive infinitesimal] By the [First-Order Compactness Theorem](/theorems/???), since every finite subset of $\Sigma$ is satisfiable, the whole theory $\Sigma$ is satisfiable. Let $M_c$ be an $\mathcal{L}_c$-structure such that \begin{align*} M_c \models \Sigma. \end{align*} Let $M$ denote the $\mathcal{L}_{\mathrm{or}}$-reduct of $M_c$, and let $a \in M$ denote the interpretation of the constant symbol $c$ in $M_c$: \begin{align*} a := c^{M_c}. \end{align*} Since $T \subset \Sigma$, we have \begin{align*} M \models T. \end{align*} By the defining property assumed for $T$, the structure $M$ is an Archimedean real closed field. On the other hand, for every $n \in \mathbb{N}$, the sentence $\varphi_n$ belongs to $\Sigma$, so $M_c \models \varphi_n$. Therefore \begin{align*} 0 < a \end{align*} and \begin{align*} n \cdot a < 1 \end{align*} for every $n \in \mathbb{N}$. [guided] We now use compactness. The [First-Order Compactness Theorem](/theorems/???) says that a first-order theory has a model whenever every finite subset of it has a model. We have just verified this finite satisfiability for $\Sigma$, so compactness gives an $\mathcal{L}_c$-structure $M_c$ such that \begin{align*} M_c \models \Sigma. \end{align*} Let $M$ be the $\mathcal{L}_{\mathrm{or}}$-reduct of $M_c$, meaning that $M$ is obtained from $M_c$ by forgetting the interpretation of the extra constant symbol $c$. Let \begin{align*} a := c^{M_c} \end{align*} be the element of the universe of $M_c$ named by $c$. Because $T \subset \Sigma$ and $M_c \models \Sigma$, every sentence of $T$ is true in $M_c$. Since the sentences of $T$ are in the smaller language $\mathcal{L}_{\mathrm{or}}$, this is exactly the assertion that the reduct $M$ satisfies $T$: \begin{align*} M \models T. \end{align*} By the assumed characterization of $T$, it follows that $M$ is an Archimedean real closed field. At the same time, the sentences $\varphi_n$ are all in $\Sigma$. Therefore $M_c \models \varphi_n$ for every $n \in \mathbb{N}$. Unpacking the definition of $\varphi_n$, we get \begin{align*} 0 < a \end{align*} and \begin{align*} n \cdot a < 1 \end{align*} for every $n \in \mathbb{N}$. Thus the element $a$ is positive and smaller than every positive integer reciprocal in the ordered-field sense. [/guided] [/step] [step:Derive the contradiction to Archimedeanness] Since $M$ is an Archimedean ordered field and $a \in M$ satisfies $0<a$, the [Archimedean property](/page/Archimedean%20Property) applied to $a$ gives an integer $n \in \mathbb{N}$ such that \begin{align*} 1 < n \cdot a. \end{align*} But the sentence $\varphi_n$ holds in $M_c$, so \begin{align*} n \cdot a < 1. \end{align*} This is impossible in an ordered field, because the strict order $<$ is asymmetric. The contradiction shows that no first-order $\mathcal{L}_{\mathrm{or}}$-theory $T$ can have exactly the Archimedean real closed fields as its models. [guided] We now use the [Archimedean property](/theorems/737) of the ordered field $M$. In an ordered field, Archimedeanness means that for every positive element $x$, there exists $n \in \mathbb{N}$ such that \begin{align*} 1 < n \cdot x. \end{align*} Apply this definition to the positive element $a \in M$. Since $0<a$, there must exist $n \in \mathbb{N}$ such that \begin{align*} 1 < n \cdot a. \end{align*} However, the same integer $n$ also indexes one of the sentences in $\Sigma$. Since $M_c \models \varphi_n$, the definition of $\varphi_n$ gives \begin{align*} n \cdot a < 1. \end{align*} Thus the ordered field $M$ would satisfy both $1 < n \cdot a$ and $n \cdot a < 1$. This contradicts asymmetry of the strict order in an ordered field. The contradiction came only from the assumption that such a theory $T$ existed. Therefore there is no first-order theory in the language of ordered rings whose models are exactly the Archimedean real closed fields. [/guided] [/step]