[proofplan] We first use compactness to replace the finite order-property configurations by one configuration indexed by the dense order $\mathbb{Q}$. The parameter side of this configuration gives a [countable set](/page/Countable%20Set) $A$. Each irrational cut of $\mathbb{Q}$ determines a finitely satisfiable partial type over $A$, and compactness extends it to a complete $1$-type. Distinct cuts force opposite decisions on some formula $\varphi(x;b_q)$, giving continuum many complete types over the countable set $A$. [/proofplan] [step:Build a rationally indexed order pattern by compactness] Let $\mathcal{L}'$ be the expansion of $\mathcal{L}$ obtained by adding, for each $q \in \mathbb{Q}$, a constant symbol $c_q$ of the sort of $x$ and a tuple $d_q$ of constant symbols of the same sorts and length as $y$. Define the $\mathcal{L}'$-theory \begin{align*} \Gamma := T \cup \{\varphi(c_q;d_r) : q,r \in \mathbb{Q},\ q < r\} \cup \{\neg \varphi(c_q;d_r) : q,r \in \mathbb{Q},\ q \geq r\}. \end{align*} We show that every finite subset of $\Gamma$ is satisfiable. Let $\Gamma_0 \subset \Gamma$ be finite. Let $F \subset \mathbb{Q}$ be the finite set of rational numbers whose constants occur in $\Gamma_0$, and enumerate it increasingly as \begin{align*} F = \{q_1,\dots,q_n\}, \qquad q_1 < \cdots < q_n. \end{align*} By the order-property hypothesis, there are elements $\alpha_1,\dots,\alpha_n$ and parameter tuples $\beta_1,\dots,\beta_n$ in some model $M \models T$ such that \begin{align*} M \models \varphi(\alpha_i;\beta_j) \quad \Longleftrightarrow \quad i < j. \end{align*} Interpret $c_{q_i}$ as $\alpha_i$ and $d_{q_i}$ as $\beta_i$ for $1 \leq i \leq n$, and interpret the remaining new constants arbitrarily. Then every instance from $\Gamma_0$ is satisfied, because $q_i < q_j$ is equivalent to $i < j$. Thus $\Gamma$ is finitely satisfiable. By the [Compactness Theorem for first-order logic](/theorems/4290) (citing a result not yet in the wiki: [Compactness Theorem](/theorems/2748) for first-order logic), there is an $\mathcal{L}'$-structure $N'$ satisfying $\Gamma$. Let $N$ be the $\mathcal{L}$-reduct of $N'$. For each $q \in \mathbb{Q}$, let \begin{align*} a_q &:= c_q^{N'}, & b_q &:= d_q^{N'}. \end{align*} Then $N \models T$ and, for all $q,r \in \mathbb{Q}$, \begin{align*} N \models \varphi(a_q;b_r) \quad \Longleftrightarrow \quad q < r. \end{align*} [guided] The order property gives finite chains, but the proof needs a single infinite chain indexed by $\mathbb{Q}$. To obtain it without choosing compatible finite configurations by hand, we add names for all desired elements and use compactness. For each $q \in \mathbb{Q}$, introduce a constant symbol $c_q$ for the object variable $x$ and a tuple $d_q$ of constant symbols matching the parameter tuple $y$. We then demand exactly the rational order pattern: \begin{align*} \Gamma := T \cup \{\varphi(c_q;d_r) : q,r \in \mathbb{Q},\ q < r\} \cup \{\neg \varphi(c_q;d_r) : q,r \in \mathbb{Q},\ q \geq r\}. \end{align*} The point of this theory is that any model of it contains elements $a_q$ and parameter tuples $b_q$ with $\varphi(a_q;b_r)$ holding exactly when $q<r$. We verify finite satisfiability. Let $\Gamma_0$ be a finite subset of $\Gamma$. Only finitely many rational indices appear in $\Gamma_0$; call this finite set $F$. Write \begin{align*} F = \{q_1,\dots,q_n\}, \qquad q_1 < \cdots < q_n. \end{align*} The order-property hypothesis applies to this integer $n$, so there are elements $\alpha_1,\dots,\alpha_n$ and parameter tuples $\beta_1,\dots,\beta_n$ in a model $M \models T$ such that \begin{align*} M \models \varphi(\alpha_i;\beta_j) \quad \Longleftrightarrow \quad i < j. \end{align*} Now interpret the constants indexed by $F$ by \begin{align*} c_{q_i}^{M'} := \alpha_i, \qquad d_{q_i}^{M'} := \beta_i \end{align*} in the corresponding expansion $M'$ of $M$, and interpret all unused new constants arbitrarily. Since the enumeration of $F$ preserves the rational order, we have $q_i<q_j$ exactly when $i<j$. Therefore every positive or negative formula from $\Gamma_0$ is satisfied. Thus every finite subset of $\Gamma$ is satisfiable. The Compactness Theorem for first-order logic (citing a result not yet in the wiki: Compactness Theorem for first-order logic) gives a model $N' \models \Gamma$. Passing to the $\mathcal{L}$-reduct $N$ gives a model of $T$. If \begin{align*} a_q := c_q^{N'}, \qquad b_q := d_q^{N'}, \end{align*} then the defining axioms of $\Gamma$ give, for all $q,r \in \mathbb{Q}$, \begin{align*} N \models \varphi(a_q;b_r) \quad \Longleftrightarrow \quad q < r. \end{align*} [/guided] [/step] [step:Choose the countable parameter set] Let $m \in \mathbb{N}$ be the length of the parameter tuple $y$. For each $q \in \mathbb{Q}$, write \begin{align*} b_q = (b_{q,1},\dots,b_{q,m}). \end{align*} Define the parameter set \begin{align*} A := \{b_{q,i} : q \in \mathbb{Q},\ 1 \leq i \leq m\} \subseteq N. \end{align*} Since $\mathbb{Q}$ is countable and $m$ is finite, $A$ is countable. Moreover, the tuples $b_q$ are pairwise distinct. If $q<r$, choose $s \in \mathbb{Q}$ with $q<s<r$. Then \begin{align*} N \models \neg\varphi(a_s;b_q) \qquad\text{and}\qquad N \models \varphi(a_s;b_r), \end{align*} so $b_q \neq b_r$. Hence $A$ is infinite, and therefore $|A|=\aleph_0$. [/step] [step:Associate a partial type to each irrational cut of $\mathbb{Q}$] For each irrational real number $\alpha \in \mathbb{R}\setminus\mathbb{Q}$, define subsets of $\mathbb{Q}$ by \begin{align*} L_\alpha &:= \{q \in \mathbb{Q} : q < \alpha\}, & U_\alpha &:= \{q \in \mathbb{Q} : q > \alpha\}. \end{align*} Define the set of $\mathcal{L}(A)$-formulas in the variable $x$ \begin{align*} p_\alpha(x) := \{\neg\varphi(x;b_q) : q \in L_\alpha\} \cup \{\varphi(x;b_q) : q \in U_\alpha\}. \end{align*} We show that $p_\alpha(x)$ is finitely satisfiable in $N$. Let $\Delta(x) \subset p_\alpha(x)$ be finite. Let $L_0 \subset L_\alpha$ and $U_0 \subset U_\alpha$ be the finite sets of rational indices that occur in $\Delta(x)$. Since $\alpha$ is irrational and $\mathbb{Q}$ is dense in $\mathbb{R}$, there exists $s \in \mathbb{Q}$ such that \begin{align*} q < s < r \end{align*} for every $q \in L_0$ and every $r \in U_0$. For this $s$, the rational order pattern gives \begin{align*} q \in L_0 &\implies s \geq q \implies N \models \neg\varphi(a_s;b_q),\\ r \in U_0 &\implies s < r \implies N \models \varphi(a_s;b_r). \end{align*} Thus $a_s$ realizes $\Delta(x)$ in $N$. Hence $p_\alpha(x)$ is finitely satisfiable, and so it is consistent with $T$. [/step] [step:Extend each cut type to a complete type over $A$] Fix $\alpha \in \mathbb{R}\setminus\mathbb{Q}$. Since $p_\alpha(x)$ is consistent with $T$, it extends to a complete $1$-type over $A$. Indeed, partially order the collection of consistent sets of $\mathcal{L}(A)$-formulas in the variable $x$ that contain $p_\alpha(x)$ by inclusion. The union of every chain is consistent, because any finite subset of such a union lies in one member of the chain. By [Zorn's lemma](/theorems/1226), there is a maximal consistent extension, denoted $P_\alpha(x)$. If $\psi(x)$ is any $\mathcal{L}(A)$-formula, maximality implies that exactly one of $\psi(x)$ and $\neg\psi(x)$ belongs to $P_\alpha(x)$; otherwise $P_\alpha(x)$ would have a proper consistent extension or would be inconsistent. Hence $P_\alpha(x) \in S_1(A)$. [/step] [step:Separate distinct cuts by a formula instance] Let $\alpha,\beta \in \mathbb{R}\setminus\mathbb{Q}$ with $\alpha<\beta$. Choose $q \in \mathbb{Q}$ such that \begin{align*} \alpha < q < \beta. \end{align*} Then $q \in U_\alpha$ and $q \in L_\beta$. Therefore \begin{align*} \varphi(x;b_q) &\in p_\alpha(x) \subseteq P_\alpha(x),\\ \neg\varphi(x;b_q) &\in p_\beta(x) \subseteq P_\beta(x). \end{align*} Thus $P_\alpha \neq P_\beta$ as complete types over $A$. The map \begin{align*} \mathbb{R}\setminus\mathbb{Q} &\to S_1(A)\\ \alpha &\mapsto P_\alpha \end{align*} is therefore injective. Since $|\mathbb{R}\setminus\mathbb{Q}|=2^{\aleph_0}$, we obtain \begin{align*} |S_1(A)| \geq 2^{\aleph_0} > \aleph_0. \end{align*} Because $|A|=\aleph_0$, this proves that $T$ is not $\aleph_0$-stable. [/step]