[proofplan] We expand the language by constants for the elements of $M$ and by one additional constant $c$. The finite satisfiability hypothesis lets us apply the [compactness theorem](/theorems/2748) to obtain a model of the full theory $\operatorname{Diag}_{\mathrm{el}}(M) \cup \Sigma(c)$. The constants naming elements of $M$ then define an elementary embedding of $M$ into the reduct of this model, because the elementary diagram records every first-order truth with parameters from $M$. Finally, the interpretation of $c$ realizes $\Sigma$, and we identify $M$ with its elementary image. [/proofplan] [step:Use compactness to realize the full expanded theory] Let $\mathcal{L}^+ := \mathcal{L}(M) \cup \{c\}$, where $c$ is a constant symbol not belonging to $\mathcal{L}(M)$. Define the $\mathcal{L}^+$-theory \begin{align*} T := \operatorname{Diag}_{\mathrm{el}}(M) \cup \Sigma(c). \end{align*} By hypothesis, every finite subset of $T$ is satisfiable. Hence, by the [compactness theorem for first-order logic](/theorems/4290) (citing a result not yet in the wiki: Compactness Theorem for First-Order Logic), there exists an $\mathcal{L}^+$-structure $N^+$ such that \begin{align*} N^+ \models T. \end{align*} [guided] We first put all the relevant information into one first-order theory. The language $\mathcal{L}(M)$ already contains a constant symbol $\bar{m}$ for each element $m \in M$, and we add one more constant symbol $c$ whose eventual interpretation will be the desired realizing element. Thus the working language is \begin{align*} \mathcal{L}^+ := \mathcal{L}(M) \cup \{c\}. \end{align*} In this language, define \begin{align*} T := \operatorname{Diag}_{\mathrm{el}}(M) \cup \Sigma(c). \end{align*} The first part of $T$ forces a model to contain an elementary copy of $M$, while the second part forces the interpretation of $c$ to satisfy all formulas in $\Sigma$. The hypothesis says precisely that every finite subset of this theory $T$ is satisfiable. The compactness theorem for first-order logic states that if every finite subset of a first-order theory is satisfiable, then the entire theory is satisfiable. Applying compactness to the $\mathcal{L}^+$-theory $T$, we obtain an $\mathcal{L}^+$-structure $N^+$ satisfying \begin{align*} N^+ \models T. \end{align*} This is the only place where compactness is used. [/guided] [/step] [step:Extract an elementary copy of $M$ from the elementary diagram] Let $N$ be the reduct of $N^+$ to the original language $\mathcal{L}$. Define a map \begin{align*} j: M &\to N \\ m &\mapsto \bar{m}^{N^+}. \end{align*} The map $j$ is injective: if $m_1,m_2 \in M$ and $m_1 \neq m_2$, then the sentence $\bar{m}_1 \neq \bar{m}_2$ belongs to $\operatorname{Diag}_{\mathrm{el}}(M)$, so $N^+ \models \bar{m}_1 \neq \bar{m}_2$. We show that $j$ is elementary. Let $\psi(y_1,\dots,y_n)$ be any $\mathcal{L}$-formula, and let $m_1,\dots,m_n \in M$. If \begin{align*} M \models \psi(m_1,\dots,m_n), \end{align*} then the $\mathcal{L}(M)$-sentence $\psi(\bar{m}_1,\dots,\bar{m}_n)$ belongs to $\operatorname{Diag}_{\mathrm{el}}(M)$, hence \begin{align*} N \models \psi(j(m_1),\dots,j(m_n)). \end{align*} If instead \begin{align*} M \models \neg \psi(m_1,\dots,m_n), \end{align*} then $\neg \psi(\bar{m}_1,\dots,\bar{m}_n)$ belongs to $\operatorname{Diag}_{\mathrm{el}}(M)$, hence \begin{align*} N \models \neg \psi(j(m_1),\dots,j(m_n)). \end{align*} Therefore, for every $\mathcal{L}$-formula $\psi(y_1,\dots,y_n)$ and every tuple $(m_1,\dots,m_n) \in M^n$, \begin{align*} M \models \psi(m_1,\dots,m_n) \iff N \models \psi(j(m_1),\dots,j(m_n)). \end{align*} Thus $j: M \to N$ is an elementary embedding. [guided] The constants $\bar{m}$ in the language are meant to name the elements of $M$. In the model $N^+$, each such constant has an interpretation $\bar{m}^{N^+}$. We use these interpretations to define \begin{align*} j: M &\to N \\ m &\mapsto \bar{m}^{N^+}, \end{align*} where $N$ is the $\mathcal{L}$-reduct of $N^+$. First, $j$ is injective. Suppose $m_1,m_2 \in M$ and $m_1 \neq m_2$. Since the elementary diagram contains every $\mathcal{L}(M)$-sentence true in $M$, the sentence \begin{align*} \bar{m}_1 \neq \bar{m}_2 \end{align*} belongs to $\operatorname{Diag}_{\mathrm{el}}(M)$. Because $N^+ \models \operatorname{Diag}_{\mathrm{el}}(M)$, we get \begin{align*} N^+ \models \bar{m}_1 \neq \bar{m}_2. \end{align*} Therefore $\bar{m}_1^{N^+} \neq \bar{m}_2^{N^+}$, so $j(m_1) \neq j(m_2)$. Now we verify elementarity. Let $\psi(y_1,\dots,y_n)$ be an arbitrary $\mathcal{L}$-formula, and let $m_1,\dots,m_n \in M$. The point of using the elementary diagram, rather than only the atomic diagram, is that it records the truth of every formula with parameters from $M$, not only atomic formulas. If \begin{align*} M \models \psi(m_1,\dots,m_n), \end{align*} then the sentence \begin{align*} \psi(\bar{m}_1,\dots,\bar{m}_n) \end{align*} is true in the natural $\mathcal{L}(M)$-expansion of $M$. Hence it belongs to $\operatorname{Diag}_{\mathrm{el}}(M)$. Since $N^+ \models \operatorname{Diag}_{\mathrm{el}}(M)$, its $\mathcal{L}$-reduct $N$ satisfies \begin{align*} N \models \psi(j(m_1),\dots,j(m_n)). \end{align*} Conversely, if \begin{align*} M \models \neg \psi(m_1,\dots,m_n), \end{align*} then \begin{align*} \neg \psi(\bar{m}_1,\dots,\bar{m}_n) \end{align*} belongs to $\operatorname{Diag}_{\mathrm{el}}(M)$, so \begin{align*} N \models \neg \psi(j(m_1),\dots,j(m_n)). \end{align*} Combining the two cases gives \begin{align*} M \models \psi(m_1,\dots,m_n) \iff N \models \psi(j(m_1),\dots,j(m_n)). \end{align*} Since $\psi$ and the tuple from $M$ were arbitrary, $j$ is an elementary embedding. [/guided] [/step] [step:Use the interpretation of $c$ to realize $\Sigma$] Define \begin{align*} a := c^{N^+} \in N. \end{align*} Let $\varphi(x) \in \Sigma(x)$. Then $\varphi(c) \in \Sigma(c) \subseteq T$, and $N^+ \models T$, so \begin{align*} N^+ \models \varphi(c). \end{align*} Equivalently, in the $\mathcal{L}$-reduct $N$, with each parameter $\bar{m}$ interpreted as $j(m)$, \begin{align*} N \models \varphi(a). \end{align*} Thus $a$ realizes every formula in $\Sigma(x)$ over the elementary copy $j[M] \subseteq N$. [/step] [step:Identify $M$ with its elementary image] Since $j: M \to N$ is an elementary embedding, its image $j[M]$ is an elementary substructure of $N$. Replacing $M$ by the isomorphic copy $j[M]$, we may regard $M$ as an elementary substructure of $N$; that is, \begin{align*} N \succeq M. \end{align*} Under this identification, the element $a = c^{N^+}$ satisfies \begin{align*} N \models \varphi(a) \end{align*} for every $\varphi(x) \in \Sigma(x)$. Therefore $N$ is an elementary extension of $M$ containing an element realizing $\Sigma(x)$, as required. [/step]