[proofplan] We enlarge $A$ to a set closed under both the function symbols of $L$ and a chosen family of Skolem witness functions for all existential formulas. The closure is obtained by countably iterating these operations, and the cardinal bound follows because there are at most $|L|+\aleph_0$ formulas in finitely many variables and only finitely many arguments are used at each stage. The resulting set is the universe of an $L$-substructure of $\mathcal{N}$, and the chosen witnesses verify the Tarski-Vaught condition, which is then proved directly to avoid an external citation. [/proofplan]

[step:Choose Skolem witnesses for existential formulas]Let \begin{align*} \kappa := |A| + |L| + \aleph_0. \end{align*} Since $\kappa$ is infinite, adjoining finitely or countably many elements to a set of cardinality at most $\kappa$ preserves the bound $\kappa$. Choose an element $b_0 \in N$, which exists because $\mathcal{N}$ is infinite. Let $\Phi$ denote the set of all $L$-formulas of the form $\varphi(y,x_1,\dots,x_n)$, where $n \in \mathbb{N} \cup \{0\}$ and the displayed variables include all free variables of $\varphi$. The usual recursive construction of first-order formulas gives \begin{align*} |\Phi| \leq |L| + \aleph_0 \leq \kappa. \end{align*} For each $\varphi(y,x_1,\dots,x_n) \in \Phi$, choose a function \begin{align*} F_\varphi: N^n &\to N \end{align*} with the following property: for every tuple $(a_1,\dots,a_n) \in N^n$, if \begin{align*} \mathcal{N} \models \exists y\,\varphi(y,a_1,\dots,a_n), \end{align*} then \begin{align*} \mathcal{N} \models \varphi(F_\varphi(a_1,\dots,a_n),a_1,\dots,a_n). \end{align*} If no such $y$ exists, define $F_\varphi(a_1,\dots,a_n) := b_0$. This choice is possible by the [axiom of choice](/page/Axiom%20of%20Choice), applied independently to the nonempty witness sets \begin{align*} \{c \in N : \mathcal{N} \models \varphi(c,a_1,\dots,a_n)\}. \end{align*}[/step]

[guided]The purpose of this step is to build, in advance, a witness-selection mechanism for every possible existential formula. Fix \begin{align*} \kappa := |A| + |L| + \aleph_0. \end{align*} This is the target size bound. Since $\kappa$ is infinite, finite powers, countable unions, and multiplication by at most $\kappa$ many choices will remain bounded by $\kappa$. We first choose one element $b_0 \in N$. This is legitimate because the universe $N$ is nonempty; indeed $\mathcal{N}$ is assumed infinite. The element $b_0$ is used only as a default value when an existential formula has no witness. Let $\Phi$ be the set of all formulas $\varphi(y,x_1,\dots,x_n)$, with $n \in \mathbb{N} \cup \{0\}$, in which the displayed variables contain all free variables. The set of first-order formulas is built from the symbols of $L$, variables, logical connectives, quantifiers, and parentheses by finite strings. Hence the number of such formulas is bounded by \begin{align*} |\Phi| \leq |L| + \aleph_0 \leq \kappa. \end{align*} For each formula $\varphi(y,x_1,\dots,x_n)$ we now define a witness function \begin{align*} F_\varphi: N^n &\to N. \end{align*} Given $(a_1,\dots,a_n) \in N^n$, if $\mathcal{N}$ satisfies $\exists y\,\varphi(y,a_1,\dots,a_n)$, then the set \begin{align*} \{c \in N : \mathcal{N} \models \varphi(c,a_1,\dots,a_n)\} \end{align*} is nonempty, so we choose one element of it and call it $F_\varphi(a_1,\dots,a_n)$. If the set is empty, we set $F_\varphi(a_1,\dots,a_n) := b_0$. Thus $F_\varphi$ is a total function on $N^n$, and whenever a witness exists in $\mathcal{N}$, $F_\varphi$ returns one.[/guided]

[step:Iterate closure under language functions and Skolem functions]Define subsets $B_i \subset N$ recursively. Set \begin{align*} B_0 := A \cup \{b_0\}. \end{align*} Given $B_i$, define $B_{i+1}$ to be the union of $B_i$ with all elements of the following two forms: \begin{align*} f^{\mathcal{N}}(a_1,\dots,a_m) \end{align*} where $f$ is an $m$-ary function symbol of $L$, $m \in \mathbb{N} \cup \{0\}$, and $(a_1,\dots,a_m) \in B_i^m$; and \begin{align*} F_\varphi(a_1,\dots,a_n) \end{align*} where $\varphi(y,x_1,\dots,x_n) \in \Phi$ and $(a_1,\dots,a_n) \in B_i^n$. Finally define \begin{align*} M := \bigcup_{i=0}^{\infty} B_i. \end{align*} Then $A \subset M$ and $b_0 \in M$, so $M$ is nonempty. We prove by induction that $|B_i| \leq \kappa$ for every $i \in \mathbb{N} \cup \{0\}$. For $i=0$, \begin{align*} |B_0| \leq |A| + 1 \leq \kappa. \end{align*} Assume $|B_i| \leq \kappa$. For each finite $m$, the set $B_i^m$ has cardinality at most $\kappa$. There are at most $|L| \leq \kappa$ function symbols and at most $|\Phi| \leq \kappa$ Skolem functions. Therefore the set of all new values added at stage $i+1$ has cardinality at most $\kappa$, and hence $|B_{i+1}| \leq \kappa$. Thus \begin{align*} |M| = \left|\bigcup_{i=0}^{\infty} B_i\right| \leq \kappa \cdot \aleph_0 = \kappa. \end{align*}[/step]

[guided]We now close $A$ under all operations that are needed for elementarity. Start with \begin{align*} B_0 := A \cup \{b_0\}. \end{align*} The element $b_0$ ensures that the eventual universe is nonempty, even when $A=\varnothing$ and the language has no constant symbols. Suppose $B_i$ has been constructed. We put into $B_{i+1}$ every value obtained from elements of $B_i$ by an actual function symbol of the language: \begin{align*} f^{\mathcal{N}}(a_1,\dots,a_m), \end{align*} where $f$ is an $m$-ary function symbol of $L$ and $(a_1,\dots,a_m) \in B_i^m$. This guarantees that the final set will be closed under the functions of the language. We also put into $B_{i+1}$ every value obtained by a Skolem witness function: \begin{align*} F_\varphi(a_1,\dots,a_n), \end{align*} where $\varphi(y,x_1,\dots,x_n) \in \Phi$ and $(a_1,\dots,a_n) \in B_i^n$. This guarantees that if an existential statement with parameters already in the construction is true in $\mathcal{N}$, then a chosen witness is added at the next stage. Define the final set by \begin{align*} M := \bigcup_{i=0}^{\infty} B_i. \end{align*} Then $A \subset M$ because $A \subset B_0$, and $M$ is nonempty because $b_0 \in B_0$. It remains to check the size bound. We prove $|B_i| \leq \kappa$ by induction. The base case is \begin{align*} |B_0| \leq |A| + 1 \leq \kappa. \end{align*} For the induction step, assume $|B_i| \leq \kappa$. If $m$ is finite, then \begin{align*} |B_i^m| \leq \kappa^m = \kappa, \end{align*} because $\kappa$ is infinite. There are at most $\kappa$ function symbols of $L$ and at most $\kappa$ formulas in $\Phi$. Therefore all values added using language functions and all values added using Skolem functions together form a set of cardinality at most $\kappa$. Hence $|B_{i+1}| \leq \kappa$. Finally, $M$ is a countable union of sets of cardinality at most $\kappa$, so \begin{align*} |M| = \left|\bigcup_{i=0}^{\infty} B_i\right| \leq \kappa \cdot \aleph_0 = \kappa. \end{align*}[/guided]

[step:Make the closed set into an $L$-substructure of $\mathcal{N}$] Let $\mathcal{M}$ be the $L$-structure with universe $M$ whose interpretations are inherited from $\mathcal{N}$: each relation symbol is restricted to $M$, each constant symbol is interpreted as its value in $\mathcal{N}$, and each function symbol is interpreted by restricting the corresponding function of $\mathcal{N}$ to tuples from $M$. This is well-defined because $M$ is closed under all function symbols of $L$. Indeed, if $f$ is an $m$-ary function symbol and $(a_1,\dots,a_m) \in M^m$, then each $a_j$ belongs to some $B_{i_j}$. Let \begin{align*} r := \max\{i_1,\dots,i_m\}. \end{align*} Then $(a_1,\dots,a_m) \in B_r^m$, so by construction \begin{align*} f^{\mathcal{N}}(a_1,\dots,a_m) \in B_{r+1} \subset M. \end{align*} Thus $\mathcal{M}$ is an $L$-substructure of $\mathcal{N}$. [/step]

[step:Verify the existential witness condition inside $M$] Let $\varphi(y,x_1,\dots,x_n) \in \Phi$ and let $(a_1,\dots,a_n) \in M^n$. Suppose \begin{align*} \mathcal{N} \models \exists y\,\varphi(y,a_1,\dots,a_n). \end{align*} Choose $r \in \mathbb{N} \cup \{0\}$ such that $a_1,\dots,a_n \in B_r$. By the defining property of $F_\varphi$, \begin{align*} \mathcal{N} \models \varphi(F_\varphi(a_1,\dots,a_n),a_1,\dots,a_n). \end{align*} By construction, \begin{align*} F_\varphi(a_1,\dots,a_n) \in B_{r+1} \subset M. \end{align*} Therefore every existential formula with parameters in $M$ that is realized in $\mathcal{N}$ has a realization in $M$. [/step]

[step:Derive elementarity from the existential witness condition]We prove that $\mathcal{M} \preccurlyeq \mathcal{N}$. It suffices to prove, by induction on formulas, that for every $L$-formula $\theta(x_1,\dots,x_n)$ and every tuple $(a_1,\dots,a_n) \in M^n$, \begin{align*} \mathcal{M} \models \theta(a_1,\dots,a_n) \quad\Longleftrightarrow\quad \mathcal{N} \models \theta(a_1,\dots,a_n). \end{align*} For atomic formulas this follows from the definition of substructure: function terms have the same values in $\mathcal{M}$ and $\mathcal{N}$ on tuples from $M$, and relation symbols are interpreted in $\mathcal{M}$ by restriction from $\mathcal{N}$. The Boolean connectives follow immediately from the induction hypothesis. It remains to handle existential quantifiers. Let $\theta(x_1,\dots,x_n)$ be the formula \begin{align*} \exists y\,\psi(y,x_1,\dots,x_n). \end{align*} Assume the induction hypothesis holds for $\psi$. If \begin{align*} \mathcal{M} \models \exists y\,\psi(y,a_1,\dots,a_n), \end{align*} then some $c \in M$ satisfies $\mathcal{M} \models \psi(c,a_1,\dots,a_n)$, so by the induction hypothesis \begin{align*} \mathcal{N} \models \psi(c,a_1,\dots,a_n), \end{align*} and hence $\mathcal{N} \models \exists y\,\psi(y,a_1,\dots,a_n)$. Conversely, suppose \begin{align*} \mathcal{N} \models \exists y\,\psi(y,a_1,\dots,a_n). \end{align*} By the existential witness condition proved above, there exists $c \in M$ such that \begin{align*} \mathcal{N} \models \psi(c,a_1,\dots,a_n). \end{align*} By the induction hypothesis applied to $\psi$, \begin{align*} \mathcal{M} \models \psi(c,a_1,\dots,a_n). \end{align*} Therefore \begin{align*} \mathcal{M} \models \exists y\,\psi(y,a_1,\dots,a_n). \end{align*} Universal quantifiers are handled by rewriting $\forall y\,\psi$ as $\neg\exists y\,\neg\psi$. Thus $\mathcal{M} \preccurlyeq \mathcal{N}$. Together with $A \subset M$ and $|M| \leq |A|+|L|+\aleph_0$, this proves the theorem.[/step]

[guided]The final point is to show that the substructure is elementary, meaning that it satisfies exactly the same first-order formulas with parameters from $M$ as $\mathcal{N}$ does. We prove the full statement by induction on formulas: for every formula $\theta(x_1,\dots,x_n)$ and every tuple $(a_1,\dots,a_n) \in M^n$, \begin{align*} \mathcal{M} \models \theta(a_1,\dots,a_n) \quad\Longleftrightarrow\quad \mathcal{N} \models \theta(a_1,\dots,a_n). \end{align*} For atomic formulas, this is exactly what it means for $\mathcal{M}$ to be a substructure of $\mathcal{N}$. Terms are evaluated the same way in both structures because the functions of $\mathcal{M}$ are restrictions of the functions of $\mathcal{N}$. Relation symbols also agree because every relation in $\mathcal{M}$ is the restriction of the corresponding relation in $\mathcal{N}$ to tuples from $M$. The induction steps for negation, conjunction, disjunction, and implication are direct consequences of the induction hypothesis. The only non-formal case is the existential quantifier. Suppose \begin{align*} \theta(x_1,\dots,x_n) \equiv \exists y\,\psi(y,x_1,\dots,x_n). \end{align*} If $\mathcal{M}$ satisfies this formula, then there is some $c \in M$ with \begin{align*} \mathcal{M} \models \psi(c,a_1,\dots,a_n). \end{align*} By the induction hypothesis for $\psi$, the same tuple satisfies $\psi$ in $\mathcal{N}$, so \begin{align*} \mathcal{N} \models \exists y\,\psi(y,a_1,\dots,a_n). \end{align*} The converse is where the Skolem closure is used. Suppose \begin{align*} \mathcal{N} \models \exists y\,\psi(y,a_1,\dots,a_n). \end{align*} The existential witness condition gives a witness $c \in M$ such that \begin{align*} \mathcal{N} \models \psi(c,a_1,\dots,a_n). \end{align*} Now apply the induction hypothesis for $\psi$ again, this time from $\mathcal{N}$ back to $\mathcal{M}$: \begin{align*} \mathcal{M} \models \psi(c,a_1,\dots,a_n). \end{align*} Hence \begin{align*} \mathcal{M} \models \exists y\,\psi(y,a_1,\dots,a_n). \end{align*} Universal quantifiers require no separate argument because $\forall y\,\psi$ is equivalent to $\neg\exists y\,\neg\psi$ in first-order logic. Therefore all formulas agree between $\mathcal{M}$ and $\mathcal{N}$ on parameters from $M$, which is precisely $\mathcal{M} \preccurlyeq \mathcal{N}$.[/guided]