[proofplan] The closure operator is obtained by restricting model-theoretic algebraic closure to the parameter-free definable strongly minimal set $D$. Extensivity, monotonicity, idempotence, and finite character are inherited directly from model-theoretic [algebraic closure](/page/Algebraic%20Closure) in the monster model. The only non-formal point is exchange: from a finite definable fibre witnessing $a \in \operatorname{acl}_{\mathfrak{M}}(A \cup \{b\})$, [strong minimality](/page/Strong%20Minimality) forces the opposite fibre over $a$ to be finite, unless $a$ would already be algebraic over $A$. [/proofplan]

[step:Inherit extensivity from algebraic closure] Let $A \subseteq D$ and let $a \in A$. Since model-theoretic algebraic closure is extensive, $a \in \operatorname{acl}_{\mathfrak{M}}(A)$. Because $a \in D$, we have \begin{align*} a \in D \cap \operatorname{acl}_{\mathfrak{M}}(A) = \operatorname{cl}(A). \end{align*} Thus $A \subseteq \operatorname{cl}(A)$, so $\operatorname{cl}$ is extensive. [/step]

[step:Inherit monotonicity from algebraic closure] Let $A,B \subseteq D$ and assume $A \subseteq B$. Since model-theoretic [algebraic closure](/page/Algebraic%20Closure) is increasing, \begin{align*} \operatorname{acl}_{\mathfrak{M}}(A) \subseteq \operatorname{acl}_{\mathfrak{M}}(B). \end{align*} Intersecting both sides with $D$ gives \begin{align*} \operatorname{cl}(A) = D \cap \operatorname{acl}_{\mathfrak{M}}(A) \subseteq D \cap \operatorname{acl}_{\mathfrak{M}}(B) = \operatorname{cl}(B). \end{align*} Thus $\operatorname{cl}$ is increasing. [/step]

[step:Inherit idempotence from algebraic closure] Let $A \subseteq D$. We prove both inclusions. Since $A \subseteq \operatorname{cl}(A)$, monotonicity gives \begin{align*} \operatorname{cl}(A) \subseteq \operatorname{cl}(\operatorname{cl}(A)). \end{align*} Conversely, because $\operatorname{cl}(A) \subseteq \operatorname{acl}_{\mathfrak{M}}(A)$, monotonicity and idempotence of model-theoretic [algebraic closure](/page/Algebraic%20Closure) give \begin{align*} \operatorname{acl}_{\mathfrak{M}}(\operatorname{cl}(A)) \subseteq \operatorname{acl}_{\mathfrak{M}}(\operatorname{acl}_{\mathfrak{M}}(A)) = \operatorname{acl}_{\mathfrak{M}}(A). \end{align*} Therefore \begin{align*} \operatorname{cl}(\operatorname{cl}(A)) = D \cap \operatorname{acl}_{\mathfrak{M}}(\operatorname{cl}(A)) \subseteq D \cap \operatorname{acl}_{\mathfrak{M}}(A) = \operatorname{cl}(A). \end{align*} Hence $\operatorname{cl}$ is idempotent. [/step]

[step:Inherit finite character from algebraic closure] Let $A \subseteq D$ and $a \in \operatorname{cl}(A)$. Then $a \in D$ and $a \in \operatorname{acl}_{\mathfrak{M}}(A)$. By finite character of model-theoretic [algebraic closure](/page/Algebraic%20Closure), there is a finite subset $A_0 \subseteq A$ such that \begin{align*} a \in \operatorname{acl}_{\mathfrak{M}}(A_0). \end{align*} Since $a \in D$, this implies \begin{align*} a \in D \cap \operatorname{acl}_{\mathfrak{M}}(A_0) = \operatorname{cl}(A_0). \end{align*} Thus $\operatorname{cl}$ has finite character. [/step]

[step:Choose a finite definable fibre witnessing $a \in \operatorname{acl}_{\mathfrak{M}}(A b)$]Let $A \subseteq D$ and let $a,b \in D$ satisfy \begin{align*} a \in \operatorname{cl}(A \cup \{b\}) \setminus \operatorname{cl}(A). \end{align*} Equivalently, \begin{align*} a \in D \cap \operatorname{acl}_{\mathfrak{M}}(A \cup \{b\}) \quad\text{and}\quad a \notin \operatorname{acl}_{\mathfrak{M}}(A). \end{align*} By finite character of [algebraic closure](/page/Algebraic%20Closure), choose a finite tuple $c$ from $A$ such that \begin{align*} a \in \operatorname{acl}_{\mathfrak{M}}(c,b) \quad\text{and}\quad a \notin \operatorname{acl}_{\mathfrak{M}}(c). \end{align*} Because $D$ is parameter-free definable, every formula defining a subset of $D$ below uses only the displayed parameters, and no hidden parameters from a definition of $D$ are added. Choose a formula $\varphi(x,y,c)$ such that $\mathfrak{M} \models \varphi(a,b,c)$ and the set \begin{align*} \{x \in D : \mathfrak{M} \models \varphi(x,b,c)\} \end{align*} is finite. Let $n \in \mathbb{N}$ be its cardinality. Define a relation \begin{align*} R \subseteq D \times D \end{align*} by \begin{align*} R(x,y) \iff \varphi(x,y,c) \quad\text{and}\quad \left|\{x' \in D : \mathfrak{M} \models \varphi(x',y,c)\}\right| \leq n. \end{align*} The condition on the fibre is first-order expressible by saying that there do not exist $n+1$ pairwise distinct elements $x'_1,\dots,x'_{n+1} \in D$ satisfying $\varphi(x'_i,y,c)$. Hence $R$ is definable over $c$. Also $R(a,b)$ holds, and every vertical fibre \begin{align*} R_y := \{x \in D : R(x,y)\} \end{align*} has cardinality at most $n$.[/step]

[guided]The purpose of this step is to replace the bare fact $a \in \operatorname{acl}_{\mathfrak{M}}(c,b)$ by a definable binary relation with uniformly finite fibres. Algebraicity means exactly that some formula with parameter $b$ isolates $a$ inside a finite definable subset of $D$. Since $D$ is parameter-free definable, restricting quantifiers and fibres to $D$ introduces no additional parameters beyond $c$ and $b$. Thus we choose $\varphi(x,y,c)$ with $\mathfrak{M} \models \varphi(a,b,c)$ and finite fibre over $b$: \begin{align*} \{x \in D : \mathfrak{M} \models \varphi(x,b,c)\}. \end{align*} Let $n$ be the size of this fibre. The formula $\varphi(x,y,c)$ may have infinite fibres for other values of $y$, so we refine it. Define $R(x,y)$ to mean that $\varphi(x,y,c)$ holds and the fibre over $y$ has at most $n$ elements in $D$. This second condition is first-order because it says: there do not exist pairwise distinct $x'_1,\dots,x'_{n+1} \in D$ such that each $\varphi(x'_i,y,c)$ holds. Therefore $R \subseteq D \times D$ is definable over the finite tuple $c$ from $A$, the pair $(a,b)$ lies in $R$, and every vertical fibre \begin{align*} R_y := \{x \in D : R(x,y)\} \end{align*} has cardinality at most $n$. This uniform finite bound is what allows strong minimality to force exchange.[/guided]

[step:Show the opposite fibre over $a$ is finite] Define, for each $x \in D$, the horizontal fibre \begin{align*} H_x := \{y \in D : R(x,y)\}. \end{align*} Also define \begin{align*} R^a := \{y \in D : R(a,y)\}. \end{align*} Thus $R^a = H_a$. We prove that $H_a$ is finite. We use the standard uniformity consequence of [strong minimality](/page/Strong%20Minimality) for the formula $R(x,y)$, with $x$ as the parameter variable and $y$ as the object variable ranging over $D$: there is an integer $N \in \mathbb{N}$ such that, for every $x \in D$, if $H_x$ is finite then $|H_x| \leq N$. To justify this, suppose no such $N$ existed. For every $m \in \mathbb{N}$, there would be $x_m \in D$ such that $H_{x_m}$ is finite and has at least $m$ elements. Hence, for each $m \in \mathbb{N}$, the partial type over $c$ asserting that $x \in D$, that there exist $m$ pairwise distinct elements of $H_x$, and that there exist $m$ pairwise distinct elements of $D \setminus H_x$ is finitely satisfiable. By the [compactness theorem](/page/Compactness%20Theorem), it is realized by some $x_* \in D$. Then $H_{x_*}$ and $D \setminus H_{x_*}$ are both infinite, so $H_{x_*}$ is an infinite coinfinite subset of $D$ definable with parameters $c$ and $x_*$. This contradicts strong minimality, which allows parameters in definable subsets of $D$. Therefore the set of parameters with cofinite horizontal fibre is definable over $c$ by \begin{align*} T := \{x \in D : \exists y_1,\dots,y_{N+1} \in D \text{ pairwise distinct and } R(x,y_i) \text{ for all } i\}. \end{align*} For each $x \in D$, strong minimality says $H_x$ is finite or cofinite in $D$. By the choice of $N$, the displayed condition is equivalent to $H_x$ being cofinite. Suppose toward a contradiction that $H_a$ is infinite. Since $H_a$ is definable with parameters from $c$ and $a$, strong minimality implies that $H_a$ is cofinite in $D$, so $a \in T$. We now show that $T$ is finite. If $T$ contained distinct elements $x_1,\dots,x_{n+1} \in D$, then each $H_{x_i}$ would be cofinite in $D$. The intersection \begin{align*} H_{x_1} \cap \cdots \cap H_{x_{n+1}} \end{align*} would also be cofinite, hence nonempty. Choose $y \in D$ in this intersection. Then $R(x_i,y)$ holds for every $i \in \{1,\dots,n+1\}$, so the vertical fibre \begin{align*} R_y := \{x \in D : R(x,y)\} \end{align*} has at least $n+1$ elements, contradicting the construction of $R$, which gives $|R_y| \leq n$ for every $y \in D$. Hence $|T| \leq n$. Thus $T$ is a finite set definable over $c$. Since $D$ is parameter-free definable and $R$ is definable over $c$, the definition of $T$ uses no parameters outside $c$. Because $a \in T$, it follows that \begin{align*} a \in \operatorname{acl}_{\mathfrak{M}}(c) \subseteq \operatorname{acl}_{\mathfrak{M}}(A), \end{align*} contradicting the choice of $c$ with $a \notin \operatorname{acl}_{\mathfrak{M}}(c)$. Therefore $H_a = R^a$ is finite. [/step]

[step:Conclude exchange from the finite opposite fibre] Since $R(a,b)$ holds, we have $b \in R^a$. The previous step shows that \begin{align*} R^a = \{y \in D : R(a,y)\} \end{align*} is finite and definable over the parameters $c$ and $a$. Therefore \begin{align*} b \in \operatorname{acl}_{\mathfrak{M}}(c,a). \end{align*} Since $c$ is a finite tuple from $A$, this gives \begin{align*} b \in \operatorname{acl}_{\mathfrak{M}}(A \cup \{a\}). \end{align*} Because $b \in D$, we obtain \begin{align*} b \in D \cap \operatorname{acl}_{\mathfrak{M}}(A \cup \{a\}) = \operatorname{cl}(A \cup \{a\}). \end{align*} This proves exchange. Together with extensivity, monotonicity, idempotence, and finite character, $\operatorname{cl}$ is a pregeometry on $D$. [/step]