Let ${\cal M}$ be an o-minimal expansion of a dense linear order $(M,\lt)$.

For $A \subseteq M$, the **definable closure** of $A$ is defined by $$\dcl(A):= \set{b \in M:\ \{b\} \text{ is } A\text{-definable}}.$$

**Exercise**

Let $A \subseteq M$.

- Prove that $\dcl(A) = \acl(A)$.
- Let $\phi(x)$ be a formula with parameters in $A$. Prove that $\phi(M)$ is infinite if and only if there exist an elementary extension $\M^*$ of $\M$ and $b \in \phi(M^*)$ such that $b \notin \dcl(A)$.

Since the boundary of any definable subset of $M$ is finite, we obtain:

**Corollary**

*Let $S \subseteq M$ be $A$-definable. Then $\bd(S) \subseteq \dcl(A)$.* $\qed$

**Lemma**

*Let $A \subseteq M$ and $a,b \in M$. Then $b \in \dcl(Aa)$ if and only if there is an $A$-definable function $g:I \into M$ such that $a \in I$ and $b = g(a)$.*

**Proof.** Assume first that $b \in \dcl(Aa)$, and let $\phi(x,y)$ be a formula with parameters from $A$ such that $\{b\} = \phi(a,M)$. So $\M \models \exists! y \phi(a,y)$, where “$\exists!y$” abbreviates “there is a unique $y$”. Thus, the set $$I:= \set{x \in M:\ \exists! y\phi(x,y)}$$ is definable over $A$ and contains $a$, and $\phi(M^2) \cap (I \times M)$ is the graph of a function $g:I \into M$ definable over $A$ such that $g(a) = b$.

Conversely, assume that $b = g(a)$ for some function $g:I \into M$ definable over $A$. Let $\phi(x,y)$ be a formula with parameters from $A$ such that $\phi(M^2)$ is the graph of $g$. Then $\{b\} = \phi(a,M)$, so $b \in \dcl(Aa)$. $\qed$

**Exercise**

Let $A \subseteq M$, and let $f:I \into M$ be $A$-definable, with $I \subseteq M$. Prove that the $a_i$ obtained by the Monotonicity Theorem for $f$ can be chosen to lie in $\dcl(A)$.

**Proposition**

*The pair $(M,\dcl)$ is a pregeometry.*

**Proof** Since $\dcl = \acl$, it suffices to establish the exchange property. Let $A \subseteq M$ and $a,b \in M$ be such that $a \in \dcl(Ab) \setminus \dcl(A)$. By the above Lemma, it suffices to show that there exists $g:J \into M$ definable over $A$, with $J \subseteq M$, such that $a \in J$ and $b = g(a)$.

Also by the above Lemma, there is an $f:I \into M$ definable over $A$ such that $I \subseteq M$, $b \in I$ and $a = f(b)$. By o-minimality and because $\bd(I) \subseteq \dcl(A)$, we may assume that $I$ is a singleton or an interval. If $I$ is a singleton, then $I=\{b\} = \bd(I) \subseteq \dcl(A)$, so $a \in \dcl(A)$, a contradiction. We therefore may assume that $I = (c,d)$ for some $c,d \in M \cup \{-\infty,+\infty\}$ with $c \lt b \lt d$.

By the Monotonicity Theorem and the above Exercise, there are $a_1, \dots, a_k \in \dcl(A)$ such that $$a_0:= c \lt a_1 \lt \cdots \lt a_k \lt a_{k+1}:= d$$ and, for $i=0, \dots, k$, the restriction of $f$ to $(a_i,a_{i+1})$ is strictly monotone. As in the previous paragraph, we must have $b \ne a_i$ for each $i$. Thus, replacing $I$ by $(a_i,a_{i+1})$ for some $i$ if necessary, we may assume that $f$ is strictly monotone.

If $f$ is constant, then $f(x) = a$ for all $x \in I$, so $\lim_{x \to c^+} f(x) = a$ as well. Hence $\{a\} = \psi(M)$, where $\psi(y)$ is the $\la(A)$-formula $$\forall y_1 y_2 x_1 \left(y_1 \lt y \lt y_2 \wedge c \lt x_1 \to \exists x (c \lt x \lt x_1 \wedge y_1 \lt f(x) \lt y_2)\right),$$ that is, $a \in \dcl(A)$, a contradiction. Therefore, $f$ must be injective; let $g:J \into M$ be the compositional inverse of $f$. Then $g$ is definable over $A$, $b \in J$ and $a = g(b)$. $\qed$

It follows that every set $A \subseteq M$ has a well-defined dimension, denoted here by $\pdim A$. More generally, for $A,B \subseteq M$, the set $B$ has a well defined **dimension over $A$**, denoted here by $\pdim(B/A)$. It is well known that in this situation, we have $$\pdim(AB) = \pdim A + \pdim(B/A).$$ For $a = (a_1, \dots, a_k) \in M^k$, we set $$\dim(a/A) := \pdim(\{a_1, \dots, a_k\}/A) \in \{0, \dots, k\}.$$

This dimension is not very useful, as long as we do not know whether it is defined in elementary extensions of $\M$, as the following example shows:

**Example**

Let $R$ be the set of all real algebraic numbers, together with the usual ordering, addition and multiplication. Then $\RR_{\textrm{alg}}:= (R,\lt,+,\cdot)$ is a real closed field, hence is o-minimal. However, for any $a \in R^n$, we have $\dim a = 0$; and this remains so for any o-minimal expansion $\R$ of $\RR_{\textrm{alg}}$ (for which we do not yet know whether any elementary extension is o-minimal).