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

We call ${\cal M}$ **o-minimal** if every definable subset of $M$ is a finite union of points and intervals.

**Examples** (without details)

- By quantifier elimination, every dense linear order without endpoints is o-minimal.
- Let ${\cal V} = (V,\lt,+,(\lambda_k)_{k \in K})$ be an ordered vector space over an ordered field $K$. By quantifier elimination, ${\cal V}$ is o-minimal.
- By Tarski’s Theorem, every real closed field is o-minimal.
- By Wilkie’s Theorem, the real exponential field, that is, the expansion of the real field by the exponential function, is o-minimal.

We will discuss examples of o-minimal structures (and the proofs of their o-minimality) later.

**Exercise**

Assume that ${\cal M}$ is o-minimal.

- Prove that every infinite definable subset of $M$ contains an interval.
- Prove that a definable subset of $M$ is finite if and only if it is discrete.
- Let $A \subseteq M^{n+1}$ be definable. Prove that the set $\{x \in M^n:\ A_x \text{ is finite}\}$ is definable.
- Prove that ${\cal M}$ is definably complete.

### Lemma

* Let $S \subseteq M$ be definable and $a \in M$. Then there exists $\epsilon \gt a$ in $M$ such that either $(a,\epsilon) \subseteq S$ or $(a,\epsilon) \subseteq M \setminus S$.*

**Proof.** If $a$ is not in the boundary ${\rm bd} (S)$ of $S$, then either $a$ is in the interior of $S$ or $a$ is in the interior of $M \setminus S$; the lemma follows in both of these cases.

So we assume that $a \in {\rm bd}(S)$. By o-minimality, ${\rm bd}(S)$ is finite, so we are in one of the following cases:

- if $a$ is an isolated point of $S$ or the right endpoint of an interval contained in $S$, then $(a,\epsilon) \subseteq M \setminus S$ for some $\epsilon \gt 0$ in $M$;
- if $a$ is the left endpoint of some interval contained in $S$, then $(a,\epsilon)

\subseteq M$ for some $\epsilon \gt 0$ in $M$.