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

Let $f:I \longrightarrow M$ be definable, with $I = (a,b)$ an interval in $M$.

We start by translating the condition “$f$ is strictly increasing” into a condition involving two subsets of $I^2$: on the one hand, we have the triangle

$\displaystyle \Delta(I):= \{(x,y) \in I^2:\ x \lt y\},$

and on the other hand, we have the set

$\displaystyle X_\lt(f):= \{(x,y) \in I^2:\ f(x) \lt f(y)\}$.

**Exercise**

Show that $f$ is strictly increasing if and only if $\Delta(I) \subseteq X_\lt(f)$.

Similarly, we define

$\displaystyle X_\gt(f):= \{(x,y) \in I^2:\ f(x) \gt f(y)\}$

and

$\displaystyle X_=(f):= \{(x,y) \in I^2:\ f(x) = f(y)\}$,

and we obtain that

- $f$ is strictly decreasing if and only if $\Delta(I) \subseteq X_\gt(f)$;
- $f$ is constant if and only if $\Delta(I) \subseteq X_=(f)$.

Therefore, the Monotonicity Lemma is equivalent to the

**Claim**

*There exist an open interval $J \subseteq I$ and $\ast \in \{\lt,=,\gt\}$ such that $\Delta(J) \subseteq X_\ast(f)$.*

This claim is a special case of what Peterzil and Starchenko call an “ordered Ramsey” theorem, which we discuss next.

Here is an attempt at a proof to the exercise:

Suppose $f$ is strictly increasing, and let $(x,y)\in\Delta(I)$. Then $x<y$, such that since $f$ is strictly increasing, $f(x)<f(y)$, and we have $(x,y)\in X_<(f)$ as desired. To show the converse, suppose $\Delta(I)\subseteq X_<(f)$, and let $x<y$. Then $(x,y)\in\Delta(I)\subseteq X_<(f)$, such that $(x,y)\in X_<(f)$, and so $f(x)<f(y)$ whenever $x<y$ as desired.

This was fairly straightforward. From the next post we can easily see the claim at the end is proven via the ordered Ramsey theorem. I think the point is to let the $S_1,S_2,…,S_k\subseteq M^2$ be the sets $X_(f)$ and $X_=(f)$, since at least one of these is true in a dense linear order. These sets are also clearly definable. It follows by the ordered Ramsey theorem that there exists $J\subseteq I$ with $\bigvee\limits_{*\in\{,=\}}J\subseteq X_*(f)$, as desired thereby proving the monotonicity lemma.

Should say $X_(f)$ in the second paragraph with $*\in\{,=\}$

Correct.