A definable function is constant or strictly monotone on a subinterval

Let $M$ be an o-minimal structure over a dense linear order without endpoints $D$, let $f : I \to B$ be a definable function of one variable, and let $u < v$ be such that the open interval $(u,v)$ is contained in the domain $I$.

Claim. There are $u_1 < v_1$ with $(u_1,v_1) \subseteq (u,v)$ such that $f$ is constant on $(u_1,v_1)$, or strictly increasing on $(u_1,v_1)$, or strictly decreasing on $(u_1,v_1)$:

This is the local core of the Monotonicity Theorem. The usual argument has two halves. First, either some fibre $f^{-1}(c)$ has nonempty interior inside $(u,v)$ — and then $f$ is constant on a subinterval — or every fibre is finite; in the latter case the image of $(u,v)$ is infinite, hence contains an interval, and taking the least preimage of each value yields a definable injection, so $f$ is injective on a subinterval. Second, an injective definable function on an interval is strictly monotone on some subinterval: for each $x$ the definable sets $\{y > x : f(y) > f(x)\}$ and $\{y > x : f(y) < f(x)\}$ partition a one-sided neighbourhood, so by o-minimality $f$ is locally strictly increasing or locally strictly decreasing to the right of every point, and one of the two behaviours persists on a whole subinterval.

Note that no hypothesis of infinitude of $I$ is needed beyond $(u,v) \subseteq I$, which already forces $I$ to be infinite, and that the conclusion is exactly what is required to produce a good window around every point of $(u_1,v_1)$.