On a subinterval the germ class of an injective definable function is constant

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 $u<v$ satisfy $(u,v) \subseteq I$. Assume $f$ is injective on $(u,v)$.

Claim. There are $u_1 < v_1$ with $(u_1,v_1) \subseteq (u,v)$ on which one of the four germ patterns of $f$ holds uniformly: below-above, above-below, above-above, or below-below.

This is the classification step in the analysis of an injective definable function of one variable. It is a pure consequence of o-minimality, and it does not yet decide which of the four classes occurs.

The argument has three parts. First, the germ dichotomy: a definable subset $A$ of the line either contains or misses an interval with endpoint $x$, on each side of $x$, because $A$ is a finite union of points and intervals. Second, for each $x$ the two definable sets

cover the punctured neighbourhoods of $x$ inside $(u,v)$, by injectivity; so the germ dichotomy assigns to each side of $x$ one of the two comparisons, and $x$ falls into exactly one of the four classes. Third, the four classes are definable subsets of the line, since each is defined by a first-order condition on $x$ built from the graph of $f$ and the order; their union is the infinite set $(u,v)$, so one of them is infinite and therefore contains an interval, which may be shrunk to lie inside $(u,v)$.

Formalization note. The four patterns are the predicates UniformBelowAbove, UniformAboveBelow, UniformAboveAbove and UniformBelowBelow of the accompanying definitions; each requires the witnessing window to lie inside the interval on which the pattern is asserted.