Lemma 3, monotone implies continuous somewhere
ProvedMonotonicity_Theorem.lemma_three2A strictly monotone definable function is continuous on a subinterval.
import Definitions.Def_Monotonicity_Theorem_Framework2
theorem Monotonicity_Theorem.lemma_three2 {R : Type} [LinearOrder R] [DenselyOrdered R] [NoMaxOrder R] [NoMinOrder R] (M : OMinimalStructure R)
{I B : Set (Power R 1)} (f : DefinableFunction M I B)
(hmono : StrictlyMonotoneOn I B f.toFun) (hI : Set.Infinite I) :
∃ a b : Endpoint R, Endpoint.lt a b ∧
(openInterval a b).Subset I ∧
(openInterval a b).Subset (ContinuousPoints I
(FunctionGraph (R := R) (m := 1) (n := 1) (A := I) (B := B) f.toFun)) := by sorryRead-back
What the Lean code literally says, in plain math · muse-spark
For any type equipped with a linear order that is densely ordered and has no maximum and no minimum, and any o-minimal structure over , and any sets and of unary tuples over , and any -definable function from to , if the underlying ordinary function of is strictly monotone on with values in and if is infinite as a set, so in particular is not empty or finite, then there exist endpoints and over , allowing infinite ends, such that is strictly less than in the endpoint order, hence excluding degenerate empty or reversed endpoint configurations, and the open interval determined by and is contained as a subset in , and that same open interval is also contained as a subset in the set of continuity points in of the function-graph, viewed as a -to--ary graph with domain and codomain , of the underlying function of , where both containments are required jointly and would hold vacuously only if the interval were empty.
Confirmed by the mission captain (proposal self-audit).