LemmaStatement: AI-adaptedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-11

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The parity of transverse ray crossings with a polygon is locally constant on its complement

Statement

For a polygon $P$ (Polygonal arcs and polygons as non-self-intersecting finite unions of line segments in $\mathbb R^2$ ) and $x\notin P$ , count the intersections modulo two of any general-position ray supplied by Every point off a polygon admits a ray meeting it transversely in finitely many nonvertex points. This parity is independent of the chosen general-position ray and is constant throughout some open neighbourhood of $x$ in the complement. Consequently it is constant on every region of $\mathbb R^2\setminus P$ (Regions of the complement of a planar set and their frontiers). The elementary alternation at successive transverse crossings follows the convention of The even and odd index maps and the alternating sequence: strictly increasing $e, o$ with $\mathbb{N}$ their disjoint union, and the unique $(s_k)$ with $s_0 = 1$ , $s_{\sigma(k)} = -s_k$ , which satisfies $|s_k| = 1$ , $s \circ e \equiv 1$ and $s \circ o \equiv -1$ .

Facts & Assumptions

Given: A polygon $P$ and a point $x\notin P$ .

[L1]

Every point off a polygon admits a ray meeting it transversely in finitely many nonvertex points (Every point off a polygon admits a ray meeting it transversely in finitely many nonvertex points).

Proof

technique · direct

1.1

Fix a general-position ray from $x$ . The finite intersection points have positive distance from every polygon vertex and from every nonincident edge; transversality also supplies a positive angle at each crossing. Taking the minimum of finitely many positive tolerances gives a ball about $x$ in which parallel translated rays retain exactly these crossings.

1.2

Rotate one general-position ray continuously to another, avoiding the finitely many exceptional directions except at isolated parameters. Crossing an edge tangentially creates or destroys two intersections, while passing a polygon vertex transfers the intersection from one incident edge to the other or changes the count by two. Thus the count modulo two never changes, so parity is independent of the chosen ray.

2.1

Steps 1.1 and 1.2 make parity locally constant on the complement. A locally constant map to the discrete set $\{0,1\}$ is constant on each connected component, hence on each complementary region.

step 1.1step 1.2∎

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 93 results over 20 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources

R. Diestel, Graph Theory, 6th ed., Chapter 4, Section 4.1 (standard reference, not scraped)