LemmaStatement: AI-adaptedProof: AI-generatedprecheck passaudited 2026-08-11

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

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

Statement

Let $P$ be a polygon (Polygonal arcs and polygons as non-self-intersecting finite unions of line segments in $R^{2}$ ) and $x \in R^{2} ∖ P$ , whose complementary regions use Regions of the complement of a planar set and their frontiers. There is a polygonal ray from $x$ that meets $P$ in finitely many points, none a vertex of $P$ , and crosses the containing edge transversely at every intersection. The finite edge list and its unions use The sum rule: a finite disjoint union is finite with $∣ A \cup B ∣ = ∣ A ∣ + ∣ B ∣$ and $∣ ⋃_{i \in I} A_{i} ∣ = \sum_{i \in I} ∣ A_{i} ∣$ , and a sum over a finite index set splits along a partition and the polygonal-path convention of Polygonal paths and polygonally connected subsets of $R^{n}$ .

Facts & Assumptions

Given: A polygon $P$ with its finite edge and vertex sets, and $x \notin P$ .

[L1]

In an Archimedean ordered field $F$ , for any $x < y$ there is a rational $q$ whose canonical image lies strictly between them (ℚ is dense in every Archimedean ordered field).

[L2]

Every complete ordered field, in particular $R$ , is Archimedean (Every complete ordered field is Archimedean).

[F1]

A polygonal path is specified by a finite list of vertices $v_{0}, \dots, v_{m} \in A$ (Polygonal paths and polygonally connected subsets of $R^{n}$ ).

Proof

technique · constructive

1.1

A ray direction is bad if its line through $x$ contains a polygon vertex or is parallel to an edge line. There are only finitely many such directions. By [L2], [L1] applies in $R$ and supplies a rational-slope direction in an open angular interval avoiding them.

L1L2F1construct

2.1

In the chosen direction the ray misses every vertex and is not parallel to any edge. It therefore meets each closed edge segment in at most one point, and every such intersection is transverse. Since the polygon has finitely many edges, the total intersection set is finite and has the required properties.

step 1.1F1discharge-construct∎

Depends on

Used by

Dependency tree · two levels

41 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

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