LemmaStatement: AI-adaptedProof: AI-generatedPipeline-generatedprecheck passjudge pass (gpt-6-luna)verified 2026-11-04 (gpt-6.1-sol)

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Transverse complementary submanifolds meet in finitely many points

Statement

Let $X$ be a Hausdorff smooth $n$ -manifold and let $C, D \subseteq X$ be compact embedded smooth submanifolds without boundary, of dimensions $p$ and $n - p$ . Assume they are transverse, meaning $T_{x} C + T_{x} D = T_{x} X$ for every $x \in C \cap D$ . Then $C \cap D$ is finite.

Statement references

Smooth manifolds and their smooth charts

Embedded smooth submanifolds with boundary

Boundary submanifolds of a boundaryless manifold have half-slice charts

Facts & Assumptions

Given: The Hausdorff smooth manifold $X$ and the compact embedded submanifolds in the statement.

[F1]

Embedded smooth submanifolds with boundary only requires the inclusion to be a smooth embedding. The published slice-chart theorem Boundary submanifolds of a boundaryless manifold have half-slice charts supplies the missing local form: because $C$ has no boundary, every $x \in C$ is an interior point of its manifold-with-boundary structure, so near $x$ it is an ordinary coordinate slice.

[F2]

The smooth inverse function theorem on manifolds states that a smooth map with invertible differential at a point restricts to a diffeomorphism between neighborhoods of that point and its image.

[F3]

In a Hausdorff space a point and a disjoint compact set, and two disjoint compact sets, have disjoint open neighbourhoods; hence every compact subset is closed, and in a compact Hausdorff space the compact subsets are exactly the closed ones states that compact subsets of a Hausdorff space are closed.

Proof

Given: Put $q = n - p$ . At every intersection point, the hypothesis is $T_{x} C + T_{x} D = T_{x} X$ .

1.1F3given

Since $D$ is compact and $X$ is Hausdorff, [F3] makes $D$ closed in $X$ . Thus $C \cap D$ is closed in the compact space $C$ , hence compact.

1.2F1F2givenalgebra

Fix $x \in C \cap D$ . By [F1], choose coordinates $(u^{1}, \dots, u^{p}, v^{1}, \dots, v^{q})$ near $x$ with $C$ given by $v = 0$ . The coordinate projection $v$ restricted to $D$ has derivative with kernel $T_{x} D \cap T_{x} C = {0}$ , because the two tangent dimensions sum to $n$ and their sum is $T_{x} X$ . Its domain and target both have dimension $q$ , so this derivative is an isomorphism. By [F2], after shrinking a neighborhood of $x$ in $D$ , the map $v ∣_{D}$ is one-to-one. The only point of that neighborhood lying in $C$ is therefore $x$ ; hence $x$ is isolated in $C \cap D$ .

2.1step 1.1step 1.2given∎

Every point of $C \cap D$ is isolated by step 1.2, so its singleton neighborhoods form an open cover of the compact subspace $C \cap D$ . Compactness gives a finite subcover, which means $C \cap D$ is finite. If $C \cap D = \emptyset$ , the conclusion holds directly.

Transverse complementary submanifolds meet in finitely many points

Statement

Statement references

Facts & Assumptions

Proof

Statement references

Depends on

Dependency tree · 2 levels

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