LemmaStatement: Literature-sourcedProof: AI-generatedPipeline-generatedjudge pass (gpt-6-luna)verified 2026-10-22 (gpt-6.1-sol)

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A smooth knot exterior is irreducible

Statement

Assume countable choice. Let $K \subset S^{3}$ be a smooth knot with a chosen closed tubular neighborhood $N (K)$ , and let $E_{K} = S^{3} ∖ int N (K)$ . Every smoothly embedded two-sphere in the interior of $E_{K}$ bounds a smoothly embedded closed three-ball contained in $E_{K}$ .

Statement references

Chosen compact knot exterior

Countable choice

Smooth Schoenflies in the three-sphere

Hatcher, Notes on Basic 3-Manifold Topology, Chapter 1, Theorem 1.1

Facts & Assumptions

Given: Countable choice, a smooth knot $K \subset S^{3}$ , its chosen closed tubular neighborhood $N (K)$ , and a smooth embedded sphere $S \subset int E_{K}$ .

[F1]

Countable choice is the axiom that every countable family of nonempty sets has a choice function (The Axiom of Countable Choice ( ${A C}_{ω}$ )).

[F2]

Under countable choice, $S$ separates $S^{3}$ into two components whose closures are smooth three-balls with common boundary $S$ (Smooth Schoenflies for an embedded two-sphere in the three-sphere, Statement).

[F3]

$E_{K} = S^{3} ∖ int N (K)$ , where $N (K)$ is connected and contains $K$ (Knot exterior, Definition).

Proof

1.1F1F2F3givenalgebra

By [F2], $S^{3} ∖ S$ has two components. The connected set $K$ is disjoint from $S$ , so it lies in one component. The connected neighborhood $N (K)$ is also disjoint from $S$ and contains $K$ , hence it lies in that same component.

2.1F2F3step 1.1∎

The closure of the other component is a smooth three-ball disjoint from $N (K)$ , so it is contained in $E_{K}$ and has boundary $S$ . This proves irreducibility of the chosen exterior.

Depends on

Used by

Dependency tree · two levels

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Sources

Allen Hatcher, Notes on Basic 3-Manifold Topology, Chapter 1, Theorem 1.1 (standard reference, not scraped)