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LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck pass
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  • AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
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A Reeb component obstructs tautness

Statement

Assume ACω. A cooriented codimension-one foliation of a closed oriented three-manifold containing a Reeb component is not taut. Thus every taut foliation is Reebless.

Facts & Assumptions

Given: The foliation and Reeb component R of the statement.

[F1]

A Reeb component is a compact saturated solid-torus region with its connected boundary torus as a leaf (Reeb components of a codimension-one foliation, The Reeb foliation of the solid torus has the boundary as a leaf).

[F2]

Tautness requires an embedded closed transversal through every leaf (Taut codimension-one foliations).

Proof

1.1F1given

Along the connected boundary torus the positive transverse direction is everywhere inward or everywhere outward: it is continuous, transverse to that leaf, and cannot change the sign of its boundary normal component. Reversing the direction chosen on a hypothetical transversal through that torus, if necessary, makes all its boundary crossings inward.

2.1F1F2step 1.1∎

Each such crossing is isolated, and the compact parameter circle has only finitely many crossings. In a boundary defining coordinate every crossing goes from outside R to inside R. A periodic curve with an entry must also have an exit; all crossings inward makes that impossible. Thus no closed transversal meets the boundary leaf, contradicting F2's condition for tautness. No accessible-set characterization or monotonicity across infinitely many interior leaves is needed.

Depends on

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Dependency tree · two levels

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Sources