Alphabeta Math
TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-14
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.

A separable reflexive Banach space without the approximation property

Statement

Assume AC. There exists a separable reflexive real Banach space B without the approximation property. Consequently B has no Schauder basis.

Facts & Assumptions

[A1]

The Axiom of Choice holds (The Axiom of Choice).

[L1]

The Walsh-block assembly produces a separable reflexive space B whose finite-rank operators satisfy Enflo's logarithmic lower bound (Enflo's Walsh-block assembly).

[L2]

That lower bound excludes every finite BAP constant (Enflo's quantitative localized-trace obstruction).

[L3]

Under AC, a reflexive space with AP has MAP (Reflexive approximation property implies metric approximation property).

[L4]

Under DC, every Schauder basis has a finite basis constant, and a space with such a basis has BAP (Coordinate functionals of a Schauder basis are bounded, A Schauder basis implies the bounded approximation property).

Proof

technique · contradiction through Grothendieck's tensor criterion

Given: AC.

1.1

Take the separable reflexive space B supplied by [L1].

A1L1
1.2

The logarithmic lower bound and [L2] show that B has no BAP.

L1L2
2.1

Rule out AP using the reflexive MAP theorem. If B had AP, its reflexivity and [L3] would give MAP, hence BAP, contradicting step 1.2. Therefore B has no AP.

L3step 1.2
3.1

Rule out a Schauder basis and close the boundary cases. If B had a Schauder basis, AC would supply DC and [L4] would give BAP, again contradicting step 1.2. Thus B has no Schauder basis. The zero-space case is irrelevant because the constructed space has a nonempty independent generator; real scalars are part of [L1].

A1L1L4step 1.2∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

29 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