Alphabeta Math
LemmaStatement: Literature-sourcedProof: Literature-sourcedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-06
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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 precompact end-limit point of a negative-gradient trajectory is critical

Statement

Every point of a nonempty precompact α- or ω-limit set of a full negative-gradient trajectory is a critical point of f.

Facts & Assumptions

Given: A full negative-gradient trajectory γ with a precompact positive or negative tail.

[F1]
[F2]

Along γ, (fγ)=gradgfg2 (A negative-gradient trajectory satisfies the energy identity).

[F3]

A noncritical point has nonzero gradient (The Riemannian gradient vanishes exactly at the critical points).

Proof

technique · direct
1.1

On a precompact positive tail, fγ is decreasing by [F2] and bounded below because f is continuous on its compact closure. It therefore has a finite limit ; every point of ω(γ) is a limit of tail values and has f=. The same argument, with increasing time reversed, applies to α(γ).

F2F1given
1.2

Let z lie in either limit set. If z were noncritical, [F3] and [F2] would give a sufficiently short positive orbit segment from z on which f strictly decreases.

F2F3assume-contra
2.1

By [F1] the entire short segment in step 1.2 remains in the same limit set, whereas step 1.1 makes f constant there. This contradiction proves that z is critical.

F1step 1.1step 1.2discharge-contradiction

Depends on

Used by

Dependency tree · two levels

9 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