Alphabeta Math
LemmaStatement: AI-adaptedProof: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-12
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.

Tail suprema land in the scale subspace

Statement

Assume AC. Let 1mk<ω and let (hξ)ξ<m be points of the scale subspace X such that hξ(n)<hη(n) for every ξ<η<m and nB with n>k. Put g(n)=supξ<mhξ(n) on B. There is hX with h(n)=g(n) at every n>k in B, hence h=g. On that tail g(n)<n and cf(g(n))=m. No condition on the cofinalities of the finitely many initial values of g is needed.

Facts & Assumptions

Given: The normalized scale (fα)α<λ defining X, with λ=ω+1, and the sequence in the statement.

[F1]

Points of X are Rudin points eventually equal to scale terms; the scale is strictly increasing in <, and admissible finite modifications preserve X (Kojman-Shelah scale subspace).

[F2]

For strictly increasing scale indices indexed by m and pointwise strictly increasing representatives in the actual product on n>km, their tail supremum is below each factor, has coordinate cofinality m, and is eventually equal to fδ for the index supremum δ<λ of cofinality m (Tail suprema and normalized scales).

[A1]

AC is assumed for the scale and cardinal-regularity suppliers (The Axiom of Choice).

Proof

1.1

Each hξ has exactly one scale index αξ with hξ=fαξ. Existence is F1. If two distinct indices worked, put the smaller first; scale strictness outside a finite set would give hξ(n)<hξ(n) at all but finitely many coordinates. The infinite B has a coordinate outside their finite union, a contradiction. For ξ<η, the supplied strict inequality on the cofinite tail excludes αηαξ: equality would imply eventual equality of the two points, while a strict reverse index inequality would imply hη<hξ. Either conflicts with the strict tail inequalities outside finitely many coordinates. Thus (αξ) is strictly increasing.

F1
2.1

Set T={nB:n>k}. For every ξ<m there is a successor ξ+1<m, since the infinite cardinal m is a limit ordinal (also its cofinality exceeds one by F3–F4). For nT, hξ(n)<hξ+1(n)n, so hξ(n)<n. Hence the restrictions hξT are members of the actual strict product, not just the inclusive-top product. They are pointwise increasing by hypothesis and eventually equal to fαξT by step 1.1. Thus all hypotheses of F2, including 1mk, hold. F2 and A1 give δ=supξαξ<λ, cf(δ)=m, and gT=fδT with g(n)<n and cf(g(n))=m on T.

step 1.1F2F3F4A1
3.1

Define h(n)=n for nB with nk, and h(n)=g(n) for n>k. On the finite prefix, F3 gives uncountable cofinalities nk; on the tail, step 2.1 gives cofinality mk. Thus every coordinate cofinality lies strictly between ω and k+1, and hXR(B). Step 2.1 and the finite prefix imply h=fδ, so F1 gives hX. It agrees with g on every required tail coordinate and differs, if at all, only on the finite prefix. QED.

step 2.1F1F3

Depends on

Used by

Dependency tree · two levels

33 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