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.

Cofinality and size of the scale subspace

Statement

Assume AC. For every bnBn there is hX with b<h pointwise. Moreover X=ω+1. No assertion that every normalized-scale class meets the Rudin space is required.

Facts & Assumptions

Given: The normalized scale defining X. Put μ=ω, λ=ω+1, and Q=nBn.

[F1]

The normalized scale (fα)α<λ is strictly increasing and cofinal in the eventual order on Q, and X consists of its eventual-equality classes intersected with XR(B) (Kojman-Shelah scale subspace).

[F2]

A strictly pointwise increasing ω1-sequence of product representatives of strictly increasing scale indices has, on n>1, a supremum below each factor of cofinality ω1, eventually equal to fδ for δ<λ (Tail suprema and normalized scales, m=k=1).

[F7]

Specified rules recurse on ordinals (Transfinite recursion).

[A1]

AC is assumed for cardinal regularity and for cardinal bounds on unions of the finite-modification classes (The Axiom of Choice).

Proof

1.1

Fix bQ. Recursively for ξ<ω1 construct indices αξ<λ and functions gξQ. Given the earlier choices, put tξ(n)=sup({b(n)+1}{gη(n)+1:η<ξ}). All entries in this supremum are below n, since that cardinal is a limit and the earlier functions are in Q. The set is countable, so F3–F4 give tξ(n)<n and tξQ. The countably many earlier indices are bounded in regular λ by F3–F4. By F1 there is an index above them all with tξ<fα: first obtain eventual domination by cofinality, and if necessary pass to a larger scale index, whose strict eventual increase preserves domination. Choose the least such index αξ and set gξ(n)=max(tξ(n),fαξ(n)). It belongs to Q, is eventually equal to fαξ, and is strictly above b and every earlier gη at every coordinate. The exceptional modification set is contained in the finite set where tξ(n)fαξ(n). These are specified rules with proved witnesses, so F7 constructs them throughout ω1. The initial stage uses only b(n)+1 and has no earlier-index obligation.

F1F3F4F7A1
1.2

Fix α<λ. Every point eventually equal to fα is determined by a finite exceptional set SB and its values in nS(n+1). There are countably many finite S, using their binary codes nS2n. For nonempty S let r=maxS; each inclusive ordinal factor has cardinality nr, since its extra top can be sent to zero, the finite ordinals shifted by one and the other values fixed. F6 bounds the finite product by r<μ; for empty S the product has one member. Thus the countable union of these possibilities has size at most μ by A1 and F6. Every class intersected with XR(B) has size at most μ, including empty classes. There are λ possible indices, so again A1 and F6 give Xλμ=λ.

F1F5F6A1
2.1

Apply F2 to (αξ,gξ)ξ<ω1. The indices strictly increase, the functions are actual members of Q, and they strictly increase pointwise by step 1.1. Since Bω{0,1}, the tail n>1 is all of B. F2 gives h(n)=supξ<ω1gξ(n)<n, cf(h(n))=ω1 at every coordinate, and h=fδ for some δ<λ. Thus hXR(B) with uniform strict bound 2, and F1 gives hX. Since g0>b and hg0, also h>b. This proves pointwise cofinality and in particular nonemptiness.

step 1.1F1F2
3.1

Every xX has a unique scale index α(x). Existence is F1; two different indices would force the same function to be eventually strictly less than itself outside finitely many coordinates, impossible on infinite B. Suppose X<λ. By F3–F4 choose β<λ strictly above all its indices; the empty case is already excluded by step 2.1. Then x<fβ for every xX, by F1 and each eventual equality. But step 2.1 applied to b=fβ gives hX with fβ<h pointwise, contradicting h<fβ. Hence Xλ by F5 and A1.

step 2.1F1F3F4F5A1
4.1

The lower bound of step 3.1 and upper bound of step 1.2 give X=λ=ω+1. Step 2.1 establishes the asserted strict pointwise cofinality. The argument only used nonempty classes when a constructed point or an already given point provided a member. QED.

step 2.1step 3.1step 1.2

Depends on

Used by

Dependency tree · two levels

39 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