Alphabeta Math
CorollaryStatement: AI-adaptedProof: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-07
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.

Lipschitz monotonicity and bi-Lipschitz invariance of dimension

Statement

Assume the Axiom of Countable Choice. A Lipschitz map f:D⊆X→Y satisfies dim⁡Hf(A)≤dim⁡HA for every A⊆D. If f is a bijection from D onto f(D) and there exist 0<a≤b<∞ with

a dX(x,y)≤dY(f(x),f(y))≤b dX(x,y)(x,y∈D),

then dim⁡Hf(A)=dim⁡HA (bi-Lipschitz invariance).

Facts & Assumptions

Given: The objects, conventions, and hypotheses in the statement above.

[F1]

Under the standing Countable Choice hypothesis, positive Lipschitz constants transfer each zero Hausdorff measure to zero; a zero-Lipschitz image is empty or a singleton. Lipschitz maps control Hausdorff measure

[F2]

Every exponent strictly above finite dimension has zero Hausdorff measure. Hausdorff dimension is the unique critical exponent

Proof

1.1

For a positive Lipschitz constant and finite d=dim⁡HA, every t>d has Ht(f(A))=0, so dim⁡Hf(A)≤d. For d=∞ the inequality is automatic. A constant map has empty or singleton image, with dimension zero from its singleton covers, so its case also holds.

F1F2
2.1

The two-sided bounds make f Lipschitz with constant b and its inverse on f(D) Lipschitz with constant 1/a. Apply the first step in both directions. Empty sets and the value zero are allowed throughout.

step 1.1given∎

Depends on

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