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.

Euclidean space and positive-volume sets have their Euclidean dimension

Statement

Assume the Axiom of Countable Choice. For each integer $n \geq 1$ , $\dim_{H} R^{n} = n$ and every subset of $R^{n}$ has dimension at most $n$ . Every $A \subseteq R^{n}$ with $λ_{n}^{*} (A) > 0$ has dimension $n$ .

Facts & Assumptions

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

[F1]

Under the standing Countable Choice hypothesis, $H^{n} (A) = c_{n} λ_{n}^{*} (A)$ for all subsets, with $0 < c_{n} < \infty$ . Euclidean Hausdorff measure is proportional to Lebesgue measure

[F2]

Under the standing Countable Choice hypothesis, dimension is monotone and countably stable. Hausdorff dimension is monotone and countably stable

[F3]

Finite positive measure at exponent $n$ gives dimension $n$ ; positive measure at $n$ rules out dimension below $n$ . Hausdorff dimension is the unique critical exponent

Proof

1.1

Every nondegenerate bounded cube has finite positive Lebesgue volume, hence finite positive $H^{n}$ and dimension $n$ . Countably many such cubes cover $R^{n}$ , so its dimension is $n$ by countable stability.

F1F2F3

2.1

Monotonicity bounds every subset above by $n$ , including the empty set. If its Lebesgue outer measure is positive, its $H^{n}$ is positive, possibly infinite, and the critical-exponent theorem bounds its dimension below by $n$ . Thus equality holds. The statements include $n = 1$ .

F1F2F3step 1.1∎

Depends on

Used by

A continuous image can raise Hausdorff dimension Counterexample
The rationals are dense but have dimension zero Example
Critical Hausdorff measure is always finite and positive False statement
Dimensions add under unions False statement
Dimension leaves the critical measure undetermined Remark

Dependency tree · two levels

14 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

Bishop–Peres §1.2 p.7 after Lemma 1.2.8; Falconer §1.2 p.8 (standard reference, not scraped)