CounterexampleConstruction: AI-adaptedVerification: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-07

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Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
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A dimension-one set can have zero length

Statement refuted

Assume the Axiom of Countable Choice. The implication “a compact subset of $R$ of Hausdorff dimension one has positive length” is false. Let $S = N_{+} ∖ {k^{2} : k \geq 1}$ . Then $A_{S}$ is compact, $\dim_{H} A_{S} = 1$ , and $H^{1} (A_{S}) = λ_{1} (A_{S}) = 0$ .

Facts & Assumptions

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

[F1]

Under the standing Countable Choice hypothesis, for every position set $S$ , $A_{S}$ is compact with dimension $lim inf a_{S} (n) / n$ ; an infinite complement implies both Lebesgue and Hausdorff one-measure zero. Digit-position density determines Hausdorff dimension

Counterexample

1.1

For the nonsquare positions, $a_{S} (n) = n - ⌊ \sqrt{n} ⌋$ . Therefore $a_{S} (n) / n \to 1$ and the dimension formula gives $\dim_{H} A_{S} = 1$ , with compactness supplied by the same theorem.

2.1

The forbidden positions include every positive square and are infinite. Thus $λ_{1} (A_{S}) = H^{1} (A_{S}) = 0$ . The set contains zero and is a nonempty witness refuting the implication.

F1step 1.1∎

Depends on

Digit-position density determines Hausdorff dimension

Used by

Critical Hausdorff measure is always finite and positive False statement

Dependency tree · two levels

10 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 Example 1.4.2, nonsquare-position specialisation (standard reference, not scraped)