Alphabeta Math
CounterexampleConstruction: AI-adaptedVerification: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-07
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An uncountable compact set can have dimension zero

Statement refuted

Assume the Axiom of Countable Choice. The implication “Hausdorff dimension zero forces countability” is false. For S={k2:k1}, AS is compact and uncountable, yet dimHAS=0 and Ht(AS)=0 for every finite t>0.

Facts & Assumptions

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

[F1]

Under the standing Countable Choice hypothesis, the binary digit set is compact with dimension equal to the lower density of allowed positions; if the positions and their complement are infinite it is uncountable. Digit-position density determines Hausdorff dimension

[F2]

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

Counterexample

1.1

Here aS(n)=n, so aS(n)/n0. The digit theorem gives compactness and dimension zero. Both the square positions and the nonsquare positions are infinite, so its uncountability conclusion applies.

F1
2.1

Every t>0 lies strictly above this dimension. Hence Ht(AS)=0 for every such finite exponent. The conclusion concerns positive exponents only; at exponent zero the uncountable set is not null.

F2step 1.1

Depends on

Used by

Dependency tree · two levels

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Sources