Alphabeta Math
False statementConstruction: AI-adaptedVerification: AI-generatedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-27
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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.

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FALSE: every nowhere dense subset of R has measure zero

Statement

False claim: every nowhere dense subset of R (Nowhere dense, meager (first category), residual, and second category subsets of R) has measure zero (Measure zero (a countable cover by intervals of total length below every ε) and content zero (a finite such cover)).

The claim is tempting because a nowhere dense set is topologically thin: its closure contains no interval at all, so it is "full of holes" everywhere. The error is to read that as a statement about total length. Holes may be plentiful and short at the same time, and the Smith-Volterra-Cantor set is built precisely so that they are.

Facts & Assumptions

[A1]

The false claim: every nowhere dense subset of R has measure zero.

[L2]

If sequences (ak), (bk) with ak≤bk cover S and all their partial total lengths are at most M, then M≥2−1; in particular S does not have measure zero (The Smith-Volterra-Cantor set is compact, perfect and nowhere dense, and does not have measure zero, claim 4).

[L3]

A set is null when for every real ε>0 it has a cover by a sequence of closed intervals with all partial total lengths at most ε (Measure zero (a countable cover by intervals of total length below every ε) and content zero (a finite such cover)).

Refutation

technique · direct
1.1

The set S is a subset of R and is nowhere dense, by [L1].

L1
1.2

S does not have measure zero: a cover witnessing nullity at ε:=4−1 would have all partial total lengths at most 4−1, and [L2] then forces 4−1≥2−1, which is false.

L2L3
2.1

So S is a nowhere dense subset of R that does not have measure zero, and the claim [A1] fails at S; the claim is therefore false.

step 1.1step 1.2A1∎

Remarks

Depends on

Used by

Dependency tree · two levels

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Sources