Alphabeta MathHow 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.
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- 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.
The Smith–Volterra–Cantor slab is compact and not Jordan measurable
Statement refuted
Every compact bounded subset of is Jordan measurable.
Facts & Assumptions
Given: The Smith-Volterra-Cantor set and .
is compact, nowhere dense, and every finite interval cover has total length at least (The Smith-Volterra-Cantor set: the same construction removing, at stage , an open middle interval of length from each of the remaining intervals, The Smith-Volterra-Cantor set is compact, perfect and nowhere dense, and does not have measure zero).
The product lower bound is If every finite interval cover of has total length at least , then every rectangle cover of has total area at least .
Counterexample
The slab is closed and bounded, hence compact by Heine-Borel in : with the Euclidean metric a subset of is compact if and only if it is closed and bounded, and the proof by bisection uses no choice principle; the same holds on the real line. Since has empty interior, has empty interior and, being closed, equals its boundary.
By [L1] and [L2], every rectangle cover of has total area at least ; its boundary therefore does not have content zero.
The boundary criterion A bounded set in is Jordan measurable iff its boundary is null, equivalently of content zero makes non-Jordan-measurable.
Depends on
- The Smith-Volterra-Cantor set: the same construction removing, at stage $n \ge 1$, an open middle interval of length $4^{-n}$ from each of the $2^{n-1}$ remaining intervals
- The Smith-Volterra-Cantor set is compact, perfect and nowhere dense, and does not have measure zero
- If every finite interval cover of $A\subseteq\mathbb{R}$ has total length at least $c$, then every rectangle cover of $A\times[0,d]$ has total area at least $cd$
- A bounded set in $\mathbb{R}^m$ is Jordan measurable iff its boundary is null, equivalently of content zero
- Heine-Borel in $\mathbb{R}^n$: with the Euclidean metric a subset of $\mathbb{R}^n$ is compact if and only if it is closed and bounded, and the proof by bisection uses no choice principle; the same holds on the real line
- Interior, closure, boundary, limit point, isolated point and dense subset of a metric space
- Jordan inner and outer content and Jordan measurable bounded sets in $\mathbb{R}^m$
Used by
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Sources
- J. Lebl, Basic Analysis, Jordan Measurable Sets (standard reference, not scraped)
- J. Lebl, Basic Analysis, Outer Measure and Null Sets (standard reference, not scraped)
- Whitman College real analysis notes (standard reference, not scraped)