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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A compact subset of need not be Jordan measurable
Statement refuted
Every compact subset of is Jordan measurable.
Facts & Assumptions
Given: The Smith--Volterra--Cantor set of The Smith-Volterra-Cantor set: the same construction removing, at stage , an open middle interval of length from each of the remaining intervals and .
No cover of by intervals has total length below (The Smith-Volterra-Cantor set is compact, perfect and nowhere dense, and does not have measure zero).
A metric-bounded set is Jordan measurable if and only if its boundary is null, equivalently has content zero (A bounded set in is Jordan measurable iff its boundary is null, equivalently of content zero).
The Smith--Volterra--Cantor set is closed, bounded, and nowhere dense (The Smith-Volterra-Cantor set is compact, perfect and nowhere dense, and does not have measure zero).
Counterexample
By [F3], the set is closed and bounded, hence compact, and has empty interior. Therefore has empty interior; being closed, it equals its boundary.
Consider any finite axis-parallel box cover of . Partition at all endpoints of the last two coordinate intervals of those boxes. For the midpoint of each nondegenerate planar cell, the first-coordinate intervals of the boxes active there cover , so [F1] makes their total length at least . Multiplying by the cell area and summing shows that the original boxes have total volume at least .
Thus does not have content zero. By [F2], the compact set is not Jordan measurable, refuting the claim.
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
- Grid partitions of a rectangle in $\mathbb{R}^m$, their cells, refinements and mesh
- Laws of finite sums and finite products
- 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
Used by
- FALSE: every compact subset of ℝ³ has Jordan volume False statement
Dependency tree · two levels
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Sources
- Sheldon Axler, Measure, Integration & Real Analysis (standard reference, not scraped)