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.
- 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.
These labels describe origin, not correctness: citations and verification chips remain separate evidence.
FALSE: every bounded plane set has Jordan area
Statement
Every bounded subset of is Jordan measurable and therefore has a Jordan area.
Facts & Assumptions
Given: The set .
A set is bounded if it is empty or is contained in some metric ball (Bounded subset, diameter, distance from a point to a set, and distance between two sets in a metric space).
The rationals and the irrationals are both dense in (Both and are dense in , and every nonempty open subset of is uncountable).
A boundary consists exactly of the points every ball about which meets both the set and its complement (Interior, closure, boundary, limit point, isolated point and dense subset of a metric space).
Every finite inner-packing sum is at most every finite outer-cover sum, and the Jordan contents are their supremum and infimum (Jordan inner and outer content and Jordan measurable bounded sets in ).
The unit square has rectangle volume (Axis-parallel rectangles in and their volume).
A metric-bounded set is Jordan measurable if and only if its boundary has content zero (A bounded set in is Jordan measurable iff its boundary is null, equivalently of content zero).
Refutation
The set lies in the unit square, hence in a sufficiently large ball about the origin, so it is bounded by [L1].
The unit square is both a one-rectangle inner family and a one-rectangle outer cover of itself, each of total volume by [L5]; the inner-versus-outer inequality in [L4] therefore forces both contents to equal , so the square does not have content zero.
By coordinatewise use of [L2], every ball centred at a point of meets and also meets its complement, including at the four sides; no point outside the closed square is adherent to . Thus [L3] gives .
By steps 2.1 and 1.2, the boundary of does not have content zero, so [L6] shows that is not Jordan measurable.
The bounded set has no Jordan area, contradicting the universal Statement.
Remarks
The same witness is developed further in The rational points of form a bounded null set that is not Jordan measurable, where its nullity is also proved. That item is not used in this refutation.
Depends on
- Bounded subset, diameter, distance from a point to a set, and distance between two sets in a metric space
- Both $\mathbb{Q}$ and $\mathbb{R} \setminus \mathbb{Q}$ are dense in $\mathbb{R}$, and every nonempty open subset of $\mathbb{R}$ is uncountable
- 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 $\mathbb{R}^m$
- Axis-parallel rectangles in $\mathbb{R}^m$ and their volume
- A bounded set in $\mathbb{R}^m$ is Jordan measurable iff its boundary is null, equivalently of content zero
Used by
Nothing in the library uses this result yet.
Dependency tree · two levels
45 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
- J. Lebl, Basic Analysis, Jordan Measurable Sets (standard reference, not scraped)
- A. Treibergs, MATH 3225 final solutions (standard reference, not scraped)