False statementConstruction: AI-adaptedVerification: AI-adaptedprecheck passaudited 2026-08-13

How 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 closed C1 field on a connected open set is exact

Statement

Every closed $C^{1}$ vector field on a connected open subset of $R^{n}$ is exact.

Facts & Assumptions

Given: The proposed implication, the punctured plane $U = R^{2} ∖ {0}$ , and the vortex field $F (x, y) = (\frac{- y}{x^{2} + y^{2}}, \frac{x}{x^{2} + y^{2}}) .$

[L1]

With coordinates indexed from $0$ , so that $F = (F_{0}, F_{1})$ , closedness means $\partial_{y} F_{0} = \partial_{x} F_{1}$ , while exactness supplies a $C^{2}$ potential with gradient $F$ (Exact and closed C1 vector fields).

[L2]

A gradient line integral is its potential's endpoint increment and is therefore zero on a closed path (The gradient theorem: the line integral of a gradient is the endpoint increment).

[L3]

Vector line integrals integrate $⟨ F (γ (t)), γ^{'} (t) ⟩$ ; sine and cosine have derivatives $\cos t$ and $- \sin t$ and satisfy $\sin^{2} t + \cos^{2} t = 1$ (Scalar line integrals with respect to arc length and vector-field line integrals, The derivatives of sine and cosine are cosine and minus sine, Parity and the Pythagorean identity for sine and cosine).

[L4]

The integral of $1$ on $[0, 2 π]$ is $2 π$ , and $π > 0$ (If $m \leq f \leq M$ on $[a, b]$ then $m (b - a) \leq L (f, P) \leq \underline{\int_{a}^{b}} f \leq \overline{\int_{a}^{b}} f \leq U (f, P) \leq M (b - a)$ for every partition $P$ ; in particular every constant function is integrable, with $\int_{a}^{b} c = c (b - a)$ , Pi as twice the smallest positive zero of cosine).

[L5]

A continuous real function on a closed interval takes every value between its endpoint values (Intermediate value theorem, by bisection with a canonical left-half rule: a continuous function on $[a, b]$ takes every value between $f (a)$ and $f (b)$ ).

[L6]

On a star-shaped open domain, closedness and exactness are equivalent for $C^{1}$ vector fields (On a star-shaped open domain, closed, exact, conservative, path-independent, and zero-loop are equivalent).

Refutation

technique · direct

1.1

Direct differentiation gives $\partial_{y} F_{0} = (y^{2} - x^{2}) / (x^{2} + y^{2})^{2} = \partial_{x} F_{1}$ , so [L1] makes $F$ closed and $C^{1}$ on the open set $U$ . Radial segments at positive radius followed by a circular arc give a piecewise- $C^{1}$ path in $U$ between any two of its points.

givenL1algebra

1.2

For $γ (t) = (\cos t, \sin t)$ on $[0, 2 π]$ , [L3] gives $F (γ (t)) = γ^{'} (t) = (- \sin t, \cos t)$ . Hence [L3] and [L4] give $\int_{γ} F \cdot d r = \int_{0}^{2 π} 1 d t = 2 π \neq 0.$

givenL3L4algebra

2.1

If $U = A \cup B$ were a separation into disjoint nonempty relatively open sets, choose $u \in A$ and $v \in B$ and let $η : [0, 1] \to U$ be the path from step 1.1. The function equal to $0$ when $η (t) \in A$ and to $1$ when $η (t) \in B$ is locally constant, hence continuous, and has endpoint values $0$ and $1$ . By [L5] it would take the value $1 / 2$ , which is impossible. Thus $U$ is connected.

step 1.1L5choosealgebra

3.1

If $F$ were exact, [L1] and [L2] would make the integral in step 1.2 zero. Thus it is not exact, and the statement is false despite steps 1.1 and 2.1. The valid correction in [L6] replaces connectedness by the stronger star-shaped hypothesis.

step 1.1step 2.1step 1.2L1L2L6∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

56 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.-B. Campesato, Poincare Lemma, sections 1 and 2 (standard reference, not scraped)