False statementConstruction: AI-adaptedVerification: AI-generatedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-31

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 connected subset of $R^{n}$ is polygonally connected

Statement

False claim: every connected subset of $R^{n}$ is polygonally connected.

The unit circle $S^{1} \subseteq R^{2}$ is connected but is not polygonally connected.

Facts & Assumptions

Given: The unit circle $S^{1} = {u \in R^{2} : ∥ u ∥_{2} = 1}$ and the points $e_{0}, - e_{0} \in S^{1}$ .

[A1]

Every connected subset of Euclidean space is polygonally connected.

[L1]

$S^{1}$ is path-connected and connected (For $n \geq 2$ , the sphere $S^{n - 1}$ is path-connected and connected).

[L2]

A polygonal path is a finite concatenation of straight segments (Polygonal paths and polygonally connected subsets of $R^{n}$ ).

[L3]

If distinct unit vectors $u, v$ are joined by a segment, its midpoint has squared Euclidean norm $∥ (u + v) / 2 ∥_{2}^{2} = 1 - ∥ u - v ∥_{2}^{2} / 4 < 1$ (The Euclidean inner product $⟨ x, y ⟩ = \sum_{k < n} x_{k} y_{k}$ on $R^{n}$ , A norm on a real vector space, the induced metric, and the dictionary with the metric axioms).

Refutation

technique · contradiction

1.1

Suppose the claim [A1] holds. Since $S^{1}$ is connected by [L1], there is a polygonal path in $S^{1}$ from $e_{0}$ to $- e_{0}$ .

A1L1assume-contra

2.1

In its finite vertex list some adjacent vertices $u, v$ are distinct, since its endpoints are distinct. The straight segment from $u$ to $v$ lies in $S^{1}$ by [L2].

L2step 1.1

3.1

But the midpoint of this segment has norm strictly less than $1$ by [L3], so it does not lie in $S^{1}$ . This contradicts step 2.1.

L3step 2.1discharge-contradiction∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

35 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

Sphere (standard reference, not scraped)
Convex set (standard reference, not scraped)

FALSE: every connected subset of RnRn is polygonally connected