LemmaStatement: AI-adaptedProof: AI-generatedSession-authored (Fable 5 assisted)precheck passjudge pass (z-ai/glm-5.2)audited 2026-07-31

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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.
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For $n\ge2$ , the punctured space $\mathbb{R}^n\setminus\{0\}$ is polygonally connected

Statement

For $n\ge2$ , $\mathbb R^n\setminus\{0\}$ is polygonally connected.

Facts & Assumptions

Given: $n\ge2$ and nonzero vectors $x,y\in\mathbb R^n$ .

[L1]

The standard unit vectors form a basis of $\mathbb R^n$ , so a one-dimensional span is a proper subspace when $n\ge2$ (The standard list $e : n \to F^{n}$ with $e_i(i) = 1_F$ and $e_i(j) = 0_F$ for $j \ne i$ is an ordered basis of $F^{n}$ ; hence $\dim_F F^{n} = n$ , and $F^{0}$ is the zero space with basis $\varnothing$ and dimension $0$ , Linear combination of a finite list, and the span $\operatorname{span}(S)$ as the smallest linear subspace containing $S$ ).

[L2]

A vector outside $\operatorname{span}\{x\}$ cannot lie on a segment from $x$ to $0$ , except at no point; the corresponding statement holds for $y$ , by the vector-space axioms (Linear combination of a finite list, and the span $\operatorname{span}(S)$ as the smallest linear subspace containing $S$ , A norm on a real vector space, the induced metric, and the dictionary with the metric axioms).

[L3]

A finite concatenation of segments contained in a subset is a polygonal path in that subset (A finite concatenation of straight segments in $\mathbb{R}^n$ is a continuous path).

Proof

technique · constructive

1.1

Choose $z\in\mathbb R^n\setminus(\operatorname{span}\{x\}\cup\operatorname{span}\{y\})$ . Such a vector exists: if the two spans differ, $x+y$ lies in neither; if they agree, [L1] gives a vector outside their common proper subspace.

L1choose

2.1

The segment from $x$ to $z$ avoids $0$ : an equality $(1-t)x+tz=0$ with $0<t\le1$ would give $z=-((1-t)/t)x\in\operatorname{span}\{x\}$ , contrary to step 1.1. The segment from $z$ to $y$ similarly avoids $0$ .

L2step 1.1

3.1

The two segments therefore form a polygonal path in $\mathbb R^n\setminus\{0\}$ from $x$ to $y$ by [L3].

L3step 2.1

4.1

Since $x,y$ were arbitrary nonzero vectors, the punctured space is polygonally connected.

step 3.1discharge-construct∎

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 111 results over 27 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources

Path-connected space (standard reference, not scraped)
Euclidean space (Wikipedia) (standard reference, not scraped)

For n≥2n\ge2n≥2, the punctured space Rn∖{0}\mathbb{R}^n\setminus\{0\}Rn∖{0} is polygonally connected