LemmaStatement: AI-adaptedProof: AI-generatedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-31

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For $n \geq 2$ , the punctured space $R^{n} ∖ {0}$ is polygonally connected

Statement

For $n \geq 2$ , $R^{n} ∖ {0}$ is polygonally connected.

Facts & Assumptions

Given: $n \geq 2$ and nonzero vectors $x, y \in R^{n}$ .

[L1]

The standard unit vectors form a basis of $R^{n}$ , so a one-dimensional span is a proper subspace when $n \geq 2$ (The standard list $e : n \to F^{n}$ with $e_{i} (i) = 1_{F}$ and $e_{i} (j) = 0_{F}$ for $j \neq i$ is an ordered basis of $F^{n}$ ; hence $\dim_{F} F^{n} = n$ , and $F^{0}$ is the zero space with basis $\emptyset$ and dimension $0$ , Linear combination of a finite list, and the span $span (S)$ as the smallest linear subspace containing $S$ ).

[L2]

A vector outside $span {x}$ 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 $span (S)$ 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 $R^{n}$ is a continuous path).

Proof

technique · constructive

1.1L1choose

Choose $z \in R^{n} ∖ (span {x} \cup span {y})$ . 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.

2.1L2step 1.1

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

3.1L3step 2.1

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

4.1step 3.1discharge-construct∎

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

For $n \geq 2$ , the punctured space $R^{n} ∖ {0}$ is polygonally connected

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources

For n≥2n≥2, the punctured space Rn∖{0}Rn∖{0} is polygonally connected

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources

For $n \geq 2$ , the punctured space $R^{n} ∖ {0}$ is polygonally connected