LemmaStatement: AI-adaptedProof: 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.

The points polygonally reachable from a fixed point form a clopen subset of every open subset of $R^{n}$

Statement

Let $U \subseteq R^{n}$ be open and let $a \in U$ . The set $R_{a}$ of points of $U$ joined to $a$ by a polygonal path (Polygonal paths and polygonally connected subsets of $R^{n}$ ) in $U$ is both open and closed in the subspace $U$ .

Facts & Assumptions

Given: An open subset $U \subseteq R^{n}$ , a point $a \in U$ , and the polygonally reachable set $R_{a} \subseteq U$ .

[L1]

Every point of an open set in a metric topology has an open metric ball contained in that set (The metric topology: a set is open when every one of its points has a ball around it inside the set; closed means open complement, Open ball, closed ball and sphere in a metric space).

[L2]

A straight segment between two points of an Euclidean ball stays in that ball, by the triangle inequality for the Euclidean norm ( $R^{n}$ as the set of functions $n \to R$ , and $d_{1}$ , $d_{2}$ , $d_{\infty}$ are metrics on it).

[L3]

A finite concatenation of such segments is a continuous polygonal path (A finite concatenation of straight segments in $R^{n}$ is a continuous path).

Proof

technique · direct

1.1

Let $y \in R_{a}$ . Choose $r > 0$ with $B (y, r) \subseteq U$ . For every $z \in B (y, r)$ , the segment from $y$ to $z$ lies in $B (y, r)$ by [L2].

L1L2choose

1.2

Let $y \in U ∖ R_{a}$ and choose $r > 0$ with $B (y, r) \subseteq U$ . If some $z \in B (y, r)$ lay in $R_{a}$ , a path from $a$ to $z$ followed by the segment from $z$ to $y$ would put $y$ in $R_{a}$ , a contradiction.

L1L2L3choose

2.1

Concatenating a polygonal path from $a$ to $y$ with that segment gives a polygonal path from $a$ to $z$ in $U$ . Thus $B (y, r) \subseteq R_{a}$ , so $R_{a}$ is open in $U$ .

L3step 1.1

3.1

Hence $B (y, r) \subseteq U ∖ R_{a}$ , so the complement is open in $U$ . Therefore $R_{a}$ is clopen in $U$ .

step 2.1step 1.2∎

Depends on

Used by

For an open subset of ℝⁿ, connectedness, path-connectedness and polygonal connectedness are equivalent Theorem

Dependency tree · two levels

32 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

Connected space (standard reference, not scraped)
Locally connected space (Wikipedia) (standard reference, not scraped)

The points polygonally reachable from a fixed point form a clopen subset of every open subset of RnRn