CorollaryStatement: AI-adaptedProof: AI-generatedprecheck passverified 2026-08-05 (claude-sonnet-5)

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.

Every connected component of an open subset of $R^{n}$ is open and polygonally connected

Statement

Every connected component of an open subset $U \subseteq R^{n}$ is open in $R^{n}$ and polygonally connected.

Facts & Assumptions

Given: An open subset $U \subseteq R^{n}$ and a connected component $C$ of $U$ .

[L1]

A Euclidean ball is path-connected, hence connected: the norm triangle inequality keeps every straight segment in the ball, the segment is continuous, and every path-connected space is connected (Open ball, closed ball and sphere in a metric space, A norm on a real vector space, the induced metric, and the dictionary with the metric axioms, A finite concatenation of straight segments in $R^{n}$ is a continuous path, Every path-connected space is connected, and every path component lies inside a component).

[L2]

A component is the largest connected subset containing each of its points (Connected components, quasicomponents, and totally disconnected spaces).

[L3]

An open connected Euclidean subset is polygonally connected (For an open subset of $R^{n}$ , connectedness, path-connectedness and polygonal connectedness are equivalent).

Proof

technique · direct

1.1L1L2choose

Let $x \in C$ . Since $U$ is open, choose $r > 0$ with $B (x, r) \subseteq U$ . The ball is connected by [L1], meets $C$ at $x$ , and so lies in $C$ by maximality of the component.

2.1step 1.1

Therefore every point of $C$ has a Euclidean ball contained in $C$ , so $C$ is open in $R^{n}$ .

3.1L3step 2.1∎

The component $C$ is connected and now open, so [L3] makes it polygonally connected.

Depends on

Dependency tree · 2 levels

Sources

Locally connected space