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

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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.

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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

1.1

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.

L1L2choose

2.1

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

step 1.1

3.1

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

L3step 2.1∎

Depends on

Used by

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

Locally connected space (standard reference, not scraped)

Every connected component of an open subset of RnRn is open and polygonally connected