CorollaryStatement: AI-adaptedProof: AI-generatedSession-authored (Fable 5 assisted)precheck 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.

$\mathbb{R}^n$ is polygonally connected, connected, locally path-connected and locally connected

Statement

For $n\ge1$ , $\mathbb{R}^n$ is polygonally connected and connected, and it is locally path-connected and locally connected.

Facts & Assumptions

Given: $\mathbb{R}^n$ with $n\ge1$ and its Euclidean topology.

[L1]

A segment $t\mapsto(1-t)x+ty$ is a continuous polygonal path in $\mathbb{R}^n$ (A finite concatenation of straight segments in $\mathbb{R}^n$ is a continuous path, Polygonal paths and polygonally connected subsets of $\mathbb{R}^n$ ).

[L2]

Euclidean balls are open and every open neighbourhood contains a Euclidean ball about its point (Open ball, closed ball and sphere in a metric space, 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).

[L3]

Every path-connected space is connected, and every locally path-connected space is locally connected (Every path-connected space is connected, and every path component lies inside a component, Locally connected and locally path-connected spaces: a neighbourhood base of open connected, respectively open path-connected, sets at every point).

[L4]

The norm triangle inequality keeps a segment joining two points of an open ball inside that ball (A norm on a real vector space, the induced metric, and the dictionary with the metric axioms).

Proof

technique · direct

1.1

For $x,y\in\mathbb{R}^n$ , the one-segment path $t\mapsto(1-t)x+ty$ joins them. Hence $\mathbb{R}^n$ is polygonally connected.

1.2

Let $x\in\mathbb{R}^n$ and let $U$ be open with $x\in U$ . Choose $r>0$ with $B(x,r)\subseteq U$ . Each pair of points in this ball is joined by its segment, which stays in the ball by [L4].

L1L2L4choose

2.1

It is path-connected and therefore connected by [L3].

L3step 1.1

3.1

Thus every open neighbourhood contains an open path-connected ball, so $\mathbb{R}^n$ is locally path-connected, and it is locally connected by [L3].

L2L3step 1.2∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 103 results over 20 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

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

Rn\mathbb{R}^nRn is polygonally connected, connected, locally path-connected and locally connected