CorollaryStatement: 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.

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

Statement

For $n \geq 1$ , $R^{n}$ is polygonally connected and connected, and it is locally path-connected and locally connected.

Facts & Assumptions

Given: $R^{n}$ with $n \geq 1$ and its Euclidean topology.

[L1]

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

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

1.2

Let $x \in R^{n}$ 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 $R^{n}$ is locally path-connected, and it is locally connected by [L3].

L2L3step 1.2∎

Depends on

Used by

Cauchy's theorem for a null-homologous cycle Corollary
FALSE: every continuous complex-valued function on a convex domain has a primitive False statement
Geodesic completeness means compactness False statement
Topological manifolds are locally compact and locally path connected Proposition
Boundary maximum modulus principle on a bounded domain Theorem
Liouville's theorem: every bounded entire function is constant Theorem
Maximum modulus principle with boundary and infinity control Theorem
The complement of a compact plane set has exactly one unbounded connected component Theorem

Dependency tree · two levels

36 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

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

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