Alphabeta Math
LemmaStatement: AI-adaptedProof: AI-generatedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-31
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  • 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.
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A finite concatenation of straight segments in Rn is a continuous path

Statement

Let v0,…,vm∈Rn and 0=t0<⋯<tm=1. The affine pieces joining vi−1 to vi define a continuous map [0,1]→Rn. If every piece lies in a subset A, the map is a polygonal path (Polygonal paths and polygonally connected subsets of Rn) in A from v0 to vm.

Facts & Assumptions

Given: Vertices v0,…,vm∈Rn and a partition 0=t0<⋯<tm=1.

[L1]

A finite family of closed sets covering a space pastes continuous restrictions to a continuous map (Continuity may be checked on any open cover, and on any finite closed cover; composites of continuous maps are continuous, claim 3).

[L2]

A map whose values lie in a subspace is continuous into that subspace exactly when its composite with the inclusion into the ambient space is continuous (Subspace topology: the traces of the open sets, its closed sets and its bases, the continuity of the inclusion, and the characteristic property of a map into a subspace).

[L3]

A map into Rn is continuous if and only if all of its coordinate functions are continuous (A vector-valued function has a limit, or is continuous, if and only if each of its components does; with the algebra of continuous vector-valued functions, clause 1).

Proof

technique · constructive
1.1

On [ti−1,ti] define γi(t):=((ti−t)/(ti−ti−1))vi−1+((t−ti−1)/(ti−ti−1))vi. Each coordinate is an affine real function of t, hence continuous.

L3construct
2.1

At every shared endpoint ti, the adjacent formulas both give vi, so the pieces define one function γ:[0,1]→Rn.

step 1.1construct
3.1

Each interval [ti−1,ti] is closed in [0,1], the finitely many intervals cover it, and each restriction of γ is continuous by step 1.1. Thus γ is continuous by [L1].

L1step 1.1step 2.1
4.1

If the pieces lie in A, then γ takes values in A. Its composite with the inclusion A↪Rn is the continuous map of step 3.1, so [L2] makes it continuous into the subspace A. Its endpoints are v0,vm, hence it is a path in A.

step 3.1L2discharge-construct∎

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