Alphabeta Math
ExampleConstruction: AI-generatedVerification: AI-generatedprecheck passaudited 2026-08-16
How statement and proof provenance work

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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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Affine interpolants with endpoints in a compact rectangle form a compact family

Example

Fix reals α≤β and γ≤δ. For (a,b) in the rectangle P=[α,β]×[γ,δ], define

fa,b(t):=(1−t)a+tb(0≤t≤1).

The family A:={fa,b:(a,b)∈P} is compact in the uniform topology on C([0,1],R). It is equicontinuous and pointwise relatively compact, and the endpoint map f↦(f(0),f(1)) identifies it homeomorphically with P.

Facts & Assumptions

Verification

technique · direct
1.1L2

By [L2], P is compact. Define Φ:P→C([0,1],R) by Φ(a,b)=fa,b.

1.2

For 0≤t≤1,

∣fa,b(t)−fa′,b′(t)∣≤∣a−a′∣+∣b−b′∣.

Thus Φ is continuous into the uniform topology of [L3]. [L3, algebra]

1.3algebra

Put M=max⁡{∣α∣,∣β∣,∣γ∣,∣δ∣}. Every member satisfies ∣fa,b(s)−fa,b(t)∣=∣b−a∣∣s−t∣≤2M∣s−t∣, so the family is equicontinuous, including the degenerate case M=0.

2.1L1step 1.1step 1.2

By [L1], A=Φ[P] is compact.

2.2L3step 1.2

The endpoint map E:A→P, E(f)=(f(0),f(1)), is continuous because uniform distance controls both endpoint differences, and E∘Φ and Φ∘E are identity maps. Hence Φ is a homeomorphism onto A.

3.1L1step 1.1∎

For fixed t, the coordinate set {fa,b(t):(a,b)∈P} is the continuous image of compact P, hence compact by [L1]. It is therefore already its compact closure, proving pointwise relative compactness.

Depends on

Used by

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Dependency tree · two levels

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