ExampleConstruction: AI-generatedVerification: AI-generatedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-16

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.

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

Example

Fix reals $α \leq β$ and $γ \leq δ$ . For $(a, b)$ in the rectangle $P = [α, β] \times [γ, δ]$ , define

$f_{a, b} (t) : = (1 - t) a + t b (0 \leq t \leq 1) .$

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

Facts & Assumptions

Given: The compact rectangle $P$ and the affine family $A$ .

[L1]

Continuous images of compact spaces are compact (A continuous image of a compact space is compact; a continuous real-valued map on a nonempty compact space attains a maximum and a minimum; and a continuous bijection from a compact space to a Hausdorff space is a homeomorphism).

[L2]

Closed bounded rectangles in $R^{2}$ are compact (Heine-Borel in $R^{n}$ : with the Euclidean metric a subset of $R^{n}$ is compact if and only if it is closed and bounded, and the proof by bisection uses no choice principle; the same holds on the real line).

[L3]

The uniform topology is induced by the uniform metric on the function space (Uniform convergence, and the topology of uniform convergence: the metric topology of the uniform metric on $Y^{X}$ and on $C (X, Y)$ ).

Verification

technique · direct

1.1

By [L2], $P$ is compact. Define $Φ : P \to C ([0, 1], R)$ by $Φ (a, b) = f_{a, b}$ .

1.2

For $0 \leq t \leq 1$ ,

$∣ f_{a, b} (t) - f_{a^{'}, b^{'}} (t) ∣ \leq ∣ a - a^{'} ∣ + ∣ b - b^{'} ∣ .$

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

1.3

Put $M = \max {∣ α ∣, ∣ β ∣, ∣ γ ∣, ∣ δ ∣}$ . Every member satisfies $∣ f_{a, b} (s) - f_{a, b} (t) ∣ = ∣ b - a ∣ ∣ s - t ∣ \leq 2 M ∣ s - t ∣$ , so the family is equicontinuous, including the degenerate case $M = 0$ .

algebra

2.1

By [L1], $A = Φ [P]$ is compact.

L1step 1.1step 1.2

2.2

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

L3step 1.2

3.1

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

L1step 1.1∎

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