CounterexampleConstruction: Literature-sourcedVerification: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-04

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.

Bounded continuous functions need not form a sheaf

Statement refuted

The assignment $U ⟼ C_{b}^{0} (U, R) : = {f : U \to R continuous and bounded}$ with the usual restriction maps is a sheaf on $R$ .

Facts & Assumptions

Given: The presheaf $C_{b}^{0} (-, R)$ on $R$ .

[F1]

Restriction of a bounded continuous function is again bounded and continuous, so this is a presheaf (A presheaf on a topological space).

[L1]

A sheaf must glue every compatible local family to a global section (A sheaf on a topological space).

Counterexample

technique · direct

1.1F1givenconstruct

For each integer $n \geq 1$ , let $U_{n} : = (- n, n)$ and let $s_{n} : U_{n} \to R$ be the identity function $s_{n} (x) = x$ . Each $s_{n}$ lies in $C_{b}^{0} (U_{n}, R)$ because $U_{n}$ is bounded.

2.1step 1.1given

If $m \leq n$ , then $s_{n} ∣_{U_{m}} = s_{m}$ , so the family $(s_{n})$ is compatible on the open cover $R = ⋃_{n \geq 1} U_{n}$ .

3.1L1step 2.1∎

Any glued section on $R$ would have to equal the identity function $x \mapsto x$ , because it agrees with each $s_{n}$ on $U_{n}$ . But $x \mapsto x$ is not bounded on $R$ , so it does not lie in $C_{b}^{0} (R, R)$ . This violates the gluing requirement in [L1]. Therefore the stated presheaf is not a sheaf.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

5 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

The Stacks Project, Sheaves on Spaces, Example 7.6 (standard reference, not scraped)