False statementConstruction: Literature-sourcedVerification: AI-adaptedSession-authored (Fable 5 assisted)precheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-21

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.

FALSE: every pointwise bounded sequence of continuous functions has a uniformly convergent subsequence

Statement

False claim: every pointwise bounded sequence of continuous real functions on a compact interval has a uniformly convergent subsequence.

Facts & Assumptions

Given: The universal claim in the Statement.

[L1]

For $f_{k} (x) = \sin ((k + 1) x)$ on $[0, π]$ , the sequence $(f_{k})$ is uniformly bounded, is not equicontinuous, and has no uniformly convergent subsequence (The sine harmonics are pointwise bounded but have no uniformly convergent subsequence).

[L2]

Sine is continuous, affine real functions are continuous, and composites of continuous real functions are continuous (The derivatives of sine and cosine are cosine and minus sine, A function differentiable at $c$ is continuous at $c$ , Sums, scalar multiples, products, absolute values, maxima, minima and quotients with nonvanishing denominator of continuous functions are continuous, as are constants, the identity and every polynomial function, claim 5, A composite of continuous functions is continuous, with no side hypothesis of the kind the composition of limits needs).

[L3]

The number $π = 2 γ$ is positive because the smallest positive zero of cosine satisfies $γ \in (0, 2)$ , and every closed bounded interval in $R$ is compact (Pi as twice the smallest positive zero of cosine, Cosine has a smallest positive zero, lying strictly between zero and two, Heine-Borel by bisection: every closed bounded interval $[a, b]$ is compact).

Refutation

technique · contradiction

1.1

Suppose, for contradiction, that every pointwise bounded sequence of continuous real functions on a compact interval has a uniformly convergent subsequence.

assume-contra

1.2

Each function in [L1] is continuous by [L2], and the sequence is uniformly bounded by [L1], hence pointwise bounded, on the compact interval $[0, π]$ from [L3].

L1L2L3

2.1

The assumed claim gives this sequence a uniformly convergent subsequence, contradicting [L1]. Therefore the claim is false; the missing Arzelà–Ascoli hypothesis is equicontinuity.

step 1.1step 1.2L1discharge-contradiction∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

52 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

John Hutchinson, Introduction to Analysis, §15.7, Remark 15.7.2 (standard reference, not scraped)