CounterexampleConstruction: Literature-sourcedVerification: AI-adaptedSession-authored (Fable 5 assisted)precheck passjudge pass (deepseek-v4-pro + claude-opus-5[1m])audited 2026-08-24

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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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Agreement accumulating only at the boundary does not force a holomorphic identity

Statement refuted

Two holomorphic functions on a complex domain that agree on a set accumulating at a boundary point must agree everywhere.

Facts & Assumptions

Given: The punctured plane $Ω = C ∖ {0}$ , the functions $f (z) = \sin (1 / z)$ and $g (z) = 0$ , the entire complex sine (Complex sine, cosine, hyperbolic sine, and hyperbolic cosine are entire with their standard derivatives, The exponential formulas, real restrictions, and trigonometric-hyperbolic dictionary over $C$ ), the complex chain and quotient rules (The chain rule for complex derivatives, Linearity, product, reciprocal, and quotient rules for complex derivatives), the plane topology dictionary ( $C = R [x] / (x^{2} + 1)$ as the Euclidean plane and as a normed real algebra: what the identification preserves), and $\sin (π / 2) = 1$ (Quarter-turn values and shifts by pi/2 and pi).

[L1]

For complex $z$ , $\sin z = 0$ exactly when $z = k π$ for some integer $k$ (The zeros of complex sine are the integer multiples of pi, and the zeros of complex cosine are the odd half-integer multiples of pi).

[L2]

If two holomorphic functions on a complex domain agree on a set with an accumulation point in the domain, then they agree everywhere on the domain (Identity theorem for holomorphic functions).

[L3]

For $n \geq 2$ , $R^{n} ∖ {0}$ is polygonally connected (For $n \geq 2$ , the punctured space $R^{n} ∖ {0}$ is polygonally connected).

Counterexample

technique · direct

1.1

The set $Ω$ is open and is connected by [L3] under the plane dictionary, so it is a complex domain (A complex domain is a nonempty connected open subset of $C$ ). The chain and quotient rules make $f$ holomorphic there, and $g$ is holomorphic as a constant.

L3givenalgebra

1.2

For every natural $k \geq 1$ , put $z_{k} = 1 / (k π)$ . Then $z_{k} \in Ω$ , [L1] gives $f (z_{k}) = \sin (k π) = 0 = g (z_{k})$ , the points are distinct, and $z_{k} \to 0$ .

L1algebra

2.1

The accumulation point $0$ is not in $Ω$ , so [L2] does not apply. Moreover, $2 / π \in Ω$ and $f (2 / π) = \sin (π / 2) = 1 \neq 0 = g (2 / π)$ . Thus the functions agree on a set accumulating only at the boundary but are not identical.

step 1.1step 1.2L2givenalgebra∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

42 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

J. Lebl, Guide to Cultivating Complex Analysis, §2.4 (standard reference, not scraped)
B. V. Shabat, Introduction to Complex Analysis, §2.3 (standard reference, not scraped)