CounterexampleConstruction: AI-adaptedVerification: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + claude-sonnet-5)audited 2026-08-17

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.

A connected complex domain need not be star-shaped

Statement refuted

Refuted claim: Every complex domain is star-shaped.

The punctured plane

$U = C ∖ {0}$

is a complex domain, but it has no star centre.

Facts & Assumptions

Given: The punctured complex plane $U = C ∖ {0}$ under the Euclidean identification of $C = R [x] / (x^{2} + 1)$ as the Euclidean plane and as a normed real algebra: what the identification preserves.

[L1]

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

[L2]

Polygonal connectedness supplies paths, and every path-connected space is connected (Polygonal paths and polygonally connected subsets of $R^{n}$ , Paths, path-connected spaces and path components, Every path-connected space is connected, and every path component lies inside a component).

[L3]

A complex domain is a nonempty connected open subset of $C$ (A complex domain is a nonempty connected open subset of $C$ ).

[L4]

A set is star-shaped with respect to $a$ only if the segment $(1 - t) a + t z$ lies in the set for every member $z$ and every $0 \leq t \leq 1$ (Star-shaped open subsets of Euclidean space).

[L5]

The complex modulus satisfies $∣ z + w ∣ \leq ∣ z ∣ + ∣ w ∣$ and vanishes exactly at zero (Conjugation is an involutive real-field automorphism, $z \overline{z} = ∣ z ∣^{2}$ , and modulus is definite, multiplicative, and subadditive).

Refutation

technique · direct

1.1givenL1L2L3L5

The set $U$ is nonempty. If $z \in U$ , the ball $B (z, ∣ z ∣ / 2)$ avoids $0$ by [L5], so $U$ is open; under $C = R^{2}$ , [L1] and [L2] make it connected. Hence [L3] makes $U$ a complex domain.

2.1step 1.1L4algebra∎

For any proposed centre $a \in U$ , the point $- a$ also lies in $U$ , but the segment from $a$ to $- a$ contains $(1 - 1 / 2) a + (1 / 2) (- a) = 0 \notin U$ . By [L4], $a$ is not a star centre, so no star centre exists and the claim is false.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

39 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

Richard Howell and John Mathews, Complex Analysis, Section 6.3 (standard reference, not scraped)