CounterexampleConstruction: AI-adaptedVerification: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-28

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 domain of holomorphy need not be convex

Statement refuted

Every domain of holomorphy in $C^{m}$ is convex.

Facts & Assumptions

Given: The domain $U = C^{2} ∖ {(z_{1}, z_{2}) : z_{1} z_{2} = 1} .$

[L1]

A domain of holomorphy is characterized by the failure of every common simultaneous extension pair (Holomorphic extension and domains of holomorphy in several variables).

[L2]

A holomorphic function on a connected open set is determined by its values on any nonempty open subset (A holomorphic function vanishing on a nonempty open subset of a domain vanishes identically).

Counterexample

technique · direct

1.1

The function $f (z_{1}, z_{2}) = 1 / (z_{1} z_{2} - 1)$ is holomorphic on $U$ . Let $p$ be a point of the removed hypersurface $z_{1} z_{2} = 1$ . If $f$ extended holomorphically to a neighborhood $V$ of $p$ , then $(z_{1} z_{2} - 1) F$ would be holomorphic on $V$ and equal to $1$ on the nonempty open set $V \cap U$ . By [L2], it would equal $1$ on all of $V$ , impossible at $p$ where $z_{1} z_{2} - 1 = 0$ . So the same function is singular at every boundary point of $U$ , and [L1] makes $U$ a domain of holomorphy.

L1L2given

2.1

The points $(2, 0)$ and $(0, 2)$ lie in $U$ , but their midpoint $(1, 1)$ lies on the removed hypersurface and therefore is not in $U$ . Hence $U$ is not convex. This refutes the statement.

step 1.1givenalgebra∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

8 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

Jiří Lebl, Tasty Bits of Several Complex Variables, §2.1 (standard reference, not scraped)
Harold P. Boas, Lecture Notes on Several Complex Variables, Example 12 (standard reference, not scraped)