Alphabeta Math
CorollaryStatement: Literature-sourcedProof: AI-generatedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-27
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 containing a Hartogs figure but not its hull is not a domain of holomorphy

Statement

Let ΩC2 be a domain. If there exist 0<r,s<1 such that

H(r,s)ΩH^(r,s)andΩH^(r,s),

then Ω is not a domain of holomorphy.

Facts & Assumptions

Given: A domain Ω with H(r,s)ΩH^(r,s) and ΩH^(r,s).

[L1]

A domain of holomorphy is defined by the nonexistence of one fixed overlap from which every holomorphic function extends farther (Holomorphic extension and domains of holomorphy in several variables).

[L2]

Every holomorphic function on H(r,s) extends uniquely to the full bidisc H^(r,s) (A holomorphic function on a Hartogs figure extends to the full bidisc).

Proof

technique · direct
1.1

Let fO(Ω). Its restriction to the open subset H(r,s) is holomorphic, so [L2] gives a holomorphic function FfO(H^(r,s)) with Ff=f on H(r,s).

givenL2
2.1

The domain H^(r,s) is not contained in Ω by hypothesis, and H(r,s) is a nonempty open subset of ΩH^(r,s). Thus the same open overlap works for every holomorphic function on Ω, namely the fixed set U1:=H(r,s) and the larger domain U2:=H^(r,s).

step 1.1given
3.1

By the definition in [L1], the existence of that common pair U1,U2 shows that Ω is not a domain of holomorphy.

L1step 2.1

Depends on

Used by

Dependency tree · two levels

10 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