Alphabeta Math
TheoremStatement: Literature-sourcedProof: 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.

Convex domains are holomorphically convex

Statement

Let ΩCm be a convex domain, and let KΩ be compact. Then

K^Ωconv(K).

In particular, Ω is holomorphically convex.

Facts & Assumptions

Given: A convex domain ΩCm and a compact set KΩ.

[L1]

A point outside a compact convex set can be strictly separated from it by the real part of a complex-linear functional (A compact convex set and an exterior point admit a complex-linear separator).

[L2]

The holomorphic hull is defined by inequalities against all holomorphic functions on Ω (Holomorphic hulls and holomorphic convexity).

[L3]

Convex subsets contain the line segment between any two of their points (A convex subset of Rm contains every line segment between two of its points).

Proof

technique · direct
1.1

Let pΩconv(K). Since conv(K) is compact and convex, [L1] gives a complex-linear functional L such that ReL(z)β<ReL(p) for every zconv(K), hence in particular for every zK. The holomorphic function h(z):=exp(L(z)) then satisfies h(z)eβ<h(p) on K. By [L2], this excludes p from K^Ω.

L1L2given
2.1

Step 1.1 proves K^Ωconv(K). Because Ω is convex, [L3] gives conv(K)Ω. In finite-dimensional Euclidean space the convex hull of a compact set is compact, so K^Ω is contained in a compact subset of Ω. Therefore K^ΩΩ, and Ω is holomorphically convex.

L3step 1.1

Depends on

Used by

Dependency tree · two levels

7 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