LemmaStatement: Literature-sourcedProof: AI-generatedPipeline-generatedprecheck 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.

Every plane domain has the canonical nested compact exhaustion by distance and radius cutoffs

Statement

Let $Ω \subseteq C$ be a plane domain, and define for $n \geq 1$ $K_{n} : = {z \in Ω : ∣ z ∣ \leq n and dist (z, \partial Ω) \geq 1 / n},$ with the convention $dist (z, \emptyset) = \infty$ when $Ω = C$ . Then each $K_{n}$ is compact, one has $K_{n} \subseteq int K_{n + 1},$ and $Ω = ⋃_{n \geq 1} K_{n} .$

Facts & Assumptions

Given: A plane domain $Ω$ and the sets $K_{n} = {z \in Ω : ∣ z ∣ \leq n and dist (z, \partial Ω) \geq 1 / n}$ .

[L1]

Closed and bounded subsets of $R^{2} ≅ C$ are compact (Heine-Borel in $R^{n}$ : with the Euclidean metric a subset of $R^{n}$ is compact if and only if it is closed and bounded, and the proof by bisection uses no choice principle; the same holds on the real line).

Proof

technique · direct

1.1L1given

Each $K_{n}$ is bounded by $∣ z ∣ \leq n$ and closed because limits preserve both the radius bound and the distance-to-boundary inequality, so [L1] makes each $K_{n}$ compact; early members are allowed to be empty.

1.2givenalgebra

If $z \in K_{n}$ , then $∣ z ∣ < n + 1$ and $dist (z, \partial Ω) > 1 / (n + 1)$ , so a small disc about $z$ stays inside $K_{n + 1}$ . Hence $K_{n} \subseteq int K_{n + 1}$ .

2.1givenchoose∎

If $z \in Ω$ , openness gives a closed disc $\overline{D} (z, r) \subseteq Ω$ ; choosing $n > ∣ z ∣$ and $1 / n < r$ puts $z$ in $K_{n}$ . Therefore $⋃_{n \geq 1} K_{n} = Ω$ .

Depends on

Used by

Dependency tree · two levels

31 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

Matthias Weber, Complex Analysis, Ch. 5 §§5.1-5.2 (standard reference, not scraped)
Elias M. Stein and Rami Shakarchi, Complex Analysis, Ch. 2 §5.2 and Ch. 8 §3.2 (standard reference, not scraped)
Sheldon Axler, Paul Bourdon, and Wade Ramey, Harmonic Function Theory, Ch. 2 (standard reference, not scraped)