TheoremStatement: 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.

The exhaustion metric induces exactly the topology of locally uniform convergence

Statement

Let $(K_{n})$ be the canonical compact exhaustion of a plane domain $Ω$ , and let $(f_{m})$ be a sequence of functions $Ω \to C$ . Then $f_{m} \to f in d_{K} ⟺ f_{m} \to f locally uniformly on Ω .$ Consequently the exhaustion metric induces exactly the topology of locally uniform convergence.

Facts & Assumptions

Given: The canonical exhaustion $(K_{n})$ , the exhaustion metric $d_{K}$ , and a sequence $f_{m} : Ω \to C$ .

[L1]

The sets $K_{n}$ are compact, nested inside successive interiors, and exhaust $Ω$ (Every plane domain has the canonical nested compact exhaustion by distance and radius cutoffs).

[L2]

On a plane domain, local uniform convergence is the same as uniform convergence on every compact subset (Locally uniform convergence on a plane domain is the already-published compact-convergence notion).

Proof

technique · direct

1.1

If $d_{K} (f_{m}, f) \to 0$ , then each summand $2^{- n} \min (1, s_{n} (f_{m}, f))$ tends to $0$ , so $f_{m} \to f$ uniformly on every $K_{n}$ .

givenalgebra

2.1

Any compact set $L \subseteq Ω$ is contained in some $K_{N}$ because the interiors of the $K_{n}$ cover $Ω$ by [L1]. Step 1.1 then gives uniform convergence on every compact $L$ , and [L2] makes the convergence locally uniform.

L1L2given

3.1

Conversely, if $f_{m} \to f$ locally uniformly, then [L2] gives uniform convergence on every $K_{n}$ , including the empty stages with zero supremum. A finite-head plus geometric-tail estimate then gives $d_{K} (f_{m}, f) \to 0$ .

L2givenalgebra∎

Depends on

Used by

Dependency tree · two levels

6 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)