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TheoremStatement: Literature-sourcedProof: AI-generatedPipeline-generatedprecheck passaudited 2026-08-28
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The exhaustion metric induces exactly the topology of locally uniform convergence

Statement

Let (Kn) be the canonical compact exhaustion of a plane domain Ω, and let (fm) be a sequence of functions Ω→C. Then fm→f in dK⟺fm→f locally uniformly on Ω. Consequently the exhaustion metric induces exactly the topology of locally uniform convergence.

Facts & Assumptions

Given: The canonical exhaustion (Kn), the exhaustion metric dK, and a sequence fm:Ω→C.

[L1]

The sets Kn 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
1.1givenalgebra

If dK(fm,f)→0, then each summand 2−nmin⁡(1,sn(fm,f)) tends to 0, so fm→f uniformly on every Kn.

2.1L1L2given

Any compact set L⊆Ω is contained in some KN because the interiors of the Kn cover Ω by [L1]. Step 1.1 then gives uniform convergence on every compact L, and [L2] makes the convergence locally uniform.

3.1L2givenalgebra∎

Conversely, if fm→f locally uniformly, then [L2] gives uniform convergence on every Kn, including the empty stages with zero supremum. A finite-head plus geometric-tail estimate then gives dK(fm,f)→0.

Depends on

Used by

Dependency tree · two levels

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Sources