Alphabeta Math
DefinitionDefinition: AI-adaptedProof: Not applicableverified 2026-09-23 (gpt-6-sol)
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.

Locally compact metric space: every point has a compact neighbourhood

Definition

Let (X,d) be a metric space (Metric space: d(x,y)=0 iff x=y, symmetry, and the triangle inequality; pseudometric and ultrametric), with balls as in Open ball, closed ball and sphere in a metric space and compact subsets as in Open cover, subcover, compact metric space, and compact subset of a metric space.

(X,d) is locally compact if for every x∈X there are a compact subset K⊆X and a real r>0 with

B(x,r)  ⊆  K.

This is the condition "every point has a compact neighbourhood", written out. Give X its metric topology Td (The metric topology: a set is open when every one of its points has a ball around it inside the set; closed means open complement), so that (X,Td) is a topological space (Metrizable space: a topological space whose topology is induced by some metric; metrizability is topological, the metric is not). A set K⊆X is a neighbourhood of x in the sense of Neighbourhood of a point and neighbourhood base, with this library's convention that a neighbourhood need not be open exactly when some open U satisfies x∈U⊆K, and by The metric topology: a set is open when every one of its points has a ball around it inside the set; closed means open complement that holds exactly when some ball B(x,r) satisfies B(x,r)⊆K. So the displayed condition says precisely that x has a compact neighbourhood, and the two readings are the same condition and not two notions.

Two conventions are fixed here.

Every compact metric space is locally compact, since K:=X and any r>0 serve at every point. The empty metric space is locally compact, the condition being vacuous.

Remarks

Depends on

Dependency tree · 2 levels

Sources