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DefinitionDefinition: AI-adaptedProof: Not applicableSession-authored (Fable 5 assisted)verified 2026-07-29 (claude-sonnet-5)
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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)(X,d) be a metric space (Metric space: d(x,y)=0d(x,y) = 0 iff x=yx = 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)(X,d) is locally compact if for every xXx \in X there are a compact subset KXK \subseteq X and a real r>0r > 0 with

B(x,r)    K.B(x,r) \;\subseteq\; K .

This is the condition "every point has a compact neighbourhood", written out. Give XX its metric topology Td\mathcal{T}_d (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)(X, \mathcal{T}_d) 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 KXK \subseteq X is a neighbourhood of xx 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 UU satisfies xUKx \in U \subseteq 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)B(x,r) satisfies B(x,r)KB(x,r) \subseteq K. So the displayed condition says precisely that xx has a compact neighbourhood, and the two readings are the same condition and not two notions.

Two conventions are fixed here, because both are live in the literature.

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

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