Locally compact metric space: every point has a compact neighbourhood

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 $x\in X$ there are a compact subset $K\subseteq X$ and a real $r>0$ with

$B(x,r)\subseteq K\mathrm{.}$

This is the condition "every point has a compact neighbourhood", written out. Give $X$ its metric topology ${\mathcal{T}}_d$ (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 $(X,{\mathcal{T}}_d)$ 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 $K\subseteq X$ 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 $x\in U\subseteq K$, 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 $B(x,r)$ satisfies $B(x,r)\subseteq K$. 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, because both are live in the literature.

• Neighbourhoods need not be open. This library's convention is Neighbourhood of a point and neighbourhood base, with this library's convention that a neighbourhood need not be open's, and it is what makes "compact neighbourhood" a useful phrase at all: a compact set is rarely open. Note that 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 uses the word neighbourhood for an open set containing the point; that narrower usage is confined to that item, and the present definition never relies on it.
• Compactness of a subset is intrinsic. $K$ compact means the metric subspace $(K,d_K)$ is a compact metric space (Open cover, subcover, compact metric space, and compact subset of a metric space, Isometry, isometric embedding, and the subspace metric on a subset); the equivalent description by families of open subsets of the ambient $X$ is A subset of a metric space is open in the subspace metric exactly when it is the trace of an open set of the ambient space, and it is compact as a metric space in its own right exactly when every family of open subsets of the ambient space covering it, indexed or not, has finitely many members covering it and is cited wherever it is used.

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.

• Dictionary: this was the metric special case of a notion this library did not yet define in general, and now does. Locally compact is ordinarily defined for an arbitrary topological space, by the same words: every point has a compact neighbourhood, compactness there being the open-cover condition for arbitrary topological spaces. General topological compactness is now available at this point in the reading order, alongside the metric notion (Open cover, subcover, compact metric space, and compact subset of a metric space) this page uses, so the definition above is stated for a metric space and for nothing else, and never claimed to be the general one.

  The agreement is immediate and not a theorem about metrics. A metric space is a topological space whose topology is ${\mathcal{T}}_d$ (Metrizable space: a topological space whose topology is induced by some metric; metrizability is topological, the metric is not), and the metric topology is the topology: the open sets of $(X,d)$ used by Open cover, subcover, compact metric space, and compact subset of a metric space and by A subset of a metric space is open in the subspace metric exactly when it is the trace of an open set of the ambient space, and it is compact as a metric space in its own right exactly when every family of open subsets of the ambient space covering it, indexed or not, has finitely many members covering it are literally the members of ${\mathcal{T}}_d$. So a subset of $X$ is compact in the metric sense exactly when it is compact in the topological sense, and "has a compact neighbourhood" means the same thing on both sides. This agreement is now recorded explicitly, exactly as Metrizable space: a topological space whose topology is induced by some metric; metrizability is topological, the metric is not records the agreement of the metric and topological notions of neighbourhood, closure, convergence and continuity: see For a metric space with its metric topology, compactness in the topological sense is compactness in the metric sense, and the two notions of compact subset coincide, which discharges the standing obligation this Remark used to record. Stating both here and there is what stops the library from acquiring two unrelated notions under one name.

• What local compactness is used for on this page. Two things, and only two: continuity of the evaluation map, and the converse half of the exponential law. The compact-open topology itself, the comparison of the three topologies, the uniform limit theorem, completeness and Dini's theorem all do without it.

• The real line is locally compact and the rationals are not. For $\mathbb{R}$ with its usual metric a closed bounded interval around a point is compact (Heine-Borel in ${\mathbb{R}}^n$: with the Euclidean metric a subset of ${\mathbb{R}}^n$ 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), which is the hypothesis discharged. For $\mathbb{Q}$ with the metric of $\mathbb{R}$ it fails at every point, and that failure is exactly what the false statement about the evaluation map on this page exploits.