Locally compact topological space: every point has a compact neighbourhood; and what this says in a metric space

A topological space $(X, \mathcal{T})$ (Topology on a set, open and closed sets, clopen sets, the closed-set axiomatisation, and the coarser/finer comparison) is locally compact when

every point of $X$ has a compact neighbourhood:

that is, for every $x \in X$ there is a neighbourhood $N$ of $x$ (Neighbourhood of a point and neighbourhood base, with this library's convention that a neighbourhood need not be open) that is a compact subset of $X$ (Open cover, subcover, and compact topological space; a compact subset is a subspace that is compact in its own right, Subspace topology: the traces of the open sets, its closed sets and its bases, the continuity of the inclusion, and the characteristic property of a map into a subspace).

A neighbourhood need not be open here, and that is what makes the condition the weak one it is meant to be: $N$ is required only to contain some open set containing $x$. Writing "compact open neighbourhood" instead would define a strictly stronger property, satisfied by no space in which a point has no compact open neighbourhood, $\mathbb{R}$ among them; and requiring the compact set merely to contain $x$ would define a property so weak that every space with a singleton has it, singletons being compact.

Every compact space is locally compact, since $X$ itself is a neighbourhood of each of its points and is a compact subset of itself. The converse fails, and $\mathbb{R}^n$ is the standard witness.

What the condition says in a metric space. Let $(X,d)$ be a metric space carrying its metric topology (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, Metrizable space: a topological space whose topology is induced by some metric; metrizability is topological, the metric is not), with balls as in Open ball, closed ball and sphere in a metric space, and let $x \in X$. Then

$x$ has a compact neighbourhood if and only if there are a real $r > 0$ and a compact $K \subseteq X$ with $B(x,r) \subseteq K$.

Both directions are immediate and are discharged here. If $N$ is a compact neighbourhood of $x$, fix an open $U$ with $x \in U \subseteq N$; 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 there is $r > 0$ with $B(x,r) \subseteq U \subseteq N$, so $K := N$ serves. Conversely, if $B(x,r) \subseteq K$ with $K$ compact, then $K$ contains the open set $B(x,r)$, which contains $x$, so $K$ is a neighbourhood of $x$ and is compact. Compactness of a subset of $(X,d)$ means the same thing read metrically and read topologically (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), so the criterion may be applied with either development's theorems.

$\mathbb{R}^n$ is locally compact for every $n \ge 1$. Give $\mathbb{R}^n$ the product topology, which is the metric topology of the Euclidean metric $d_2$ ($\mathbb{R}^n$ as the set of functions $n \to \mathbb{R}$, and $d_1$, $d_2$, $d_\infty$ are metrics on it, A subset of $\mathbb{R}^n$ with the product topology is compact exactly when it is closed and bounded, the product topology being the Euclidean metric topology). For $p \in \mathbb{R}^n$ the set

$Q_p \;:=\; \{\, x \in \mathbb{R}^n : d_2(x,p) \le 1 \,\}$

is closed, being the complement of the union of the open balls $B(y, d_2(y,p) - 1)$ over the points $y$ with $d_2(y,p) > 1$, and it is bounded (Bounded subset, diameter, distance from a point to a set, and distance between two sets in a metric space), lying inside $B(p, 2)$; so $Q_p$ is compact by A subset of $\mathbb{R}^n$ with the product topology is compact exactly when it is closed and bounded, the product topology being the Euclidean metric topology. It contains the open ball $B(p,1)$, which contains $p$, so it is a compact neighbourhood of $p$. The space $\mathbb{R}^n$ is not compact, so local compactness is strictly weaker than compactness.

Local compactness is a local condition and compactness is not. The definition quantifies over points and asks for something in a neighbourhood of each; nothing is asserted about covers of the whole space. That is why a locally compact space may be as large as one likes, and why the two properties separate.

Where the extra strength is needed. For an arbitrary space, "every point has a compact neighbourhood" does not by itself give a base of compact neighbourhoods at each point, nor an open set with compact closure around each compact set. Both of those do follow once the space is also Hausdorff, and that is In a locally compact Hausdorff space every point has a neighbourhood base of compact sets, every open subspace and every closed subspace is locally compact, every open set around a point contains an open set with compact closure inside it, and every compact set sits inside an open set with compact closure; several authors build the stronger condition into the definition and then note the agreement in the Hausdorff case. This library takes the weak definition and proves the strengthening under the hypothesis that licenses it.

Local compactness is not hereditary, unlike metrizability. A subspace of a locally compact space need not be locally compact, and FALSE: every subspace of a locally compact space is locally compact records the failure with a witness; what does survive is heredity along open and along closed subspaces of a locally compact Hausdorff space (In a locally compact Hausdorff space every point has a neighbourhood base of compact sets, every open subspace and every closed subspace is locally compact, every open set around a point contains an open set with compact closure inside it, and every compact set sits inside an open set with compact closure).