Many of the most important spaces in analysis and geometry are not [compact](/page/Compact%20Space) — $\mathbb{R}^n$, smooth manifolds, locally compact groups — yet they still support a surprisingly rich theory of integration, function spaces, and duality. The Extreme Value Theorem fails on $\mathbb{R}$ (the function $f(x) = x$ has no maximum), the space $C(\mathbb{R})$ of all continuous functions has no useful norm, and an arbitrary open cover of $\mathbb{R}$ need not admit a finite subcover. At first glance, it appears that without compactness, the powerful machinery of analysis — Riesz representation, partitions of unity, spectral theory — should be unavailable.

Yet $\mathbb{R}^n$ is not entirely without compactness structure. Every point $x \in \mathbb{R}^n$ sits inside a compact neighbourhood: the closed ball $\overline{B}(x, 1)$ is compact by the Heine-Borel theorem, and it contains the open ball $B(x, 1)$ as an open neighbourhood of $x$. This means that *locally* — near each individual point — the full force of compactness is available. The space fails to be compact only because these local compact pieces cannot be assembled into a single finite cover. This local-versus-global distinction turns out to be the key: a space that is "compact near every point" admits much of the theory that global compactness provides, as long as the arguments are structured to work locally first and then assembled by carefully controlled global constructions.

The contrast with infinite-dimensional spaces is instructive.

[example: No Point in $\ell^2$ Has a Compact Neighbourhood] Consider the Hilbert space $\ell^2 = \ell^2(\mathbb{N})$ of square-summable real sequences with norm $\|x\|_{\ell^2} = (\sum_{k=1}^\infty x_k^2)^{1/2}$. We claim that no point $x_0 \in \ell^2$ has a compact neighbourhood in the norm topology. Suppose, for contradiction, that $K \subset \ell^2$ is a compact set containing an open neighbourhood $B(x_0, r)$ for some $r > 0$. Consider the sequence of vectors $y_k = x_0 + (r/2) e_k$, where $e_k$ is the $k$-th standard basis vector. Each $y_k$ lies in $B(x_0, r) \subset K$, so $\{y_k\}_{k=1}^\infty$ is a sequence in the compact set $K$ and must have a convergent subsequence. But for $j \neq k$, \begin{align*} \|y_j - y_k\|_{\ell^2} = \frac{r}{2}\|e_j - e_k\|_{\ell^2} = \frac{r}{2}\sqrt{2}, \end{align*} so the sequence has no Cauchy subsequence. This contradicts the compactness of $K$. The same argument applies to any infinite-dimensional normed space: the closed unit ball is never compact (by [Riesz's characterisation of finite-dimensional spaces](/page/Compact%20Space)), so no point has a compact neighbourhood in the norm topology. Infinite-dimensional normed spaces are never locally compact. This is why the theory of locally compact spaces is predominantly a theory of *finite-dimensional* or *discrete* phenomena — and why functional analysis on infinite-dimensional spaces requires fundamentally different tools (weak topologies, compact operators, compact embeddings). [/example]

## Definition

The observations above suggest that the right weakening of compactness is to require compact neighbourhoods only *locally*, at each individual point. The challenge is to formulate this in a way that is strong enough to support the constructions of analysis — partitions of unity, exhaustion by compact sets, the one-point compactification — without requiring global compactness. There are several candidate definitions in the literature, and they diverge in the absence of the [Hausdorff](/page/Hausdorff%20Space) condition.

[definition: Locally Compact Space] A [topological space](/page/Topology) $(X, \tau)$ is **locally compact** if every point $x \in X$ has a compact neighbourhood: there exist an open set $U \in \tau$ and a [compact](/page/Compact%20Space) set $K \subset X$ such that \begin{align*} x \in U \subset K. \end{align*} [/definition]

Several remarks on this definition are in order.

[remark: Equivalent Formulations in the Hausdorff Case] When $(X, \tau)$ is Hausdorff, the following conditions are all equivalent to local compactness: 1. Every point has a compact neighbourhood. 2. Every point has an open neighbourhood whose closure is compact. 3. Every point has a neighbourhood base consisting of compact sets. 4. For every point $x \in X$ and every open set $V$ containing $x$, there exists an open set $W$ with $x \in W \subset \overline{W} \subset V$ and $\overline{W}$ compact. Condition (4) is the strongest and most useful: it says that compact neighbourhoods can be found *inside any prescribed open neighbourhood*, not merely somewhere around the point. The equivalence uses the Hausdorff property in an essential way — specifically, the fact that compact subsets of Hausdorff spaces are closed (so compact neighbourhoods automatically have closed, hence compact, closures contained in open neighbourhoods). In non-Hausdorff spaces, these conditions diverge, and authors adopt different conventions. We follow the convention above (condition 1) as the default and state additional hypotheses when needed. [/remark]

Since all four conditions coincide in the Hausdorff setting, and since the non-Hausdorff case is rarely encountered in analysis, we adopt the following standing assumption.

[remark: Convention on Hausdorff] Throughout this page, all locally compact spaces are assumed Hausdorff unless explicitly stated otherwise. This is standard in analysis, measure theory, and harmonic analysis. The combination "locally compact Hausdorff" (abbreviated **LCH**) is the default setting for the Riesz representation theorem, the theory of Radon measures, and abstract harmonic analysis. [/remark]

The following examples establish the basic landscape of locally compact spaces.

[example: Euclidean Spaces and Manifolds] The space $\mathbb{R}^n$ is locally compact: for each $x \in \mathbb{R}^n$, the closed ball $\overline{B}(x, 1)$ is a compact neighbourhood (compact by the Heine-Borel theorem). More generally, every topological [manifold](/page/Smooth%20Manifold) $M$ of dimension $n$ is locally compact, since every point has a neighbourhood homeomorphic to an open subset of $\mathbb{R}^n$, and open subsets of $\mathbb{R}^n$ are locally compact (as we show below). Every compact space is locally compact (take $K = X$ for any point). Every discrete space is locally compact (every singleton $\{x\}$ is a compact neighbourhood of $x$). [/example]

[example: $\mathbb{Q}$ Is Not Locally Compact] The rational numbers $\mathbb{Q}$ with the subspace topology inherited from $\mathbb{R}$ are **not** locally compact. To see this, suppose $K \subset \mathbb{Q}$ is compact and contains an open neighbourhood $U$ of some point $q \in \mathbb{Q}$. Since $U$ is open in the subspace topology, $U$ contains an interval $(q - \varepsilon, q + \varepsilon) \cap \mathbb{Q}$ for some $\varepsilon > 0$. In particular, $U$ contains a sequence of rationals converging to some irrational number $\alpha \in (q - \varepsilon, q + \varepsilon)$. Since $K$ is compact and Hausdorff (as a subspace of $\mathbb{R}$, hence metrizable), $K$ is closed in $\mathbb{R}$. But then $K$ must contain $\alpha$, which contradicts $K \subset \mathbb{Q}$. The failure of local compactness for $\mathbb{Q}$ reflects a fundamental topological deficiency: $\mathbb{Q}$ is not a [Baire space](/page/Baire%20Category%20Theorem), and its topology is too "fragmented" to support the constructions that local compactness enables. This example shows that being a dense subspace of a locally compact space does not guarantee local compactness — local compactness is not inherited by arbitrary subspaces. [/example]

## Subspaces of Locally Compact Spaces

A natural question is: which subspaces of a locally compact Hausdorff space inherit local compactness? The answer determines, for instance, which subsets of $\mathbb{R}^n$ are locally compact — and the example of $\mathbb{Q}$ above shows that the answer is not "all of them." The correct characterisation involves a topological condition that captures the spaces that are "well-embedded" enough to preserve the local compactness structure.

[quotetheorem:1059]

The locally closed condition $Y = U \cap C$ (with $U$ open and $C$ closed in $X$) includes two important special cases: open subspaces ($C = X$) and closed subspaces ($U = X$). Both of these inherit local compactness, but for different reasons.

For **open subspaces**: if $Y \subset X$ is open and $x \in Y$, then by local compactness of $X$ (using the strong form, condition (4) in the remark above), there exists an open set $W$ with $x \in W \subset \overline{W} \subset Y$ and $\overline{W}$ compact. So $\overline{W}$ is a compact neighbourhood of $x$ in $Y$.

For **closed subspaces**: if $Y \subset X$ is closed and $x \in Y$, take a compact neighbourhood $K$ of $x$ in $X$. Then $K \cap Y$ is a closed subset of a compact set, hence compact, and it contains an open neighbourhood of $x$ relative to $Y$.