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 $Y$. The key negative example is that an arbitrary subspace need not be locally compact, as $\mathbb{Q} \subset \mathbb{R}$ demonstrates. The rationals are neither open nor closed in $\mathbb{R}$, and indeed $\mathbb{Q}$ is not locally closed in $\mathbb{R}$: for any open set $U$ and closed set $C$ in $\mathbb{R}$, the intersection $U \cap C$ is either empty, or an open subset of $C$, or an $F_\sigma$ set — but $\mathbb{Q}$ is dense and co-dense, so it cannot be expressed as such an intersection. ## The One-Point Compactification A space that is locally compact but not compact is "almost" compact — it only fails compactness because it is not "closed off." The natural question is whether there is a canonical way to add a single point to make it compact. This construction is fundamental: it provides the right framework for understanding functions that "vanish at infinity," it connects locally compact spaces to compact ones, and it underlies the theory of $C_0(X)$. The idea is simple: adjoin a new point $\infty$ to $X$ and declare that the open sets around $\infty$ are the complements of compact subsets of $X$. This forces every open cover to eventually "use" a set containing $\infty$, which swallows everything outside a compact set — and compact sets in $X$ can be covered by finitely many of the remaining sets. [definition: One-Point Compactification] Let $(X, \tau)$ be a topological space. The **one-point compactification** (or **Alexandroff compactification**) of $X$ is the space $X^+ = X \cup \{\infty\}$, where $\infty \notin X$, equipped with the topology $\tau^+$ consisting of: 1. All open sets $U \in \tau$ (i.e., the original open sets of $X$), and 2. All sets of the form $(X \setminus K) \cup \{\infty\}$, where $K \subset X$ is compact. The topology $\tau^+$ is the unique topology on $X^+$ that makes $X^+$ compact and restricts to $\tau$ on $X$. [/definition] The verification that $\tau^+$ is indeed a topology requires checking closure under unions and finite intersections. For sets of type (2): the intersection $(X \setminus K_1) \cup \{\infty\}$ and $(X \setminus K_2) \cup \{\infty\}$ equals $(X \setminus (K_1 \cup K_2)) \cup \{\infty\}$, which is of the same form since $K_1 \cup K_2$ is compact. The union of a family of such sets corresponds to intersecting compact sets, but the intersection of compact sets in a Hausdorff space is compact, so the result is again of the correct form. The compactness of $X^+$ follows from the construction: given an open cover $\{U_\alpha\}$ of $X^+$, at least one set $U_{\alpha_0}$ contains $\infty$, so $U_{\alpha_0} \supset (X \setminus K) \cup \{\infty\}$ for some compact $K \subset X$. The remaining sets $\{U_\alpha \cap X\}$ cover $K$, which is compact, so finitely many suffice. The critical question is: *when is $X^+$ Hausdorff?* This is where local compactness enters as a necessary and sufficient condition. [quotetheorem:1060] The "only if" direction is instructive: if $X^+$ is Hausdorff, then $X$, as a subspace of a Hausdorff space, is Hausdorff. To see that $X$ is locally compact, fix $x \in X$. Since $X^+$ is Hausdorff and compact (hence regular), there exist disjoint open sets in $X^+$ separating $x$ from $\infty$. The open set containing $\infty$ has the form $(X \setminus K) \cup \{\infty\}$ for some compact $K$. The open set containing $x$ is then contained in $K$, so $K$ is a compact neighbourhood of $x$ in $X$. The "if" direction requires separating a point $x \in X$ from $\infty$ in $X^+$. By local compactness, $x$ has an open neighbourhood $U$ with $\overline{U}$ compact. Then $U$ and $(X \setminus \overline{U}) \cup \{\infty\}$ are disjoint open sets in $X^+$ separating $x$ from $\infty$. [example: Classical One-Point Compactifications] **The real line.** The one-point compactification $\mathbb{R}^+$ of $\mathbb{R}$ is homeomorphic to the circle $S^1$. The homeomorphism can be constructed via stereographic projection: identify $S^1$ with the unit circle in $\mathbb{R}^2$ centred at the origin, let $N = (0, 1)$ be the "north pole," and define \begin{align*} \varphi: S^1 \setminus \{N\} &\to \mathbb{R} \\ (x, y) &\mapsto \frac{x}{1 - y}. \end{align*} This is a homeomorphism $S^1 \setminus \{N\} \cong \mathbb{R}$, and under this identification, the point $N$ plays the role of $\infty$. As a point on $S^1$ moves toward $N$ from either direction, its image under $\varphi$ diverges to $+\infty$ or $-\infty$ — the two "ends" of $\mathbb{R}$ are identified in the compactification. **Euclidean space.** Similarly, the one-point compactification $(\mathbb{R}^n)^+$ is homeomorphic to the $n$-sphere $S^n$, via stereographic projection from the north pole. This identifies $\mathbb{R}^n$ with $S^n \setminus \{N\}$ and adds the single point at infinity. **A discrete space.** If $X$ is an infinite discrete space (every subset is open), then $X^+$ has the topology where open sets are either subsets of $X$ or cofinite subsets of $X^+$. Points of $X$ are isolated, and $\infty$ has a neighbourhood base consisting of cofinite subsets. This is the one-point compactification of a discrete space. [/example] The one-point compactification provides a clean characterisation of locally compact Hausdorff spaces: they are precisely the spaces that arise as open subspaces of compact Hausdorff spaces. [quotetheorem:1061] One direction is immediate: an open subspace of a compact Hausdorff space is locally compact and Hausdorff (as established in the subspace section above). For the converse, $X$ embeds as an open subspace of its one-point compactification $X^+$, which is compact Hausdorff by the previous theorem. ## Exhaustion by Compact Sets On a compact space, arguments that require finiteness (covering arguments, partition-of-unity arguments) work directly. On a locally compact space, the strategy is to reduce to compact subsets and then assemble the local results. The tool that makes this reduction systematic is an **exhaustion** — a nested sequence of compact sets that fills out the space. The basic question is: can every locally compact Hausdorff space be written as a countable union of compact sets, arranged so that each is contained in the interior of the next? The answer is yes, provided the space satisfies a mild countability condition. [definition: Sigma-Compact Space] A topological space $(X, \tau)$ is **$\sigma$-compact** if it can be written as a countable union of compact subsets: \begin{align*} X = \bigcup_{j=1}^\infty K_j, \end{align*} where each $K_j \subset X$ is compact. [/definition] Every compact space is $\sigma$-compact (take $K_1 = X$). The space $\mathbb{R}^n$ is $\sigma$-compact: $\mathbb{R}^n = \bigcup_{j=1}^\infty \overline{B}(0, j)$. More generally, every [second-countable](/page/Separable%20Space) locally compact Hausdorff space is $\sigma$-compact (given a countable base, each basic open set has compact closure, and countably many of these cover the space). In particular, every locally compact metrizable separable space is $\sigma$-compact. The following theorem gives the exhaustion in its most useful form. [quotetheorem:1062] The nesting condition $K_j \subset \operatorname{int}(K_{j+1})$ is essential — it provides the "room" needed to construct cut-off functions, partitions of unity, and mollification arguments. Without the nesting, the compact sets might meet along their boundaries with no open buffer between them. In PDE theory on $\mathbb{R}^n$, the standard exhaustion $K_j = \overline{B}(0, j)$ satisfies $K_j \subset B(0, j+1) = \operatorname{int}(K_{j+1})$, and this is the starting point for cut-off function arguments, approximation of Sobolev functions, and passage from local to global estimates. [example: Exhaustion of $\mathbb{R}^n$ and Cut-Off Functions] Take $K_j = \overline{B}(0, j)$. Then $K_j \subset B(0, j + 1) = \operatorname{int}(K_{j+1})$ and $\mathbb{R}^n = \bigcup_{j=1}^\infty K_j$. For each $j$, define the **annular region** $A_j = K_{j+1} \setminus \operatorname{int}(K_{j-1})$ (with $K_0 = \varnothing$). The annuli $A_j$ provide an open cover of $\mathbb{R}^n$ with controlled overlap — each point lies in at most three consecutive annuli. For each $j$, one constructs a smooth **cut-off function** $\eta_j \in C_c^\infty(\mathbb{R}^n)$ satisfying: \begin{align*} 0 \le \eta_j \le 1, \quad \eta_j \equiv 1 \text{ on } K_j, \quad \operatorname{supp}(\eta_j) \subset \operatorname{int}(K_{j+1}). \end{align*} Such functions exist by mollification: take the indicator $\mathbb{1}_{K_j}$, convolve with a smooth mollifier $\rho_\varepsilon$ for sufficiently small $\varepsilon > 0$, and the result is smooth with support in a small enlargement of $K_j$. These cut-off functions are the fundamental tool for localisation in PDE theory: multiplying a function $u$ by $\eta_j$ produces a compactly supported function that agrees with $u$ on $K_j$, allowing the application of results that require compact support (Sobolev inequalities, integration by parts, density arguments). [/example] ## Functions Vanishing at Infinity and $C_0(X)$ On a compact Hausdorff space $K$, the space $C(K)$ of all continuous real-valued functions is a Banach space under the supremum norm $\|f\|_\infty = \sup_{x \in K} |f(x)|$, and it plays a central role in functional analysis through the Riesz representation theorem and the Stone-Weierstrass theorem. On a locally compact but non-compact space $X$, the space $C(X)$ of all continuous functions is too large — it is not a normed space in any natural way (unbounded continuous functions exist, and there is no natural norm). The space $C_b(X)$ of bounded continuous functions is a Banach space, but it is too large for duality theory: its dual is the space of finitely additive measures, not the Radon measures that arise in analysis. The correct replacement is to consider functions that are "negligible at infinity" — that become arbitrarily small outside compact sets. [definition: Functions Vanishing at Infinity] Let $X$ be a locally compact Hausdorff space. A continuous function $f: X \to \mathbb{R}$ **vanishes at infinity** if for every $\varepsilon > 0$, there exists a compact set $K \subset X$ such that $|f(x)| < \varepsilon$ for all $x \in X \setminus K$. The space of all continuous functions vanishing at infinity is denoted $C_0(X)$. It is equipped with the supremum norm: \begin{align*} \|f\|_\infty = \sup_{x \in X} |f(x)|. \end{align*} [/definition] The condition "vanishing at infinity" is the precise formulation of "the function becomes negligible far from any compact set." In the one-point compactification $X^+$, a function $f \in C_0(X)$ extends continuously to $X^+$ by setting $f(\infty) = 0$, and conversely, every continuous function on $X^+$ that vanishes at $\infty$ restricts to a function in $C_0(X)$. This gives an isometric identification: \begin{align*} C_0(X) \cong \{g \in C(X^+) : g(\infty) = 0\}. \end{align*} When $X$ is compact, every continuous function automatically vanishes at infinity (take $K = X$), so $C_0(X) = C(X)$. The space $C_c(X)$ of continuous functions with **compact support** — those $f \in C(X)$ for which $\{x \in X : f(x) \neq 0\}$ has compact closure — is a dense subspace of $C_0(X)$. This density is crucial: many constructions (integration, convolution, approximation) are first carried out on $C_c(X)$ and then extended to $C_0(X)$ by continuity. [quotetheorem:1102] The completeness of $C_0(X)$ follows from the completeness of $C(X^+)$: the subspace $\{g \in C(X^+) : g(\infty) = 0\}$ is closed in $C(X^+)$ (since evaluation at $\infty$ is continuous), hence complete. The density of $C_c(X)$ uses the exhaustion by compact sets: given $f \in C_0(X)$ and $\varepsilon > 0$, choose a compact $K$ with $|f| < \varepsilon$ outside $K$, then multiply $f$ by a continuous cut-off function that equals $1$ on $K$ and has compact support. The result is in $C_c(X)$ and is within $\varepsilon$ of $f$ in the supremum norm. [example: $C_0(\mathbb{R})$ and Its Elements] On $\mathbb{R}$, the function $f(x) = e^{-x^2}$ belongs to $C_0(\mathbb{R})$: for any $\varepsilon > 0$, choose $R$ with $e^{-R^2} < \varepsilon$, and then $|f(x)| < \varepsilon$ for $|x| > R$, so $K = [-R, R]$ works. More generally, any continuous function with $\lim_{|x| \to \infty} f(x) = 0$ belongs to $C_0(\mathbb{R})$. The function $g(x) = \sin(x)/(1 + x^2)$ is in $C_0(\mathbb{R})$: $|g(x)| \le 1/(1 + x^2) \to 0$ as $|x| \to \infty$. The constant function $h(x) = 1$ is in $C_b(\mathbb{R})$ but **not** in $C_0(\mathbb{R})$: for $\varepsilon = 1/2$, no compact $K$ satisfies $|h(x)| < 1/2$ for all $x \notin K$. The function $p(x) = x$ is not even in $C_b(\mathbb{R})$, let alone $C_0(\mathbb{R})$. [/example] ## The Riesz Representation Theorem for Locally Compact Spaces One of the most far-reaching consequences of local compactness is the **Riesz-Markov-Kakutani representation theorem**, which identifies positive linear functionals on $C_c(X)$ with Radon measures. This theorem is the foundation of modern measure theory on locally compact spaces: it explains why Radon measures — rather than arbitrary Borel measures — are the "correct" measures in analysis, and it provides the link between the functional-analytic viewpoint (linear functionals) and the measure-theoretic viewpoint (integration). The question the theorem answers is: if $\Lambda: C_c(X) \to \mathbb{R}$ is a positive linear functional (meaning $\Lambda(f) \ge 0$ whenever $f \ge 0$), must $\Lambda$ be given by integration against some measure? And if so, what properties must the measure satisfy? In the compact case, the answer is clean: every positive linear functional on $C(K)$ is integration against a finite Borel measure. In the locally compact case, the measure may be infinite (as for Lebesgue measure on $\mathbb{R}$), and additional regularity conditions on the measure are required. [definition: Radon Measure] Let $X$ be a locally compact Hausdorff space and let $\mathcal{B}(X)$ denote the Borel $\sigma$-algebra on $X$. A **Radon measure** on $X$ is a Borel measure $\mu: \mathcal{B}(X) \to [0, +\infty]$ satisfying: 1. **Locally finite**: $\mu(K) < \infty$ for every compact $K \subset X$. 2. **Outer regular**: For every Borel set $E \in \mathcal{B}(X)$, \begin{align*} \mu(E) = \inf\{\mu(U) : U \supset E, \, U \text{ open}\}. \end{align*} 3. **Inner regular on open sets**: For every open set $U \subset X$, \begin{align*} \mu(U) = \sup\{\mu(K) : K \subset U, \, K \text{ compact}\}. \end{align*} [/definition] The regularity conditions are the key: they ensure that the measure is determined by its values on compact sets (inner regularity) and approximable from outside by open sets (outer regularity). Without these, Borel measures on a locally compact space can be pathological — for instance, there exist non-regular Borel measures on locally compact spaces that are not $\sigma$-compact. [quotetheorem:976] The uniqueness relies on the density of $C_c(X)$ in $L^1(X, \mu)$ and on the regularity of $\mu$. Two Radon measures that agree on $C_c(X)$ must agree on all open sets (by inner regularity), hence on all Borel sets (by outer regularity). This is where the locally compact Hausdorff structure is essential: it provides enough continuous compactly supported functions (via Urysohn's lemma for LCH spaces) to separate measures. The theorem does not extend to spaces that are not locally compact. On an infinite-dimensional Banach space $X$ with the norm topology, the space $C_c(X)$ may contain only the zero function (since the only compact sets have empty interior), and the Riesz representation framework collapses. ## Locally Compact Groups and Haar Measure The combination of local compactness with algebraic structure — specifically, the structure of a topological group — produces one of the most powerful objects in mathematics: the **Haar measure**, an invariant measure that provides the foundation for harmonic analysis and representation theory. A **topological group** is a group $G$ equipped with a topology making the group operations continuous: multiplication $G \times G \to G$ and inversion $G \to G$ are both continuous. The question is: when does a topological group carry a nonzero measure $\mu$ that is invariant under left translation, i.e., $\mu(gE) = \mu(E)$ for all $g \in G$ and all measurable $E$? On a compact group, the counting measure (normalised) or the Riesz representation theorem immediately yields such a measure. On a general topological group, there is no reason for such a measure to exist. But on a locally compact group, the Haar measure exists and is essentially unique. [quotetheorem:1063] The local compactness hypothesis is essential for both existence and uniqueness. The proof of existence (due to Haar, with refinements by Weil and Cartan) constructs the measure by approximating it with averages over compact neighbourhoods — an approach that requires compact neighbourhoods to exist. The uniqueness proof uses the Riesz representation theorem, which itself requires local compactness. [example: Haar Measures on Classical Groups] **$(\mathbb{R}^n, +)$.** The additive group of Euclidean space is locally compact, and the Lebesgue measure $\mathcal{L}^n$ is the (left and right) Haar measure, unique up to scalar multiples. **$(\mathbb{R}_{> 0}, \times)$.** The multiplicative group of positive reals is locally compact. The Haar measure is $d\mu(x) = dx/x$, where $dx$ denotes Lebesgue measure. To verify invariance: for $a > 0$, the substitution $y = ax$ gives \begin{align*} \mu(aE) = \int_{aE} \frac{dy}{y} = \int_E \frac{a \, dx}{ax} = \int_E \frac{dx}{x} = \mu(E). \end{align*} **$\operatorname{GL}_n(\mathbb{R})$.** The general linear group, as an open subset of $\mathbb{R}^{n^2}$, is locally compact. The left Haar measure is \begin{align*} d\mu(A) = \frac{d\mathcal{L}^{n^2}(A)}{|\det A|^n}, \end{align*} where $d\mathcal{L}^{n^2}(A)$ denotes Lebesgue measure on $\mathbb{R}^{n^2}$ (identifying $n \times n$ matrices with $\mathbb{R}^{n^2}$). The power $n$ in the denominator arises from the Jacobian of the left-multiplication map $B \mapsto AB$. **Finite groups.** Every finite group, with the discrete topology, is compact (hence locally compact). The Haar measure is the counting measure: $\mu(\{g\}) = 1$ for every $g \in G$. [/example] ## Products of Locally Compact Spaces A basic stability question for any topological property is whether it is preserved under products. For compactness, Tychonoff's theorem gives a strong affirmative answer: arbitrary products of compact spaces are compact. For local compactness, the situation is more delicate. The fundamental difficulty is that the [product topology](/page/Product%20Topology) on an infinite product is very coarse — basic open sets are "cylinders" that constrain only finitely many coordinates. A compact neighbourhood in the product would need to control all coordinates simultaneously, but a basic open set only constrains finitely many. This conflict means that infinite products of locally compact spaces need not be locally compact. [quotetheorem:1064] The argument is constructive: given $(x_1, \ldots, x_N) \in X_1 \times \cdots \times X_N$, choose compact neighbourhoods $K_i$ of $x_i$ in $X_i$ (with open sets $U_i$ satisfying $x_i \in U_i \subset K_i$). Then $K_1 \times \cdots \times K_N$ is compact (by [Tychonoff's theorem](/page/Compact%20Space) for finite products), and $U_1 \times \cdots \times U_N$ is an open neighbourhood of the point contained in the product of the compact sets. For infinite products, the situation changes dramatically. [quotetheorem:1103] This theorem reveals a sharp difference between compactness and local compactness. Tychonoff's theorem asserts that an arbitrary product of compact spaces is compact — no finiteness restriction on the index set is needed. For local compactness, the "all but finitely many must be compact" condition is essential: the non-compact factors must be finite in number. [example: $\mathbb{R}^{\mathbb{N}}$ Is Not Locally Compact] Consider the countably infinite product $\mathbb{R}^{\mathbb{N}} = \prod_{k=1}^\infty \mathbb{R}$ with the product topology. Each factor $\mathbb{R}$ is locally compact but not compact. By the theorem above, $\mathbb{R}^{\mathbb{N}}$ is **not** locally compact. To see this directly, suppose $K \subset \mathbb{R}^{\mathbb{N}}$ is compact and contains a basic open neighbourhood of the origin. A basic open set containing the origin has the form $U = \prod_{k=1}^\infty U_k$, where $U_k$ is open in $\mathbb{R}$ with $0 \in U_k$, and $U_k = \mathbb{R}$ for all but finitely many $k$. In particular, there exists an index $k_0$ such that $U_{k_0} = \mathbb{R}$. The projection $\pi_{k_0}: \mathbb{R}^{\mathbb{N}} \to \mathbb{R}$ is continuous, and $\pi_{k_0}(K) \supset \pi_{k_0}(U) = \mathbb{R}$. Since continuous images of compact sets are compact, $\pi_{k_0}(K)$ would need to be compact. But $\pi_{k_0}(K) = \mathbb{R}$ is not compact. Contradiction. This failure is not merely a technical nuisance — it reflects a genuine difference between finite-dimensional and infinite-dimensional product spaces. The space $\mathbb{R}^{\mathbb{N}}$ is metrizable (it is homeomorphic to $\ell^2$ with a different, weaker topology), [Polish](/page/Polish%20Space), and $\sigma$-compact, but it is not locally compact. The function space $C([0,1])$ with the compact-open topology is another important example of a metrizable space that fails local compactness. [/example] ## Urysohn-Type Results and Partitions of Unity One of the most important features of locally compact Hausdorff spaces is the availability of continuous functions that separate compact sets from their complements. In a general Hausdorff space, the topology may not have enough continuous functions to separate points from closed sets — this is why the Urysohn lemma requires normality. But in a locally compact Hausdorff space, the local compactness provides a substitute: compact sets and points can always be separated by open sets with compact closure, and Urysohn's lemma can be applied within these compact neighbourhoods. The following result is the local version of Urysohn's lemma tailored to locally compact Hausdorff spaces. It is used constantly in analysis — every construction of cut-off functions, mollifiers, and partitions of unity on $\mathbb{R}^n$ or on manifolds is an instance of this result. [quotetheorem:1065] The idea is to use the strong form of local compactness (condition (4) in the remark after the definition) to find an open set $V$ with $K \subset V \subset \overline{V} \subset U$ and $\overline{V}$ compact. Since $\overline{V}$ is a compact Hausdorff space — hence normal — Urysohn's lemma for normal spaces provides a continuous function $g: \overline{V} \to [0,1]$ with $g \equiv 1$ on $K$ and $g \equiv 0$ on $\overline{V} \setminus V$. Extending $g$ by zero outside $\overline{V}$ gives the desired function $f$. This result enables the construction of **partitions of unity** on locally compact Hausdorff spaces, which are the tool for passing from local constructions to global ones. [quotetheorem:1066] Partitions of unity are the primary mechanism for globalising local constructions. Given a property or construction defined locally on each $U_\alpha$ (a local section of a vector bundle, a local solution to a PDE, a local definition of a measure), the partition of unity $\{\psi_\alpha\}$ allows these local pieces to be "blended" into a global object: $\sum_\alpha \psi_\alpha \cdot (\text{local object on } U_\alpha)$. The local finiteness ensures that this sum is well-defined, and the compact supports ensure that each piece has controlled behaviour. ## Standard Techniques for Working with Locally Compact Spaces The following techniques recur throughout analysis and topology whenever locally compact spaces arise. Mastering them is essential for applying the theory effectively. ### Localisation via Cut-Off Functions The basic strategy: given a problem on a non-compact locally compact space $X$, multiply by a cut-off function to reduce to a problem on a compact set, solve the compact problem, and then take limits as the cut-off exhausts $X$. [example: Proving an Integral Identity on $\mathbb{R}^n$ by Localisation] Suppose $u, v \in C_c^\infty(\mathbb{R}^n)$ and we wish to establish the identity \begin{align*} \int_{\mathbb{R}^n} u \, \Delta v \, d\mathcal{L}^n = \int_{\mathbb{R}^n} v \, \Delta u \, d\mathcal{L}^n. \end{align*} Since both $u$ and $v$ have compact support, there exists $R > 0$ with $\operatorname{supp}(u) \cup \operatorname{supp}(v) \subset B(0, R)$. The identity reduces to an integral over the compact set $\overline{B}(0, R)$. On this bounded domain, integration by parts (Green's identity) applies directly: the boundary terms vanish because $u$ and $v$ vanish on $\partial B(0, R)$. For functions not in $C_c^\infty$ but in a Sobolev space $W^{2,2}(\mathbb{R}^n)$, the same identity is established by inserting a cut-off function $\eta_R \in C_c^\infty(\mathbb{R}^n)$ with $\eta_R \equiv 1$ on $B(0, R)$, applying the identity to $\eta_R u$ and $v$ (which is valid since $\eta_R u$ has compact support), and then sending $R \to \infty$ using dominated convergence. [/example] ### Exhaustion Arguments When a result holds on compact sets and one needs it on the full (non-compact) space, the strategy is to apply the result to each compact set $K_j$ in an exhaustion and then pass to the limit. [example: Measurability via Exhaustion] Let $X$ be a $\sigma$-compact locally compact Hausdorff space with Radon measure $\mu$, and let $f: X \to \mathbb{R}$ be a function that is measurable when restricted to each $K_j$ in an exhaustion. Then $f$ is measurable on $X$. Take an exhaustion $K_1 \subset K_2 \subset \cdots$ with $X = \bigcup_{j=1}^\infty K_j$. For any $a \in \mathbb{R}$, \begin{align*} \{x \in X : f(x) > a\} = \bigcup_{j=1}^\infty \{x \in K_j : f(x) > a\}. \end{align*} Each set $\{x \in K_j : f(x) > a\}$ is measurable (since $f|_{K_j}$ is measurable), and a countable union of measurable sets is measurable. [/example] ### The One-Point Compactification Trick Some problems on locally compact spaces become simpler when reformulated on the compact space $X^+$. The idea is to work with $C(X^+)$ instead of $C_0(X)$, exploit the compactness of $X^+$, and then restrict back to $X$. [example: Dual of $C_0(X)$ via Compactification] To identify the dual space $C_0(X)^*$, one uses the identification $C_0(X) \cong \{g \in C(X^+) : g(\infty) = 0\}$. Since $X^+$ is compact Hausdorff, the Riesz representation theorem for compact spaces identifies $C(X^+)^*$ with the space of signed Radon measures on $X^+$. The subspace $C_0(X)$ corresponds (via the Hahn-Banach theorem) to measures $\nu$ on $X^+$ satisfying $\nu(\{\infty\}) = 0$, which are exactly the signed Radon measures on $X$ (extended by zero at infinity). This yields the identification \begin{align*} C_0(X)^* \cong \mathcal{M}(X), \end{align*} where $\mathcal{M}(X)$ denotes the space of finite signed Radon measures on $X$. [/example] ## References 1. Munkres, J. R., *Topology* (2000). 2. Folland, G. B., *Real Analysis: Modern Techniques and Their Applications* (1999). 3. Rudin, W., *Real and Complex Analysis* (1987). 4. Conway, J. B., *A Course in Functional Analysis* (1990). 5. Hewitt, E. and Ross, K. A., *Abstract Harmonic Analysis, Vol. I* (1963). 6. Nachbin, L., *The Haar Integral* (1965). 7. Kelley, J. L., *General Topology* (1955). 8. Bourbaki, N., *General Topology, Chapters 1-4* (1995).