From Compact Domains to Locally Compact Spaces
On a compact metric domain, the Arzelà–Ascoli theorem gives a practical test for relative compactness of a family of continuous functions: the family must be equicontinuous and its values at each point must be relatively compact. A noncompact domain has no single compact set on which to apply that theorem. A compact exhaustion lets us apply it successively on larger compact pieces, then make the resulting subsequences agree by a diagonal argument.
Let \(X\) be a locally compact metric space with a compact exhaustion \((K_n)_{n\in\mathbb{N}}\) satisfying \(K_n\subseteq\operatorname{int}K_{n+1}\) and \(\bigcup_{n\in\mathbb{N}}K_n=X\). We take \(X\ne\varnothing\), and the exhaustion sets to be nonempty. Let \(\mathbb{R}^m\) have its standard Euclidean topology and metric. Write \(C(X,\mathbb{R}^m)\) for the set of continuous maps from \(X\) to \(\mathbb{R}^m\).
The topology on \(C(X,\mathbb{R}^m)\) in this tutorial is the compact-open topology: its subbasis consists of the sets \([K,U]=\{f:f(K)\subseteq U\}\), where \(K\subseteq X\) is compact and \(U\subseteq\mathbb{R}^m\) is open. For this locally compact metric domain and metric target, this topology is equivalently the topology of uniform convergence on each compact subset. In particular, convergence means uniform convergence on each \(K_n\), not necessarily uniform convergence on all of \(X\).
For clarity, the exhaustion also gives a metric for this topology. Define \(p_n(f,g)=\sup_{x\in K_n}\|f(x)-g(x)\|\), which is finite because the functions are continuous and \(K_n\) is compact, and put
This is a metric: symmetry and the triangle inequality follow from those of the Euclidean metric and \(\min\{1,a+b\}\leq\min\{1,a\}+\min\{1,b\}\); if \(d_c(f,g)=0\), then \(f=g\) on every \(K_n\), hence on \(X\). Convergence in this metric is equivalent to uniform convergence on every \(K_n\): each summand tending to zero forces \(p_n\to0\), and conversely convergence of each \(p_n\), together with the bounded tail \(\sum_{n>N}2^{-n}\), makes the whole sum tend to zero. Every compact \(K\subseteq X\) is contained in some \(K_N\): the increasing open sets \(\operatorname{int}K_n\) cover \(X\), so finitely many cover \(K\), and the largest index among them suffices. Thus convergence on the exhaustion is convergence on every compact set. The compact-open/uniform-on-compacts identification follows by comparing neighborhoods: uniform closeness on a compact set ensures its image stays in a prescribed open set, and compactness gives finitely many local neighborhoods to control a uniform bound there.
The Local Arzelà–Ascoli Criterion
- \(\mathcal{F}\) is relatively compact in the compact-open topology;
- \(\mathcal{F}\) is equicontinuous on every compact subset of \(X\) and pointwise relatively compact.
Proof. First suppose \(\mathcal{F}\) is relatively compact, and let \(\overline{\mathcal{F}}\) denote its compact closure. For any compact \(K\subseteq X\), restriction \(f\mapsto f|_K\) is continuous into \(C(K,\mathbb{R}^m)\) with its uniform topology. Therefore the restrictions of \(\overline{\mathcal{F}}\) form a compact family there. The compact-domain Arzelà–Ascoli theorem says that a compact family in \(C(K,\mathbb{R}^m)\) is equicontinuous and has relatively compact values at each point. Applying this to each \(K\) gives equicontinuity of \(\mathcal{F}\) on every compact subset. Applying the pointwise conclusion at each \(x\in X\), with \(K=\{x\}\), gives pointwise relative compactness.
Conversely, suppose \(\mathcal{F}\) is equicontinuous on every compact subset and pointwise relatively compact. On each \(K_n\), its restrictions are equicontinuous, and at every \(x\in K_n\) their values have compact closure. The compact-domain Arzelà–Ascoli theorem therefore says that these restrictions are relatively compact in \(C(K_n,\mathbb{R}^m)\), with the uniform topology.
Take any sequence \((f_j)\) in \(\mathcal{F}\). Relative compactness on \(K_1\) gives a subsequence whose restrictions converge uniformly on \(K_1\). From that subsequence choose one converging uniformly on \(K_2\); continue in this way. Taking the diagonal subsequence gives a single subsequence \((f_{j_k})\) whose restrictions converge uniformly on every \(K_n\). Denote the limit on \(K_n\) by \(g_n\). If \(m\leq n\), the limits \(g_m\) and \(g_n|_{K_m}\) agree, because they are pointwise limits of the same subsequence on \(K_m\). Thus they define one function \(f:X\to\mathbb{R}^m\). This function is continuous: each point lies in some \(\operatorname{int}K_n\), and on that open set \(f\) agrees with the continuous function \(g_n\). The subsequence converges uniformly to \(f\) on every \(K_n\), hence converges to \(f\) in the compact-open topology.
We have shown that every sequence in \(\mathcal{F}\) has a subsequence converging in \(C(X,\mathbb{R}^m)\). The same is true of every sequence in its closure: approximate its \(k\)-th term by a point of \(\mathcal{F}\) within \(1/k\) in the metric \(d_c\), and use the subsequence result just proved. The closure is therefore sequentially compact. In metric spaces sequential compactness implies compactness: otherwise one can first construct a sequence with pairwise distances bounded below if the space is not totally bounded; and, once total boundedness holds, a sequentially compact metric space has a Lebesgue number for each open cover, so a finite net gives a finite subcover. Since \(d_c\) metrizes the compact-open topology, \(\overline{\mathcal{F}}\) is compact. This proves relative compactness. \(\square\)
The compact-domain theorem used here is the standard Arzelà–Ascoli theorem for compact metric domains: equicontinuity and pointwise relative compactness are equivalent to relative compactness in \(C(K,\mathbb{R}^m)\). The work specific to a noncompact domain is coordinating the compact-domain limits so that they define one continuous function on all of \(X\).
Applying the Criterion
Worked Example: A Compact Family on the Real Line
Let \(X=\mathbb{R}\) with its standard topology, take \(K_n=[-n,n]\), and consider \(\mathcal{F}=\{f_a:a\in[-1,1]\}\), where \(f_a(x)=a\sin x\). These \(K_n\) are compact, their interiors are nested, and their union is \(\mathbb{R}\).
For any \(x,y\in\mathbb{R}\) and \(a\in[-1,1]\), the mean value theorem applied to sine gives \(\lvert f_a(x)-f_a(y)\rvert=\lvert a\rvert\lvert\sin x-\sin y\rvert\leq\lvert x-y\rvert\). Thus, given \(\varepsilon>0\), choosing \(\delta=\varepsilon\) works on every \(K_n\) for every member of the family. At a fixed \(x\), the value set is \(\{a\sin x:a\in[-1,1]\}=[-|\sin x|,|\sin x|]\), a compact interval (including the singleton \(\{0\}\) when \(\sin x=0\)). Both conditions of the theorem hold, so \(\mathcal{F}\) is relatively compact in the compact-open topology. In fact, it is compact: the map \(a\mapsto f_a\) is continuous in that topology, since on \(K_n\) the uniform difference between \(f_a\) and \(f_b\) is at most \(|a-b|\).
Check: at \(x=\pi/2\), the values are \(f_a(\pi/2)=a\), so the value set is exactly \([-1,1]\); it is indeed compact and bounded as the pointwise condition requires.
Worked Example: Convergence on Compact Sets Without Global Uniform Convergence
On the same exhaustion of \(\mathbb{R}\), let \(\mathcal{G}=\{g_n:n\in\mathbb{N}\}\cup\{0\}\), where \(g_n(x)=\sin(x/n)\) and \(0\) denotes the zero function. The inequality \(\lvert\sin u-\sin v\rvert\leq\lvert u-v\rvert\) gives \(\lvert g_n(x)-g_n(y)\rvert\leq |x-y|/n\leq |x-y|\). The zero function also satisfies this bound, so the family is equicontinuous on every compact subset. At any fixed \(x\), all its values lie in \([-1,1]\); this closed bounded interval is compact in \(\mathbb{R}\). Hence \(\mathcal{G}\) is relatively compact.
More specifically, on \(K_N=[-N,N]\), for each \(x\in K_N\) we have \(|g_n(x)|=|\sin(x/n)|\leq |x|/n\leq N/n\). Therefore \(\sup_{x\in K_N}|g_n(x)|\leq N/n\), which tends to zero for each fixed \(N\). Thus \(g_n\) converges to zero in the compact-open topology.
Check: on all of \(\mathbb{R}\), the supremum is different: \(|g_n(n\pi/2)|=|\sin(\pi/2)|=1\), while \(|g_n(x)|\leq1\) for every \(x\). Therefore \(\sup_{x\in\mathbb{R}}|g_n(x)|=1\) for every \(n\), so the convergence is not uniform on \(\mathbb{R}\).
Worked Example: Pointwise Boundedness Does Not Replace Equicontinuity
On the compact domain \(K=[0,1]\), consider \(h_n(x)=\sin(nx)\), for \(n\in\mathbb{N}\). For every fixed \(x\), the values \(\{h_n(x):n\in\mathbb{N}\}\) lie in \([-1,1]\), so they are relatively compact. But the family is not equicontinuous at \(0\). If it were, for \(\varepsilon=1/2\) there would be a \(\delta>0\) such that \(|h_n(x)-h_n(0)|<1/2\) whenever \(x\in[0,1]\) and \(x<\delta\), for every \(n\).
Choose an integer \(n\geq2\) large enough that \(\pi/(2n)<\delta\), and set \(x=\pi/(2n)\). Then \(0\leq x\leq1\), \(x<\delta\), \(h_n(0)=\sin(0)=0\), and \(h_n(x)=\sin(n\pi/(2n))=\sin(\pi/2)=1\). Hence \(|h_n(x)-h_n(0)|=1\), contradicting the required bound \(1/2\). By the compact-domain Arzelà–Ascoli theorem, the family is not relatively compact in \(C([0,1],\mathbb{R})\).
Worked Example: An Equicontinuous Family That Is Not Relatively Compact
On \(X=\mathbb{R}\), let \(c_n(x)=n\) for \(n\in\mathbb{N}\), and let \(\mathcal{C}=\{c_n:n\in\mathbb{N}\}\). For every \(x,y\), \(|c_n(x)-c_n(y)|=|n-n|=0\), so this family is equicontinuous on every compact subset. At \(x=0\), its value set is \(\{1,2,3,\ldots\}\), which is unbounded and therefore has no compact closure in \(\mathbb{R}\). The theorem says \(\mathcal{C}\) is not relatively compact.
This failure can also be checked directly. Evaluation at \(0\), the map \(f\mapsto f(0)\), is continuous for the compact-open topology. A compact closure of \(\mathcal{C}\) would have compact image under evaluation, but that image would contain every positive integer, which cannot lie in a compact subset of \(\mathbb{R}\).
What the Local Criterion Does—and Does Not—Say
The criterion is useful when behavior at infinity is not controlled uniformly. It asks for the Arzelà–Ascoli conditions on each compact piece, not a single estimate that handles every point of \(X\) at once. In the example \(g_n(x)=\sin(x/n)\), every fixed compact interval sees convergence to zero, while the whole real line still contains points where the function has absolute value one.
Common pitfall. It is incorrect to infer global uniform convergence from uniform convergence on every \(K_n\). For \(g_n(x)=\sin(x/n)\), the estimate on \(K_N\) is \(\sup_{x\in K_N}|g_n(x)|\leq N/n\), which tends to zero when \(N\) is fixed. But choosing \(x=n\pi/2\) gives \(|g_n(x)|=1\), so the global supremum is exactly \(1\), not a quantity tending to zero. The exhaustion criterion concerns compact-open relative compactness; it does not assert compactness in a topology requiring uniform control over the entire noncompact domain.
Check Your Understanding
Use the compact-open topology and distinguish conditions on each compact subset from conditions on the whole domain.
- State the two conditions in the compact-exhaustion Arzelà–Ascoli criterion.
- Why does pointwise relative compactness alone fail for the family \(h_n(x)=\sin(nx)\) on \([0,1]\)?
- For \(g_n(x)=\sin(x/n)\) on \(\mathbb{R}\), show that \(\sup_{x\in[-N,N]}|g_n(x)|\leq N/n\) for fixed \(N\), but \(\sup_{x\in\mathbb{R}}|g_n(x)|=1\).
- Explain why the diagonal subsequence in the theorem gives one limit function on all of \(X\), rather than unrelated limits on the different \(K_n\).
- For the constant functions \(c_n(x)=n\), identify exactly which condition of the criterion fails.