A Sequential Proof Of Arzela-Ascoli

From Pointwise Selection to Uniform Convergence

For a sequence of continuous functions, compactness of the range values at each individual point can produce a subsequence that converges pointwise. Pointwise convergence alone does not control what happens between those points. Equicontinuity supplies that missing control: nearby domain points have nearby function values, uniformly across the family. The sequential proof below combines these two ingredients to obtain uniform convergence and then checks that the limit is continuous.

Let \((K,d)\) be a compact metric space, with \(K\) allowed to be empty, and give \(\mathbb{R}^m\) its Euclidean norm \(\|\cdot\|\). Write \(C(K,\mathbb{R}^m)\) for the set of continuous maps from \(K\) to \(\mathbb{R}^m\). A family \(\mathcal{F}\subseteq C(K,\mathbb{R}^m)\) is equicontinuous if for every \(x\in K\) and every \(\varepsilon>0\), there is a \(\delta>0\) such that, for every \(f\in\mathcal{F}\) and every \(y\in K\), \(d(x,y)<\delta\) implies \(\|f(x)-f(y)\|<\varepsilon\). The choice of \(\delta\) may depend on \(x\) and \(\varepsilon\), but not on \(f\).

For nonempty \(K\), the uniform metric on \(C(K,\mathbb{R}^m)\) is \(d_\infty(f,g)=\sup_{x\in K}\|f(x)-g(x)\|\). This supremum is finite: the function \(x\mapsto\|f(x)-g(x)\|\) is continuous on compact \(K\), so it is bounded. The nonnegativity, symmetry, and triangle inequality follow from those of the Euclidean norm; if \(d_\infty(f,g)=0\), then \(\|f(x)-g(x)\|=0\) for every \(x\), so \(f=g\). Thus \(d_\infty\) is a metric. When \(K=\varnothing\), there is exactly one map from \(K\) to \(\mathbb{R}^m\); define its distance from itself to be \(0\). This is a metric on that singleton function space.

A Countable Dense Set and a Diagonal Choice

The selection argument needs only countably many points at first. Compactness provides such a set in every nonempty compact metric space.

Proof. For each \(n\in\mathbb{N}\), the collection of open balls \(B_d(x,1/n)\), for \(x\in K\), covers \(K\). By compactness, choose a finite subcover with centers \(x_{n,1},\ldots,x_{n,r_n}\). Let \(D\) be the union of these finite sets of centers over all \(n\). It is countable, being a countable union of finite sets. To show it is dense, take \(x\in K\) and an open neighborhood \(U\) of \(x\). There is \(r>0\) such that \(B_d(x,r)\subseteq U\). Choose \(n\) with \(1/n<r\). The finite subcover at level \(n\) contains a ball \(B_d(x_{n,j},1/n)\) that contains \(x\); hence \(d(x,x_{n,j})<1/n<r\), so \(x_{n,j}\in U\). Thus every open neighborhood of every point meets \(D\), which proves density. \(\square\)

We can list the elements of \(D\) as \(a_1,a_2,\ldots\). If \(D\) is finite, repeat its elements in a sequence so that every element still occurs. At the first point, pointwise relative compactness allows a subsequence whose values converge at \(a_1\). From that subsequence select one whose values converge at \(a_2\), and continue. The diagonal subsequence then converges at every listed point.

The Sequential Arzelà–Ascoli Criterion

Proof. If \(K=\varnothing\), its function space consists of one map, so every sequence of functions is constant and converges uniformly to that map. Assume now that \(K\ne\varnothing\), and take any sequence \((f_n)\) in \(\mathcal{F}\). Let \(D=\{a_1,a_2,\ldots\}\) be a countable dense subset as in the lemma. At \(a_1\), the sequence \((f_n(a_1))\) lies in a set with compact closure in \(\mathbb{R}^m\), so it has a convergent subsequence. From that subsequence select one whose values converge at \(a_2\), and repeat for each \(a_j\). Here we use the standard fact that every sequence in a compact subset of \(\mathbb{R}^m\) has a convergent subsequence. Write \(I_j\subseteq\mathbb{N}\) for the nested infinite sets of indices after stage \(j\), so \(I_{j+1}\subseteq I_j\) and the values converge at \(a_j\) along indices in \(I_j\). Choose indices \(n_k\in I_k\) with \(n_1<n_2<\cdots\), which is possible because each \(I_k\) is infinite. For each fixed \(j\), all \(n_k\) with \(k\ge j\) belong to \(I_j\). Thus \((f_{n_k}(a_j))_k\) converges for every \(j\).

We prove that this diagonal subsequence is uniformly Cauchy. Fix \(\varepsilon>0\). By equicontinuity, for each \(x\in K\) choose \(\delta_x>0\) so that \(d(y,x)<\delta_x\) implies \(\|f(y)-f(x)\|<\varepsilon/6\) for all \(f\in\mathcal{F}\). The open balls \(B_d(x,\delta_x)\), for \(x\in K\), cover \(K\). Choose a finite subcover with centers \(x_1,\ldots,x_s\). For each \(i\), density of \(D\) lets us choose \(a_{j_i}\in D\) with \(d(a_{j_i},x_i)<\delta_{x_i}\). Since the subsequence values converge at each of these finitely many dense points, there is \(N\) such that, whenever \(k,l\ge N\), \(\|f_{n_k}(a_{j_i})-f_{n_l}(a_{j_i})\|<\varepsilon/3\) for every \(i=1,\ldots,s\).

Now take any \(x\in K\). It lies in \(B_d(x_i,\delta_{x_i})\) for some \(i\), so both \(d(x,x_i)<\delta_{x_i}\) and \(d(a_{j_i},x_i)<\delta_{x_i}\). Equicontinuity at the center \(x_i\), applied separately to these two points, gives \(\|f_{n_k}(x)-f_{n_k}(a_{j_i})\|\leq\|f_{n_k}(x)-f_{n_k}(x_i)\|+\|f_{n_k}(x_i)-f_{n_k}(a_{j_i})\|<\varepsilon/6+\varepsilon/6=\varepsilon/3\). The same reasoning gives \(\|f_{n_l}(x)-f_{n_l}(a_{j_i})\|<\varepsilon/3\). The triangle inequality and the choice of \(N\) now yield, for all \(k,l\ge N\), \[ \|f_{n_k}(x)-f_{n_l}(x)\| \leq \|f_{n_k}(x)-f_{n_k}(a_{j_i})\| +\|f_{n_k}(a_{j_i})-f_{n_l}(a_{j_i})\| +\|f_{n_l}(a_{j_i})-f_{n_l}(x)\| <\varepsilon. \] The index \(N\) works for every \(x\in K\), so \((f_{n_k})\) is uniformly Cauchy.

For each \(x\), the sequence \((f_{n_k}(x))\) is Cauchy in the complete space \(\mathbb{R}^m\), and therefore has a limit; define \(f(x)=\lim_{k\to\infty}f_{n_k}(x)\). The uniform Cauchy estimate implies uniform convergence to \(f\): given \(\varepsilon>0\), apply that estimate with \(\varepsilon/2\), obtaining \(N\) such that \(\|f_{n_k}(x)-f_{n_l}(x)\|<\varepsilon/2\) for all \(x\) and \(k,l\ge N\). Fix \(k\ge N\) and let \(l\to\infty\); continuity of the Euclidean norm gives \(\|f_{n_k}(x)-f(x)\|\leq\varepsilon/2<\varepsilon\) for every \(x\). Thus \(d_\infty(f_{n_k},f)\to0\).

It remains to verify that \(f\) is continuous. Let \(x\in K\) and \(\varepsilon>0\). Equicontinuity at \(x\) gives \(\delta>0\) such that \(d(x,y)<\delta\) implies \(\|g(x)-g(y)\|<\varepsilon/3\) for every \(g\in\mathcal{F}\). Choose \(k\) with \(\sup_{z\in K}\|f_{n_k}(z)-f(z)\|<\varepsilon/3\), using the uniform convergence just proved. For \(d(x,y)<\delta\), the triangle inequality gives \(\|f(x)-f(y)\|\leq\|f(x)-f_{n_k}(x)\|+\|f_{n_k}(x)-f_{n_k}(y)\|+\|f_{n_k}(y)-f(y)\|<\varepsilon\). This is continuity at \(x\). Since \(x\) was arbitrary, \(f\in C(K,\mathbb{R}^m)\), completing the proof. \(\square\)

Examples and a Common Pitfall

Common pitfall. A wrong argument is: “At each \(x\), the values \(f_n(x)\) have a convergent subsequence, so a subsequence converges uniformly.” Pointwise subsequences can depend on \(x\), and even pointwise convergence of the whole sequence does not imply uniform convergence. For \(h_n(x)=x^n\), the pointwise limit above exists, but the uniform error has supremum \(1\) for every \(n\), not a quantity tending to zero. The diagonal construction first secures convergence on a countable dense set; the finite-cover estimate and equicontinuity are the separate steps that make this convergence uniform on all of \(K\).