Arzela-Ascoli Theorem For Compact Metric Domains

Compactness for Families of Functions

A family of continuous functions can have bounded values at every point and still fail to be compact in a metric that measures uniform error. The missing ingredient is control of how rapidly the functions can change as the input varies. The Arzelà–Ascoli theorem makes this distinction precise: on a compact metric domain, equicontinuity together with pointwise relative compactness is exactly the condition for relative compactness in the uniform metric, provided the target is complete.

Let \((K,d_K)\) be a compact metric space and \((Y,d_Y)\) a metric space. Write \(C(K,Y)\) for the set of continuous functions from \(K\) to \(Y\). For \(f,g\in C(K,Y)\), define the uniform metric by \[ d_\infty(f,g)=\sup_{x\in K}d_Y(f(x),g(x)). \] The supremum is finite: \(f(K)\) and \(g(K)\) are compact, hence bounded in \(Y\). If \(K\) is empty, there is just one function from \(K\) to \(Y\), and \(C(K,Y)\) is a one-point metric space with distance zero.

Equicontinuity requires one \(\delta\) to work for every function in the family. Pointwise relative compactness, by contrast, concerns the values at one fixed input at a time. The theorem combines these different kinds of control.

Why Relative Compactness Gives the Two Conditions

We first prove the forward direction. Suppose the uniform closure \(H=\overline{F}\) is compact. Fix \(x\in K\). The evaluation map \(e_x:C(K,Y)\to Y\), defined by \(e_x(f)=f(x)\), is continuous, because \[ d_Y(e_x(f),e_x(g))=d_Y(f(x),g(x))\leq d_\infty(f,g). \] Thus \(e_x(H)\) is compact. It contains \(F(x)\), so the closure of \(F(x)\) is a closed subset of the compact set \(e_x(H)\), and is compact. Here we use the standard fact that compact subsets of a metric space are closed.

To show equicontinuity, fix \(\varepsilon>0\). Compactness of \(H\) gives finitely many functions \(g_1,\ldots,g_m\in H\) such that every \(f\in H\) satisfies \(d_\infty(f,g_i)<\varepsilon/3\) for some \(i\). Each \(g_i\) is uniformly continuous on the compact metric space \(K\). Taking the minimum of the finitely many resulting positive distance thresholds gives a single \(\delta>0\) such that \(d_K(x,y)<\delta\) implies \(d_Y(g_i(x),g_i(y))<\varepsilon/3\) for every \(i\). For \(f\in H\), choose a corresponding \(g_i\). Then \[ d_Y(f(x),f(y)) \leq d_Y(f(x),g_i(x))+d_Y(g_i(x),g_i(y))+d_Y(g_i(y),f(y)) <\varepsilon. \] This proves equicontinuity of \(H\), and hence of \(F\).

Why the Two Conditions Give Relative Compactness

For the converse, we build finite uniform approximations to \(F\). A subset of a metric space is called totally bounded if, for every positive radius, finitely many open balls of that radius cover it. We will show that \(F\) is totally bounded, then use completeness.

Assume \(F\) is equicontinuous and pointwise relatively compact, and fix \(\varepsilon>0\). Choose \(\delta>0\) from equicontinuity for the error \(\varepsilon/3\). Compactness of \(K\) gives points \(x_1,\ldots,x_m\in K\) such that the balls \(B(x_i,\delta)\) cover \(K\). For each \(i\), pointwise relative compactness implies that \(F(x_i)\) has a finite \(\varepsilon/6\)-net: finitely many points of \(Y\) such that every element of \(F(x_i)\) is within distance \(\varepsilon/6\) of one of them.

Assign to each \(f\in F\), at each \(x_i\), one of the finitely many net points within \(\varepsilon/6\) of \(f(x_i)\). There are only finitely many possible lists of assignments. For each list that occurs, choose one function \(g\in F\) with that list. If \(f\) and \(g\) have the same list, then at every \(x_i\), \(d_Y(f(x_i),g(x_i))<2(\varepsilon/6)=\varepsilon/3\). For any \(x\in K\), choose \(i\) with \(d_K(x,x_i)<\delta\). Equicontinuity applies to both functions, so \[ d_Y(f(x),g(x)) \leq d_Y(f(x),f(x_i))+d_Y(f(x_i),g(x_i))+d_Y(g(x_i),g(x)) <\frac{\varepsilon}{3}+\frac{\varepsilon}{3}+\frac{\varepsilon}{3} =\varepsilon. \] Thus these finitely many chosen functions give an \(\varepsilon\)-cover of \(F\). The family \(F\) is totally bounded.

The closure of a totally bounded set is totally bounded: cover \(F\) by finitely many balls of radius \(\varepsilon/2\); the corresponding closed balls of radius \(\varepsilon/2\) contain its closure, and are contained in the open balls with the same centers and radius \(\varepsilon\). It remains to verify completeness and the metric compactness fact needed here.

Proof. Let \((f_n)\) be Cauchy in \(d_\infty\). For each \(x\in K\), the inequality \(d_Y(f_n(x),f_m(x))\leq d_\infty(f_n,f_m)\) shows that \((f_n(x))\) is Cauchy in \(Y\). Completeness of \(Y\) gives a limit \(f(x)\in Y\). To prove uniform convergence, let \(\varepsilon>0\). There is \(N\) such that \(d_\infty(f_n,f_m)<\varepsilon/2\) whenever \(n,m\geq N\). Fix \(n\geq N\) and \(x\in K\), and let \(m\) tend to infinity in \(d_Y(f_n(x),f_m(x))<\varepsilon/2\). Continuity of the metric gives \(d_Y(f_n(x),f(x))\leq\varepsilon/2\). This bound holds for every \(x\), so \(d_\infty(f_n,f)\leq\varepsilon/2<\varepsilon\). Thus \(f_n\) converges uniformly to \(f\).

It remains to check that \(f\) is continuous. Fix \(x\in K\) and \(\varepsilon>0\). Choose \(n\) with \(d_\infty(f_n,f)<\varepsilon/3\). Continuity of \(f_n\) at \(x\) gives \(\delta>0\) such that \(d_K(x,y)<\delta\) implies \(d_Y(f_n(x),f_n(y))<\varepsilon/3\). Then \(d_Y(f(x),f(y))<\varepsilon/3+\varepsilon/3+\varepsilon/3=\varepsilon\). So \(f\) is continuous and belongs to \(C(K,Y)\). Every Cauchy sequence therefore converges in \(C(K,Y)\). \(\square\)

We also use the metric-space fact that a complete, totally bounded metric space is compact. One direct proof is as follows. Given an open cover with no finite subcover, cover the space by finitely many closed balls of radius \(1/4\). At least one intersection of such a ball with the space cannot be covered by finitely many members of the open cover. Repeat inside that intersection, using finite covers by closed balls with radii tending to zero, and each time retain an intersection that still has no finite subcover. This gives nested nonempty closed sets with diameters tending to zero. Choose one point from each set; the nested property and shrinking diameters make the chosen points Cauchy. Completeness gives a limit, which belongs to every closed set. Some member of the original open cover contains a ball about that limit, so a sufficiently small one of the nested sets lies inside that member. That set then has a finite subcover, a contradiction.

Now \(C(K,Y)\) is complete by the lemma, so its closed subset \(\overline{F}\) is complete. We have shown \(\overline{F}\) is totally bounded, and the metric-space fact makes it compact. Hence \(F\) is relatively compact. If \(K\) is empty, \(C(K,Y)\) is a singleton, and both conditions and the conclusion hold directly. This completes the theorem.

Using the Criterion

A Common Pitfall

Pointwise relative compactness is not a substitute for equicontinuity. In the last example, every fixed input produces values in the compact interval \([0,1]\); that fact alone does not control the functions uniformly as the input changes. A related error is to allow the distance threshold \(\delta\) to depend on the function. Equicontinuity requires one \(\delta\) for all members of \(F\) at the chosen error tolerance. The calculation with \(f_n(x)=x^n\) shows why that shared threshold matters: the functions can change by \(1/2\) across input distances tending to zero.

The assumptions have distinct roles. Compactness of \(K\) supplies finitely many input locations that control all of the domain. Completeness of \(Y\) ensures that uniform Cauchy limits take values in \(Y\), so the closure remains complete. The conclusion concerns compactness in the uniform metric, not merely compactness of each set of values \(F(x)\).