Compactness Controls an Entire Family
Continuity gives a local guarantee for one function at a time: near each point of its domain, sufficiently close inputs have close outputs. For a family of functions, the required input tolerance might depend on which function is chosen. Compactness in the uniform metric rules out that variation becoming uncontrollable. A finite uniform net reduces the family to finitely many representative functions, and continuity of those representatives supplies a single tolerance that works for every member.
Let \((K,d_K)\) be a nonempty compact metric space and let \((Y,d_Y)\) be a metric space. Write \(C(K,Y)\) for the set of continuous functions from \(K\) to \(Y\). A continuous image of a compact space is compact, hence bounded in a metric space; therefore the following supremum is finite for \(f,g\in C(K,Y)\).
We give the compactness argument explicitly. A finite \(r\)-net in a metric space \(M\) is a finite set of points such that every point of \(M\) is at distance less than \(r\) from at least one of them. The centers of the net below can be chosen from the compact family itself.
Proof. For each \(f\in\mathcal{F}\), take the open ball \(B_M(f,r)\). These balls cover \(\mathcal{F}\), since each \(f\) belongs to its own ball. Compactness gives a finite subcover \(B_M(f_1,r),\ldots,B_M(f_N,r)\), with every \(f_i\in\mathcal{F}\). Thus each member of \(\mathcal{F}\) is at distance less than \(r\) from one of these finitely many centers, which is the required finite net. \(\square\)
Continuity on a Compact Metric Domain
The finite net will consist of continuous functions. We first verify that each such function has a uniform continuity bound on \(K\). The argument also explains why compactness of the domain is relevant: continuity provides a radius at each point, and a finite subcover lets us reduce those radii to finitely many.
Proof. Fix \(\eta>0\). For each \(a\in K\), continuity at \(a\) gives \(r_a>0\) such that \(d_K(a,z)<r_a\) implies \(d_Y(g(a),g(z))<\eta/2\). The balls \(B_{d_K}(a,r_a/2)\), for \(a\in K\), cover \(K\). By compactness, choose a finite subcover with centers \(a_1,\ldots,a_m\). Set \(\rho=\min_{1\leq j\leq m}(r_{a_j}/2)>0\). Given \(u,v\in K\) with \(d_K(u,v)<\rho\), choose \(j\) with \(d_K(u,a_j)<r_{a_j}/2\). Then \(d_K(v,a_j)\leq d_K(v,u)+d_K(u,a_j)<\rho+r_{a_j}/2\leq r_{a_j}\). Both \(u\) and \(v\) are within \(r_{a_j}\) of \(a_j\), so the choice of \(r_{a_j}\) and the triangle inequality give \(d_Y(g(u),g(v))\leq d_Y(g(u),g(a_j))+d_Y(g(a_j),g(v))<\eta/2+\eta/2=\eta\). This proves the assertion. \(\square\)
The Compact-Family Theorem
Proof. If \(\mathcal{F}=\varnothing\), the definition of uniform equicontinuity holds vacuously, so suppose \(\mathcal{F}\ne\varnothing\). Fix \(\varepsilon>0\). By the finite-net lemma, choose \(g_1,\ldots,g_N\in\mathcal{F}\) such that for every \(f\in\mathcal{F}\), some \(g_i\) satisfies \(d_\infty(f,g_i)<\varepsilon/3\). Each \(g_i\) is continuous on compact \(K\). Apply the preceding lemma to each \(g_i\), with \(\eta=\varepsilon/3\), to obtain a positive number \(\rho_i\). Since there are finitely many indices, \(\delta=\min_{1\leq i\leq N}\rho_i>0\).
Take any \(f\in\mathcal{F}\) and \(x,y\in K\) with \(d_K(x,y)<\delta\). Choose \(g_i\) with \(d_\infty(f,g_i)<\varepsilon/3\). By the definition of the supremum metric, \(d_Y(f(x),g_i(x))<\varepsilon/3\) and \(d_Y(f(y),g_i(y))<\varepsilon/3\). Also, \(d_K(x,y)<\delta\leq\rho_i\), so the uniform continuity bound for \(g_i\) gives \(d_Y(g_i(x),g_i(y))<\varepsilon/3\). The triangle inequality now yields \(d_Y(f(x),f(y))\leq d_Y(f(x),g_i(x))+d_Y(g_i(x),g_i(y))+d_Y(g_i(y),f(y))<\varepsilon\). The chosen \(\delta\) does not depend on \(f,x,\) or \(y\). This is uniform equicontinuity. \(\square\)
The proof has two separate compactness inputs. Compactness of \(\mathcal{F}\) in the uniform metric gives finitely many representatives that approximate every function uniformly. Compactness of \(K\) gives a uniform continuity bound for each representative. The finite minimum of those bounds works for the whole family.
Worked Example: A Finite Family on the Unit Interval
Let \(K=[0,1]\) with its usual metric, \(Y=\mathbb{R}\) with its usual metric, and \(\mathcal{F}=\{f_1,f_2\}\), where \(f_1(x)=x^2\) and \(f_2(x)=2x\). The family is compact in the uniform metric because every finite metric space is compact: from any open cover, choose one set covering each of its finitely many points.
For \(x,y\in[0,1]\), \(|f_1(x)-f_1(y)|=|x-y||x+y|\leq 2|x-y|\), since \(0\leq x+y\leq2\). Also, \(|f_2(x)-f_2(y)|=2|x-y|\). Thus, for any \(\varepsilon>0\), take \(\delta=\varepsilon/2\). If \(|x-y|<\delta\), both output differences are less than \(2\delta=\varepsilon\). For instance, when \(\varepsilon=0.4\), \(\delta=0.2\), and \(|x-y|<0.2\) gives both differences less than \(2(0.2)=0.4\). This directly verifies the common bound for the two functions.
Check: at \(x=1\) and \(y=0.9\), the input distance is \(0.1<0.2\). The output differences are \(|1^2-0.9^2|=|1-0.81|=0.19<0.4\) and \(|2-1.8|=0.2<0.4\), as required.
Worked Example: A Compact Infinite Family
Again let \(K=[0,1]\) and \(Y=\mathbb{R}\), and consider \(\mathcal{F}=\{f_a:a\in[0,1]\}\), where \(f_a(x)=ax\). First check compactness in the uniform metric. For \(a,b\in[0,1]\), \(d_\infty(f_a,f_b)=\sup_{x\in[0,1]}|ax-bx|=|a-b|\sup_{x\in[0,1]}x=|a-b|\). The final supremum is \(1\), attained at \(x=1\). Thus \(a\mapsto f_a\) preserves distances from \([0,1]\) to \(\mathcal{F}\), and in particular it is continuous and maps onto \(\mathcal{F}\). Since \([0,1]\) is compact, its continuous image \(\mathcal{F}\) is compact.
There is also a direct common estimate. For \(a\in[0,1]\) and \(x,y\in[0,1]\), \(|f_a(x)-f_a(y)|=|a||x-y|\leq|x-y|\). Given \(\varepsilon>0\), take \(\delta=\varepsilon\). Then \(|x-y|<\delta\) implies \(|f_a(x)-f_a(y)|<\varepsilon\) for every \(a\). For the specific choice \(\varepsilon=0.3\), \(\delta=0.3\), and \(|x-y|<0.3\), the difference is at most \(|x-y|<0.3\).
Check: with \(a=0.6\), \(x=0.8\), and \(y=0.5\), the input distance is \(0.3\), which is not strictly less than \(\delta=0.3\); this pair is therefore not covered by the implication. If instead \(y=0.51\), the input distance is \(0.29<0.3\), and the output difference is \(|0.6(0.8)-0.6(0.51)|=|0.48-0.306|=0.174<0.3\).
Why Compactness of the Family Matters
Compactness of the domain alone does not make every family of continuous functions uniformly equicontinuous. The next example gives a concrete failure. It also guards against a common mistaken argument: choosing a continuity radius separately for each function does not produce one radius for the entire family.
Worked Example: Powers Near an Endpoint
Take \(K=[0,1]\), \(Y=\mathbb{R}\), and \(f_n(x)=x^n\) for \(n\in\mathbb{N}\). Each \(f_n\) is continuous. We show that \(\mathcal{G}=\{f_n:n\in\mathbb{N}\}\) is not uniformly equicontinuous. Fix \(\varepsilon=1/2\), and let \(\delta>0\) be any proposed common input tolerance. Set \(h=\min(\delta/2,1/2)\), so \(0<h\leq1/2\) and \(h<\delta\). The number \(q=1-h\) satisfies \(0<q<1\), so \(q^n\) tends to \(0\). Choose \(n\) such that \(q^n<1/2\). For \(x=1\) and \(y=1-h\), the input distance is \(h<\delta\), whereas \(|f_n(1)-f_n(1-h)|=|1-(1-h)^n|=1-q^n>1/2=\varepsilon\). Therefore no \(\delta\) works for all \(n\).
This family is indeed not compact in the uniform metric. The sequence \((f_n)\) has no uniformly convergent subsequence: any such subsequence would converge pointwise to \(g(x)=0\) for \(0\leq x<1\) and \(g(1)=1\), which is discontinuous. A uniform limit of continuous functions is continuous: given a point \(x\) and a tolerance \(\eta>0\), choose one function within \(\eta/3\) of the limit uniformly, then use its continuity at \(x\) with tolerance \(\eta/3\); the triangle inequality bounds the limit's change by less than \(\eta\). Thus no subsequence can converge uniformly. By the standard theorem that compact metric spaces are sequentially compact, \(\mathcal{G}\) cannot be compact.
Common pitfall. For each fixed \(n\), continuity of \(x^n\) on \([0,1]\) gives a tolerance that works for that function. It is incorrect to conclude that one tolerance works for all \(n\) merely because the domain is compact. The calculation above uses, for every proposed \(\delta\), a function \(f_n\) and two inputs less than \(\delta\) apart whose outputs differ by more than \(1/2\). What makes the theorem work is compactness of the family in the uniform metric, not just compactness of \(K\).
The theorem is useful whenever convergence or compactness is studied in a function space: a compact family cannot develop arbitrarily sharp changes at smaller and smaller input scales. Its conclusion is a uniform statement across the family, stronger than continuity of each member separately. It does not assert that all continuous functions on \(K\) form a compact family; the powers example shows why such a claim would fail.
Check Your Understanding
For each question, keep track of which compactness assumption is being used and whether the tolerance must be common to the entire family.
- In the theorem, where is compactness of \(\mathcal{F}\) used, and where is compactness of \(K\) used?
- For \(\mathcal{F}=\{x\mapsto 3x,\ x\mapsto x^2\}\) on \([0,1]\), find a common \(\delta\) for a given \(\varepsilon>0\).
- Show that the family of constant functions \(f_a(x)=a\), \(a\in[0,1]\), is compact in the uniform metric and uniformly equicontinuous.
- In the powers example, explain why the inputs \(1\) and \(1-h\) satisfy the strict condition \(d_K(x,y)<\delta\).
- Is a compact family required to be finite for the theorem to apply? Use the family \(f_a(x)=ax\) to justify your answer.