The Supremum Metric and Uniform Topology

Measuring the Largest Separation Between Maps

Pointwise convergence measures what happens at each input separately. Uniform closeness asks for one bound on the distance between two maps that works at every input. A natural way to measure that closeness is to take the supremum of all their pointwise distances. This tutorial constructs that metric for bounded maps into an arbitrary metric space and proves that its metric topology is the uniform topology.

Let \(X\) be a nonempty set, and let \((Y,d)\) be a nonempty metric space. No topology on \(X\) is needed: the maps considered here need not be continuous. A subset \(E\subseteq Y\) is bounded if there are \(y_0\in Y\) and \(R\ge 0\) such that \(d(y,y_0)\le R\) for every \(y\in E\). Write \(\mathcal{B}(X,Y)\) for the set of maps \(f:X\to Y\) whose images \(f(X)\) are bounded.

Before using the supremum, we must know it is finite. Choose \(y_0,z_0\in Y\) and finite \(R_f,R_g\ge0\) such that \(d(f(x),y_0)\le R_f\) and \(d(g(x),z_0)\le R_g\) for all \(x\in X\). The triangle inequality gives, for every \(x\in X\),

Thus the set of pointwise distances is nonempty and bounded above, so its supremum exists as a finite real number. The bound depends on the maps and on the chosen centers; it need not be attained at any input.

The Supremum Distance Is a Metric

Proof. For each \(x\in X\), \(d(f(x),g(x))\ge0\), so \(d_\infty(f,g)\ge0\). Symmetry of \(d\) gives \(d(f(x),g(x))=d(g(x),f(x))\) for every \(x\), and therefore \(d_\infty(f,g)=d_\infty(g,f)\). If \(f=g\), then every pointwise distance is zero, so \(d_\infty(f,g)=0\). Conversely, suppose \(d_\infty(f,g)=0\). Each pointwise distance is nonnegative and bounded above by its supremum, so for every \(x\in X\), \(0\le d(f(x),g(x))\le0\). The metric property of \(d\) gives \(f(x)=g(x)\) for every \(x\), hence \(f=g\).

It remains to prove the triangle inequality. Let \(f,g,h\in\mathcal{B}(X,Y)\). For every \(x\in X\),

The right side is an upper bound for all the distances \(d(f(x),h(x))\). Taking their supremum gives \(d_\infty(f,h)\le d_\infty(f,g)+d_\infty(g,h)\). All metric axioms hold, so \(d_\infty\) is a metric. \(\square\)

If \(X=\varnothing\), there is exactly one map from \(X\) to \(Y\). In that special case, define its distance to itself to be zero; the resulting one-point metric space has the expected trivial topology. If \(Y\) has just one point, \(\mathcal{B}(X,Y)\) also has just one point, and the metric axioms still apply.

Uniform Neighborhoods and the Induced Topology

For \(f\in\mathcal{B}(X,Y)\) and \(\varepsilon>0\), the open ball of radius \(\varepsilon\) in the supremum metric is

Membership means that the supremum of the distances is strictly less than \(\varepsilon\), so in particular \(d(f(x),g(x))<\varepsilon\) for every \(x\in X\). Conversely, a pointwise bound \(d(f(x),g(x))<\varepsilon\) for every \(x\) alone only implies \(d_\infty(f,g)\le\varepsilon\); the supremum could equal \(\varepsilon\). To describe open neighborhoods exactly, one can use the stricter condition \(d_\infty(f,g)<\varepsilon\). This distinction between all individual values being strictly below a threshold and their supremum being strictly below it matters when the supremum is not attained.

Proof. If \(g\in B_{d_\infty}(f,\varepsilon)\), set \(r=d_\infty(f,g)\). Then \(r<\varepsilon\), and for every \(x\in X\), \(d(f(x),g(x))\le r<\varepsilon\). Thus the single bound \(r\) works at every input. Conversely, if there is a number \(r<\varepsilon\) such that \(d(f(x),g(x))\le r\) for every \(x\), then \(r\) is an upper bound for all the pointwise distances. Hence \(d_\infty(f,g)\le r<\varepsilon\), so \(g\in B_{d_\infty}(f,\varepsilon)\). \(\square\)

This proposition identifies the metric topology with the uniform topology: its basic neighborhoods constrain all inputs at once, rather than restricting a finite selection of inputs as in the pointwise topology. No topology on the domain \(X\) is involved. If one starts with a specified collection of continuous maps and restricts \(d_\infty\) to that collection, the same formulas describe its induced metric and its uniform neighborhoods, provided the maps in the collection are bounded.

Metric Convergence Is Uniform Convergence

For maps \(f_n:X\to Y\) and \(f:X\to Y\), uniform convergence to \(f\) means that for every \(\varepsilon>0\), there is \(N\in\mathbb{N}\) such that for every \(n\ge N\) and every \(x\in X\), \(d(f_n(x),f(x))<\varepsilon\). This definition requires one index \(N\) to work simultaneously for all inputs.

Proof. Suppose first that \(d_\infty(f_n,f)\to0\). Given \(\varepsilon>0\), metric convergence supplies \(N\) such that \(n\ge N\) implies \(d_\infty(f_n,f)<\varepsilon\). For each \(x\in X\), the pointwise distance is bounded by its supremum, so \(d(f_n(x),f(x))\le d_\infty(f_n,f)<\varepsilon\). This is uniform convergence.

Conversely, suppose \(f_n\) converges uniformly to \(f\), and let \(\varepsilon>0\). Apply uniform convergence with the positive tolerance \(\varepsilon/2\). There is \(N\) such that for \(n\ge N\) and every \(x\in X\), \(d(f_n(x),f(x))<\varepsilon/2\). Therefore \(\varepsilon/2\) is an upper bound for all these distances, and \(d_\infty(f_n,f)\le\varepsilon/2<\varepsilon\). This proves convergence to zero in the supremum metric. The use of \(\varepsilon/2\) ensures a strict bound on the supremum even if the pointwise distances approach \(\varepsilon/2\) without attaining it. \(\square\)

The boundedness assumption makes every distance \(d_\infty(f_n,f)\) finite, so the metric convergence statement is meaningful in \(\mathcal{B}(X,Y)\). Neither the theorem nor the metric construction asserts that a uniform limit is continuous; that issue requires separate hypotheses and results.

A Common Pitfall

Common mistake: Replacing a supremum by a maximum. In the example \(d(f(k),g(k))=\frac{k}{k+1}\), one might write “the largest distance is \(1\), attained as \(k\to\infty\).” This is not correct: \(k\) is always a finite natural number, and \(\frac{k}{k+1}<1\) for every such \(k\), so there is no input where the distance equals \(1\). The correct statement is that the distances have supremum \(1\), as verified by showing that \(1\) is an upper bound and every \(c<1\) fails to be one. A maximum exists only when the supremum is itself one of the values; the first worked example had a maximum, while this one does not.

The main practical benefit of \(d_\infty\) is that it turns a condition involving every input into one numerical distance. Once the supremum metric is available, uniform convergence is ordinary convergence in a metric space, and uniform neighborhoods are exactly its open balls.