Compactness and Uniform Continuity

From Local Continuity to One Global Scale

Continuity at a point allows the input distance required for a given output tolerance to depend on that point. Uniform continuity is stronger: one input distance works at every point of the domain. Compactness is what lets us pass from many local choices to one global choice. In this tutorial, we use the Lebesgue Number Lemma to prove the Heine-Cantor theorem, and then see how compactness also ensures that continuous real-valued functions attain their extreme values.

Recall that \(f:X\to Y\), between metric spaces \((X,d_X)\) and \((Y,d_Y)\), is uniformly continuous if for every \(\varepsilon>0\) there is a \(\delta>0\) such that for all \(x,y\in X\), \(d_X(x,y)<\delta\) implies \(d_Y(f(x),f(y))<\varepsilon\). The same \(\delta\) must work regardless of where \(x\) and \(y\) lie. By contrast, continuity at each \(x\) allows the corresponding \(\delta\) to depend on \(x\).

Earlier in the course, the Complete and Totally Bounded Metric Spaces Are Compact theorem and the result that a continuous map from a complete, totally bounded metric space is uniformly continuous established a route to this conclusion. The compactness proof below gives a different useful perspective: it identifies how an open cover of the domain supplies a single scale for all points.

The Heine-Cantor Theorem

Fix an output tolerance \(\varepsilon>0\). Continuity at each point \(z\in X\) gives a radius \(r_z>0\) such that points within \(r_z\) of \(z\) have images within \(\varepsilon/3\) of \(f(z)\). The balls with half these radii cover \(X\). Compactness supplies a finite subcover, and the Lebesgue Number Lemma supplies a positive scale for that cover. The triangle inequality then compares the images of any two sufficiently close points to the image of one common center.

Proof. If \(X\) is empty, the definition of uniform continuity holds vacuously, so suppose \(X\) is nonempty. Let \(\varepsilon>0\). For each \(z\in X\), continuity at \(z\) gives \(r_z>0\) such that \(d_X(w,z)<r_z\) implies \(d_Y(f(w),f(z))<\varepsilon/3\). The collection of open balls \(\{B_{r_z/2}(z):z\in X\}\) covers \(X\). By compactness, finitely many of these balls cover \(X\). The Lebesgue Number Lemma gives a number \(\lambda>0\) such that every subset of \(X\) with diameter less than \(\lambda\) is contained in one member of this finite cover.

Set \(\delta=\lambda\). Take any \(x,y\in X\) with \(d_X(x,y)<\delta\). The set \(\{x,y\}\) has diameter less than \(\lambda\), so it is contained in one of the covering balls, say \(B_{r_z/2}(z)\). Thus \(d_X(x,z)<r_z/2<r_z\) and \(d_X(y,z)<r_z/2<r_z\). The choice of \(r_z\) gives

The chosen \(\delta\) depends on \(\varepsilon\) and the cover, not on \(x\) or \(y\). This is uniform continuity. \(\square\)

The factor of \(1/3\) is simply convenient: it leaves room to apply the triangle inequality through the common center \(f(z)\). The crucial step is not the particular fraction but finding one Lebesgue number for a cover built from local continuity neighborhoods. Continuity supplies those neighborhoods; compactness makes a global scale possible.

Worked Examples: Uniform Control in Practice

Compactness and Attainment of Extreme Values

Uniform continuity is not the only useful consequence of compactness for continuous functions. For real-valued functions, compactness also ensures that the largest and smallest values are actually reached. This is stronger than merely saying the values are bounded: a bounded set of values can have a supremum or infimum that it never attains.

Proof. By the theorem that Continuous Images of Compact Sets Are Compact, \(f[K]\) is compact in \(\mathbb{R}\). Compact subsets of metric spaces are bounded, so \(f[K]\) is nonempty and bounded. Let \(M=\sup f[K]\). For each positive integer \(n\), the definition of supremum gives \(t_n\in f[K]\) such that \(M-1/n<t_n\leq M\). Consequently, \(t_n\to M\). Compact subsets of the Hausdorff space \(\mathbb{R}\) are closed, so \(M\in f[K]\). Hence \(M=f(b)\) for some \(b\in K\).

Likewise, let \(m=\inf f[K]\). For each positive integer \(n\), choose \(s_n\in f[K]\) with \(m\leq s_n<m+1/n\). Then \(s_n\to m\), and closedness of \(f[K]\) implies \(m\in f[K]\). Thus \(m=f(a)\) for some \(a\in K\). Since \(m\) and \(M\) are the infimum and supremum of the image, \(f(a)\leq f(x)\leq f(b)\) for all \(x\in K\). \(\square\)

What Compactness Adds—and What It Does Not

The Heine-Cantor theorem does not say that every continuous function on every metric space is uniformly continuous. Continuity controls behavior near each point separately; without compactness, those local controls may become less effective as the point moves through the domain. The sequence example on \(\mathbb{R}\) makes that failure explicit. On a compact domain, finite coverage and a Lebesgue number prevent the local scales from degenerating without a global bound.

A second distinction is between uniform continuity and a Lipschitz estimate. A Lipschitz bound such as \(|f(x)-f(y)|\leq L|x-y|\) is a particularly strong way to prove uniform continuity, as in the polynomial example. The Heine-Cantor theorem guarantees uniform continuity for continuous maps on compact metric spaces, but it does not promise one global Lipschitz constant. The square-root estimate illustrates a different modulus of control: the output distance is bounded by the square root of the input distance.

When applying the theorem, check both hypotheses: the domain must be compact and the function must be continuous on that domain. A restriction to a compact subset can be uniformly continuous even when the same formula is not uniformly continuous on a larger, noncompact domain. For real-valued functions, the Extreme Value Theorem adds another conclusion: boundedness of the image becomes actual attainment of its upper and lower bounds.