Uniform Continuity and Cauchy Sequences

Uniform Continuity Meets Cauchy Sequences

Uniform continuity controls pairs of inputs using one choice of \(\delta\), regardless of where the inputs lie in the domain. Cauchy sequences are built from pairs of terms that eventually become arbitrarily close. These ideas fit together: a uniformly continuous function sends a Cauchy sequence to a Cauchy sequence.

The previous tutorial concerned extending a function to a missing endpoint when it has a finite limit there. Uniform continuity gives a broader way to find such limits. In fact, a uniformly continuous function on any subset of \(\mathbb{R}\) has a unique uniformly continuous extension to the closure of that subset. The closure need not be compact or bounded.

Recall that a sequence \((x_n)\) is Cauchy if, for every \(\varepsilon>0\), there is an \(N\) such that \(m,n\geq N\) implies \(|x_m-x_n|<\varepsilon\). Its terms need not have a limit in their domain. The key fact we will use is the earlier Theorem (Uniform Continuity Preserves Cauchy Sequences): if \(f\) is uniformly continuous and \((x_n)\) is Cauchy in its domain, then \((f(x_n))\) is Cauchy. Since \(\mathbb{R}\) is complete, every such image sequence has a real limit.

Extending to the Closure

Proof. For each \(x\in\overline{E}\), choose a sequence \((x_n)\) in \(E\) such that \(x_n\to x\). Such a sequence exists by the definition of closure. The sequence \((x_n)\) is Cauchy, so the earlier theorem says that \((f(x_n))\) is Cauchy. Completeness of \(\mathbb{R}\) gives a real limit. We would like to define \(F(x)\) to be this limit; first we must check that it does not depend on the sequence chosen.

Suppose also that \(y_n\in E\) and \(y_n\to x\). Then \(|x_n-y_n|\leq |x_n-x|+|y_n-x|\to0\). Given \(\varepsilon>0\), uniform continuity of \(f\) supplies \(\delta>0\) such that \(u,v\in E\) and \(|u-v|<\delta\) imply \(|f(u)-f(v)|<\varepsilon\). For all sufficiently large \(n\), \(|x_n-y_n|<\delta\), so \(|f(x_n)-f(y_n)|<\varepsilon\). The two image sequences converge, and therefore the distance between their limits is at most \(\varepsilon\). Since this holds for every \(\varepsilon>0\), their limits are equal. Thus the definition

is independent of the approximating sequence. If \(x\in E\), we may use the constant sequence \(x_n=x\), which gives \(F(x)=f(x)\).

We next prove that \(F\) is uniformly continuous. Fix \(\varepsilon>0\). By uniform continuity of \(f\), choose \(\delta>0\) such that \(u,v\in E\) and \(|u-v|<\delta\) imply \(|f(u)-f(v)|<\varepsilon/2\). Let \(x,y\in\overline{E}\) satisfy \(|x-y|<\delta/2\). Choose sequences \(x_n,y_n\in E\) with \(x_n\to x\) and \(y_n\to y\). For all sufficiently large \(n\), both \(|x_n-x|<\delta/8\) and \(|y_n-y|<\delta/8\). Hence

It follows that \(|f(x_n)-f(y_n)|<\varepsilon/2\) for all sufficiently large \(n\). Taking limits gives \(|F(x)-F(y)|\leq\varepsilon/2<\varepsilon\). Thus \(\delta/2\) works for \(F\), which proves its uniform continuity.

Finally, suppose \(G:\overline{E}\to\mathbb{R}\) is continuous and agrees with \(f\) on \(E\). For any \(x\in\overline{E}\), choose \(x_n\in E\) with \(x_n\to x\). Continuity gives \(G(x_n)\to G(x)\), while \(G(x_n)=f(x_n)\to F(x)\) by the definition of \(F\). Uniqueness of limits gives \(G(x)=F(x)\). This holds for every \(x\in\overline{E}\), so the extension is unique. \(\square\)

The construction has two separate ingredients. Uniform continuity ensures that the output values along an approximating sequence are Cauchy and that different approximating sequences give the same limit. Completeness of the codomain ensures that these Cauchy sequences actually converge. Compactness of \(\overline{E}\) is not needed.

Worked Examples

What Cauchy-Sequence Preservation Does Not Tell Us

Uniform continuity guarantees that Cauchy input sequences have Cauchy image sequences. The converse is not true: a function may send every Cauchy sequence in its domain to a Cauchy sequence without being uniformly continuous. The square function on \(\mathbb{R}\) gives a useful example.

Proof. Let \((x_n)\) be any Cauchy sequence in \(\mathbb{R}\). Every Cauchy sequence in \(\mathbb{R}\) is bounded, so there is an \(M\geq0\) with \(|x_n|\leq M\) for every \(n\). For any \(m,n\),

Given \(\varepsilon>0\), if \(M>0\), the Cauchy property gives an \(N\) such that \(m,n\geq N\) implies \(|x_n-x_m|<\varepsilon/(2M)\), and then \(|x_n^2-x_m^2|<\varepsilon\). If \(M=0\), every \(x_n=0\), so the image sequence is constant and is Cauchy. In either case \((x_n^2)\) is Cauchy, and completeness of \(\mathbb{R}\) makes it convergent.

To see that \(f\) is not uniformly continuous, take \(x_n=n\) and \(y_n=n+1/n\). Their input distances satisfy \(|x_n-y_n|=1/n\to0\), while

Thus arbitrarily close inputs can have outputs at least \(2\) apart, which contradicts uniform continuity. \(\square\)

There is no conflict between this example and the extension theorem. Each individual Cauchy sequence in \(\mathbb{R}\) is bounded, so squaring behaves in a controlled way along that sequence. Uniform continuity requires one \(\delta\) to work for all pairs in the entire domain, including pairs far from the origin. A Cauchy sequence tests only the behavior of its own tail; it does not test all distant regions at once.

How to Use the Connection

When a uniformly continuous function is defined on a domain with missing limit points, Cauchy sequences provide a systematic route to extending it. To construct the value at a point of the closure, approximate that point by domain points and take the limit of their function values. The proof above explains why the limit exists, why it is independent of the approximation, and why the extended function remains uniformly continuous.

A useful diagnostic follows in the other direction: if a function sends some Cauchy sequence in its domain to a sequence that is not Cauchy, then the function cannot be uniformly continuous. But passing this test for every Cauchy sequence is not enough to establish uniform continuity, as the square function demonstrates.