Proof Using Cauchy Sequences

Using Compactness to Control an Inverse

Uniform continuity controls how much outputs can change when inputs are close. A related question reverses the direction: if two function values are close, must their inputs be close? For an injective function, this asks whether its inverse behaves continuously, or even uniformly continuously.

The previous tutorial established that uniform continuity preserves Cauchy sequences and used Cauchy sequences to construct extensions. Here we use a different sequence argument. Compactness lets us extract convergent subsequences from inputs that might otherwise escape or behave unpredictably. Continuity identifies the limits of their function values, and injectivity then forces the input limits to agree.

The argument is useful because it does not require a formula for the inverse. It proves a general result: if \(K\) is compact and \(f:K\to\mathbb{R}\) is continuous and injective, then the inverse function from \(f(K)\) back to \(K\) is uniformly continuous. We first isolate the sequence argument that drives the proof.

A Sequence Lemma for Injective Functions

Proof. Suppose the conclusion fails. Then there is an \(\varepsilon_0>0\) and a subsequence of indices \(n_j\) such that

for every \(j\). By sequential compactness of \(K\), a result established earlier in the course, the sequence \((x_{n_j})\) has a subsequence converging to some \(x\in K\). Along that subsequence, the corresponding \(z\)-terms are still in \(K\), so they have a further subsequence converging to some \(z\in K\). Denote the resulting common subsequence of indices by \(j_\ell\). Then

Continuity gives \(f(x_{n_{j_\ell}})\to f(x)\) and \(f(z_{n_{j_\ell}})\to f(z)\). Since \(|f(x_n)-f(z_n)|\to0\), the difference of these two limits is zero:

Thus \(f(x)=f(z)\), and injectivity implies \(x=z\). But taking limits in \(|x_{n_{j_\ell}}-z_{n_{j_\ell}}|\geq\varepsilon_0\) gives \(|x-z|\geq\varepsilon_0\), contradicting \(x=z\). The contradiction proves that \(|x_n-z_n|\to0\). \(\square\)

This proof has a useful Cauchy-sequence interpretation. If the input distances did not tend to zero, we could select pairs separated by a fixed positive amount. Compactness would give convergent subsequences of both members of those pairs. Every convergent sequence is Cauchy, so continuity would make the corresponding function values converge, while their differences tend to zero. The two limiting inputs would have to be equal by injectivity, despite remaining separated. That is the contradiction.

The Inverse Is Uniformly Continuous

Proof. Suppose, to the contrary, that \(f^{-1}\) is not uniformly continuous on \(f(K)\). By the definition of uniform continuity, there is an \(\varepsilon_0>0\) such that for every \(\delta>0\), there are \(u,v\in f(K)\) satisfying

For each positive integer \(n\), use \(\delta=1/n\) to choose \(u_n,v_n\in f(K)\) with these properties. Set \(x_n=f^{-1}(u_n)\) and \(z_n=f^{-1}(v_n)\). Then \(x_n,z_n\in K\), and

The lemma says that \(|x_n-z_n|\to0\), contradicting the second inequality. Therefore \(f^{-1}\) is uniformly continuous on \(f(K)\). \(\square\)

The proof uses compactness where an arbitrary sequence of inputs is replaced by convergent subsequences. It uses continuity to pass from limits of inputs to limits of function values, and injectivity to identify inputs with equal limiting function values. The failure of uniform continuity supplies pairs whose outputs approach one another while their inverse values stay separated; the lemma rules out precisely that behavior.

Worked Examples

Why Compactness Matters

Compactness is not a decorative hypothesis. Without it, a continuous injective function may have an inverse that is not uniformly continuous. Consider \(f:\mathbb{R}\to(0,\infty)\), \(f(x)=e^x\). The exponential function is continuous and strictly increasing, so its inverse is \(f^{-1}(u)=\ln u\). Take \(u_n=1/n\) and \(v_n=2/n\). Both inputs belong to \((0,\infty)\), and

Thus \(\ln\) is not uniformly continuous on \((0,\infty)\). In the sequence proof, the corresponding inputs to \(f\) are \(-\ln n\) and \(\ln(2/n)\). They do not stay in a compact set: both move without bound as \(n\) increases. There is no compactness argument that supplies convergent subsequences in the domain.

This illustrates the precise role of the hypothesis. Compactness prevents the input pairs in a proposed counterexample from escaping all bounded regions. Once subsequences converge, continuity and injectivity force their limits to coincide. On a noncompact domain, close function values can occur at inputs that have no convergent subsequences in the domain, so this proof no longer applies.

A common mistake is to infer uniform continuity of an inverse merely from continuity and injectivity of the original function. The example \(e^x\) shows why that inference fails on a noncompact domain. Another mistake is to assume that a formula for the inverse is necessary. The compactness proof applies equally well to functions whose inverses have no simple expression.

The Cauchy-sequence viewpoint helps organize the contradiction: compactness gives convergent, and therefore Cauchy, subsequences; continuity carries their limits through \(f\); injectivity prevents distinct input limits from sharing the same output. This method proves a uniform estimate qualitatively, even when no explicit modulus of continuity is available.