The Cauchy Property

Comparing Terms Instead of Guessing a Limit

In the previous tutorial, the upper and lower limits of a bounded sequence were used to decide whether the sequence converges. There is another way to approach convergence: instead of comparing terms with a proposed limit, compare terms with one another. If all terms sufficiently far out in a sequence are close to each other, then the sequence has the Cauchy property.

This comparison is useful because it does not require us to know the limit in advance. It also highlights a key feature of the real numbers: a sequence whose terms eventually become arbitrarily close to one another does converge to a real number. We will first establish the Cauchy property as a necessary consequence of convergence, then use boundedness and Bolzano-Weierstrass to prove the converse for real sequences.

The Cauchy Property

A convergence statement compares \(a_n\) with a fixed number \(L\). The Cauchy property instead compares \(a_m\) with \(a_n\), requiring both indices to be sufficiently large. The indices need not be adjacent; the condition applies to every pair of terms in the tail.

The order of the quantifiers matters. We first choose an arbitrary tolerance \(\varepsilon\), then find one index \(N\), and that same \(N\) must work for every pair \(m,n\geq N\). It is not enough that each term be close to the next one. The requirement concerns all pairs of terms in the tail, including terms whose indices are far apart.

Convergence Implies the Cauchy Property

Proof. Suppose \(a_n\to L\), and let \(\varepsilon>0\). By the definition of convergence, there exists \(N\in\mathbb{N}_0\) such that \(|a_k-L|<\varepsilon/2\) whenever \(k\geq N\). For any \(m,n\geq N\), the Triangle Inequality gives

Thus this \(N\) works for every pair \(m,n\geq N\), so the sequence has the Cauchy property. \(\square\)

The proof explains why the tolerance is split in half: each term is within \(\varepsilon/2\) of the same limit, so the two terms are within \(\varepsilon\) of each other. More generally, the triangle inequality lets us compare two terms by inserting an intermediate point. Here that point is the limit.

The Cauchy Property Forces Boundedness

A sequence with the Cauchy property cannot spread out without bound. To see why, use the property just once, with a fixed tolerance such as \(1\). All terms in the resulting tail lie within distance \(1\) of one particular term. The finitely many earlier terms can then be bounded separately.

Proof. Apply the Cauchy property with \(\varepsilon=1\). There is an \(N\in\mathbb{N}_0\) such that \(|a_n-a_N|<1\) whenever \(n\geq N\). Thus, for every \(n\geq N\),

Let \(K=\max\{|a_0|,|a_1|,\ldots,|a_N|\}+1\). This maximum exists because it is the maximum of a nonempty finite set. For \(0\leq n\leq N\), we have \(|a_n|<K\). For \(n>N\), the tail estimate gives \(|a_n|<1+|a_N|\leq K\). Hence \(|a_n|\leq K\) for every \(n\), and the sequence is bounded. \(\square\)

This proof also covers \(N=0\): the finite set used to define \(K\) then consists just of \(|a_0|\). The point is not that the tail is bounded by a universal constant, but that a single tail term, \(a_N\), provides a fixed reference for every later term.

A Convergent Subsequence Determines the Whole Sequence

Boundedness alone does not imply convergence. The Bolzano-Weierstrass Theorem, established earlier, says that a bounded real sequence has a convergent subsequence. For a sequence with the Cauchy property, that subsequential limit controls not just the selected terms, but every sufficiently late term.

Proof. Let \(\varepsilon>0\). By the Cauchy property, there is an \(N_1\in\mathbb{N}_0\) such that \(|a_m-a_n|<\varepsilon/2\) whenever \(m,n\geq N_1\). Write the convergent subsequence as \((a_{n_k})\), where its indices are strictly increasing. Since \(a_{n_k}\to L\), there is a \(K_1\) such that \(|a_{n_k}-L|<\varepsilon/2\) for every \(k\geq K_1\).

Choose \(k\geq K_1\) large enough that \(n_k\geq N_1\). Such a choice is possible because the indices of a subsequence increase without bound. For any \(n\geq N_1\), both \(n\) and \(n_k\) are at least \(N_1\), so

The same chosen \(k\) works for every \(n\geq N_1\). This is precisely the definition of \(a_n\to L\), proving the theorem. \(\square\)

The Cauchy condition is doing essential work here. Convergence of one subsequence only tells us about its selected terms. The Cauchy property ensures that every sufficiently late term is close to a selected term that is already close to \(L\).

Why the Real Number System Matters

These results together give a convergence criterion for real sequences. A sequence with the Cauchy property is bounded, so Bolzano-Weierstrass supplies a convergent subsequence. The theorem just proved then shows that the whole sequence converges. In the other direction, every convergent sequence has the Cauchy property. Therefore, for real sequences, the Cauchy property is equivalent to convergence.

For the reverse direction, the proof is exactly the chain of results just established: Cauchy property implies boundedness; boundedness and Bolzano-Weierstrass give a convergent subsequence; and a Cauchy sequence with a convergent subsequence converges to the same limit. The forward direction is the theorem that every convergent sequence has the Cauchy property. The existence of a limit in the real numbers is important: the corresponding statement need not hold for sequences whose terms are restricted to an incomplete number system, such as the rational numbers.

A common pitfall is to check only successive differences, such as \(|a_{n+1}-a_n|\to0\). The Cauchy property is stronger: it requires \(|a_m-a_n|\) to be small for every pair of sufficiently late indices. Small steps alone do not guarantee that a sequence stays within a small distance over a long stretch. When testing the property, keep both indices in view and verify the required bound uniformly across the entire tail.