Pairwise Control in a Tail
Convergence asks whether the terms of a sequence get close to one fixed point. The Cauchy condition asks a different question: do all pairs of terms become close to each other when their indices are sufficiently large? It is useful precisely because it can be stated without first specifying a candidate limit. The distinction is in the quantifiers: for a Cauchy sequence, both indices may vary throughout the tail.
Throughout, \((X,d)\) is a metric space and \((x_n)\) is a sequence in \(X\). Recall that \(x_n\to x\) means that for every \(\varepsilon>0\), there is \(N\in\mathbb{N}\) such that \(d(x_n,x)<\varepsilon\) whenever \(n\ge N\). The point \(x\) is fixed before the tail is tested.
In particular, it is not enough to control \(d(x_n,x_{n+1})\), or to show that each term is close to some point that is allowed to change with \(n\). The Cauchy test compares arbitrary terms \(x_m\) and \(x_n\) from the same tail. It uses the strict inequality \(d(x_m,x_n)<\varepsilon\); estimates that give a non-strict bound are still useful if that bound is itself strictly less than \(\varepsilon\).
A Basic Estimate for Cauchy Sequences
Proof. Apply the Cauchy condition with \(\varepsilon=1\). There is \(N\in\mathbb{N}\) such that \(d(x_n,x_N)<1\) for every \(n\ge N\). For the finitely many earlier indices, let \(M=0\) if \(N=1\), and otherwise let \(M\) be the maximum of the finite set \(\{d(x_i,x_N):1\le i<N\}\). Set \(R=\max\{M,1\}\). If \(n<N\), then \(d(x_n,x_N)\le M\le R\). If \(n\ge N\), then \(d(x_n,x_N)<1\le R\). Thus every term lies in the closed ball of radius \(R\) about \(x_N\), meaning \(d(x_n,x_N)\le R\) for every \(n\). This is boundedness. \(\square\)
The non-strict bound in the conclusion matters. The finite maximum \(M\) might be attained, so an earlier term can have distance exactly \(M\) from \(x_N\). We therefore use a closed-ball bound, not the claim that all terms lie in the open ball of radius \(M\). Equivalently, one could use the open ball of radius \(R+1\), which contains every term because \(d(x_n,x_N)\le R<R+1\).
Proof. Let \(\varepsilon>0\). Since \(a_n\to0\), there is \(N\in\mathbb{N}\) such that \(a_k<\varepsilon/2\) for every \(k\ge N\). For any \(m,n\ge N\), the assumed estimate gives \(d(x_m,x_n)\le a_m+a_n<\varepsilon/2+\varepsilon/2=\varepsilon\). This is exactly the Cauchy condition. \(\square\)
This estimate is convenient when each term can be compared to a common reference point, but it does not require that reference point to be a limit. The triangle inequality supplies the pairwise estimate; the tail of the sequence \((a_n)\) then controls all pairs at once.
Worked Example: A Geometric Tail in the Real Line
Give a Cauchy estimate for \(x_n=1-2^{-n}\) in \(\mathbb{R}\) with its standard metric \(d(x,y)=|x-y|\).
Take indices \(m>n\). Subtracting the terms gives \(|x_m-x_n|=|(1-2^{-m})-(1-2^{-n})|=2^{-n}-2^{-m}<2^{-n}\). If \(m,n\ge N\), the same calculation with the smaller index first gives \(|x_m-x_n|<2^{-N}\) when the indices differ; when \(m=n\), the distance is \(0\). Thus, for any \(\varepsilon>0\), choose \(N\) so that \(2^{-N}<\varepsilon\). Then every pair \(m,n\ge N\) has \(|x_m-x_n|<\varepsilon\), proving the Cauchy condition.
Check with a specific tolerance: for \(\varepsilon=1/16\), choose \(N=5\), since \(2^{-5}=1/32<1/16\). For the tail pair \(n=5,m=6\), the distance is \(|x_6-x_5|=|(1-1/64)-(1-1/32)|=1/32-1/64=1/64<1/16\). The general bound handles every pair in the tail, not only this checked pair.
Worked Example: A Sequence in the Euclidean Plane
Let \(X=\mathbb{R}^2\) with its Euclidean metric \(d(u,v)=\|u-v\|\), and set \(x_n=(1/n,(-1)^n/n)\). Verify the Cauchy condition with an explicit estimate.
First compute the distance from \(x_n\) to the origin: \(\|x_n\|=\sqrt{(1/n)^2+((-1)^n/n)^2}=\sqrt{2}/n\), since \(((-1)^n)^2=1\). By the triangle inequality, for any \(m,n\), \(\|x_m-x_n\|\le\|x_m\|+\|x_n\|=\sqrt{2}/m+\sqrt{2}/n\). If \(m,n\ge N\), this is at most \(2\sqrt{2}/N\). Given \(\varepsilon>0\), choose \(N\in\mathbb{N}\) with \(N>2\sqrt{2}/\varepsilon\). Then \(2\sqrt{2}/N<\varepsilon\), so the sequence is Cauchy.
Check with a specific tolerance and a specific pair: take \(\varepsilon=1/2\) and \(N=6\). The uniform estimate is \(2\sqrt{2}/6=\sqrt{2}/3<1/2\), because both sides are positive and \((\sqrt{2}/3)^2=2/9<1/4=(1/2)^2\). For the pair \(m=7,n=6\), direct subtraction gives \(x_7-x_6=(1/7-1/6,-1/7-1/6)=(-1/42,-13/42)\). Its distance is \(\sqrt{(1/42)^2+(13/42)^2}=\sqrt{170}/42<1/2\), since \(4\cdot170=680<1764=42^2\). This direct check agrees with the bound, which applies to every pair of indices at least \(6\).
What the Discrete Metric Makes Visible
On a set \(X\), the discrete metric is \(d(x,y)=0\) when \(x=y\) and \(d(x,y)=1\) when \(x\ne y\). It satisfies the metric axioms: distances are nonnegative and symmetric, zero distance occurs exactly for equal points, and if \(x\ne z\), then at least one of \(x\ne y\) or \(y\ne z\) holds, so \(d(x,z)=1\le d(x,y)+d(y,z)\). For this metric, the Cauchy condition has a particularly direct form.
Proof. Suppose \((x_n)\) is Cauchy. Apply the definition with \(\varepsilon=1/2\). For some \(N\), every \(m,n\ge N\) satisfies \(d(x_m,x_n)<1/2\). The only possible discrete-metric distance less than \(1/2\) is \(0\), so \(x_m=x_n\) for all \(m,n\ge N\). Taking \(p=x_N\) shows the sequence is eventually constant. Conversely, suppose \(x_n=p\) for all \(n\ge N\). For any \(\varepsilon>0\) and any \(m,n\ge N\), \(d(x_m,x_n)=d(p,p)=0<\varepsilon\). Hence the sequence is Cauchy. \(\square\)
Worked Example: An Alternating Sequence in the Discrete Metric
Take \(X=\{a,b\}\) with the discrete metric and define \(x_n=a\) for even \(n\), \(x_n=b\) for odd \(n\). This sequence is not Cauchy.
Use the fixed tolerance \(\varepsilon=1/2\). For any proposed tail index \(N\), choose \(n\ge N\) with \(n\) even and set \(m=n+1\). Then \(m\ge N\), \(x_n=a\), and \(x_m=b\), so \(d(x_m,x_n)=d(b,a)=1\). Since \(1\not<1/2\), this pair violates the Cauchy requirement. The choice of an even \(n\ge N\) is always possible: if \(N\) is even take \(n=N\), and if \(N\) is odd take \(n=N+1\). The failure therefore occurs in every tail.
Check: the distances between consecutive terms are all \(1\), not less than \(1/2\), in agreement with the failure. The proposition gives the same conclusion because the sequence does not become constant on any tail.
A Common Pitfall: Successive Terms Are Not Enough
A frequent error is to check only the distance between consecutive terms and infer that the sequence is Cauchy. The Cauchy definition requires control of every pair in a tail; small successive steps can accumulate into a substantial distance.
Worked Example: Small Increments but No Cauchy Tail
In \(\mathbb{R}\) with the standard metric, let \(x_n=\sum_{k=1}^{n}1/k\). Consecutive terms satisfy \(|x_{n+1}-x_n|=1/(n+1)\), which tends to \(0\). Nevertheless, the sequence is not Cauchy.
To check this, fix any proposed tail index \(N\), and choose \(n\ge N\). Set \(m=4n\), so \(m\ge n\ge N\). The distance between these terms is \[ |x_{4n}-x_n|=\sum_{k=n+1}^{4n}\frac1k. \] There are \(4n-(n+1)+1=3n\) summands. For each index \(k\) in this sum, \(k\le4n\), so \(1/k\ge1/(4n)\). Therefore \[ |x_{4n}-x_n|\ge 3n\cdot\frac1{4n}=\frac34>\frac12. \] For the fixed tolerance \(\varepsilon=1/2\), every proposed tail contains a pair whose distance is greater than \(\varepsilon\). Thus no \(N\) can satisfy the Cauchy condition. The correction is to estimate arbitrary pairs \(m,n\) in the tail, rather than only neighboring pairs.
Check of the count and inequality: the indices \(n+1,n+2,\ldots,4n\) number \(3n\), and each reciprocal is at least \(1/(4n)\). Hence the lower bound is exactly \((3n)/(4n)=3/4\), which is strictly larger than \(1/2\).
Reading the Two Definitions Side by Side
| Property | What must be close | What is fixed |
|---|---|---|
| Convergence to \(x\) | \(x_n\) and \(x\) | The candidate point \(x\) |
| Cauchy condition | Every pair \(x_m,x_n\) in a sufficiently late tail | The tolerance \(\varepsilon\); both indices vary in the tail |
For computations, begin by fixing an arbitrary \(\varepsilon>0\), then choose a tail index, and finally verify the distance inequality for arbitrary indices in that tail. A successful estimate must account for both indices and must produce a strict upper bound below \(\varepsilon\). The boundedness proposition is one consequence of this pairwise control; the examples show how the same definition can be applied in different metrics.
Check Your Understanding
In each question, state the metric being used and pay attention to whether the required inequality is strict.
- Write the full quantifiers in the definition of a Cauchy sequence, and explain how they differ from the quantifiers in convergence to a fixed point.
- For \(x_n=2+1/n\) in the standard metric on \(\mathbb{R}\), find an estimate for \(|x_m-x_n|\) when \(m,n\ge N\), and use it to verify the Cauchy condition.
- Prove that every eventually constant sequence is Cauchy in any metric space.
- In a discrete metric space, can a Cauchy sequence take two different values arbitrarily far out in the sequence? Justify your answer from the definition.
- For the sequence \(x_n=\sum_{k=1}^{n}1/k\), explain why \(1/(n+1)\to0\) does not establish the Cauchy condition.