Computing Cauchy Sequences in Familiar Metrics

What the Cauchy Criterion Tests

A sequence is Cauchy when its terms eventually become arbitrarily close to one another. Unlike convergence, this definition does not ask for a candidate limit or even require that a limit point be present in the space. That distinction matters when comparing a metric space with a larger space containing it.

For a metric space \((X,d)\), convergence \(x_n\to x\) means that for every \(\varepsilon>0\), there is \(N\in\mathbb{N}\) such that \(d(x_n,x)<\varepsilon\) whenever \(n\ge N\). The Cauchy condition compares two terms of the sequence instead.

Definition: A sequence \((x_n)\) in a metric space \((X,d)\) is Cauchy if for every \(\varepsilon>0\), there is \(N\in\mathbb{N}\) such that \(d(x_n,x_m)<\varepsilon\) whenever \(n,m\ge N\). It converges in \(X\) if there is a point \(x\in X\) such that \(x_n\to x\).

The index \(N\) may depend on \(\varepsilon\), but once chosen it must work for every pair \(n,m\ge N\). A metric space is called complete if every Cauchy sequence in it converges to a point of that space. Here we will calculate the Cauchy condition directly in several metrics and keep track of which space contains a limit.

The Usual Metric on the Real Line

In \(\mathbb{R}\) with its usual metric \(d(x,y)=|x-y|\), the Cauchy condition is the familiar numerical test: for every \(\varepsilon>0\), all sufficiently late terms differ by less than \(\varepsilon\). The test does not itself name a limit, though a known candidate can make the estimates easier.

Worked Example: A Cauchy Sequence in the Usual Metric

Consider \(a_n=2+\frac{1}{n}\) in \(\mathbb{R}\). For any \(n,m\in\mathbb{N}\), the triangle inequality gives

|a_n-a_m| =\left|\left(2+\frac{1}{n}\right)-\left(2+\frac{1}{m}\right)\right| =\left|\frac{1}{n}-\frac{1}{m}\right| \le \frac{1}{n}+\frac{1}{m}.

Given \(\varepsilon>0\), choose \(N\in\mathbb{N}\) with \(N>2/\varepsilon\). If \(n,m\ge N\), then \(1/n\le 1/N\) and \(1/m\le 1/N\), so

|a_n-a_m|\le\frac{2}{N}<\varepsilon.

Thus \((a_n)\) is Cauchy. It also converges to \(2\), since \(|a_n-2|=1/n\), which is less than \(\varepsilon\) whenever \(n>1/\varepsilon\). Check: \(a_1=2+1=3\), and the formula gives \(|a_1-2|=1\), as expected. The Cauchy estimate concerns pairs of late terms; the convergence estimate concerns distance to the specified point \(2\).

In a complete space, the Cauchy condition guarantees convergence in that space. In an incomplete space, the same calculation may show that terms settle toward a point that is missing. Two examples of that distinction follow below.

The Discrete Metric

For any nonempty set \(X\), the discrete metric is \(d(x,y)=0\) when \(x=y\), and \(d(x,y)=1\) when \(x\ne y\). Its Cauchy sequences have a particularly rigid form: they must eventually stop changing.

Theorem: A sequence in a space with the discrete metric is Cauchy if and only if it is eventually constant, meaning that there are \(N\in\mathbb{N}\) and \(a\in X\) such that \(x_n=a\) for every \(n\ge N\).

Proof. Suppose \((x_n)\) is Cauchy. Apply the definition with \(\varepsilon=1\). There is \(N\) such that \(d(x_n,x_m)<1\) for all \(n,m\ge N\). In the discrete metric, the distance between distinct points is \(1\), not less than \(1\). Therefore \(x_n=x_m\) for every \(n,m\ge N\), so the sequence is eventually constant.

Conversely, suppose \(x_n=a\) for every \(n\ge N\). For any \(\varepsilon>0\) and any \(n,m\ge N\), \(d(x_n,x_m)=d(a,a)=0<\varepsilon\). Hence the sequence is Cauchy. \(\square\)

Worked Example: Alternating and Eventually Constant Sequences

Let \(X=\{a,b\}\) with the discrete metric. Define \(x_n=a\) for even \(n\) and \(x_n=b\) for odd \(n\). For any proposed index \(N\), there are an even \(n\ge N\) and an odd \(m\ge N\). Thus \(x_n\ne x_m\) and \(d(x_n,x_m)=1\). Taking \(\varepsilon=1/2\), the required inequality \(d(x_n,x_m)<1/2\) fails for every \(N\), so this sequence is not Cauchy.

Now define \(y_1=b\), \(y_2=b\), and \(y_n=a\) for every \(n\ge3\). For all \(n,m\ge3\), \(y_n=y_m=a\), and \(d(y_n,y_m)=0\). Since \(0<\varepsilon\) for every \(\varepsilon>0\), the sequence is Cauchy. It converges in \(X\) to \(a\), because \(d(y_n,a)=0\) for all \(n\ge3\). Check: the first two terms do not affect either eventual condition; the index \(N=3\) works for every positive \(\varepsilon\).

The Sup Metric on Finite-Dimensional Real Space

For \(x=(x_1,\ldots,x_k)\) and \(y=(y_1,\ldots,y_k)\) in \(\mathbb{R}^k\), the sup metric is

d_\infty(x,y)=\max_{1\le i\le k}|x_i-y_i|.

The maximum is over the finitely many coordinate differences. Consequently, \(d_\infty(x,y)<\varepsilon\) exactly when every coordinate difference has absolute value less than \(\varepsilon\). This gives a coordinate-by-coordinate way to test the Cauchy condition.

Theorem: A sequence \((x_n)\) in \(\mathbb{R}^k\) is Cauchy for \(d_\infty\) if and only if each coordinate sequence is Cauchy in \(\mathbb{R}\).

Proof. Write \(x_n=(x_{n,1},\ldots,x_{n,k})\). First suppose \((x_n)\) is Cauchy for \(d_\infty\). For each coordinate \(i\) and every \(n,m\), the definition of the maximum gives \(|x_{n,i}-x_{m,i}|\le d_\infty(x_n,x_m)\). Given \(\varepsilon>0\), the Cauchy condition supplies an \(N\) for which \(d_\infty(x_n,x_m)<\varepsilon\) whenever \(n,m\ge N\). The same \(N\) then gives \(|x_{n,i}-x_{m,i}|<\varepsilon\) for each \(i\), so every coordinate sequence is Cauchy.

Conversely, suppose each coordinate sequence is Cauchy. Fix \(\varepsilon>0\). For each \(i\in\{1,\ldots,k\}\), choose \(N_i\) such that \(|x_{n,i}-x_{m,i}|<\varepsilon\) whenever \(n,m\ge N_i\). There are only finitely many coordinates, so \(N=\max\{N_1,\ldots,N_k\}\) is an integer. For \(n,m\ge N\), every coordinate difference is less than \(\varepsilon\). Their maximum is therefore also less than \(\varepsilon\), giving \(d_\infty(x_n,x_m)<\varepsilon\). Thus \((x_n)\) is Cauchy. \(\square\)

Worked Example: Computing in the Sup Metric

In \(\mathbb{R}^2\) with \(d_\infty\), let \(x_n=(1+\frac{1}{n},-2+\frac{3}{n})\). Its coordinate differences from \((1,-2)\) are \(1/n\) and \(3/n\), so

d_\infty(x_n,(1,-2))=\max\left\{\frac{1}{n},\frac{3}{n}\right\}=\frac{3}{n}.

For the Cauchy test, the triangle inequality in each coordinate gives

d_\infty(x_n,x_m) =\max\left\{\left|\frac{1}{n}-\frac{1}{m}\right|, 3\left|\frac{1}{n}-\frac{1}{m}\right|\right\} \le \frac{3}{n}+\frac{3}{m}.

Given \(\varepsilon>0\), choose \(N>6/\varepsilon\). For \(n,m\ge N\), the last expression is at most \(6/N<\varepsilon\). Thus the sequence is Cauchy, and the distance formula above also shows \(x_n\to(1,-2)\). Check: \(x_2=(3/2,-1/2)\), \(x_4=(5/4,-5/4)\), and \(d_\infty(x_2,x_4)=\max\{1/4,3/4\}=3/4\). The coordinate formula gives the same value: the first-coordinate difference is \(1/4\), and the second-coordinate difference is \(3/4\).

The usual Euclidean metric on \(\mathbb{R}^k\) is \(d_2(x,y)=\sqrt{\sum_{i=1}^k|x_i-y_i|^2}\). For every pair \(x,y\), each coordinate difference is at most \(d_2(x,y)\), and the sum of \(k\) squared coordinate differences is at most \(k\,d_\infty(x,y)^2\). Hence \(d_\infty(x,y)\le d_2(x,y)\le\sqrt{k}\,d_\infty(x,y)\). These inequalities show directly that the Cauchy condition in either metric implies the Cauchy condition in the other: for a desired bound \(\varepsilon\), use \(\varepsilon\) in the first inequality and \(\varepsilon/\sqrt{k}\) in the second.

A Cauchy Sequence May Have No Limit in Its Space

When \(Y\subseteq X\) has the subspace metric inherited from \(X\), distances between points of \(Y\) are the same whether calculated in \(Y\) or \(X\). Thus a sequence in \(Y\) is Cauchy in \(Y\) exactly when it is Cauchy in \(X\). But a limit in \(X\) need not belong to \(Y\). If it does not, it cannot be a limit in \(Y\).

Uniqueness of limits: In a metric space, a sequence has at most one limit. Indeed, if \(x_n\to x\) and \(x_n\to y\), then for every \(n\), \(d(x,y)\le d(x,x_n)+d(x_n,y)\). Given any \(\varepsilon>0\), both terms on the right are less than \(\varepsilon/2\) for all sufficiently large \(n\), so \(d(x,y)<\varepsilon\). This holds for every \(\varepsilon>0\), forcing \(d(x,y)=0\), and therefore \(x=y\).

Worked Example: Rational Approximations to \(\sqrt{2}\)

Give \(\mathbb{Q}\) the metric inherited from \(\mathbb{R}\). Define \(q_n=\lfloor 10^n\sqrt{2}\rfloor/10^n\), where \(\lfloor t\rfloor\) is the greatest integer less than or equal to \(t\). Each \(q_n\) is rational. From the definition of the floor function,

q_n\le\sqrt{2}<q_n+10^{-n}, \qquad 0\le\sqrt{2}-q_n<10^{-n}.

For orientation, \(1.414<\sqrt{2}<1.415\), since \(1.414^2=1.999396<2\) and \(1.415^2=2.002225>2\). These inequalities give \(q_1=1.4\), \(q_2=1.41\), and \(q_3=1.414\). For any \(n,m\), the bounds around \(\sqrt{2}\) give

|q_n-q_m| \le |q_n-\sqrt{2}|+|q_m-\sqrt{2}| <10^{-n}+10^{-m}.

Given \(\varepsilon>0\), choose \(N\in\mathbb{N}\) so that \(2\cdot10^{-N}<\varepsilon\). If \(n,m\ge N\), then \(10^{-n}\le10^{-N}\) and \(10^{-m}\le10^{-N}\), so \(|q_n-q_m|<2\cdot10^{-N}<\varepsilon\). Thus \((q_n)\) is Cauchy in \(\mathbb{Q}\), and the same estimate \(0\le\sqrt{2}-q_n<10^{-n}\) shows that it converges to \(\sqrt{2}\) in \(\mathbb{R}\).

It does not converge in \(\mathbb{Q}\). If it converged in \(\mathbb{Q}\) to some \(q\in\mathbb{Q}\), it would also converge to \(q\) in \(\mathbb{R}\), because the metric is inherited. In \(\mathbb{R}\) it already converges to \(\sqrt{2}\); uniqueness of limits would give \(q=\sqrt{2}\), which is not rational. Check: the first three terms are the rational numbers \(14/10\), \(141/100\), and \(1414/1000\); each lies below \(\sqrt{2}\) and within \(10^{-n}\) of it at its corresponding index \(n\).

Worked Example: A Limit Missing from an Open Interval

Give \(Y=(0,1)\) the metric inherited from \(\mathbb{R}\), and let \(y_n=1/(n+1)\). For \(n,m\ge N\),

|y_n-y_m| \le \frac{1}{n+1}+\frac{1}{m+1} \le \frac{2}{N+1}.

For any \(\varepsilon>0\), choose \(N\) so that \(2/(N+1)<\varepsilon\). Then \((y_n)\) is Cauchy in \(Y\). In the ambient real line, \(|y_n-0|=1/(n+1)\to0\), so \(y_n\to0\) in \(\mathbb{R}\). But \(0\notin Y\). If the sequence converged to some \(y\in Y\), it would also converge to \(y\) in \(\mathbb{R}\); uniqueness of limits would force \(y=0\), contradicting \(y\in(0,1)\). Thus it is Cauchy in \(Y\) and converges in \(\mathbb{R}\), but it does not converge in \(Y\). Check: \(y_1=1/2\), \(y_2=1/3\), and both belong to \((0,1)\); the excluded point is the limit, not a term of the sequence.

A Common Mistake and a Practical Check

Common pitfall. A Cauchy sequence in a subspace is sometimes treated as if it must converge in that subspace. For the sequence \(y_n=1/(n+1)\) in \((0,1)\), the calculation \(y_n\to0\) is correct in \(\mathbb{R}\), but the claim “therefore it converges in \((0,1)\)” is wrong: \(0\notin(0,1)\). The correction is to name the space with every convergence claim. The sequence is Cauchy in \((0,1)\), converges to \(0\) in \(\mathbb{R}\), and has no limit in \((0,1)\).

For a Cauchy calculation, first write the metric formula, then bound the distance between two arbitrary terms with indices \(n,m\ge N\). For a convergence claim, specify a candidate point and verify its distance from \(x_n\) tends to zero. In the discrete metric, test whether the sequence is eventually constant. In the finite-dimensional sup metric, test each coordinate and take a single index large enough for all finitely many coordinates.

Check Your Understanding

In each question, keep the metric space in view as well as the distance calculation.

Show directly from the Cauchy definition that \(a_n=5-\frac{2}{n}\) is Cauchy in \(\mathbb{R}\).
In a discrete metric space, can a Cauchy sequence take infinitely many different values? Justify your answer from the definition.
For \(x_n=(1/n,1/n^2)\) in \(\mathbb{R}^2\), use the sup metric to calculate \(d_\infty(x_n,(0,0))\) and show that \((x_n)\) is Cauchy.
Let \(z_n=1/(n+2)\) in \((0,1)\). State its limit in \(\mathbb{R}\), and explain whether it converges in \((0,1)\).
What part of the argument for the sup-metric theorem uses the fact that \(\mathbb{R}^k\) has finitely many coordinates?