Cauchy Sequences — Introduction to Analysis

Convergence describes the asymptotic behavior of a sequence by relating its terms to a point of the ambient space. This raises a natural question. How much of this asymptotic behavior can be recognized from the terms of the sequence themselves, without first specifying or establishing the existence of such a point? The concept of Cauchyness¹ isolates precisely this internal aspect of convergence. Instead of comparing the terms $$x_n$$ with a separate point $$x$,$ it compares sufficiently late terms with one another and asks whether they eventually become arbitrarily close.

Definition 3.3.1. Let $$M$$ be a metric space and let $$(x_n)$$ be a sequence in $$M$.$ The sequence $$(x_n)$$ is called a Cauchy sequence if for any $$\varepsilon >0$,$ there exists $n_0\in \mathbb Z^{+}$ such that

\[m,n\ge n_0\ \Longrightarrow \ d(x_m,x_n)<\varepsilon.\]

The Cauchy condition can be viewed geometrically as requiring that the tails of the sequence eventually become arbitrarily small as sets. To make this precise, we introduce the diameter of a set.

Definition 3.3.2. Let $$A$$ be a nonempty subset of a metric space $$M$,$ and let $$S$$ be the set of all reals of the form $$d(x,y)$$ where $$x,y\in A$.$ The supremum of $$S$$ is called the diameter of $$A$,$ denoted by $$\operatorname {diam}{A}$.$

Theorem 3.3.3. Let $$(x_n)$$ be a sequence in a metric space $$M$,$ and let $E_k = \{x_n : n\ge k\}.$ Then, $$(x_n)$$ is a Cauchy sequence if and only if

\[\lim\limits_{k\to \infty} {\operatorname {diam}{E_k}} = 0.\]

Proof. Let $$a_k = \operatorname {diam}{E_k}$.$ Suppose $$(x_n)$$ is a Cauchy sequence, and let $\varepsilon>0$ be given. Then, there is $n_0\in \mathbb Z^{+}$ such that $m,n\ge n_0$ implies $$d(x_m,x_n)<\varepsilon /2$.$ Hence the set of the reals of the form $$d(x_m,x_n)$$ is bounded above, and the supremum is $$a_{n_0}$.$ Therefore, we have $$a_{n_0} <\varepsilon$.$ Since $$E_{k+1}\subseteq E_k$,$ we have $$a_{k+1}\le a_k$,$ so $0\le a_k\le a_{n_0}<\varepsilon$ for every $$k\ge n_0$.$ Hence $$a_k\to 0$.$ Conversely, suppose $$a_k\to 0$,$ and let $\varepsilon>0$ be given. Then, there is $k_0\in \mathbb Z^{+}$ such that $$a_{k_0}<\varepsilon$.$ That is, if $m,n\ge k_0$ then

\[d(x_m,x_n)\le a_{k_0}<\varepsilon.\]

Thus $$(x_n)$$ is a Cauchy sequence. $\square$

The preceding results focus on describing Cauchyness itself. We now ask whether this internal asymptotic behavior of a sequence is enough to produce convergence. Every convergent sequence is necessarily Cauchy, but the converse requires something from the ambient space. Compactness provides one condition under which this can be guaranteed. To show that concisely, we first prove the following lemma.

Lemma 3.3.4. Let $$M$$ be a metric space. Then the following hold:

Let $$A$$ be a nonempty subset of $$M$.$ Then, $$\operatorname {diam} {\overline A} = \operatorname {diam} {A}$.$
Let $$(K_n)$$ be a sequence of nonempty compact sets in $$M$$ such that $K_{n+1}\subseteq K_n$ for $$n=1,2,3\ldots$.$ If
\[\lim\limits_{n\to \infty} {\operatorname {diam} {K_n}} = 0,\]
then $\bigcap_{n=1}^{\infty} K_n$ consists of one point.

Proof. (1) Let $\varepsilon>0$ be given, and let $$x,y\in \overline A$.$ Then there exist points $x',y'\in A$ such that $d(x,x')<\varepsilon$ and $$d(y,y')<\varepsilon$.$ Hence, we have

\[\begin{aligned} d(x,y) &\le d(x,x') + d(x',y') + d(y',y)\\ &< d(x',y') + 2\varepsilon\\ &\le \operatorname {diam}{A} + 2\varepsilon. \end{aligned}\]

It follows that $$d(x,y)\le\operatorname {diam} {A}$.$ Therefore $$\operatorname {diam}{\overline A} \le \operatorname {diam} {A}$.$ Since $$A\subseteq \overline A$,$ it is clear that $$\operatorname {diam} {A} \le \operatorname {diam} {\overline A}$.$ Consequently, $$\operatorname {diam} {\overline A} = \operatorname {diam} {A}$.$ (2) By Corollary 2.2.12, $\bigcap_{n=1}^{\infty} K_n$ is nonempty. If $x,y \in \bigcap_{n=1}^{\infty} K_n$ with $$x\ne y$,$ then $\operatorname {diam}{K_n}\ge d(x,y)>0$ for every $$n\in \mathbb Z^{+}$.$ This contradicts $$\lim\limits_{n\to \infty} {\operatorname {diam} {K_n}} = 0$.$ Therefore $\bigcap_{n=1}^{\infty} K_n$ consists of exactly one point. $\square$

Theorem 3.3.5. Let $$(x_n)$$ be a sequence in a metric space $$M$.$

If $$(x_n)$$ converges in $$M$,$ then $$(x_n)$$ is a Cauchy sequence.
If $$M$$ is compact and $$(x_n)$$ is a Cauchy sequence, then $$(x_n)$$ converges in $$M$.$

Proof. (1) Suppose $$(x_n)$$ converges to $x\in M$ and let $\varepsilon>0$ be given. Then, there is $n_0\in \mathbb Z^{+}$ such that $n\ge n_0$ implies $$d(x_n,x)<\varepsilon /2$.$ By the triangle inequality, it follows that

\[m,n\ge n_0 \ \Longrightarrow \ d(x_m,x_n)< \varepsilon.\]

Hence $$(x_n)$$ is a Cauchy sequence. (2) Let $$E_k = \{x_n : n\ge k\}$.$ Then Theorem 3.3.3 and Lemma 3.3.4(1) give

\[\lim\limits_{k\to \infty} {\operatorname {diam}{\overline {E_k}}} = 0.\]

Since $$M$$ is compact, by Theorem 2.2.7, $\overline {E_k}$ is compact. Since $$E_{k+1}\subseteq E_k$,$ we also have $$\overline {E_{k+1}} \subseteq \overline {E_k}$.$ Lemma 3.3.4(2) now shows that there is a unique point $$y\in \bigcap_{k=1}^{\infty}{\overline{E_k}}$.$ Let $\varepsilon >0$ be given. Then, there is $k_0\in \mathbb Z^{+}$ such that $k\ge k_0$ implies $$\operatorname {diam}{\overline {E_k}}<\varepsilon$.$ Since $$y\in \overline {E_k}$,$ if $z\in \overline {E_k}$ then $$d(z,y)<\varepsilon$;$ hence $z\in E_k$ implies $$d(z,y)<\varepsilon$.$ Since $n\ge k$ implies $$x_n\in E_k$,$ it is concluded that $n\ge k_0$ implies $$d(x_n,y)<\varepsilon$;$ this says that $$x_n\to y$.$ $\square$

A Euclidean space $\mathbb R^k$ is not compact, so Theorem 3.3.5 cannot be applied to it directly. However, every Cauchy sequence is bounded and is therefore contained in a compact $$k$-$ cell. It leads to the following corollary, which is often called the Cauchy criterion for convergence.

Corollary 3.3.6. Let $(\mathbf x_n)$ be a sequence in $$\mathbb R^k$.$ Then, $(\mathbf x_n)$ converges in $\mathbb R^k$ if and only if $(\mathbf x_n)$ is a Cauchy sequence.

Proof. The forward implication follows from Theorem 3.3.5(1). Conversely, suppose $(\mathbf x_n)$ is Cauchy. Let $$E_m = \{\mathbf x_n : n\ge m\}$.$ Then we have $\operatorname {diam}{E_{m_0}}<1$ for some $$m_0\in \mathbb Z^{+}$.$ Since the range is the union of a bounded tail and a finite set, it is bounded. Thus $\{\mathbf x_n : n\ge 1\}\subseteq I$ for some $$k$-$ cell $$I$.$ Since $$I$$ is compact, Theorem 3.3.5 completes the proof. $\square$

The argument in $\mathbb R^k$ reveals a distinction that is more fundamental than compactness itself. The Cauchy condition concerns the asymptotic relation among the terms of a sequence, whereas the assertion that every such sequence has a limit is a property of the ambient metric space. This motivates separating these two roles: a Cauchy sequence has the internal behavior required of a convergent sequence, and a complete space is one in which this behavior is always realized by an actual limit.

Definition 3.3.7. Let $$M$$ be a metric space. If every Cauchy sequence in $$M$$ converges in $$M$,$ then $$M$$ is said to be complete.

In the term Cauchyness, “Cauchy” refers to the French mathematician Augustin-Louis Cauchy (1789–1857). He was one of the major figures in the development of rigorous mathematical analysis, making fundamental contributions to limits, continuity, infinite series, and differential equations. In mathematical English, his surname is usually pronounced /ˈkoʊʃi/. ↩