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3.3.  Cauchy Sequences

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 \(n_0\in \mathbb N^{+}\) such that

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

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 \(\operatorname {diam}{A}\).

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

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

Proof. Let \(a_k = \operatorname {diam}{A_k}\). Suppose \((x_n)\) is a Cauchy sequence, and let \(\varepsilon>0\) be given. Then, there is \(n_0\in \mathbb N^{+}\) 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 \(a_{n_0} <\varepsilon\), which implies \(a_k\to 0\). Conversely, suppose \(a_k\to 0\), and let \(\varepsilon>0\) be given. Then, there is \(k_0\in \mathbb N^{+}\) such that \(a_{k_0}<\varepsilon\). That is, if \(m,n\ge k_0\) then

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

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

Theorem 3.3.4. Let \(M\) be a metric space. Then the following hold.

  1. Let \(A\) be a subset of \(M\). Then, \(\operatorname {diam} {\overline A} = \operatorname {diam} {A}\).

  2. Let \((K_n)\) be a sequence of 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 \(d(x,y)<\operatorname {diam} {A}\). 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)\(\square\)

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

  1. If \((x_n)\) converges in \(M\), then \((x_n)\) is a Cauchy sequence.

Proof. (1) Suppose \((x_n)\) converges to \(x\in M\) and let \(\varepsilon>0\) be given. Then, there is \(n_0\in \mathbb N^{+}\) 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.\(\square\)

References

  1. Rudin, W. (1976). Principles of mathematical analysis (3rd ed.). McGraw-Hill Education.