Cauchy Sequence

Definition

Let \(\{p_n\}_{n=1}^\infty\) be a sequence in $\mathbb{R}$. The sequence is a Cauchy sequence if $\forall \epsilon >0, \exists N\in \mathbb{N}$ such that \(\begin{align} n,m > N \Rightarrow |p_n - p_m| < \epsilon. \end{align}\)

Theorem1

• Every convergent sequence is a Cauchy sequence.
• Every Cauchy sequence is bounded.

<proof> Let \(\{ p_n\}\) be a convergent sequence such that $p_n \rightarrow p \text{ as } n\rightarrow \infty$. Let \(\epsilon >0\) be given. There exists $N\in \mathbb{N}$ such that \(\begin{align}n\geq N \Rightarrow |p_n -p| <\frac{\epsilon}{2} \end{align}\) By triangular inequality, for $n,m \geq N$ \(\begin{align} \begin{split} |p_n - p_m| &\leq |p_n - p| + |p_m -p|\\ &< \frac{\epsilon}{2} + \frac{\epsilon}{2} \\ &= \epsilon \end{split} \end{align}\)

\(\therefore \{p_n\}\) is a Cauchy sequence.

• Let \(\{p_n\}\) be a Cauchy sequence. There exists $N \in \mathbb{N}$ such that \(\begin{align} n\geq N \Rightarrow |p_n - p_m| < 1. \end{align}\) That is $|p_n|-|p_m| \leq |p_n -p_m| < 1.$ Thus, $|p_n| < 1 + |p_m| \text{ for } n \geq N.$ \(\begin{align} M :=\max\{|p_1|, \ldots, |p_{N-1}|, |p_N|+1\} \end{align}\)

\[\therefore |p_n| \leq M \text{ for all } n\in\mathbb{N}.\] \[\tag*{$\square$}\]

Theorem2

\(\{p_n\}_{n=1}^\infty\) be a Cauchy sequence having convergent subsequence. Then the sequence \(\{p_n\}\) converges.

<proof> Let \(\{p_{n_k}\}\) be a convergent subsequence of \(\{p_n\} \text{ such that } p_{n_k} \rightarrow p \text{ as } k\rightarrow \infty.\) Let \(\epsilon >0\) be given.
Since \(\{p_n\}\) is a Cauchy sequence, there exists \(n_0 \in \mathbb{N} \text{ such that } n,m \geq n_0 \Rightarrow |p_n -p_m| <\frac{\epsilon}{2}.\)
Since \(\{p_{n_k}\}\) is a convergent subsequence, there is \(k_0 \in \mathbb{N} \text{ such that } k\geq k_0 \Rightarrow |p_{n_k} - p| < \frac{\epsilon}{2}.\)
Put \(N :=\{n_0, k_0\}\). If \(n\geq N\), \(\begin{align} \begin{split} |p_n -p| &= |p_n-p_N +p_N -p|\\ &\leq |p_n-p_N| + |p_N - p| \\ &< \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon \end{split} \end{align}\)

\[\therefore \lim_{n\to \infty} p_n = p\] \[\tag*{$\square$}\]

Theorem 3

Every Cauchy sequence converges in $\mathbb{R}$.

<proof> Let \(\{p_n\}\) be a Cauchy sequence. Then \(\{p_n\}\) is bounded. By Bolzano-Weierstrass theorem, \(\{p_n\}\) has a convergent subsequence. By the theorem2, \(\{p_n\}\) converges. \(\tag*{$\square$}\)

Remark

We say that $\mathbb{R}$ is complete because theorem2 holds. In general, if $(X,d)$ is metric space, then we say that \((X,d)\) is complete if every Cauchy sequence in $X$ converges in $X$.

Theorem 4

If every Cauchy sequence in $\mathbb{R}$ converges, then every nonempty subsuet of $\mathbb{R}$ that is bounded above has a supremum.

<proof>

Let \(A\) be a bounded nonempty subset of \(\mathbb{R}\). If \(\lvert A\rvert\) is finite, it is trivial case. Otherwise, suppose that \(\lvert A\rvert\) is infinite. Let \(b_1\) be an upper bound of \(A\). Then, there is \(a_1 \in A\) such that \(a_1 < b_1\). Define \(m_1 :=\frac{a_1+b_1}{2}.\) If \(m_1\) is an upper bound of \(A\), put \(b_2 :=m_1\) and \(a_2 :=a_1\). Otherwise, \(a_2 := m_1, b_2 :=b1.\) Repeat this process: If \(m_n:=\frac{a_n+b_n}{2}\) is an upper bound of \(A\), put \(b_{n+1} := m_n, a_{n+1} := a_n.\) Otherwise, \(b_{n+1} := b_n, a_{n+1} :=m_n\).

Then \(\{a_n\}, \{b_n\}\) are bounded and monotone increasing and decreasing sequence, respectively. By monotone convergence theorem,

\[\begin{align*} \lim_{n\to\infty} a_n &= a \\ \lim_{n\to\infty} b_n &= b \end{align*}\]

Let \(\epsilon>0\) be given. Since \(\lvert b_n-a_n\rvert = \frac{b_1 -a_1}{2^{n-1}}\), \(\vert b_n -a_n\rvert < \epsilon \text{ if } n \geq N \text{ with } N\in \mathbb{N} \text{ such that }\frac{b_1-a_1}{2^{N-1}} < \epsilon.\) Therefore, \(\lim_{n\to\infty}(b_n-a_n) =0.\) i.e. \(\{b_n -a_n\}\) is a Cauchy sequence, by theorem 1.

Suppose \(m\geq n\). Then \(\lvert a_m - a_n\rvert\leq \lvert b_n - a_n\rvert < \epsilon \text{ for } n\geq N\) such that \(\frac{b_1-a_1}{2^{N-1}} < \epsilon.\) Similarly, \(\lvert b_m-b_n\rvert\leq \lvert a_n-b_n\rvert < \epsilon.\) Thus, \(\{a_n\}_{n=1}^\infty, \{b_n\}_{n=1}^\infty\) are Cauchy sequence.

\[\therefore b=\lim_{n\to\infty}b_n = \lim_{n\to\infty}(b_n-a_n) + \lim_{n\to\infty}a_n = \lim_{n\to\infty}a_n=a.\]

Now, we want to show that \(\lim_{n\to\infty}b_n =b\) is the supremum of $A$. Since $b_n$ is an upper bound of $A$, $\forall x \in A, x\leq b_n \text{ for all }n\in\mathbb{N}.$ We claim that \(\lim_{n\to\infty}b_n \geq x \text{ for all } x\in A.\)Suppose \(\lim_{n\to\infty}b_n < x.\) Take \(\epsilon_0 := x- b >0\). Since \(b_n \rightarrow b \text{ as } n \rightarrow \infty\), there is \(n_0 \in \mathbb{N} \text{ such that } n\geq n_0 \Rightarrow b-\epsilon_0 < b_n < b+\epsilon_0 = x\), which implies that \(b_n <x \text{ for } n\geq N.\) It contradicts to the assumption. Therefore, \(\lim_{n\to\infty}b_n \geq x.\) i.e. $b$ is an upper bound of $A$.

Lastly, we want to show that $b$ is the least upper bound of $A$. Suppose that there is another upper bound $M$ of $A$ such that $M<b=a.$ Since $a_n \leq M \text{ for all } n\in \mathbb{N}$, \(\lim_{n\to\infty}a_n = a \leq M.\) Moreover, $a_n \leq a \text{ for all } n \in \mathbb{N}$. Therefore \(a_n\leq a \leq M < b\), but it contradicts to the assumption that $a=b.$

\[\therefore b=\sup A.\] \[\tag*{$\square$}\]

Definition

Let $E$ be a subset of metric space $X$. We define diameter of the set $E$ as

\[\begin{align*} \text{diam}E= \sup \{d(p,q)\mid p,q\in E \} \end{align*}\]

Remark

Suppose that $(p_n)^\infty_{n=1}$ is a sequence. Let \(E_N := \{ p_N, p_{N+1},\ldots \}\). Then the sequence \((p_n)_{n=1}^\infty\) is a Cauchy sequence if and only if \(\displaystyle{\lim_{N\to\infty}\text{diam}E_N}=0\)

Given any $\epsilon >0$, there is $M\in\mathbb{N}$ such that $N\geq M \Rightarrow \text{diam}E_N=\sup_{n,m \geq M} d(p_n, p_m)<\epsilon$, which is equivalent to say that the sequence is Cauchy.

Theorem 5

Let $E$ be a subset of metric space $X$. Then diam$\overline{E} = E$.

<Proof> Pick any $p,q\in E$. For any $\epsilon >0$, there are $p^\prime, q^\prime\in E$ such that

\[\begin{align*} d(p,p^\prime) < \frac{\epsilon}{2}, \quad d(q,q^\prime) < \frac{\epsilon}{2} \end{align*}\]

If $p\in E^\prime$, there is $p^\prime \in N^\prime_{\frac{\epsilon}{2}}(p)\cap E\neq \emptyset$. Otherwise we can take $p^\prime :=p$.

With triangular inequality,

\[\begin{align*} d(p,q) &\leq d(p, p^\prime) + d(p^\prime, q) \\ &\leq d(p,p^\prime) + d(q^\prime, q) + d(p^\prime, q^\prime) \\ &\leq d(p^\prime,q^\prime) + \epsilon \end{align*}\]

By the definition of diameter of $E$,

\[\begin{align*} d(p,q) \leq d(p^\prime, q^\prime) + \epsilon \leq \text{diam}E + \epsilon \end{align*}\]

Since $\text{diam}\overline{E}$ is the least upper bound of \(\{d(p,q)\mid p,q\in \overline{E}\}\), $d(p,q) \leq \text{diam}\overline{E}\leq \text{diam}E + \epsilon$. Since the choice of $\epsilon$ is arbitrary, $\text{diam}\overline{E} \leq \text{diam} E$. It is trivial that $\text{diam}\overline{E} \geq \text{diam}E$.

$\therefore \text{diam}\overline{E} = \text{diam}E$.

\[\tag*{$\square$}\]

Theorem 6

Let \((K_n)_{n=1}^\infty\) be a sequence of nested sets in metric space \(X\). If \(K_n\) is compact and \(\displaystyle{\lim_{n\to\infty} \text{diam}(K_n)=0}\), then \(\bigcap_{n=1}^\infty K_n\) consist of exactly one point.

<Proof>

Since $K_n$ are nested compact sets, every finite intersection of $K_n$ is not empty. By previous Theorem, we know that $K:=\bigcap_{n=1}^\infty K_n$ is not empty. Suppose that there are two distinct points $p_1, p_2 \in \bigcap_{n=1}^\infty K_n$.

Since $d(p_1, p_2)>0, \text{diam} K >0$. But for each $n\in\mathbb{N}, K_n \supset K$, so that $\text{diam}K_n \geq \text{diam}K >0$. This contradicts the assumption that $\displaystyle{\lim_{n\to\infty} \text{diam}(K_n)=0}$.

\[\tag*{$\square$}\]

Definition

Let $X$ be a metric space. If every Cauchy sequence converges in $X$, then $X$ is complete.

Theorem 7

If $X$ is a compact metric space then $X$ is complete

<Proof>

Let \((p_n)_{n=1}^\infty\) be a Cauchy sequence and let \(E_N=\{p_N, p_{N+1}, \ldots \}\). Consider \((\overline{E}_N)_{N=1}^\infty\). We know that \(\text{diam}\overline{E} = \text{diam}E\), so \(\lim_{N\to\infty}\text{diam}\overline{E}_N=0\).

Since \(\overline{E}_N\) is closed and subset of compact \(X\), \(\overline{E}_N\) is compact by previous Theorem 3.2.5. Moreover \((\overline{E}_N)_{N=1}^\infty\) is nested. By previous theorem, there is a unique point \(p\in \bigcap_{N=1}^\infty \overline{E}_N\).

We want to show that $\lim_{n\to\infty}p_n = p$.

Let $\epsilon >0$ be given. Choose $M\in\mathbb{N}$ such that $N\geq M \Rightarrow \text{diam}\overline{E}_N=\text{diam}E_N < \epsilon$. Then if $n\geq M, d(p_n, p) < \text{diam}E_n <\epsilon$.

$\therefore \lim_{n\to\infty}p_n =p$

\[\tag*{$\square$}\]

Corollary

$\mathbb{R}^k$ is complete

<Proof>

Let \((p_n)_{n=1}^\infty\) be a Cauchy sequence in \(\mathbb{R}^k\). Since \(\lim_{N\to\infty}\text{diam}E_N=0\), we can choose \(M\in\mathbb{N}\) such that \(\text{diam}E_M < 1\). Then \(d(p_n, p_m) <1\) for all \(n,m \geq M\). Thus \(E_M\) is bounded. Furthermore, \(\{p_1, \ldots, p_{M-1}\}\) is bounded and thus \((p_n)_{n=1}^\infty\)is bounded.

Thus, \(S = \{p_n\}_{n=1}^\infty \cup \overline{ \{p_n\}}_{n=1}^\infty\) bounded and closed., which is compact. In other words, \((p_n)_{n=1}^\infty\) is a Cauchy sequence in compact set \(S\). Therefore, \((p_n)_{n=1}^\infty\) converges in \(S\).

$\therefore \mathbb{R}^k$ is complete.

\[\tag*{$\square$}\]