# Cauchy Sequence

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## 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.

$\tag*{\square}$

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