# Lecture 17

## Review

Given a sequence $(a_n)$ in $\mathbb{R}$, let $E_n=\{a_k:k\geq n\}$. Calculate $diam E_1$, $diam E_2$, $diam E_3$... for the following sequences:

1. $a_n=0$: $E_n=\{0\}$, $$diam E_1=0, diam E_2=0, diam E_3=0, \ldots$$
2. $a_n=n$: $E_n=\{n\}$, $$diam E_1=\infty, diam E_2=\infty, diam E_3=\infty, \ldots$$
3. $a_n=(-1)^n$: $E_n=\{-1,1\}$, $$diam E_1=2, diam E_2=2, diam E_3=2, \ldots$$
4. $a_n=1/n$: $E_n=\{1/n,1/(n+1),\dots\}$, $$diam E_1=\frac{1}{2}, diam E_2=\frac{1}{3}, diam E_3=\frac{1}{4}, \ldots$$
5. $a_n=\frac{(-1)^n}{n}$: $E_n=\{-1/n,1/n,\dots\}$, $$diam E_1=\frac{1}{1}+\frac{1}{2}, diam E_2=\frac{1}{2}+\frac{1}{3}, diam E_3=\frac{1}{3}+\frac{1}{4}, \ldots$$

## New materials

### Cauchy sequence

#### Theorem 3.11

(b) If $X$ is a compact metric space, then every Cauchy sequence $(p_n)$ in $X$ converges.

(c) In $\mathbb{R}^k$, every Cauchy sequence $(p_n)$ converges.

Proof:

(b) Let $E_N=\{p_n:n\geq N\}$. Since $(p_n)$ is Cauchy, $\lim_{N\to\infty} diam E_N=0$. By **Theorem 3.10 (a)**, $\lim_{N\to\infty} diam \overline{E_N}=0$.

Since $X$ is compact, and $\overline{E_N}$ is closed, by **Theorem 2.35**, $\overline{E_N}$ is compact.

Since $E_1\supset E_2\supset E_3\supset\cdots$, $\overline{E_1}\supset \overline{E_2}\supset \overline{E_3}\supset\cdots$. By **Theorem 3.10(b)**, $\exists p\in X$ such that $p\in\bigcap_{N=1}^{\infty}\overline{E_N}$.

We claim that $(p_n)$ converges to $p$. Let  $\epsilon>0$, there exists $N_0$ such that $\forall N\geq N_0$, $diam \overline{E_N}<\epsilon$.

For any $n\geq N_0$, $p_n\in \overline{E_{N_0}}$.

So $d(p_n,p)\leq diam \overline{E_{N_0}}<\epsilon$, by definition of diameter.

Therefore, $(p_n)$ converges to $p$.

(c) Let $(p_n)$ be a Cauchy sequence in $\mathbb{R}^k$.

By **Theorem 3.9**, $(p_n)$ is bounded. So $\exists R>0$ such that $p_n\in B(0,R)$ for all $n$. Moreover $p_n\in \overline{B(0,R)}$. and $\overline{B(0,R)}$ is closed and bounded. Thus by **Theorem 2.41**, $\overline{B(0,R)}$ is compact.

Note that **Theorem 2.41** only works for $\mathbb{R}^k$.

So by (b), $(p_n)$ converges to some $p\in \overline{B(0,R)}$.

QED

#### Definition 3.12

Let $X$ be a metric space. We say $X$ is **complete** if every Cauchy sequence in $X$ converges.

**Theorem 3.11(b)** can also be rephrased as:

$X$ is a compact metric space $\implies$ $X$ is complete.

**Theorem 3.11(c)** can also be rephrased as:

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

> Note: completeness is a property of the "universe" $X$, not a property of any particular sequence in $X$.

$\mathbb{Q}$ is not complete. $\{3,3.1,3.14,3.141,3.1415,\dots\}$ is a Cauchy sequence in $\mathbb{Q}$ but it does not converge in $\mathbb{Q}$.

Fact: If $X$ is complete and $E$ is a closed subset of $X$, then $E$ is complete.

#### Definition 3.13

A sequence $(s_n)$ of real numbers is said to be

- **monotone increasing** if $s_n\leq s_{n+1}$ for all $n$.
- **monotone decreasing** if $s_n\geq s_{n+1}$ for all $n$.
- **strictly monotone increasing** if $s_n<s_{n+1}$ for all $n$.
- **strictly monotone decreasing** if $s_n>s_{n+1}$ for all $n$.
- **monotone** if it is either monotone increasing or monotone decreasing.

Example:

1. $s_n=1/n$ is strictly monotone decreasing.
2. $s_n=(-1)^n$ is neither monotone increasing nor monotone decreasing.

#### Theorem 3.14

Suppose $(s_n)$ is monotonic. Then $(s_n)$ converges $\iff$ $(s_n)$ is bounded.

Proof:

If $(s_n)$ is monotonic and bounded, then by previous result, $(s_n)$ converges.

If $(s_n)$ is monotonic and converges, then by **Theorem 3.2(c)**, $(s_n)$ is bounded.

QED

### Upper and lower limits

#### Definition 3.15 (Divergence to $\infty$ or $-\infty$)

Let $(s_n)$ be a sequence of real numbers with the following properties:

For every real number $M$ there is an integer $N$ such that $n\geq N$ implies $s_n>M$. We then write $s_n\to\infty$

For every real number $M$ there is an integer $N$ such that $n\geq N$ implies $s_n<M$. We then write $s_n\to-\infty$.

_for every real number, we can find a element in the sequence that is greater than or less than it_

#### Definition 3.16

Let $(s_n)$ be a sequence of real numbers.

Let $E_n=\{x\in[-\infty,\infty]:\exists \textup{ subsequence } (s_{n_k})\textup{ of } (s_n)\textup{ such that } s_{n_k}\to x\}$.

Let $S^*=\sup E_n$, $S_*=\inf E_n$.

We define $$\limsup_{n\to\infty} s_n=S^*$$ and $$\liminf_{n\to\infty} s_n=S_*$$

Informally, $S^*$ is the largest possible value that a subsequence of $(s_n)$ can converge to.

Example:

$s_n=(-1)^n$, $E_n=\{-1,1\}$, $S^*=\sup E_n=1$, $S_*=\inf E_n=-1$. and $\lim_{n\to\infty} s_n$ does not exist.

One advantage of $\limsup$ and $\liminf$ is that they always exist (they may be $\infty$ or $-\infty$), even if the sequence does not converge.