# Lecture 9

## Review

1. Let $X=\mathbb{R}$ (and as usual, let $d(x,y)=|x-y|$). What is the set $B_1(0)$
    $B_1(0)=(-1,1)$
2. Let $X=[0,5]$ (and let $d$ be as usual). What is the set $B_1(0)$.
    $B_1(0)=[0,1)$
3. Let $X=\mathbb{R}$ and let $E=[0,2)$. Is $E$ open?
    No, $0$ is not a interior point.
4. Let $X=[0,5]$ and let $E=[0,2)$. Is $E$ open?
    Yes, $0$ is a interior point, we can set radius to $1$ and all the points of $B_1(0)\subset E$

## Continue on new materials

### Metric space

#### Theorem 2.27

If $X$ is a metric spae and $E\subset X$, then.

1. $\bar{E}$ is closed.
2. $E=\bar{E}$ if and only if $E$ is closed.
    Proof: $E=\bar{E}\iff E=E\cup E'\iff E'\subset E\iff E$ is closed.
3. $\bar{E}\subset F$ for every closed set $F\subset X$ such that $E\subset F$.
    $F\subset X$ closed, and $E'\subset F'\subset F$, so $\bar{E}=E\cup E'\subset F$

#### Theorem 2.28

$E\subset \mathbb{R}$ non-$\phi$ and bounded above, $\sup E\in \bar{E}$

Proof:  

Let $y\sup E$, To show $y\in \bar{E}$, we need to show $\forall h>0$, $B_h(y)\cap E\neq \phi$

Let $h>0$. Since $y-h$ is not an upper bound of $E$, $\exists x\in E$ such that $x>y-h$.

Since $y$ is an upper bound of $E$, $x\leq y$. So $x\in B_n(y)\cap E$, so $B_h(y)\cap E\neq \phi$.

QED

#### Remark 2.29

Let $(X,d)$ be a metric space, $E\subset X$. "$E$ is open" is short for "$E$ is open in $X$"/$E$ is open relative to $X$.

This means $\forall p\in E$, $\exists r>0\{q\in X:d(p,q)<r\}\subset E$.

Suppose $E\subset Y\subset X$. $\exists r>0\{q\in Y:d(p,q)<r\}\subset E$.

Example: $X=\mathbb{R}$, $Y=[0,5]$, $E=[0,2)$, $E$ is open in $Y$, $E$ is not open on $X$.

#### Theorem 2.30

Let $E\subset Y\subset X$,then $E$ is open in $Y\iff \exists G$ open in $X$ such that $E=Y\cap G$

Proof:

Observe. If $p\in Y, r>0$, then $\{q\in Y:d(p,q)<r\}=\{q\in X:d(p,q)<r\}$ (ball in $X$ and ball in $Y$).

$\impliedby$

Suppose $\exists G$ open in $X$ such that $E=Y\cap G$. Let $p\in E$. Then $p\in G$ so $\exists r>0$ such that $\{q\in X:d(p,q)<r\}\subset G$ intersect both sides with $Y$ to get $\{q\in Y:d(p,q)<r\}\subset G\cap Y=E$

$\implies$ 

Suppose $E$ is open in $Y$. Then $\forall p\in E,\exists r_p>0$ such that $\{q\in Y:d(p,q)<r_p\}\subset E$

Let $V_p=\{q\in X:d(p,q)<r_p\}$, $G=\bigcup_{p\in E}V_p$.

Each $V_p$ is open in $X$ so $G$ is open in $X$. We need to show $E=G\cap Y$.

To show $E\subset G\cap Y$. $p\in E\implies p\in V_p\implies p\in G$. So $E\subset G$. Also, by assumption, $E\subset Y$.

To show $G\cap Y\subset E$.

$G\cap Y=\left(\bigcup_{p\in E}V_p\right)\cap Y=\bigcup_{p\in E}(V_p\cap Y)\subset E$

QED

### Compact sets

#### Definition 2.31/2.32

Let $(X,d)$ be a metric space $K\subset X$.

Let $A$ be an index set. Suppose $\forall \alpha\in A$, $G_\alpha$ is an open set (in $X$) and suppose $K\subset \bigcup_{\alpha\in E}G_\alpha$. Then we say $\{G_\alpha:\alpha\in A\}$ is an **open cover** of $K$.

Let $\{G_\alpha\}_{\alpha \in A}$ be an open cover of $K$. Suppose $\exists \alpha_1,...,\alpha_n\in A$ such that $K\subset\bigcup_{i=1}^n G_{\alpha_i}$. Then we say $\{G_{\alpha_i}\}_{i=1}^n$ is a **finite subcover** of $\{G_\alpha\}_{\alpha\in A}$ of $K$.

#### Definition

We say $K$ is **compact** if every open cover of $K$ contains a finite subcover.

i.e. $\forall$ (This is important) open cover $\{G_\alpha\}_{\alpha\in A}$ of $K$, $\exists$ finite subcover $\{G_{\alpha_i}\}_{i=1}^n$ of $\{G_\alpha\}_{\alpha\in A}$ of $K$.

Examples: $X=\mathbb{R}$.

1.$K=\mathbb{R}$. $\mathbb{R}$ is not compact.

as we can build an infinite open cover $\bigcup_{i\in Z} (i,i+2)$ and it does not have a finite subcover because:

Suppose we consider the sub-collection $\{n_i,n_i+2:i=1,..,k\}$, Then $N+3$ is not in the union, where $N=max\{n_1,...,n_k\}$.

QED
