# Lecture 8

## Review

Let $(X,d)$ be a metric space. Recall that $B_r(x)=\{z\in X:d(x,z)<r\}$.

Let $x,y\in X$ and let $r=\frac{1}{2}d(x,y)$. What do you think is true about $B_r(x)\cap B_r(y)$? Can you prove it?

It should be empty. Proof any point cannot be in two balls at the same time. (By triangle inequality or contradiction)

### Metric space defs

1. $p\in X,r>0$, $B_r(p)=\{q\in X:d(p,q)<r\}$, also called **neighborhood**.
2. $p$ is a **limit point** of $E(p\in E')$ if $\forall r>0$, $(B_s(p)\cap E)\backslash \{p\}\neq \phi$
3. If $p\in E$ and $p$ is not a limit point of $E$, then $p$ is called an **isolated point** of $E$.
4. $E$ is **closed** if $E'\subset E$
5. $p$ is a **interior point** of $E(p\in E^{\circ})$ if $\exists r>0$ such that $B_r(p)\subset E$.

## New materials

### Metric space

#### Theorem 2.20

$p\in E'\implies \forall r>0,B_r(p)\cap E$ is infinite.

<details>
<summary>Proof</summary>

We will prove the contrapositive.

want to prove $\exists r>0$ such that $B_r(p)\cap E$ is finite $\implies p\notin E'$ ($\exists s>0$ such that $(B_s(p)\cap E)\backslash \{p\}=\phi$)

Suppose $\exists r>0$ such that $B_r(p)\cap E$ is finite

let $B_s(p)\cap E)\backslash \{p\}={q_1,...,q_n}$ 

- If $n=0$, then $B_s(p)\cap E)\backslash \{p\}=\phi$, so $p\in E'$
- If $n\geq 1$, then let $s=min\{d(p,q_m):1\leq m\leq n\}$

    Each $d(p,q_m)$ is positive and the set is finite, so $s>0$.

Then $(B_s(p)\cap E)\backslash \{p\}=\phi$, so $p\notin E$

</details>

#### Theorem 2.22 De Morgan's law

$$
\left(\bigcup_a E_a\right)^c=\bigcap_a(E^c_a)
$$

$E^c=X\backslash E$

Proof:

$x\in \cup_{a\in A} E_x\iff \exists a\in A$ such that $x\in E_a$

So $x\in \left(\bigcup_a E_a\right)^c\iff \forall a\in A, x\notin  E_a\iff \forall a\in A,x\in E_a^c\iff \bigcap_a(E^c_a)$

#### Theorem 2.23

$E$ is open $\iff$ $E^c$ is closed.

> Warning: $E$ is open $\cancel{\iff}$ $E$ is closed.
> $E$ is closed $\cancel{\iff}$ $E$ is open.
>
> Example:  
>$\phi$, $\R$ is both open and closed. "clopen set"  
>$[0,1)$ is not open and not closed. bad...

<details>
<summary>Proof</summary>

$\impliedby$ Suppose $E^c$ is closed. Let $x\in E$, so $x\notin E^c$

$E^c$ is closed and $x\notin E^c\implies x\notin (E^c)'\implies \exists r >0$ such that $(B_r(x)\cap E^c)\backslash \{x\}=\phi$

So $\phi=(B_r(x)\cap E^c)\backslash \{x\}=B_r(x)\cap E^c$

So $B_r(x)\in E$

$\implies$

Suppose $E$ is open

$$
\begin{aligned}
    x\in (E^c)'&\implies \forall r>0, (B_r(x)\cap E^c)\backslash \{x\}\neq \phi\\
    &\implies \forall r>0, (B_r(x)\cap E^c)\neq \phi\\
    &\implies \forall r>0, B-r(x)\notin E\\
    &\implies x\notin  E^{\circ}\\
    &\implies x\notin E\\
    &\implies x\in E^c
\end{aligned}
$$

So $(E^c)'\subset E^c$

</details>

#### Theorem 2.24

##### An arbitrary union of open sets is open

<details>
<summary>Proof</summary>

Suppose $\forall \alpha, G_\alpha$ is open. Let $x\in \bigcup _{\alpha} G_\alpha$. Then $\exists \alpha_0$ such that $x\in G_{\alpha_0}$. Since $G_{\alpha_0}$ is open, $\exists r>0$ such that $B_r(x)\subset G_{\alpha_0}$ Then $B_r(x)\subset G_{\alpha_0}\subset \bigcup_{\alpha} G_\alpha$

</details>

##### A finite intersection of open set is open

<details>
<summary>Proof</summary>

Suppose $\forall i\in \{1,...,n\}$, $G_i$ is open.

Let $x\in \bigcap^n_{i=1}G_i$, then $\forall i\in \{1,..,n\}$ and $G_i$ is open, so $\exists r_i>0$, such that $B_{r_i}(x)\subset G_i$

Let $r=min\{r_1,...,r_n\}$. Then $\forall i\in \{1,...,n\}$. $B_r(x)\subset B_{r_i}(x)\subset G_i$. So $B_r(x)\subset \bigcup_{i=1}^n G_i$

</details>

The other two can be proved by **Theorem 2.22,2.23**

#### Definition 2.26

The closure $\bar{E}=E\cup E'$

Remark: Using the definition of $E'$, we have, $\bar{E}=\{p\in X,\forall r>0,B_r(p)\cap E\neq \phi\}$

#### Definition 2.27

$\bar {E}$ is closed.

<details>
<summary>Proof</summary>

We will show $\bar{E}^c$ is open.

Suppose $p\in \bar{E}^c$. Then by remark, $\exists r>0$ such that $B_r(p)\cap E=\phi$ (a)

Furthermore,, we claim $B_r(p)\cap E'=\phi$ (b)

Suppose for contradiction that $\exists q\in B_r(p)\cap E'$ By **Theorem 2.19**, $\exists s>0$ such that $B_s(q)\subset B_r(p)$

Since $q\in E',(B_s(q)\cap E)\backslash \{q\}\neq \phi$. This implies $B_r(p)\cap E=\phi$, which contradicts with (a)

This proves (b)

So $\bar{E}^c$ is open

</details>