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

$p\in X,r>0$ , $B_r(p)=\{q\in X:d(p,q)<r\}$ , also called neighborhood.
$p$ is a limit point of $E(p\in E')$ if $\forall r>0$ , $(B_s(p)\cap E)\backslash \{p\}\neq \phi$
If $p\in E$ and $p$ is not a limit point of $E$ , then $p$ is called an isolated point of $E$ .
$E$ is closed if $E'\subset E$
$p$ 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.

Proof

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 $p\notin E$

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…

Proof

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

Theorem 2.24

An arbitrary union of open sets is open

Proof

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 $B_r(x)\subset G_{\alpha_0}\subset \bigcup_{\alpha} G_\alpha$

A finite intersection of open set is open

Proof

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 $B_r(x)\subset \bigcup_{i=1}^n G_i$

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.

Proof

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