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content/Math4111/Math4111_L8.md

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+# 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.
+
+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 $p\notin E$
+
+QED
+
+#### 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 $(E^c)'\subset E^c$
+
+QED
+
+#### 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 $B_r(x)\subset G_{\alpha_0}\subset \bigcup_{\alpha} G_\alpha$
+
+QED
+
+##### 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 $B_r(x)\subset \bigcup_{i=1}^n G_i$
+
+QED
+
+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
+
+QED