Chapter 9 Continuous function on R

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Definition 9.1.8 (Adherent points)

$X \subseteq \mathbf{R}$, $x \in \mathbf{R}$, $x$ is an adherent point of $X \Leftrightarrow $ $\forall \epsilon >0, \exists x' \in X, \left|x-y\right|< \epsilon$

Definition 9.1.10 (Closure)

$x \in \overline{X} \Leftrightarrow x$ is an adherent point of $X$

Lemma 9.1.11

$X \subseteq \overline{X}$

prove: $x \in X \Rightarrow \forall \epsilon>0,\left|x-x\right|=0< \epsilon \Rightarrow x \in \overline{X}$
$\overline{X \cup Y}= \overline{X} \cup \overline {Y}$

for sake of contradiction, let $x \in \overline{X \cup Y} , x \notin \overline{X} \cup \overline {Y} \Rightarrow \exists \epsilon_1, \forall x' \in X, \left|x-x'\right|> \epsilon_1, \exists \epsilon_2, \forall x' \in Y, \left|x-x'\right|> \epsilon_2 $ let $\epsilon = \min(\epsilon_1, \epsilon_2), $ then $\forall x' \in X \cup Y, \left|x-x'\right| > \epsilon \Rightarrow x $ is not an adherent point of $\overline{X \cup Y}$

therefore $x \in \overline{X \cup Y} \Rightarrow x \in \overline{X} \cup \overline {Y}$

$x \in \overline{X} \cup \overline{Y} \Rightarrow x \in \overline{X} \lor x \in \overline{Y}$, if $x \in \overline{X}, \forall \epsilon>0, \exists x' \in X, \left|x-x'\right| \le \epsilon$ since $x' \in X \Rightarrow x' \in X \cup Y$ ,

$\Rightarrow x \in \overline{X \cup Y} $, similarly, $x \in \overline{Y} \Rightarrow x \in \overline{X \cup Y}$

$\overline{\overline{X}} = \overline{X}$

by above $\overline{X} \subseteq \overline{\overline{X}}$ , $x \in \overline{\overline{X}} \Rightarrow \forall \epsilon/2 >0, \exists x_1 \in \overline{X}, \left|x-x_1\right| \le \epsilon/2, x_1 \in \overline{X} \Rightarrow \exists x_2 \in X, \left|x_1 -x_2\right| \le \epsilon$ , by triangle inequality, $\left|x-x_2\right|=\left|x-x_1+x_1 -x_2\right| \le \epsilon \Rightarrow \overline{\overline{X}} \subseteq \overline{X}$
$\overline{X \cap Y} \subseteq \overline{X} \cap \overline{Y}$

$x \in \overline{X \cap Y} \Rightarrow \forall \epsilon>0, \exists p \in X \cap Y, \left|x-p\right| \le \epsilon$, $p \in X \cap Y \Rightarrow p \in X \land p \in Y \Rightarrow x \in \overline{X} \land x \in \overline{Y} \Rightarrow x \in \overline{X} \cap \overline{Y}$
$X \subseteq Y \Rightarrow \overline{X} \subseteq \overline{Y}$

$x \in \overline{X} \Rightarrow \forall \epsilon >0, \exists x' \in X, \left|x -x'\right|\le \epsilon , X \subseteq Y \Rightarrow x' \in Y \Rightarrow x \in \overline{Y}$

Lemma 9.1.12 (Closures of intervals)

closure of $(a,b)$ is $[a,b]$

Let $I=(a,b)$
1. $a<x<b \Rightarrow x \in I \Rightarrow x \in I \Rightarrow x \in \overline{I}$
2. $x=a \Rightarrow \forall \epsilon>0, \exists x'=a+\min(\epsilon,(b-a)/2)$ , $a<x'<b$ and $\left|x-x'\right|\le \epsilon$
  
  similarly $b$ is an adherent point of $I$
3. If $x<a$, let $\epsilon=(a-x)/2$, $a<x' \Rightarrow x'-x>a-x \Rightarrow \left|x'-x\right| \ge \epsilon$
  
  similarly, $x>b \Rightarrow$ $x$ is not an adherent point of $I$
closure of $(a,b]$ is $[a,b]$

let $I=(a,b]$

similarly if $a<x \le b$, then $x \in \overline{I}$, if $x<a$, $x \notin \overline{I}$

If $x > b$, let $\epsilon=(x-b)/2$ $\quad x' \le b \Rightarrow x-x' \ge x-b > (x-b)/2 $

$\overline{I}= [a,b]$
closure of $[a,b)$ is $[a,b]$
closure of $[a,b]$ is $[a,b]$

let $I=(a,b), J=[a,b]$, $J= \overline{I} \Rightarrow \overline{J}=\overline{\overline{I}}=\overline{I}=J$

Lemma 9.1.13

$\overline{\mathbf{N}} = \mathbf{N}$
1. $x \in \overline{\mathbf{N}} \Rightarrow x \ge 0$
if $x<0$, let $\epsilon = -x/2$, $\forall n \in \mathbf{N}, n \ge 0 \Rightarrow \left|n-x\right|=n-x \ge -x >-x/2$
1. $x \in \overline{\mathrm{N}} \Rightarrow x \in \mathbf{N}$
  
  $x \in \overline{\mathbf{N}}, x \ge 0 \Rightarrow \exists N \in \mathbf{N}, N \le x <N+1$ , If $x \ne N, \mathrm{let} ; 2\epsilon = \min(x-N,N+1-x)$, Therefore $\forall n \le N, n \ge N+1, \left|n-x\right| > \epsilon $
$\overline{\mathbf{Z}} = \mathbf{Z} $
1. $x \in \mathbf{Z} \Rightarrow x \in \mathbf{R} \Rightarrow \exists N \in \mathbf{Z}, N \le x <N+1$ , similarly, if $x \in N$, let $2\epsilon= \min(x-N,N+1-x)$,
  
  $\forall x' \in \mathbf{Z}, \left|x'-x\right|> \epsilon$
$\overline{\mathbf{Q}} = \mathbf{R}$

$\forall \epsilon >0, x \in \mathbf{R}, $ since $x < x+\epsilon$, there must exists some rational $r, x <r<x+\epsilon \Rightarrow - \epsilon <x-r <0 \Rightarrow \left|x-r\right| \le \epsilon \Rightarrow \mathbf{R} \subseteq \overline{\mathbf{Q}}$

Lemma 9.1.14

$X \subseteq \mathbf{R},x $ is an adherent point of $X \Leftrightarrow \exists (a_n){n=0}^{\infty}, a_n \in X , \lim{n \rightarrow \infty}a_n =x$

Prove:

$\Rightarrow :$ since $1/n>0, x$ is an adherent point of $X$, the set $E_n =\left{x' \in X:\left|x-x'\right| < 1/n \right}$ is not empty

let $a_n \in E_n$, therefore $\forall \epsilon>0, \exists 1/N_\epsilon < \epsilon, \forall n \ge N_\epsilon,$

$\left|a_n -x\right| \le 1/N_\epsilon \le \epsilon$

$\Leftarrow :$ If $\lim_{n \rightarrow }a_n=x, a_n \in X$, $\forall \epsilon>0, \exists N_{\epsilon/2} ,\forall n \ge N_{\epsilon/2}, \left|a_n-x\right| \le \epsilon/2 < \epsilon \Rightarrow \exists x'=a_{N_{\epsilon/2}}, \left|x-x'\right|<\epsilon$

Definition 9.1.15

$X$ is closed $\Leftrightarrow X=\overline{X}$

Corollary 9.1.17

$X = \overline{X} \Rightarrow $ If $a_n \in X, \lim_{n \rightarrow \infty}a_n=x$ then $x \in X$

$a_n \in X, \lim_{n \rightarrow \infty}a_n = x \Rightarrow x$ is an adherent point of $X \Rightarrow x \in \overline{X} \Rightarrow x \in X$

Definition 9.1.18 Limit points

$x$ is an limit point of $X \Leftrightarrow x$ is an adherent point of $X\setminus \left{x\right}$