Real Analysis

Arun Ram
Department of Mathematics and Statistics
University of Melbourne
Parkville, VIC 3010 Australia
aram@unimelb.edu.au

Last update: 19 July 2014

Lecture 28

The real numbers $ℝ$

The real numbers $ℝ$ is the set $$R={ℝ}_{\le 0}\cup {ℝ}_{\ge 0},$$ where $$\begin{array}{ccc}{\displaystyle {ℝ}_{\ge 0}}& =& {\displaystyle \left\{a_{\ell }{a}_{\ell -1}\cdots a_1{a}_0\text{.}{a}_{-1}{a}_{-2}{a}_{-3}\cdots \hspace{0.17em}\right|\hspace{0.17em}\ell \in {\mathbb{Z}}_{\ge 0},a_i\in \{0,1,\dots ,9\}\}}\\ {\displaystyle {ℝ}_{\le 0}}& =& {\displaystyle \{-a_{\ell }{a}_{\ell -1}\cdots a_1{a}_0\text{.}{a}_{-1}{a}_{-2}\cdots \hspace{0.17em}|\hspace{0.17em}\ell \in {\mathbb{Z}}_{\ge 0},a_i\in \{0,1,\dots ,9\}\}}\end{array}$$ and $$a=b\text{if}a-b=0.000\dots \text{.}$$ Define a relation $\le$ on $ℝ$ by $$x\le y\text{if}y-x\in {ℝ}_{\ge 0}\text{.}$$ Define the absolute value on $ℝ$ $$\begin{array}{cccc}|·|:& ℝ& ⟶& {ℝ}_{\ge 0}\\ & x& ⟼& \left|x\right|\end{array}\text{by}\left|x\right|=\{\begin{array}{cc}x,& \text{if}\hspace{0.17em}x\in {ℝ}_{\ge 0},\\ 0,& \text{if}\hspace{0.17em}x=0,\\ -x,& \text{if}\hspace{0.17em}x\in {ℝ}_{\le 0}\text{.}\end{array}$$ Define a distance on $ℝ,$ $d:ℝ\times ℝ\to ℝ,$ by $$d(x,y)=|y-x|\text{.}$$ Let $\epsilon \in {ℝ}_{>0}$ and $x\in ℝ\text{.}$ The $\epsilon \text{-ball}$ at $x$ is $$B_{\epsilon }\left(x\right)=\{y\in ℝ\hspace{0.17em}|\hspace{0.17em}d(x,y)<\epsilon \}\text{.}$$ Let $E$ be a subset of $ℝ\text{.}$ The set $E$ is open if $E$ is a union of $\epsilon \text{-balls.}$

The Euclidean space ${ℝ}^n$

The Euclidean space ${ℝ}^n$ is the set $${ℝ}^n=\left\{(x_1,x_2,\dots ,x_n)\hspace{0.17em}\right|\hspace{0.17em}{x}_1,\dots ,x_n\in ℝ\}\text{.}$$ Define the absolute value on ${ℝ}^n,$ $$\begin{array}{cccc}|·|:& {ℝ}^n& ⟶& {ℝ}_{\ge 0}\\ & x& ⟼& \left|x\right|\end{array}\text{by}\left|x\right|=\text{sup}(\pm \sqrt{x_1^2+\cdots +x_n^2})$$ if $x=(x_1,\dots ,x_n)\text{.}$

The distance on ${ℝ}^n$ is $d:{ℝ}^n\times {ℝ}^n\to {ℝ}_{\ge 0}$ given by $$d(x,y)=|y-x|\text{.}$$ Let $\epsilon \in {ℝ}_{\ge 0}$ and $x\in {ℝ}^n\text{.}$ The $\epsilon \text{-ball}$ at $x$ is $$B_{\epsilon }\left(x\right)=\{y\in {ℝ}^n\hspace{0.17em}|\hspace{0.17em}d(x,y)<\epsilon \}\text{.}$$ Let $E$ be a subset of ${ℝ}^n\text{.}$ The set $E$ is open if $E$ is a union of $\epsilon \text{-balls.}$

A metric space is a set $X$ with a function $d:X\times X\to {ℝ}_{\ge 0}$ such that

(a)	if $x\in X$ then $d(x,x)=0,$
(b)	if $x,y\in X$ and $d(x,y)=0$ then $x=y,$
(c)	if $x,y\in X$ then $d(x,y)=d(y,x),$
(d)	if $x,y,z\in X$ then $d(x,z)\le d(x,y)+d(y,z)\text{.}$

Let $X$ be a metric space. Let $x\in X\text{.}$ Let $\epsilon \in {ℝ}_{>0}\text{.}$ The $\epsilon \text{-ball}$ at $X$ is $$B_{\epsilon }\left(X\right)=\{y\in X\hspace{0.17em}|\hspace{0.17em}d(y,x)<\epsilon \}\text{.}$$

Let $C=\{f:ℝ\to ℝ\hspace{0.17em}|\hspace{0.17em}f\hspace{0.17em}\text{is continuous and bounded}\}\text{.}$ Define $\Vert \hspace{0.17em}\Vert :C\to {ℝ}_{\ge 0}$ by $$\Vert f\Vert =\underset{x\in ℝ}{\text{sup}}\left|f\left(x\right)\right|\text{.}$$ Define a distance $d:C\times C\to {ℝ}_{\ge 0}$ by $$d(f,g)=|g-f|$$ where $(g-f)\left(x\right)=g\left(x\right)-f\left(x\right)$ for $x\in ℝ\text{.}$ Then $C$ is a metric space.

Let $X$ be a metric space and let $E$ be a subset of $X\text{.}$

The set $E$ is open if $E$ is a union of $\epsilon \text{-balls.}$

$E=\{x\in ℝ\hspace{0.17em}|\hspace{0.17em}|x+2|\le 2\hspace{0.17em}\text{or}\hspace{0.17em}\left|x\right|>1\}\text{.}$ So $$\begin{array}{ccc}{\displaystyle E}& =& {\displaystyle \{x\in ℝ\hspace{0.17em}|\hspace{0.17em}|x+2|\le 2\}\cup \{x\in ℝ\hspace{0.17em}|\hspace{0.17em}\left|x\right|>1\}}\\ & =& {\displaystyle \{x\in ℝ\hspace{0.17em}|\hspace{0.17em}x\ge -2\hspace{0.17em}\text{and}\hspace{0.17em}x+2\le 2\}}\\ & & {\displaystyle \cup \{x\in ℝ\hspace{0.17em}|\hspace{0.17em}x<-2\hspace{0.17em}\text{and}\hspace{0.17em}-(x+2)\le 2\}}\\ & & {\displaystyle \cup \{x\in ℝ\hspace{0.17em}|\hspace{0.17em}x>1\}\cup \{x\in ℝ\hspace{0.17em}|\hspace{0.17em}x<-1\}}\\ & =& {\displaystyle \{x\in ℝ\hspace{0.17em}|\hspace{0.17em}x\ge -2\hspace{0.17em}\text{and}\hspace{0.17em}x\le 0\}}\\ & & {\displaystyle \cup \{x\in ℝ\hspace{0.17em}|\hspace{0.17em}x<-2\hspace{0.17em}\text{and}\hspace{0.17em}x+2\ge -2\}}\\ & =& {\displaystyle \cup (1,\infty )\cup (-\infty ,-1)}\\ & =& {\displaystyle [-2,0]\cup \{x\in ℝ\hspace{0.17em}|\hspace{0.17em}x<-2\hspace{0.17em}\text{and}\hspace{0.17em}x\ge -4\}}\\ & & {\displaystyle \cup (1,\infty )\cup (-\infty ,-1)}\\ & =& {\displaystyle [-2,0]\cup [-4,-2)\cup (1,\infty )\cup (-\infty ,-1)\text{.}}\end{array}$$ $$\begin{array}{c} -4 -3 -2 -1 0 1 2 \end{array}$$ So $E=(-\infty ,0]\cup (1,\infty )\text{.}$

$E$ is not open in $ℝ\text{.}$

$E$ is not bounded above and not bounded below.

Notes and References

These are notes from a 2010 course on Real Analysis 620-295. This page comes from 100514Lect28.pdf and was given on 14 May 2010.

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