The Real numbers $ℝ$

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

Last updates: 29 July 2014

The real numbers $ℝ$

The real numbers $ℝ$ is the set $$ℝ={ℝ}_{\le 0}\cup {ℝ}_{\ge 0}$$ where $${ℝ}_{\ge 0}=\{a_l{a}_{l-1}\dots a_1{a}_0.a_{-1}{a}_{-2}{a}_{-3}\dots |l\in {\mathbb{Z}}_{\ge 0},a_i\in \{0,1,\dots ,9\}\}$$ $${ℝ}_{\le 0}=\{-a_l{a}_{l-1}\dots a_1{a}_0.a_{-1}{a}_{-2}{a}_{-3}\dots |l\in {\mathbb{Z}}_{\ge 0},a_i\in \{0,1,\dots ,9\}\}$$ and $$a=b\text{if}a-b=0.00000\dots ,$$ where addition and multiplication are given by

1. the $m^{\mathrm{th}}$ decimal place of $a+b$ is the same as the $m^{\mathrm{th}}$ decimal place of $$a_{\le 2m}+b_{\le 2m}\text{in}\mathbb{Q}\times \mathbb{Q}\to \mathbb{Q},\text{and}$$
2. the $m^{\mathrm{th}}$ decimal place of $ab$ is the same as the $m^{\mathrm{th}}$ decimal place of $$a_{\le 2m}{b}_{\le 2m}\text{in}\mathbb{Q}\times \mathbb{Q}\to \mathbb{Q}.$$

THESE OPERATIONS ARE REALLY MUCKED UP AND NEED TO BE FIXED.

Define a relation $\le$ on $ℝ$ by $$x\le y\text{if}y-x\in {ℝ}_{\ge 0}.$$

Define the absolute value on $ℝ$ $$\begin{array}{rrcl}\left|\phantom{T}\right|:& ℝ& ⟶& {ℝ}_{\ge 0}\\ & x& ⟼& \left|x\right|\end{array}\text{by}\left|x\right|=\{\begin{array}{cc}x,& \text{if}x>0,\\ 0,& \text{if}x=0,\\ -x,& \text{if}x<0.\end{array}$$

Define the distance on $ℝ$ $$d:ℝ\times ℝ⟶{ℝ}_{\ge 0}\text{by}d(x,y)=|y-x|.$$

Let $\epsilon \in {ℝ}_{>0}$. The $\epsilon$-ball at $x$ is $${ℬ}_{\epsilon }\left(x\right)=\{y\in ℝ\left|d\right(x,y)<\epsilon \}.$$

Let $E$ be a subset of $ℝ$. The set $E$ is open if

$E\text{is a union of}\epsilon \text{-balls.}$

(1.1)

1. The set $ℝ$ with the operations of addition, multiplication, the order $\le$ and open sets as in (1.1) is an ordered field and a topological field.
2. The set $\mathbb{Q}$ is a ordered topological subfield of $ℝ$.

Let $J\subseteq ℝ$ with $J\ne \varnothing \text{.}$ The subset $J$ is connected if and only if $J$ is an interval.

		Proof.
	$\Rightarrow \text{:}$ Assume $J$ is not an interval. Let $x,y\in J$ and $z\in ℝ$ with $$x<z<y,x,y\in J\text{and}z\notin J\text{.}$$ Let $A=(-\infty ,z)\cap J$ and $B=(z,\infty )\cap J\text{.}$ Then $A$ and $B$ are open subsets of $J$ and $$A\ne \varnothing ,B\ne \varnothing ,A\cap B=\varnothing \text{and}A\cup B=J\text{.}$$ So $J$ is not connected. $\Leftarrow \text{:}$ Assume $J$ is an interval. To show: $J$ is connected. Proof by contradiction. Assume $J$ is not connected. Let $A\subseteq J$ and $B\subseteq J$ be open subsets of $J$ such that $$A\cap B=\varnothing ,A\ne \varnothing ,B\ne \varnothing \text{and}A\cup B=J\text{.}$$ Then $f:J\to \{0,1\}$ given by $$f\left(z\right)=\{\begin{array}{cc}0,& \text{if}\hspace{0.17em}z\in A,\\ 1,& \text{if}\hspace{0.17em}z\in B\end{array}$$ is a continuous surjective function. Let $x_1,y_1\in J$ with $f\left(x_1\right)=0$ and $f\left(y_1\right)=1\text{.}$ Switching $A$ and $B$ if necessary we may assume that $x_1<y_1\text{.}$ Construct sequences $x_1,x_2,\dots$ and $y_1,y_2,\dots$ by $$\begin{array}{c}{\displaystyle x_{i+1}=\frac{x_i+y_i}{2}\text{and}{y}_{i+1}=y_i,\text{if}f\left(\frac{x_i+y_i}{2}\right)=0,}\\ {\displaystyle x_{i+1}=x_i\text{and}{y}_{i+1}=\frac{x_i+y_i}{2},\text{if}f\left(\frac{x_i+y_i}{2}\right)=1\text{.}}\end{array}$$ By induction, $x_i\in J$ and $y_i\in J,$ and, since $J$ is an interval, $\frac{x_i+y_i}{2}\in J$ so that $f\left(\frac{x_i+y_i}{2}\right)$ is defined and $$x_{i+1}\in J\text{and}{y}_{i+1}\in J\text{.}$$ Also, $$f\left(x_{i+1}\right)=0,f\left(y_{i+1}\right)=1,x_i\le x_{i+1}<y_{i+1}\le y_i\text{and}|x_{i+1}-y_{i+1}|\le \frac{1}{2}|x_i-y_i|$$ so that $$|x_{i+1}-y_{i+1}|\le \frac{1}{2^i}|x_1-y_1|\text{.}$$ Since $ℝ$ is complete and the sequence $x_1,x_2,\dots$ is increasing and bounded by $y_1,$ $\underset{n\to \infty }{\text{lim}}{x}_n$ exists in $ℝ\text{.}$ Since $ℝ$ is complete and the sequence $y_1,y_2,\dots$ is decreasing and bounded by $x_1,$ $\underset{n\to \infty }{\text{lim}}{y}_n$ exists in $ℝ\text{.}$ Since $\underset{n\to \infty }{\text{lim}}|x_n-y_n|=0$ then $\underset{n\to \infty }{\text{lim}}{x}_n=\underset{n\to \infty }{\text{lim}}{y}_n\text{.}$ Let $$z=\underset{n\to \infty }{\text{lim}}{x}_n=\underset{n\to \infty }{\text{lim}}{y}_n\text{.}$$ Since $x_1\le x_2\le \cdots \le x_n<y_n\le y_{n-1}\le \cdots \le y_1$ for $n\in {\mathbb{Z}}_{>0}$ then $$x_1<z<y_1\text{.}$$ Since $J$ is an interval, $z\in J\text{.}$ Since $f$ is continuous, $$0=\underset{n\to \infty }{\text{lim}}f\left(x_n\right)=f\left(z\right)=\underset{n\to \infty }{\text{lim}}f\left(y_n\right)=1\text{.}$$ This is a contradiction. So $J$ is connected. $\square$

Notes and References

Decimal expansions (real numbers) are introduced to a child to read and write numerical values. Long division (the Eucidean algorithm) is the algorithm that converts rational numbers to real numbers. The integers $\mathbb{Z}$ is the free group on one generator, the rationals $\mathbb{Q}$ is the field of fractions of $\mathbb{Z}$, and the real numbers $ℝ$ is a completion of $\mathbb{Q}$ (a decimal expansion is no different than a Cauchy sequence of rational numbers). Thus, all three number systems are universal objects (in the sense of category theory).

The proof of the theorem that connected sets in $ℝ$ are intervals follows the proof given in the course notes of J. Hyam Rubinstein for Metric and Hilbert spaces at the University of Melbourne. This proof does not differ substantially from the proof in [Bou, Gen Top. Ch. IV §2 No. 5 Theorem 4] but is organised to be more self contained.

References

[BouTop] N. Bourbaki, General Topology, Chapter IV, Springer-Verlag, Berlin 1989. MR?????

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