1.4: Real Numbers

Let $C$ be the set of all Cauchy sequences of rational numbers. We define a relation on $C$ as follows: If $\left\{a_{i}\right\}_{i \in I}$ and $\left\{b_{j}\right\}_{j \in J}$ are Cauchy sequences in $\mathbb{Q}$, then $\left\{a_{i}\right\}_{i \in I} \sim\left\{b_{j}\right\}_{j \in J},$ which we will write more simply as $a_{i} \sim b_{i},$ if for every rational number $\epsilon>0,$ there exists an integer $N$ such that

$$\left|a_{i}-b_{i}\right|<\epsilon$$

whenever $i>N .$ This relation is clearly reflexive and symmetric. To show that it is also transitive, and hence an equivalence relation, suppose $a_{i} \sim b_{i}$ and $b_{i} \sim c_{i} .$ Given $\epsilon \in \mathbb{Q}^{+},$ choose $N$ so that

$$\left|a_{i}-b_{i}\right|<\frac{\epsilon}{2}$$

for all $i>N$ and $M$ so that

$$\left|b_{i}-c_{i}\right|<\frac{\epsilon}{2}$$

for all $i>M .$ Let $L$ be the larger of $N$ and $M .$ Then, for all $i>L$,

$$\left|a_{i}-c_{i}\right| \leq\left|a_{i}-b_{i}\right|+\left|b_{i}-c_{i}\right|<\frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon.$$

Hence $a_{i} \sim c_{i}$.

Definition: set of equivalence classes

Using the equivalence relation just defined, we call the set of equivalence classes of $C$ the real numbers, denoted $\mathbb{R}$.

Note that if $a \in \mathbb{Q},$ we may identify $a$ with the equivalence class of the sequence $\left\{b_{i}\right\}_{i=1}^{\infty}$ where $b_{i}=a, i=1,2,3, \ldots,$ and thus consider $\mathbb{Q}$ to be a subset of $\mathbb{R} .$

Exercise $\PageIndex{1}$

Suppose $\left\{a_{i}\right\}_{i \in I}$ and $\left\{b_{i}\right\}_{i \in J}$ are sequences in $\mathbb{Q}$ with

$$\lim _{i \rightarrow \infty} a_{i}=\lim _{i \rightarrow \infty} b_{i}.$$

Show that $a_{i} \sim b_{i}$.

1.4.1 Field Properties

Suppose $\left\{a_{i}\right\}_{i \in I}$ and $\left\{b_{j}\right\}_{j \in J}$ are both Cauchy sequences of rational numbers. Let $K=I \cap J$ and define a new sequence $\left\{s_{k}\right\}_{k \in K}$ by setting $s_{k}=a_{k}+b_{k}$. Given any rational $\epsilon>0,$ choose integers $N$ and $M$ such that

$$\left|a_{i}-a_{j}\right|<\frac{\epsilon}{2}$$

for all $i, j>N$ and

$$\left|b_{i}-b_{j}\right|<\frac{\epsilon}{2}$$

for all $i, j>M .$ If $L$ is the larger of $N$ and $M,$ then, for all $i, j>L$,

$$\left|s_{i}-s_{j}\right|=\left|\left(a_{i}-a_{j}\right)+\left(b_{i}-b_{j}\right)\right| \leq\left|a_{i}-a_{j}\right|+\left|b_{i}-b_{j}\right|<\frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon,$$

showing that $\left\{s_{i}\right\}_{k \in K}$ is also a Cauchy sequence. Moreover, suppose $a_{i} \sim c_{i}$ and $b_{i} \sim d_{i} .$ Given $\epsilon \in \mathbb{Q}^{+},$ choose $N$ so that

$$\left|a_{i}-c_{i}\right|<\frac{\epsilon}{2}$$

for all $i>N$ and choose $M$ so that

$$\left|b_{i}-d_{i}\right|<\frac{\epsilon}{2}$$

for all $i>M .$ If $L$ is the larger of $N$ and $M,$ then, for all $i>L$,

$$\left|\left(a_{i}+b_{i}\right)-\left(c_{i}+d_{i}\right)\right| \leq\left|a_{i}-c_{i}\right|+\left|b_{i}-d_{i}\right|<\frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon.$$

Hence $a_{i}+b_{i} \sim c_{i}+d_{i} .$ Thus if $u, v \in \mathbb{R},$ with $u$ being the equivalence class of $\left\{a_{i}\right\}_{i \in I}$ and $v$ being the equivalence class of $\left\{b_{j}\right\}_{j \in J},$ then we may unambiguously define $u+v$ to be the equivalence class of $\left\{a_{i}+b_{i}\right\}_{i \in K},$ where $K=I \cap J$.

Suppose $\left\{a_{i}\right\}_{i \in I}$ and $\left\{b_{j}\right\}_{j \in J}$ are both Cauchy sequences of rational numbers.

Let $K=I \cap J$ and define a new sequence $\left\{p_{k}\right\}_{k \in K}$ by setting $p_{k}=a_{k} b_{k} .$ Let $B>0$ be an upper bound for the set $\left\{\left|a_{i}\right|: i \in I\right\} \cup\left\{\left|b_{j}\right|: j \in J\right\} .$ Given $\epsilon>0$ choose integers $N$ and $M$ such that

$$\left|a_{i}-a_{j}\right|<\frac{\epsilon}{2 B}$$

for all $i, j>N$ and

$$\left|b_{i}-b_{j}\right|<\frac{\epsilon}{2 B}$$

for all $i, j>M .$ If $L$ is the larger of $N$ and $M,$ then, for all $i, j>L$,

$$\begin{aligned}\left|p_{i}-p_{j}\right| &=\left|a_{i} b_{i}-a_{j} b_{j}\right| \\ &=\left|a_{i} b_{i}-a_{j} b_{i}+a_{j} b_{i}-a_{j} b_{j}\right| \\ &=\left|b_{i}\left(a_{i}-a_{j}\right)+a_{j}\left(b_{i}-b_{j} \right)\right|\\ & \leq\left|b_{i}\left(a_{i}-a_{j}\right)\right|+\left|a_{j}\left(b_{i}-b_{j}\right)\right| \\ &=\left|b_{i}\right|\left|a_{i}-a_{j}\right|+\left|a_{j}\right|\left|b_{i}-b_{j}\right| \\ &<B \frac{\epsilon}{2 B}+B \frac{\epsilon}{2 B} \\ &=\epsilon \end{aligned}$$

Hence $\left\{p_{k}\right\}_{k \in K}$ is a Cauchy sequence.

Now suppose $\left\{c_{i}\right\}_{i \in H}$ and $\left\{d_{i}\right\}_{i \in G}$ are Cauchy sequences with $a_{i} \sim c_{i}$ and $b_{i} \sim d_{i} .$ Let $B>0$ be an upper bound for the set $\left\{\left|b_{j}\right|: j \in J\right\} \cup\left\{\left|c_{i}\right|: i \in H\right\}$. Given $\epsilon>0,$ choose integers $N$ and $M$ such that

$$\left|a_{i}-c_{i}\right|<\frac{\epsilon}{2 B}$$

for all $i>N$ and

$$\left|b_{i}-d_{i}\right|<\frac{\epsilon}{2 B}$$

for all $i>M .$ If $L$ is the larger of $N$ and $M,$ then, for all $i>L$,

$$\begin{aligned}\left|a_{i} b_{i}-c_{i} d_{i}\right| &=\left|a_{i} b_{i}-b_{i} c_{i}+b_{i} c_{i}-c_{i} d_{i}\right| \\ &=\left|b_{i}\left(a_{i}-c_{i}\right)+c_{i}\left(b_{i}-d_{i} \right)\right|\\ & \leq\left|b_{i}\left(a_{i}-c_{i}\right)\right|+\left|c_{i}\left(b_{i}-d_{i}\right)\right| \\ &=\left|b_{i}\right|\left|a_{i}-c_{i}\right|+\left|c_{i}\right|\left|b_{i}-d_{i}\right| \\ &<B \frac{\epsilon}{2 B}+B \frac{\epsilon}{2 B} \\ &=\epsilon. \end{aligned}$$

Hence $a_{i} b_{i} \sim c_{i} d_{i} .$ Thus if $u, v \in \mathbb{R},$ with $u$ being the equivalence class of $\left\{a_{i}\right\}_{i \in I}$ and $v$ being the equivalence class of $\left\{b_{j}\right\}_{j \in J},$ then we may unambiguously define $u v$ to be the equivalence class of $\left\{a_{i} b_{i}\right\}_{i \in K},$ where $K=I \cap J .$

If $u \in \mathbb{R},$ we define $-u=(-1) u .$ Note that if $\left\{a_{i}\right\}_{i \in I}$ is a Cauchy sequence of rational numbers in the equivalence class of $u,$ then $\left\{-a_{i}\right\}_{i \in I}$ is a Cauchy sequence in the equivalence class of $-u .$

We will say that a sequence $\left\{a_{i}\right\}_{i \in I}$ is bounded away from 0 if there exists a rational number $\delta>0$ and an integer $N$ such that $\left|a_{i}\right|>\delta$ for all $i>N .$ It should be clear that any sequence which converges to 0 is not bounded away from $0 .$ Moreover, as a consequence of the next exercise, any Cauchy sequence which does not converge to 0 must be bounded away from $0 .$

Exercise $\PageIndex{2}$

Suppose $\left\{a_{i}\right\}_{i \in I}$ is a Cauchy sequence which is not bounded away from 0. Show that the sequence converges and $\lim _{i \rightarrow \infty} a_{i}=0$.

Exercise $\PageIndex{3}$

Suppose $\left\{a_{i}\right\}_{i \in I}$ is a Cauchy sequence which is bounded away from 0 and $a_{i} \sim b_{i} .$ Show that $\left\{b_{j}\right\}_{j \in J}$ is also bounded away from $0 .$

Now suppose $\left\{a_{i}\right\}_{i \in I}$ is a Cauchy sequence which is bounded away from 0 and choose $\delta>0$ and $N$ so that $\left|a_{i}\right|>\delta$ for all $i>N .$ Define a new sequence $\left\{c_{i}\right\}_{i=N+1}^{\infty}+1$ by setting

$$c_{i}=\frac{1}{a_{i}}, i=N+1, N+2, \ldots$$

Given $\epsilon>0,$ choose $M$ so that

$$\left|a_{i}-a_{j}\right|<\epsilon \delta^{2}$$

for all $i, j>M .$ Let $L$ be the larger of $N$ and $M .$ Then, for all $i, j>L,$ we have

$$\begin{aligned}\left|c_{i}-c_{j}\right| &=\left|\frac{1}{a_{i}}-\frac{1}{a_{j}}\right| \\ &=\left|\frac{a_{j}-a_{i}}{a_{i} a_{j}}\right| \\ &=\frac{\left|a_{j}-a_{i}\right|}{\left|a_{i} a_{j}\right|} \\ &<\frac{\epsilon \delta^{2}}{\delta^{2}} \\ &=\epsilon. \end{aligned}$$

Hence $\left\{c_{i}\right\}_{i=N+1}^{\infty}$ is a Cauchy sequence.

Now suppose $\left\{b_{j}\right\}_{j \in J}$ is a Cauchy sequence with $a_{i} \sim b_{i} .$ By Exercise 1.4.3 we know that $\left\{b_{j}\right\}_{j \in J}$ is also bounded away from $0,$ so choose $\gamma>0$ and $K$ such that $\left|b_{j}\right|>\gamma$ for all $j>K .$ Given $\epsilon>0,$ choose $P$ so that

$$\left|a_{i}-b_{i}\right|<\epsilon \delta \gamma .$$

for all $i>P .$ Let $S$ be the larger of $N, K,$ and $P .$ Then, for all $i, j>S,$ we have

$$\begin{aligned}\left|\frac{1}{a_{i}}-\frac{1}{b_{i}}\right| &=\left|\frac{b_{i}-a_{i}}{a_{i} b_{i}}\right| \\ &=\frac{\left|b_{i}-a_{i}\right|}{\left|a_{i} b_{i}\right|} \\ &<\frac{\epsilon \delta \gamma}{\delta \gamma} \\ &=\epsilon . \end{aligned}$$

Hence $\frac{1}{a_{i}} \sim \frac{1}{b_{i}} .$ Thus if $u \neq 0$ is a real number which is the equivalence class of $\left\{a_{i}\right\}_{i \in I}(\text { necessarily bounded away from } 0),$ then we may define

$$a^{-1}=\frac{1}{a}$$

to be the equivalence class of

$$\left\{\frac{1}{a_{i}}\right\}_{i=N+1}^{\infty},$$

where $N$ has been chosen so that $\left|a_{i}\right|>\delta$ for all $i>N$ and some $\delta>0 .$

It follows immediately from these definitions that $\mathbb{R}$ is a field. That is:

1. $a+b=b+a$ for all $a, b \in \mathbb{R}$;

2. $(a+b)+c=a+(b+c)$ for all $a, b, c \in \mathbb{R}$;

3. $a b=b a$ for all $a, b \in \mathbb{R}$;

4. $(a b) c=a(b c)$ for all $a, b, c \in \mathbb{R}$;

5. $a(b+c)=a b+a c$ for all $a, b, c \in \mathbb{R}$;

6. $a+0=a$ for all $a \in \mathbb{R}$;

7. $a+(-a)=0$ for all $a \in \mathbb{R}$;

8. $1 a=a$ for all $a \in \mathbb{R}$;

9. if $a \in \mathbb{R}, a \neq 0,$ then $a a^{-1}=1$.

1.4.2 Order and Metric Properties

Definition

Given $u \in \mathbb{R},$ we say that $u$ is positive, written $u>0,$ if $u$ is the equivalence class of a Cauchy sequence $\left\{a_{i}\right\}_{i \in I}$ for which there exists a rational number $\epsilon>0$ and an integer $N$ such that $a_{i}>\epsilon$ for every $i>N .$ A real number $u \in \mathbb{R}$ is said to be negative if $-u>0 .$ We let $\mathbb{R}^{+}$ denote the set of all positive real numbers.

Exercise $\PageIndex{4}$

Show that if $u \in \mathbb{R},$ then one and only one of the following is true: $(\mathrm{a}) u>0,(\mathrm{b}) u<0,$ or $(\mathrm{c}) u=0$.

Exercise $\PageIndex{5}$

Show that if $a, b \in \mathbb{R}^{+},$ then $a+b \in \mathbb{R}^{+}$.

Definition

Given real numbers $u$ and $v,$ we say $u$ is greater than $v$, written $u>v,$ or, equivalently, $v$ is less than $u,$ written, $v<u,$ if $u-v>0 .$ We write $u \geq v,$ or, equivalently, $v \leq u,$ to indicate that $u$ is either greater than or equal to $v .$ We say that $u$ is nonnegative if $u \geq 0$.

Exercise $\PageIndex{6}$

Show that $\mathbb{R}$ is an ordered field, that is, verify the following:

a. For any $a, b \in \mathbb{R},$ one and only one of the following must hold: $(i) a<b,$ (ii) $a=b,(\text { iii) } a>b$.

b. If $a, b, c \in \mathbb{R}$ with $a<b$ and $b<c,$ then $a<c$.

c. If $a, b, c \in \mathbb{R}$ with $a<b,$ then $a+c<b+c$.

d. If $a, b \in \mathbb{R}$ with $a>0$ and $b>0,$ then $a b>0$.

Exercise $\PageIndex{7}$

Show that if $a, b \in \mathbb{R}$ with $a>0$ and $b<0,$ then $a b<0$.

Exercise $\PageIndex{8}$

Show that if $a, b, c \in \mathbb{R}$ with $a<b,$ then $a c<b c$ if $c>0$ and $a c>b c$ if $c<0$.

Exercise $\PageIndex{9}$

Show that if $a, b \in \mathbb{R}$ with $a<b,$ then for any real number $\lambda$ with $0<\lambda<1, a<\lambda a+(1-\lambda) b<b$.

Definition

For any $a \in \mathbb{R},$ we call

$$|a|=\left\{\begin{array}{cc}{a,} & {\text { if } a \geq 0,} \\ {-a,} & {\text { if } a<0,}\end{array}\right.$$

the absolute value of $a$.

Exercise $\PageIndex{10}$

Show that for any $a \in \mathbb{R},-|a| \leq a \leq|a|$.

Proposition $\PageIndex{1}$

For any $a, b \in \mathbb{R},|a+b| \leq|a|+|b|$.

Proof

If $a+b \geq 0,$ then

$$|a|+|b|-|a+b|=|a|+|b|-a-b=(|a|-a)+(|b|-b).$$

Both of the terms on the right are nonnegative by Exercise $1.4 .10 .$ Hence the sum is nonnegative and the proposition follows. If $a+b<0,$ then

$$|a|+|b|-|a+b|=|a|+|b|+a+b=(|a|+a)+(|b|+b).$$

Again, both of the terms on the right are nonnegative by Exercise $1.4 .10 .$ Hence the sum is nonnegative and the proposition follows. $\quad Q.E.D.$

It is now easy to show that the absolute value function satisfies

1. $|a-b| \geq 0$ for all $a, b \in \mathbb{R},$ with $|a-b|=0$ if and only if $a=b,$

2. $|a-b|=|b-a|$ for all $a, b \in \mathbb{R}$,

3. $|a-b| \leq|a-c|+|c-b|$ for all $a, b, c \in \mathbb{R} .$

These properties show that the function

$$d(a, b)=|a-b|$$

is a metric, and we will call $|a-b|$ the distance from $a$ to $b$.

Proposition $\PageIndex{2}$

Given $a \in \mathbb{R}^{+},$ there exist $r, s \in \mathbb{Q}$ such that $0<r<a<s$.

Proof

Let $\{u\}_{i \in I}$ be a Cauchy sequence in the equivalence class of $a .$ Since $a>0,$ there exists a rational $\epsilon>0$ and an integer $N$ such that $u_{i}>\epsilon$ for all $i>N .$ Let $r=\frac{\epsilon}{2} .$ Then $u_{i}-r>\frac{\epsilon}{2}$ for every $i>N,$ so $a-r>0,$ that is, $0<r<a .$

Now choose an integer $M$ so that $\left|u_{i}-u_{j}\right|<1$ for all $i, j>M .$ Let $s=u_{M+1}+2 .$ Then

$$s-u_{i}=u_{M+1}+2-u_{i}>1$$

for all $i>M .$ Hence $s>a$. $\quad$ Q.E.D.

Proposition $\PageIndex{3}$

$\mathbb{R}$ is an archimedean ordered field.

Proof

Given real numbers $a$ and $b$ with $0<a<b,$ let $r$ and $s$ be rational numbers for which $0<r<a<b<s .$ since $\mathbb{Q}$ is a an archimedean field, there exists an integer $n$ such that $n r>s .$ Hence

$$n a>n r>s>b.$$

Q.E.D.

Proposition $\PageIndex{4}$

Given $a, b \in \mathbb{R}$ with $a<b,$ there exists $r \in \mathbb{Q}$ such that $a<r<b$.

Proof

Let $\{u\}_{i \in I}$ be a Cauchy sequence in the equivalence class of $a$ and let $\{v\}_{j \in J}$ be in the equivalence class of $b .$ Since $b-a>0,$ there exists a rational $\epsilon>0$ and an integer $N$ such that $v_{i}-u_{i}>\epsilon$ for all $i>N .$ Now choose an integer $M$ so that $\left|u_{i}-u_{j}\right|<\frac{e}{4}$ for all $i, j>M .$ Let $r=u_{M+1}+\frac{\epsilon}{2} .$ Then

$$\begin{aligned} r-u_{i} &=u_{M+1}+\frac{\epsilon}{2}-u_{i} \\ &=\frac{\epsilon}{2}-\left(u_{i}-u_{M+1}\right) \\ &>\frac{\epsilon}{2}-\frac{\epsilon}{4} \\ &=\frac{\epsilon}{4} \end{aligned}$$

for all $i>M$ and

$$\begin{aligned} v_{i}-r &=v_{i}-u_{M+1}-\frac{\epsilon}{2} \\ &=\left(v_{i}-u_{i}\right)-\left(u_{M+1}-u_{i}\right)-\frac{\epsilon}{2} \\ &>\epsilon-\frac{\epsilon}{4}-\frac{\epsilon}{2} \\ &=\frac{\epsilon}{4} \end{aligned}$$

for all $i$ larger than the larger of $N$ and $M .$ Hence $a<r<b . \quad$ Q.E.D.

1.4.3 Upper and Lower Bounds

Definition

Let $A \subset \mathbb{R}$. If $s \in \mathbb{R}$ is such that $s \geq a$ for every $a \in A,$ then we call $s$ an upper bound for $A$. If $s$ is an upper bound for $A$ with the property that $s \leq t$ whenever $t$ is an upper bound for $A,$ then we call $s$ the supremum, or least upper bound, of $A,$ denoted $s=\sup A$. Similarly, if $r \in \mathbb{R}$ is such that $r \leq a$ for every $a \in A,$ then we call $r$ a lower bound for $A .$ If $r$ is a lower bound for $A$ with the property that $r \geq t$ whenever $t$ is a lower bound for $A,$ then we call $r$ the infimum, or greatest lower bound, of $A,$ denoted $r=\inf A .$

Theorem $\PageIndex{5}$

Suppose $A \subset \mathbb{R}, A \neq \emptyset,$ has an upper bound. Then sup $A$ exists.

Proof

Let $a \in A$ and let $b$ be an upper bound for $A .$ Define sequences $\left\{a_{i}\right\}_{i=1}^{\infty}$ and $\left\{b_{i}\right\}_{i=1}^{\infty}$ as follows: Let $a_{1}=a$ and $b_{1}=b .$ For $i>1,$ let

$$c=\frac{a_{i-1}+b_{i-1}}{2}.$$

If $c$ is an upper bound for $A,$ let $a_{i}=a_{i-1}$ and let $b_{i}=c_{i}$ otherwise, let $a_{i}=c$ and $b_{i}=b_{i-1} .$ Then

$$\left|b_{i}-a_{i}\right|=\frac{|b-a|}{2^{i-1}}$$

for $i=1,2,3, \ldots$ Now, for $i=1,2,3, \ldots,$ let $r_{i}$ be a rational number such that $a_{i}<r_{i}<b_{i} .$ Given any $\epsilon>0,$ we may choose $N$ so that

$$2^{N}>\frac{|b-a|}{\epsilon}.$$

Then, whenever $i>N$ and $j>N$,

$$\left|r_{i}-r_{j}\right|<\left|b_{N+1}-a_{N+1}\right|=\frac{|b-a|}{2^{N}}<\epsilon.$$

Hence $\left\{r_{i}\right\}_{i=1}^{\infty}$ is a Cauchy sequence. Let $s \in \mathbb{R}$ be the equivalence class of $\left\{r_{i}\right\}_{i=1}^{\infty} .$ Note that, for $i=1,2,3, \ldots, a_{i} \leq s \leq b_{i}$.

Now if $s$ is not an upper bound for $A,$ then there exists $a \in A$ with $a>s .$ Let $\delta=a-s$ and choose an integer $N$ such that

$$2^{N}>\frac{|b-a|}{\delta}.$$

Then

$$b_{N+1} \leq s+\frac{|b-a|}{2^{N}}<s+\delta=a.$$

But, by construction, $b_{N+1}$ is an upper bound for $A .$ Thus s must be an upper bound for $A .$

Now suppose $t$ is another upper bound for $A$ and $t<s .$ Let $\delta=s-t$ and choose an integer $N$ such that

$$2^{N}>\frac{|b-a|}{\delta}.$$

Then

$$a_{N+1} \geq s-\frac{|b-a|}{2^{N}}>s-\delta=t,$$

which implies that $a_{N+1}$ is an upper bound for $A .$ But, by construction, $a_{N+1}$ is not an upper bound for $A$. Hence $s$ must be the least upper bound for $A$, that is, $s=\sup A .$ $\quad$ Q.E.D.

Exercise $\PageIndex{11}$

Show that if $A \subset \mathbb{R}$ is nonempty and has a lower bound, then inf $A$ exists. (Hint: You may wish to first show that inf $A=-\sup (-A),$ where $-A=\{x:-x \in A\}) .$