
# 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 $$-A=\{x:-x \in A\}) .$$