[Lecture 1] R is Complete

We have $\mathbb{N}$, $\mathbb{Z}$ and $\mathbb{Q}$. $\mathbb{Q}$ is a field. The field axiom is as follows:

Field Axiom

We say that $\mathbb{F}$ is a field if there are two operations $+$ and $\cdot$ such that $\mathbb{F},+,\cdot$ satisfies

$a + b, ab \in \mathbb{F}$
$a + (b + c) = (a + b) + c$
$a + b = b + a, ab = ba$
$a(b+c) = ab + ac$
...

Another axiom that we need is the order axiom as follows:

Order Axiom

An order relation $<$ for the field $\mathbb{F}$ has the following properties

(Trichotomy) For $a,b \in \mathbb{F}$, $a < b$ or $b < a$ or $a = b$ holds
(Transitive). $a < b$ and $b < c$ implies that $a < c$.
$a < b$ and $c \in \mathbb{F}$ implies $a + b < b + c$.
If $c > 0$ and $a < b$, then $ac < bc$.
If $c < 0$ and $a < b$, then $ac > bc$.

Based on this we get a couple of other theorems that we will need later:

Theorem

If $a \neq 0$, then $a^2 > 0$.
$-1 < 0 < 1$
If $0 < a < 1$, then $a^2 < a$. If $a > 1$, then $a^2 > a$.

Proof

(a) If $a \neq 0$, then by the order axiom (1), either $a < 0$ or $a > 0$.
Case 1: Suppose that $a > 0$, then by the order axiom (4)

$$ \begin{align*} a &> 0 \\ a \cdot a &> 0 \cdot a \\ a^2 &> 0 \end{align*} $$

Case 2: Suppose that $a < 0$, then by the order axiom (5)

$$ \begin{align*} a &< 0 \\ a \cdot a &> 0 \cdot a \\ a^2 &> 0 \end{align*} $$

(b) By (a), we know that $1 \neq 0$. Then

$$ \begin{align*} 1^2 = 1 > 0 \end{align*} $$

Next, to show that $-1 < 0$, observe that

$$ \begin{align*} 1 &> 0 \\ -1 \cdot 1 &< -1 \cdot 0 \quad \text{(By the order axiom (5))} \\ -1 &< 0 \\ \end{align*} $$

Finally, since $-1 < 0$ and $0 < 1$, then $-1 < 0 < 1$.

$$ \begin{align*} a &< 1 \\ a \cdot a &< a \\ a^2 &< a \end{align*} $$

If $a > 1$, then by the order axiom (4)

$$ \begin{align*} a &> 1 \\ a \cdot a &> a \\ a^2 &> a \end{align*} $$

We have one more theorem as follows

Theorem

There is no integer $n$ such that $0 < n < 1$.

Proof

Suppose that $n > 0$ is a positive integer, then we can construct $n$ as the sum of $n$ 1’s added together as follows

$$ \begin{align*} n = 1 + 1 + \ldots + 1 \quad \text{($n$ 1s added together)} \end{align*} $$

But we know from the previous theorem that if $a \neq 0$, then $a^2 > 0$ and from this we saw that $1 > 0$. By the order axiom (3)

$$ \begin{align*} 1 + 1 &> 1 \\ 1 + 1 + 1 &> 1 + 1 \\ &\cdots \\ 1 + n &> n > n - 1 > \ldots > 1 > 0. \end{align*} $$

Therefore, $n \geq 1$. So $n$ is at least $1$ if it’s a positive integer.
If $n < 0$, then we can use the order axiom (5) to get

$$ \begin{align*} n &< 0 \\ -n &> 0 \end{align*} $$

Then, the first case applies,

We know that $\mathbb{Q}$ is an ordered field. Things were good until we discovered that $\sqrt{2}$ can’t be rational. So this implies that $\mathbb{Q}$ has gaps …

Completeness Axiom

If a set $E$ has an upper bound and $E$ is non-empty, then the least upper bound denoted by $s = \sup E$ is the supremum of $E$. It means that for all $\epsilon > 0$, we can find $x \in E$ such that $$ \begin{align*} s - \epsilon < x \leq s \end{align*} $$

So now the Order Axiom together with the Field Axiom and the Completeness Axiom, we can finally get $\mathbb{R}$.

Theorem (Dedekind)

There is a unique field that satisfies the order axiom and the completeness axiom and contains $\mathbb{Q}$. This field is our real number system $\mathbb{R}$.

References

Lecture Notes by Professor Chun Kit Lai