Real numbers/Ordering axioms/Archimedes/Intervals/Introduction/2/Section

Definition

A field ${}K$ is called an ordered field if there is a relation ${}>$ (larger than) between the elements of ${}K$ fulfilling the following properties (${}a\geq b$ means ${}a>b$ or ${}a=b$)

1. For two elements ${}a,b\in K$ we have either ${}a>b$ or ${}a=b$ or ${}b>a$.
2. From ${}a\geq b$ and ${}b\geq c$ one may deduce ${}a\geq c$ (for any ${}a,b,c\in K$).
3. ${}a\geq b$ implies ${}a+c\geq b+c$ (for any ${}a,b,c\in K$).
4. From ${}a\geq 0$ and ${}b\geq 0$ one may deduce ${}ab\geq 0$ (for any ${}a,b\in K$).

Lemma

In an ordered field the following properties hold.

1. ${}1\geq 0$.
2. ${}a\geq 0$ holds if and only if ${}-a\leq 0$ holds.
3. ${}a\geq b$ holds if and only if ${}a-b\geq 0$ holds.
4. ${}a\geq b$ holds if and only if ${}-a\leq -b$ holds.
5. ${}a\geq b$ and ${}c\geq d$ imply ${}a+c\geq b+d$.
6. ${}a\geq b$ and ${}c\geq 0$ imply ${}ac\geq bc$.
7. ${}a\geq b$ and ${}c\leq 0$ imply ${}ac\leq bc$.
8. ${}a\geq b\geq 0$ and ${}c\geq d\geq 0$ imply ${}ac\geq bd$.
9. ${}a\geq 0$ and ${}b\leq 0$ imply ${}ab\leq 0$.
10. ${}a\leq 0$ and ${}b\leq 0$ imply ${}ab\geq 0$.

Proof

See Exercise.

$\Box$

Lemma

In an ordered field the following properties holds.

1. From ${}x>0$ one can deduce ${}x^{-1}>0$.
2. From ${}x<0$ one can deduce ${}x^{-1}<0$.
3. For ${}x>0$ we have ${}x\geq 1$ if and only if ${}x^{-1}\leq 1$.
4. From ${}x\geq y>0$ one can deduce ${}x^{-1}\leq y^{-1}$.
5. For positive elements ${}x,y$ the relation ${}x\geq y$ is equivalent with ${}{\frac {x}{y}}\geq 1$.

Proof

See Exercise.

$\Box$

Definition

Let ${}K$ be an ordered field. ${}K$ is called Archimedean, if the following Archimedean axiom holds, i.e. if for every ${}x\in K$ there exists a natural number ${}n$ such that

${}n\geq x\,.$

Lemma

1. For ${}x,y\in \mathbb {R}$ with ${}x>0$ there exists ${}n\in \mathbb {N}$ such that ${}nx\geq y$.
2. For ${}x>0$ there exists a natural number ${}n$ such that ${}{\frac {1}{n}}<x$.
3. For two real numbers ${}x<y$ there exists a rational number ${}n/k$ (with ${}n\in \mathbb {Z}$, ${}k\in \mathbb {N} _{+}$) such that

   ${}x<{\frac {n}{k}}<y\,.$

Proof

(1). We consider ${}y/x$. Because of the Archimedean axiom there exists some natural number ${}n$ with ${}n\geq y/x$. Since ${}x$ is positive, due to fact also ${}nx\geq y$ holds. For (2) and (3) see exercise.

$\Box$

Definition

For real numbers ${}a,b$, ${}a\leq b$, we call

1. ${}[a,b]={\left\{x\in \mathbb {R} \mid x\geq a{\text{ and }}x\leq b\right\}}$ the closed interval.
2. ${}]a,b[={\left\{x\in \mathbb {R} \mid x>a{\text{ and }}x<b\right\}}$ the open interval.
3. ${}]a,b]={\left\{x\in \mathbb {R} \mid x>a{\text{ and }}x\leq b\right\}}$ the half-open interval (closed on the right).
4. ${}[a,b[={\left\{x\in \mathbb {R} \mid x\geq a{\text{ and }}x<b\right\}}$ the half-open interval (closed on the left).