1 - Ordered Sets and Fields

Field

A set $F$ , where $(F, +)$ is an abelian group with identity $0$ and $(F - 0, \cdot)$ is an abelian group with identity $1$ , with a distributive law $x \cdot (y + z) = x \cdot y + x \cdot z$ for each $x, y, z \in F$ is called a field.

We can write $n x$ for $x + x \dots_{n t i m e s} + x$ , $x^{n}$ for $x \cdot x \dots_{n t i m e s} \cdot x$ , $\frac{1}{x}$ for $x^{- 1}$ .
And, it follows that $- 1 \cdot x = - x$ . (where $- 1$ is in additive inverse of the multiplicative identity). And we also get all the usual associative, commutative and distributive properties we are familiar with.

Ordered Set

A set $S$ with a relation $<$ is an ordered set if:

(Trichotomy) $\forall x, y \in S$ exactly one of $x = y, x < y, y < x$ is true.
(transitivity) $\forall x, y, z \in S$ if $x < y$ and $y < z$ , then $x < z$ .

Supremum, infimum

if $S$ is an ordered set and $E \subset S$ , $E$ is said to be bounded above if there exists $α \in S$ such that for all $e \in E, α \geq e$ .
if $γ$ is an upper bound of $E$ such that for all upper bounds $α$ of $E$ , $γ \leq α$ , then $γ$ is the supremum of $E$ (least upper bound) and we write $γ = sup E$ .
Similarly $F \subset S$ is bounded below if there exists $β \in S$ such that for all $f \in F, β \leq f$ .
If $δ$ is a lower bound of $F$ such that for all lower bounds $β$ of $F$ , $β \leq δ$ then $δ$ is the infimum of $F$ (greatest lower bound) and we write $δ = inf F$ .

Least upper bound property

An ordered set $S$ has the least upper bound property (or supremum property) if for every $E \subset S$ that is bounded above, $sup E \in S$ .
Similarly, $S$ has the greatest lower bound property if every subset that is bonded below has an infimum in $S$ .

The next theorem will show that Least upper bound property implies greatest lower bound property (and vice versa if you wish).

Supremum property implies infimum property

Let $S$ be an ordered set with the Supremum property, and $H \subset S$ be bounded below. Define $L (H) = {l : l is a lower bound of H}$ . Then, $inf H$ exists in $S$ and is equal to $sup L (H)$ .

$* p r o o f * :$ Firstly any $h \in H$ is an upper bound for $L (H)$ as for any $l \in L (H)$ , $l \leq h$ (definition of lower bound). Hence $sup L (H) = l^{*} \in S$ .

Suppose $l^{*}$ is not a lower bound of $H$ , then there exists some $v \in H$ for which $v < l^{*}$ . But, this means that $v$ is an upper bound for $L (H)$ , smaller than the least upper bound $l^{*}$ , resulting in a contradiction.

Therefore, $l^{*}$ is a lower bound of $H$ . moreover, since $l^{*} = sup L (H)$ for any other lower bound $l$ of $H$ , $l \leq l^{*}$ , hence $l^{*} = inf H$ .

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An ordered field is a field, whose set is ordered, and the order behaves "nice" with addition and multiplication.

Ordered Field

let $F, +, \cdot$ be a field an $<$ an order on $F$ , then $F$ is an ordered field if:

$y < z$ implies $x + y < x + z$ for any $x, y, z \in F$ - (OF1)
$x > 0, y > 0$ implies $x y > 0$ . - (OF2)

Properties of ordered fields

let $F$ be an ordered field. Then:

$x > 0 ⟺ - x < 0$
$x > 0$ and $y < z$ implies $x y < x z$
$x < 0$ and $y < z$ implies $x y > x z$
$x \neq 0$ implies $x^{2} > 0$
$0 < x < y$ implies $0 < 1 / y < 1 / x$

$* p r o o f * :$

Use (OF1) on $x > 0$ to get $x - x > - x$ and hence $0 > - x$ .
Use (OF2) on $x > 0$ and $z - y > 0$ to get (distributive) $x z - x y > 0$ and hence $x z > x y$ .
Similar to 2
if $x > 0$ , use (OF2) to conclude $x^{2} > 0$ . if $x < 0$ , it follows that $- x > 0$ and using (OF2) with field axioms we have $(- 1)^{2} x^{2} > 0$ hence $x^{2} > 0$ .
Since $x, y > 0$ both $x^{- 1}, y^{- 1} > 0$ use (OF2) twice.

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