Lecture 2

Ordered sets, least upper bounds and fields.

Warm up

(a) The statements says: $\forall a\in A, \exists s\in a$ such that $s\geq 7$ .

The negation is $\exist a\in A,\forall s\in a$ , such that $s<7$ .

Let $S$ be a set. An order on $S$ is a relation satisfying:

"trichotomy". If $x,y\in S$ , then exactly on eof the these statements are hold: $x<y,x=y,x>y$ .
"transitivity". If $x,y,z\in S$ , then $x<y,y\implies x<z$ .

An ordered set is a set with order.

Let $S$ be an ordered set and let $E\subset S$ ( $E$ is the "universe", all the element you can ever think of...)

$\beta\in S$ is an upper bound of $E$ if $\forall x\in E,x\leq \beta$
$E$ is bounded above if $\exist \beta \in S$ such that $\beta$ is the upper bound of $E$ .

$S=\mathbb{Q}, E=\{1,2,3\}$ $S = Q, E = {1, 2, 3}$ ( $E$ $E$ is bounded above)
- 3,4,3.5 are all upper bounds of $E$ .
- 2,2.5 is not upper bounds of $E$ .
- $\pi$ is not upper bound of $S$ since $\pi\notin S$
$S=\mathbb{Q}, E=\{x\in \mathbb{Q}:0<x<1\}$ $S = Q, E = {x \in Q : 0 < x < 1}$ ( $E$ $E$ is bounded above)
- The upper bound is $1$ .
$S=\mathbb{Q}, E=\{x\in \mathbb{Q}:0<x\}$ $S = Q, E = {x \in Q : 0 < x}$ ( $E$ $E$ is not bounded above)
- Sad
$S=\mathbb{Q}, E=\phi$ $S = Q, E = ϕ$ .
- $\beta\in S$ is an upper bound of $E$ if $\forall x\in E,x\leq \beta\equiv\beta\in S$ is not an upper bound of $E$ if $\exists x\in E,x> \beta$
- So this statement is true for any rational numbers since $\cancel{\exist} a\in E$ such that $x>\beta$ .

Least upper bound, LUB, supremum, SUP

Let $S$ be an ordered set and $E\subset S$ . We say $\alpha\in S$ is the LUB of $E$ if

$\alpha$ is the UB of $E$ . ( $\forall x\in E,x\leq \alpha$ )
if $\gamma<\alpha$ , then $\gamma$ is not UB of $E$ . ( $\forall \gamma <\alpha, \exist x\in E$ such that $x>\gamma$ )

Uniqueness of upper bounds.

If $\alpha$ and $\beta$ are LUBs of $E$ , then $\alpha=\beta$ .

Proof:

Suppose for contradiction $\alpha$ and $\beta$ are both LUB of $E$ , then $\alpha\neq\beta$

WLOG $\alpha>\beta$ and $\beta>\alpha$ .

EOP

We write $SupE$ to denote the LUB of $E$ .

This also applies to $GLB$ (greatest lower bound) and infinum of $E$

$S=\mathbb{Q}, E=\{1,2,3\}$ $S = Q, E = {1, 2, 3}$ ( $E$ $E$ is bounded above)
- $SupE=3$ , $Inf E=1$
$S=\mathbb{Q}, E=\{x\in \mathbb{Q}:0<x<1\}$ $S = Q, E = {x \in Q : 0 < x < 1}$ ( $E$ $E$ is bounded above)
- $SupE=3$ , $Inf E=1$

$SupE$ and $Inf E=1$ don't have to $\in E$

$S=\mathbb{Q}, E=\{x\in \mathbb{Q}:0<x\}$ $S = Q, E = {x \in Q : 0 < x}$ ( $E$ $E$ is not bounded above)
- $SupE=\infty$ or not defined, $Inf E=0$
$S=\mathbb{Q}, E=\phi$ $S = Q, E = ϕ$ .
- $SupE=-\infty$ or not defined, $Inf E=\infty$ or not defined, we don't put $\infty$ in $\mathbb{Q}$

Important example

$S=\mathbb{Q}, A=\{p\leq \mathbb{Q}:p>0, p^2<2\}$ $S = Q, A = {p \leq Q : p > 0, p^{2} < 2}$ .
- $A$ is not empty and bounded above. However, $Sup A$ des not exists.

If $S=\mathbb{R}, A=\{p\leq \mathbb{Q}:p>0, p^2<2\}$ . * $A$ is not empty and bounded above. However, $Sup A=\sqrt{2}$ .

if $\forall E\subset S$ that tis non-empty and bounded above, $\exist Sup E\in S$ .

S has greatest lower bound property (GLBP) if $\exist E\subset S$ that is non-empty and bounded below, $\exists Inf E\in S$

$\mathbb{Q}$ does not have LUBP and GLBP.

Let $S$ be an ordered set. Then $S$ has the LUBP $\iff$ $S$ has the GLBP

Proof:

Let $S$ be a set with LUBP. (we want to show $S$ has GLBP)