# 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 $s<7$.

## Ordered sets

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

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

An ordered set is a set with order.

### Definition 1.7

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

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

#### Example

1. $S=\mathbb{Q}, E=\{1,2,3\}$ ($E$ 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 $\pi\notin S$
2. $S=\mathbb{Q}, E=\{x\in \mathbb{Q}:0<x<1\}$  ($E$ is bounded above)
    * The upper bound is $1$.
3. $S=\mathbb{Q}, E=\{x\in \mathbb{Q}:0<x\}$ ($E$ is not bounded above)
    * Sad
4. $S=\mathbb{Q}, E=\phi$.
    * $\beta\in S$ 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 $x>\beta$.

### Definition 1.8

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 

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

#### Lemma (Uniqueness of upper bounds)

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

<details>
<summary>Proof</summary>

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

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

</details>

We write $\sup E$ to denote the LUB of $E$.

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

Example:

1. $S=\mathbb{Q}, E=\{1,2,3\}$ ($E$ is bounded above)
    * $\sup E=3$, $\inf E=1$
2. $S=\mathbb{Q}, E=\{x\in \mathbb{Q}:0<x<1\}$  ($E$ is bounded above)
    * $\sup E=3$, $\inf E=1$

$\sup E$ and $\inf E=1$ don't have to $\in E$

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

Important example

5. $S=\mathbb{Q}, A=\{p\leq \mathbb{Q}:p>0, p^2<2\}$.
    * $A$ 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, $\sup A=\sqrt{2}$.

#### Least upper bound property (LUBP)

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

#### Greatest upper bound property (GLBP)

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.

#### Theorem 1.11

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)