# Lecture 3

## Review

Let $S=\mathbb{Z}$.

1. Let $E=\{x\in S:x>0,x^2<5\}$. What are $sup\ E$ and $\inf\ E$?
  
    $sup\ E=2,inf\ E=1$

2. Can you find a subset $E\subset S$ which is bounded above but not bounded below?

    $E=\{x\in S:x<0\}$

3. Does $S$ have the least upper bound property?

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

4. Does $S$ have the greatest lower bound property?

    Yes, $\forall E\subset S$ that tis non-empty and bounded below, $\exist \inf E\in S$.

## Continue

### LUBP (The least upper bound property)

Proof that $LUBP\implies GLBP$.

<details>
<summary>Proof</summary>

Let $S$ be an ordered set with LUBP. Let $B<S$ be non-empty and bounded below.

Let $L=y\in S:y$ is a lower bound of $B$. From the picture, we expect $\sup L=\inf B$ First we'll show $\sup L$ exists.

1. To show $L\neq \phi$.

    $B$ is bounded below $\implies L\neq\phi$
2. To show $L$ id bounded above.

    $B$ is not empty $\implies \exists x\in B\implies x$ is a upper bound of $L$.

3. Since $S$ has the least upper bound property, $\sup L$ exists (in $S$).

Let's say $\alpha=sup\ L$. We claim that $\alpha=inf\ B$. We need to show $2$ things.

1. To show $\alpha$ is a lower bound of $B$, $\forall \gamma\in B,\alpha\leq \gamma$.

    Let $\gamma\in B$, then $\gamma$ is an upper bound of $L$.

    Since $\alpha$ is the least upper bound of $L$, $\alpha\leq \gamma$.

2. To show $\alpha$ is the greatest lower bound of $B$, $\forall \beta>\alpha,\beta$ is not a lower bound of $B$.

    Let $\beta>\alpha$. Since $\alpha$ is an upper bound of $L$, $\beta\notin L$.

    By definition of $L$, $\beta$ is not a lower bound of $B$.

Thus $\alpha=inf\ B$

</details>

### Field

|                | addition                                                    | multiplication                                                 |
| -------------- | ----------------------------------------------------------- | -------------------------------------------------------------- |
| closure        | $\checkmark$                                                | $\checkmark$                                                   |
| commutativity  | $\checkmark$                                                | $\checkmark$                                                   |
| associativity  | $\checkmark$                                                | $\checkmark$                                                   |
| identity       | $\checkmark$ (denoted $0$)                                  | $\checkmark$ (denoted $1$)                                     |
| inverses       | $\checkmark$ (denoted $-x$)                                 | $\checkmark$ (exists when $x\neq 0$ denoted $1/x$ or $x^{-1}$) |
| distributivity | $\checkmark$ (distributive of multiplication over addition) ||

Examples: $\mathbb{Q},\mathbb{R},\mathbb{C}$

Non-examples: $\mathbb{N}$ fails A4,A5,M5, $\mathbb{Z}$ fails M5

Another example of field: $\mathbb{Z}/5\mathbb{Z}=\{1,2,3,4,5\}$, $\forall a,b\in \mathbb{Z}/5\mathbb{Z}$, $a+b=(a+b)\mod 5$, $a\cdot b=(a\cdot b)\mod 5$

Some properties of fields: see Proposition 1.14,1.15,1.16

Remark:

1. It's more helpful if you try to prove these yourselves. The proofs are "straightforward".
2. For this course, it's not important to remember which properties are axioms, etc.

Example of proof:

#### 1.14(a) $x+y=x+z\implies y=z$

Proof:

$x+y=x+z$,

$(-x)+(x+y)=(-x)+(x+z)$,

by A3, $(-x+x)+(y)=(-x+x)+(z)$,

$0+y=0+z$,

$y=z$.

Chain of equalities.

#### 1.16(a) $\forall x\in \mathbb{F}, 0x=0$

1. A4, where 0 is defined.
2. Since $0$ is defined in the addition, identity. The proposition says something about multiplication by 0. The only proposition that relates the addition and multiplication is Distributive law.

$0x=(0+0)x=0x+0x$, cancel $0x$ on both side we have $0x=0$.

### Ordered Field (1.17)

An _ordered field_ is a _field_ $F$ which is also an _ordered set_, such that

1. $x+y<x+z$ if $x,y,\in F$ and $y<z$,
2. $xy>0$ if $x\in F,y\in F,x>0$ and $y>0$.

#### Prop 1.18

If $x>0$ and $y<z$, then $xy<yz$.

Proof: $y<z\implies 0<z-y$, $x(z-y)>0\implies xz>xy$

We define $\mathbb{R}$ to be the unique ordered field with $LUBP$. (The existence and uniqueness are discussed in the appendix of this chapter).

#### Theorem 1.20

1. (Archimedean property) If $x,y\in \mathbb{R}$ and $x>0$, then $\exists n\in \mathbb{N}$ such that $nx>y$.
2. ($\mathbb{Q}$ is dense in $\mathbb{R}$) If $x,y\in \mathbb{R}$ and $x<y$, then $\exists p\in \mathbb{Q}$ such that $x<p<y$.
