NoteNextra-origin/content/Math4111/Math4111_L3.md

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.

Proof

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

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.