NoteNextra-origin/pages/Math4111/Math4111_L2.md

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.

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

Example

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

SupE 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)
   • SupE=\infty or not defined, Inf E=0
4. S=\mathbb{Q}, E=\phi.
   • SupE=-\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)