Real Analysis

Arun Ram
Department of Mathematics and Statistics
University of Melbourne
Parkville, VIC 3010 Australia
aram@unimelb.edu.au

Last update: 19 July 2014

Lecture 27

Let S be a set.

A relation on S is a subset Γ of S×S.

S = {α,β,γ}, S×S = { (α,α), (α,β), (α,γ) (β,α), (β,β), (β,γ) (γ,α), (γ,β), (γ,γ) } , Γ = { (β,β), (β,γ), (γ,α) } .

A partial order on S is a relation Γ on S such that

(a) if x,y,zS and (x,y)Γ and (y,z)Γ then (x,z)Γ,
(b) if x,yS and (x,y)Γ and (y,x)Γ then x=y.

A total order on S is a relation Γ on S such that

(a) if x,y,zS and (x,y)Γ and (y,z)Γ then (x,z)Γ,
(b) if x,yS and (x,y)Γ and (y,x)Γ then x=y,
(c) if x,yS then (x,y)Γ or (y,x)Γ.

If Γ is a partial order on S write xy if (x,y) in Γ.

(a partial order that is not a total order.) Let E={α,β,γ} and let S={subsets ofE}= { , {α}, {β}, {γ} {α,β}, {α,γ}, {β,γ} {α,β,γ} } . Let Γ be the relation on S given by Γ={(A,B)|AB}. In other words, inclusion is a partial order on S since

(a) if A,B,CS and AB and BC then AC, and
(b) if A,B,S and AB and BA then A=B.
{α,β,γ} {α,β} {α,γ} {β,γ} {α} {β} {γ} Inclusion is not a total order since X={α} and Y={β} are in S and XY and YX.

Let S be a set with a partial order Γ. Write xy if (x,y)Γ.

Let A be a subset of S.

An upper bound of A is an element bS such that if aA then ab.

A lower bound of A is an element S such that if aA then a.

A maximum of A is an element MA such that there does not exist aA such that aM (i.e. if AA then Ma).

A minimum of A is an element mA such that if aA then ma.

A supremum of A is an element sS such that

(a) s is an upper bound of A, and
(b) if b is an upper bound of A then bs.

An infimum of A is an element iS such that

(a) i is a lower bound of A, and
(b) if is a lower bound of A then i.

If A={{α,β},{β,γ},{β}} then {α,β} and {β,γ} are both maximums of A and supA={α,β,γ}.

Prove that Card(>0)Card((0,1]) where (0,1]={x|0<x1}.

Proof.

Proof by contradiction.

Assume f: >0 (0,1] k rk is a bijection r1 = 0.r11r12r13r14 r2 = 0.r21r22r23r24 r3 = 0.r31r32r33r34 r4 = 0.r41r42r43r44 Let s=0.s1s2s3s4s5s6 with s1r11, s2r22, s3r33, Then s(0,1] and does not appear in the sequence (r1,r2,r3,). So f is not surjective. This is a contradiction to f being bijective. So Card(>0)Card((0,1]).

Let S be a set and let Γ be a partial order on S. Write xyif(x,y) Γ. Let a,bS. Then let [a,b] = { xS|axb } , (a,b] = { xS|a <xandxb } , [a,b) = { xS|ax andx<b } , (a,b) = { xS|a<x andx<b } , [a,) = { xS|ax } , (a,) = {xS|a<x}. (-,a] = {xS|xa}, (-,a) = {xS|x<a} where x<y means xy and xy. These sets are intervals in S.

Notes and References

These are notes from a 2010 course on Real Analysis 620-295. This page comes from 100512Lect27.pdf and was given on 12 May 2010.

page history