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,z∈S and (x,y)∈Γ and (y,z)∈Γ then (x,z)∈Γ,
(b) if x,y∈S 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,z∈S and (x,y)∈Γ and (y,z)∈Γ then (x,z)∈Γ,
(b) if x,y∈S and (x,y)∈Γ and (y,x)∈Γ then x=y,
(c) if x,y∈S then (x,y)∈Γ or (y,x)∈Γ.

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

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

(a) if A,B,C∈S and A⊆B and B⊆C then A⊆C, and
(b) if A,B,∈S and A⊆B and B⊆A 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 X⊈Y and Y⊈X.

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

Let A be a subset of S.

An upper bound of A is an element b∈S such that if a∈A then a≤b.

A lower bound of A is an element ℓ∈S such that if a∈A then ℓ≤a.

A maximum of A is an element M∈A such that there does not exist a∈A such that a≥M (i.e. if A∈A then M≰a).

A minimum of A is an element m∈A such that if a∈A then m≱a.

A supremum of A is an element s∈S such that

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

An infimum of A is an element i∈S such that

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

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

Prove that Card(ℤ>0)≠Card((0,1]ℝ) where (0,1]ℝ={x∈ℝ | 0<x≤1}.

Proof.

Proof by contradiction.

Assume f: ℤ>0 ⟶ (0,1]ℝ k ⟼ rk is a bijection r1 = 0.r11r12r13r14… r2 = 0.r21r22r23r24… r3 = 0.r31r32r33r34… r4 = 0.r41r42r43r44… ⋮ Let s=0.s1s2s3s4s5s6 with s1≠r11, s2≠r22, s3≠r33,… Then s∈(0,1]ℝ 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(ℤ>0)≠Card((0,1]ℝ).

□

Let S be a set and let Γ be a partial order on S. Write x≤yif(x,y) ∈Γ. Let a,b∈S. Then let [a,b] = { x∈S | a≤x≤b } , (a,b] = { x∈S | a <x and x≤b } , [a,b) = { x∈S | a≤x  and x<b } , (a,b) = { x∈S | a<x  and x<b } , [a,∞) = { x∈S | a≤x } , (a,∞) = {x∈S | a<x}. (-∞,a] = {x∈S | x≤a}, (-∞,a) = {x∈S | x<a} where x<y means x≤y and x≠y. 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.

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