Chapter 1: The Real Numbers - Tarleton State University · 2021. 2. 1. · Chapter 1: The Real...

Chapter 1: TheReal Numbers

PWhite

Discussion

SomePreliminaries

The Axiom ofCompleteness

Consequences ofCompleteness

Cantor’s Theorem

Epilogue

Chapter 1: The Real Numbers

Peter W. [email protected]

Initial development byKeith E. Emmert

Department of MathematicsTarleton State University

Spring 2021 / Advanced Analysis


PWhite

Discussion

SomePreliminaries



Cantor’s Theorem

Epilogue

Overview

Discussion: The Irrationality of√

2

Some Preliminaries

The Axiom of Completeness

Consequences of Completeness

Cantor’s Theorem

Epilogue


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Discussion

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Cantor’s Theorem

Epilogue

The Big Pythagorean OopsieI The Pythagoreans worshiped integers and fractions

of integers (rational numbers).I They believed that everything could be connected to

an integer or rational number.I Around 500 BC, the Pythagoreans discovered the

irrationality of√

2.I The wheels came off the bus.I Real Analysis began and is often driven by the study

of things that caused the “wheels to come off thebus.”


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Basic Definitions

Definition 1I The natural numbers are defined by

N = {1,2,3, . . .}.I The integers are defined by

Z = {. . . ,−3,−2,−1,0,1,2,3, . . .}.I The rational numbers are defined by

Q =

{pq| p,q ∈ Z, q 6= 0

}.

Theorem 2There is no rational number whose square is 2.Proof:


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Questions and ObservationsI Thus

√2 6∈ Q. Where does

√2 “belong?”

I We define irrational numbers as those that are notrational.

I Fact: n√

p with p prime is irrational.I Note that n

√p is a root of f (x) = xn − p.

I Are there other irrational numbers that are notalgebraic roots of polynomials with rationalcoefficients?

I The real numbers could be defined as Q with it’s“holes” filled in.

I More about these things later.


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Cantor’s Theorem

Epilogue

Overview


2

Some Preliminaries



Cantor’s Theorem

Epilogue


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Cantor’s Theorem

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Sets

Definition 3I A set is a collection of any objects, called elements.I If an element x is a member of a set A we write

x ∈ A. If x is not an element of A we write x 6∈ A.I The empty set, ∅, is the set that contains no

elements.I The union of two sets A and B is defined by

A ∪ B = {x | x ∈ A or x ∈ B (or both)}.

I The intersection of two sets A and B is defined by

A ∩ B = {x | x ∈ A and x ∈ B}.


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Sets

Definition 4I We say A is a subset of B, A ⊆ B, if

x ∈ A =⇒ x ∈ B.

I Two sets A and B are equal, A = B, if A ⊆ B andB ⊆ A.

I A is a proper subset of B if A ⊆ B and there existsy ∈ B such that y 6∈ A.

I The complement of a set A ⊆ R, Ac , is defined by

Ac = {x ∈ R | x 6∈ A}.

Remark 5The definition of complement is dependent upon yourchoice of “universe” (for us, R).


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Tiny Examples

Example 6Let E represent the even numbers and O the oddnumbers. Let S = {r ∈ Q | |r2| < 2}.I Z = E ∪O.I E ∩O = ∅.I E = Z ∩ E .I Z ∩ S = {−1,0,1}.I Let A be a set. ∅ ⊆ A.I What is ∅c?I What is Rc?I What is Qc?I What is Zc?


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Tiny Examples

Definition 7Let A1,A2,A3, . . . ,An, . . . be an infinite collection of sets.I A countable union of sets is

∞⋃n=1

An =⋃n∈N

An = A1 ∪ A2 ∪ · · · ∪ An ∪ · · · .

I A countable intersection of sets is∞⋂

n=1

An =⋂n∈N

An = A1 ∩ A2 ∩ · · · ∩ An ∩ · · · .

I A (descending) nested chain of sets satisfies thecondition

A1 ⊇ A2 ⊇ · · · ⊇ An ⊇ · · · .


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Example

Example 8For each n ∈ N, define An = {n,n + 1,n + 2, . . .}. Verifythe following statements.I A1 ⊇ A2 ⊇ · · · ⊇ An ⊇ · · · .I⋂n∈N

An = ∅.

I⋃n∈N

An = A1.


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De Morgan’s Laws

Theorem 9Let A and B be any sets. Then the following statementsare true:

1. (A ∪ B)c = Ac ∩ Bc .2. (A ∩ B)c = Ac ∪ Bc .

The proof is an exercise.


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Functions

Definition 10Let A and B be two sets. A function, f , from A to B is arule or mapping that takes each element x ∈ A andassociates with it a single element of B. Our notation is

f : A 7→ B.

For x ∈ A, the expression f (x) is used to represent theelement of B associated with x by f .

The set A is called the domain of f and the set B is calledthe codomain. The range or image of f is the set{y ∈ B | y = f (x) for some x ∈ A}.

Remark 11If f : A 7→ B, then the range of f is a subset of thecodomain, B, and need not equal B.


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Function Examples

Example 12I f : R 7→ R be defined by f (x) = x2. Then the range of

f is [0,∞) ( R.I Dirichlet’s Function: g : R 7→ R and is defined by

g(x) =

{1, x ∈ Q0, x 6∈ Q.

The range if g is the set {0,1}.

Remark 13When looking for counterexamples of statements aboutfunctions, try to remember Dirichlet’s Function.


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Preimages of Functions

Definition 14Let f be a function satisfying f : D 7→ R and chooseB ⊆ R. Define the preimage of f , f−1(B) to be the set

{x ∈ D | f (x) ∈ B}.

Example 15Let the function f : [0,4] 7→ R be given by f (x) = x2. Findthe following preimages.I f−1([0,1])I f−1([−1,1])I f−1(R)


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Some Preimage Theory

Theorem 16Let D ⊆ R, suppose f : D 7→ R is a function and A,B ⊆ Rbe two sets. Prove the following statementsI f−1(A ∩ B) = f−1(A) ∩ f−1(B).

I f−1(A ∪ B) = f−1(A) ∪ f−1(B).

Proof: (Pick One)


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Absolute Value Function

Definition 17The absolute value of a number x ∈ R is given by thefunction

|x | =

{x , x ≥ 0−x , x < 0.

Remark 18For all a,b, c ∈ R,

1. |ab| = |a||b|2. |a + b| ≤ |a|+ |b|. (The Triangle Inequality)3. Common use of the Triangle Inequality:

|a− b| = |(a− c) + (c − b)| ≤ |a− c|+ |c − b|.


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Logic and Proofs

Some Types of Proofs:I Direct Proof – Begin with a true statement (often the

hypothesis of your theorem) and use logicaldeductions to arrive at the conclusion of thestatement.

I Proof by Contradiction – Assume the hypothesis andthe negation of the conclusion. The goal is to derivea contradiction of some known statement.

I Contrapositive Proof – Assume the negation of theconclusion and use logical deductions to arrive at thenegation of the hypothesis.


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Logic and Proofs

So, if you’re trying to prove A =⇒ B, then your typicaloptions are:I Direct Proof: A =⇒ B. (Start with A and derive B).I Proof by Contradiction: A ∩ ¬B. (Start with A AND¬B and break mathematics.)

I Contrapositive Proof: ¬B =⇒ ¬A. (Start with ¬B andderive ¬A.)

Remark 19Be careful when you negate a statements containing“none,” “at least one,” “for all,” and “there exists.”I The negation of “for all” is “there exists.”I The negation of “there exists” is “for all.”I The negation of “at least one” is “none.”I The negation of “none” is “at least one.”

Remember to use De Morgan’s Laws when negatingstatements about sets.


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Proofs by Contradiction

Theorem 20If m is irrational, then

√1 + m is irrational.

Proof:


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Example

Example 21Two real numbers a and b are equal if and only if forevery real number ε > 0, it follows that |a− b| < ε.Proof:


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Induction

Fundamental Principle of InductionLet S ⊆ N. If

1. 1 ∈ S2. n ∈ S =⇒ n + 1 ∈ S

Then S = N.

The Typical UsageGiven a statement P(n), where n ≥ n0 and n0 ∈ N ∪ {0}.

1. Base Case: P(n0) is true.2. Induction Hypothesis: Assume P(k) is true for

some k ≥ n0.3. Inductive Step: Show that P(k) =⇒ P(k + 1).

Remark 22When using induction, I want to see all three steps clearlylabeled.


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Induction Example

Example 23Prove that for every n ≥ 4, n! > 2n.


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Example

Example 24Prove that for any n ≥ 2, the number√

1 +

√1 +√

1 + · · ·

is always irrational. (There are n radicals and n 1’s.)


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Homework

Pages 11 – 13Problems: 1.2.3, 1.2.5, 1.2.6, 1.2.8, 1.2.10, 1.2.11, 1.2.12


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Overview


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Cantor’s Theorem

Epilogue


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Cantor’s Theorem

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What is R?I The real numbers contains the rational numbers.I It “fills in” all the gaps of the rational numbers.I It is a field.I It is ordered – that is, if a 6= b, then either a < b or

a > b.I The real numbers satisfies the Axiom of

Completeness.

Axiom of CompletenessEvery nonempty set of real numbers that is boundedabove has a least upper bound.

Remark 25The Axiom of Completeness is an axiom. It is afundamental building block. Thus we assume that it istrue and can’t actually prove that it is true.


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Least Upper and Greatest Lower Bounds

Definition 26I A set A ⊆ R is bounded above if there exists a

number b ∈ R such that a ≤ b for all a ∈ R. Thenumber b is called an upper bound for A.

I A set A ⊆ R is bounded below if there exists anumber l ∈ R such that l ≤ a for all a ∈ R. Thenumber l is called an lower bound for A.

I The supremum or least upper bound for a setA ⊆ R, s = sup(A), satisfies the following

1. s is an upper bound for A.2. if b is any upper bound for A, then s ≤ b.

I The infimum or greatest lower bound for a setA ⊆ R, g = inf(A), satisfies the following

1. g is a lower bound for A.2. if b is any lower bound for A, then b ≤ g.


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Maximum and Minimum

Definition 27I The real number a0 is a maximum of the set A if a0

is an element of A and a0 ≥ a for all a ∈ A.I The number a1 is a minimum of the set A if a1 is an

element of A and a1 ≤ b for all b ∈ A.

Remark 28I Supremum and infimum (when they exist) are

unique. (Easy proof).I A set can have a supremum and yet have no

maximum.I A set can have an infimum and yet have no minimum.


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Supremum and Infimum

Example 29What are the (intuitive) infimum and supremum, as wellas maximum and minimum for the following sets?

1. (0,1]2. {1,2,3, . . .}3. Q4.{1

n | n ∈ N}

Example 30Let S = {r ∈ Q | r2 < 2}. Answer the following questionsabout S.

1. Explain why sup(S) exists.2. Is sup(S) ∈ Q?3. Does inf(S) exist? Is it in Q?4. Is there a maximum? Minimum?


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Tiny Theory

Lemma 31Assume s ∈ R is an upper bound for a set A ⊆ R. Then,s = sup(A) if and only if, for every choice of ε > 0, thereexists an element a ∈ A satisfying s − ε < a.Proof:


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It all holds for Infimums

Remark 32I The Axiom of Completeness does not explicitly say

anything about infimums and their existence.I However, infimums exist for any set that is bounded

below.I So, any result about supremums has an analogous

result for infimums.


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Homework

Pages: 18 – 20Problems: 1.3.3, 1.3.4, 1.3.5, 1.3.8, 1.3.9, 1.3.11


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Discussion

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Cantor’s Theorem

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Overview


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Cantor’s Theorem

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Application of the Axiom of Choice

Theorem 33 (Nested Interval Property)For each n ∈ N, assume we are given a closed intervalIn = [an,bn]. Assume also that

I1 ⊇ I2 ⊇ I3 ⊇ · · · ⊇ In ⊇ · · · .

Then,∞⋂

n=1

In 6= ∅.

Proof:I


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The Archimedean Property

Theorem 34 (The Archimedean Property)I Given any number x ∈ R, there exists an n ∈ N

satisfying n > x.I Given any real number y > 0, there exists an n ∈ N

satisfying1n< y .


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Density of Q in R

Theorem 35 (Density of Q in R)For every two real numbers a and b with a < b, thereexists a rational number r satisfying a < r < b.Proof:


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Density of the Irrational Numbers in R

Corollary 36Given any two real numbers a < b, there exists anirrational number t satisfying a < t < b.

Remark 37I We have now shown that for any two rational

numbers, there exists an irrational number betweenthem. (The previous corollary).

I We have also shown that for any two irrationalnumbers, there exists a rational number betweenthem. (The previous theorem).

I These two statements will be disturbing when we talkabout the “size” of the rational and irrationalnumbers.


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The Existence of Square Roots

Theorem 38There exists a real number α ∈ R satisfying α2 = 2.Proof:


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Cardinality of Sets

Definition 39The cardinality of a set A, |A| = card(A), is the size of aset.

Definition 40Let f : A 7→ B be a function.I f is one-to-one, 1-1, if a1 6= a2 in A implies that

f (a1) 6= f (a2) in B.I f is onto if, given any b ∈ B, there exits a ∈ A such

that f (a) = b.I The sets A and B have the same cardinality,

A ∼ B, if f is 1-1 and onto.


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Example

Example 41I Let E be the set of even numbers. Show that E ∼ N

using f (n) = 2n.

I Show that N ∼ Z using f (n) =

{n−1

2 , n odd−n

2 , n even.

I Show that f : (−1,1) 7→ R given by f (x) =x

x2 − 1is

1-1 and onto. Hence, (−1,1) ∼ R.


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Equivalence Relations

Definition 42A relation, R, is an equivalence relation if

1. ARA.2. ARB =⇒ BRA.3. ARB and BRC =⇒ ARC.

Example 43Show that the countable relation ∼ is an equivalencerelation between sets.


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Countable Sets

Definition 44I A set A is countable if N ∼ A.I An infinite set that is not countable is called an

uncountable set.

Remark 45The sets E = even numbers, O = odd numbers, N (duh!)and Z are countable.

Theorem 461. The set Q is countable.2. The set R is uncountable.

Proof:


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Remark

Theorem 47If A ⊆ B and B is countable, then A is either countable,finite, or empty.

Theorem 481. If A1,A2,A3, . . . ,Am are each countable sets, then

the union⋃m

n=1 An is countable.2. If An is countable for all n ∈ N, then

⋃∞n=1 An is

countable.

Remark 49I A previous theorem states that Q is countable and R

is uncountable. Hence, the irrationals must beuncountable and have a larger cardinality than Q.

I But, for any two irrationals, there is a rationalbetween them. The infinity concept is unintuitive.


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Homework

Page: 24Problems: 1.4.1, 1.4.3, 1.4.8Pages: 30 – 31Problems: 1.5.3, 1.5.4, 1.5.6, 1.5.9


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Overview


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Cantor’s Diagonalization Method

Theorem 50The open interval (0,1) = {x ∈ R | 0 < x < 1} isuncountable.Proof: Note: The proof uses Cantor’s DiagonalizationMethod.


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Power Sets

Definition 51Given a set A, the power set of A, P(A), is the set of allsubsets of A.

Example 52Let A = {a,b, c}.

1. Find P(A).2. If A has n elements, then |P(A)| = 2n.3. Notice that it is trivial to construct a 1− 1 function

f : A 7→ P(A). However, it is impossible to find anonto function.


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Cantor on Power Sets

Theorem 53 (Cantor’s Theorem)Given any set A, there does not exist a functionf : A 7→ P(A) that is onto.Proof:


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Homework

Pages: 32 – 35Problems: 1.6.1, 1.6.4, 1.6.10


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Something OddI Let ℵ0 = |N|. This is the first cardinal number (the

smallest infinity).I Let c = |R|. This is the second cardinal number.I We’ve shown that ℵ0 < c.I The Question: Does there exist A ⊂ R such thatℵ0 < |A| < c?

I Cantor’s Continuum Hypothesis: There is no A ⊂ Rsuch that ℵ0 < |A| < c.

I Kurt Gödel: Using the agreed upon axioms of settheory, the continuum hypothesis could not bedisproven.

I Paul Cohen: Using the same axioms, the continuumhypothesis could not be proven.

I So, we are free to accept or not the continuumhypothesis without developing any logicalcontradictions.

Chapter 1: The Real Numbers - Tarleton State University · 2021. 2. 1. · Chapter 1: The Real...

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