Section 1.2 { Preliminariesmath341.cardon.byu.edu/Math-341-Lecture-Notes/341...Chapter 1 { The Real...

§1.2: Preliminaries

Math 341 – Lecture Notes on Chapter 1

What is Math 341 about?

In this class we will study the major theorems of calculus from a theoretical point of view. Amajor goal of the class is to improve your ability to read and write mathematical proofs.

Section 1.2 – Preliminaries

N = {1, 2, 3, . . .} the set of natural numbers

Z = {. . . ,−3,−2,−1, 0, 1, 2, 3, . . .} the set of integers

Q =

{a

b: a, b,∈ Z and b 6= 0

}the set of rational numbers

N ( Z ( Q

Remark

To solve interesting problems we often need a system of numbers larger than Q.

As an illustration, we’ll prove that√2 /∈ Q.

Chapter 1 – The Real Numbers 1 / 1


Theorem 1.1.1

There is no rational number whose square is 2.

Proof.

Suppose, by way of contradiction, that there exists a rational number

r =a

bwhere a, b ∈ Z and b 6= 0

such thatr2 = 2.

Furthermore, by cancelling common factors, we may assume that ab

is a reduced fraction. Then(ab

)2= 2

a2 = 2b2.

This implies that a is even (since the square of an odd number is odd and since 2b2 is even).Write

a = 2c.

(Continued on next page.)



Proof (continued).

We already know that c ∈ Z. Then

a2 = 2b2

(2c)2 = 2b2

4c2 = 2b2

2c2 = b2.

By the same reason that a was even, we conclude that b is even.

Therefore, both a and b are divisible by 2 contradicting the assumption that ab

was a reducedfraction. This is a contradiction.

Thus, there is no r ∈ Q with r2 = 2.



Note

Q has a natural ordering. Given r, s ∈ Q, exactly one of the following holds:

r < s, r = s, r > s.

Ways to think about the real numbers

1 Limits of sequences of rational numbers.

2 The decimal representation.

3 The base-2 (binary) representation.

4 The real numbers are the one-dimensional number line.

5 The complex numbers C = {x+ iy : x, y ∈ R and i2 = −1} are equivalent to thetwo-dimensional Cartesian plane where x+ iy ↔ (x, y).

N ( Z ( Q ( R ( C

Decimals

If you fully understand the decimal representation of the real numbers (as convergent infiniteseries), then you are 90% of the way to fully understanding R.



We’ll review a few definitions and notation:

Some Definitions †1 A set is a collection of objects.

2 If the object x belongs to the set S, we write

x ∈ S.

If x does not belong to S, we writex 6∈ S.

3 The union of sets A and B is

A ∪B := {x : x ∈ A or x ∈ B}.

4 the intersection of sets A and B is

A ∩B := {x : x ∈ A and x ∈ B}.

5 The set A is a subset of the set B if every element of A is also an element of B. We writeA ⊆ B.

6 The sets A and B are equal if A ⊆ B and B ⊆ A. We write A = B.



Example

A = {1, 2, 3, 4} and B = {2, 3, 4, 5}x ∈ A ∩B ⇐⇒ x ∈ {2, 3, 4}x ∈ A ∪B ⇐⇒ x ∈ {1, 2, 3, 4, 5}

Example

For each positive integer n, let

An = {n, n+ 1, n+ 2, . . .}.

These sets form a nested sequence:

A1 ) A2 ) A3 ) A4 ) · · ·

What is ∪∞n=1An? Answer: A1.

What is ∩∞n=1An? Answer: ∅.



Complements of Sets

If U is a set containing all of the sets in a particular discussion, we call U the universal set. IfA ⊆ U , then the complement of A in U is defined to be

Ac := {x ∈ U : x /∈ A}.

De Morgan’s Laws

For sets A ⊆ U and B ⊆ U , we have

1 (A ∩B)c = Ac ∪Bc, and

2 (A ∪B)c = Ac ∩Bc.

Proof.

For Part (1),

x ∈ (A ∩B)c ⇔ x 6∈ A ∩B⇔ (x 6∈ A) or (x 6∈ B)

⇔ (x ∈ Ac) or (x ∈ Bc)

⇔ x ∈ Ac ∪Bc.

Hence, (A ∩B)c = Ac ∪Bc. The proof of the other equality is similar.



Functions

Not all functions are defined by ‘nice’ algebraic formulas. For example, consider the functionf : R→ R defined by

f(x) =

{x if x ∈ Q,

0 if x ∈ R\Q.

Definition of Absolute Value

The absolute value function is

|x| :={x if x ≥ 0,

−x if x < 0.

Note that we could also define |x| by the formula

|x| :=√x2.



Three Properties of the Absolute Value

For any a, b ∈ R, we have

1 |ab| = |a| |b|2 |a+ b| ≤ |a|+ |b| (Triangle Inequality)

3

∣∣|a| − |b|∣∣ ≤ |a+ b| (Variant of Triangle Inequality)

Proof.

The (relatively easy) proofs are in the homework.

The Triangle Inequality is the MOST IMPORTANT Inequality in Mathematics

The triangle inequality ∣∣|a| − |b|∣∣ ≤ |a+ b| ≤ |a|+ |b|.



This course frequently requires reasoning involving an arbitrary ε > 0.

Theorem 1.2.6

Two real numbers a and b are equal if and only if for every real ε > 0 it follows that |a− b| < ε.

Proof.

(⇒) Assume a = b. Then |a− b| = 0 < ε for every real ε > 0.

(⇐) Assume that, for every real ε > 0, if follows that |a− b| < ε. We’ll show that a = b.

If, to the contrary, a 6= b, then |a− b| > 0. Then the choice ε = |a− b| gives

0 < |a− b| < ε = |a− b|,

which is false. Hence, a = b.



Example †Let y1 = 6, and for each n ∈ N define

yn+1 =2yn − 6

3.

1 Use induction to prove that yn > −6 for all n ∈ N.

2 Use another induction argument to show that the sequence (y1, y2, y3, . . .) is decreasing.

y1 = 6 y4 − 22/9 = −2.44444 y7 = −1202/243 = −4.9465y2 = 2 y5 = −98/27 = −3.62963 y8 = −3862/729 = −5.29767y3 = −2/3 = −0.666667 y6 = −358/81 = −4.41975 y9 = −12098/2187 = −5.53178

Proof of (1).

Base Case: y1 = 6 > −6. Statement (1) is true if n = 1.

Induction Step: Suppose yk > −6 for some k ∈ N. Then, by the induction hypothesis,

yk+1 =2yk − 6

3>

2(−6)− 6

3= −6.

So, by induction yn > −6 for all n ∈ N.



Proof of (2).

Base Case: y2 = 2y1−63

= 2·6−63

= 2 < y1. So, we have y1 > y2.

Induction Step: Assume for some k ∈ N that yk > yk+1. We’ll show that yk+1 > yk+2.

yk > yk+1

2yk − 6

3>

2yk+1 − 6

3

yk+1 > yk+2.

By induction, Statement (2) is true for all n ∈ N.

Question

In the example we proved thaty1 > y2 > y3 > · · · > −6.

We wonder if the values yn “approach” some specific real number as n→∞.

In a later section, the Monotone Convergence Theorem will answer this question.



End of Section 1.2Next: Section 1.3 – Axiom of Completeness


§1.3: Axiom of Completeness

Section 1.3 – Axiom of Completeness• For now we’ll assume we understand R fairly well.

• Last time we saw that√2 /∈ Q. So,

√2 is somehow “filling a gap” in the rational numbers.

• Today we’ll study the Axiom of Completeness which makes more precise the intuitive ideathat “R fills in the gaps in Q”.

Definitions

1 A set A ⊆ R is bounded above if there exists a number b ∈ R such that a ≤ b for all a ∈ A.The number b is call an upper bound for A.

2 A set A ⊆ R is bounded below if there exists a number c ∈ R such that c ≤ a for all a ∈ A.We call c a lower bound for A.

3 A real number s is the least upper bound for A or the supremum of A and we writes = sup(A) if the following two conditions hold:

(i) s is an upper bound for A.(ii) If b is any upper bound for A, then s ≤ b.

4 A real number t is the greatest lower bound or infimum of A, denoted t = inf(A) if thefollowing two conditions hold:

(i) t is a lower bound for A.(ii) If c is any lower bound for A, then c ≤ t.



Question

Can a set A have two different suprema?

Answer: NO. If s1 and s2 are both least upper bounds on A, then s1 ≤ s2 and s2 ≤ s1 and sos1 = s2.

Now we state an axiom about the real numbers.

Axiom of Completeness

Every nonempty set of real number that is bounded above has a least upper bound.

Lemma

The Axiom of Completeness implies that every nonempty set of real numbers that is boundedbelow has a greatest lower bound.

Example

A = { 1n: n ∈ N} = {1, 1

2, 13, 14, . . .}.

sup(A) = 1 ∈ Ainf(A) = 0 6∈ A



Example

B = [0, 2) = {x ∈ R : 0 ≤ x < 2}

inf(B) = 0 ∈ Bsup(B) = 2 6∈ B.

Example

N = {1, 2, 3, . . .}.

inf(N) = 1 ∈ Nsup(N) does not exist since N is not bounded above.

Example

S = {r ∈ Q : r2 ≤ 2}

inf(S) = −√2 6∈ S (since −

√2 is not rational)

sup(S) =√2 6∈ S (since

√2 is not rational)



Example

Let A ⊆ R be nonempty and bounded above. Let c ∈ R and define

c+A := {c+ a : a ∈ A}.

Prove that sup(c+A) = c+ sup(A).



Proof.

Set s = sup(A). Recall that s = sup(A) is determined by the following two conditions:

(a) s is an upper bound for A.

(b) If b is any upper bound for A, then s ≤ b.Then a ≤ s for all a ∈ A. Hence c+ a ≤ c+ s for all a ∈ A and so c+ s is an upper bound onthe set c+A. So,

sup(c+A) ≤ c+ s = c+ sup(A).

Next, let b be any upper bound on c+A. Then for all a ∈ A,

c+ a ≤ b

and soa ≤ b− c.

Hence, b− c is an upper bound for A. Because s is the least upper bound on A,

s ≤ b− c

and soc+ sup(A) = c+ s ≤ c+ (b− c) = b.

Hence, sup(c+A) = c+ sup(A).



Lemma 1.3.8

Assume s ∈ R is an upper bound for a nonempty set A ⊆ R. Then s = sup(A) if and only if, forevery ε > 0, there exists an element a ∈ A satisfying s− ε < a ≤ s.

xs− ε s

a ∈ A

Proof.

(⇒) Assume s = sup(A). Let ε > 0.

Since s− ε < s, the number s− ε is not an upper bound on A. Thus there exists some a ∈ Asuch that s− ε < a ≤ s, as desired.

(Proof of ⇐ on next page.)



Lemma 1.3.8

Assume s ∈ R is an upper bound for a nonempty set A ⊆ R. Then s = sup(A) if and only if, forevery ε > 0, there exists an element a ∈ A satisfying s− ε < a ≤ s.

Proof (continued).

(⇐) Conversely, assume s is an upper bound for A such that for every ε > 0 there exists somea ∈ A with s− ε < a ≤ s. We’ll show s = sup(A).

If, to the contrary, s 6= sup(A), then sup(A) < s. Consider the choice

ε =s− sup(A)

2.

Then

s− ε = s−(s− sup(A)

2

)=s

2+

sup(A)

2

>sup(A)

2+

sup(A)

2

= sup(A)

Since s− ε > sup(A), there is no a ∈ A satisfying s− ε < a ≤ s, contrary to the hypothesis.This is a contradiction. So, s = sup(A).



End of Section 1.3Next: Section 1.4 – Consequences of Completeness


§1.4: Consequences of Completeness

Section 1.4 – Consequences of Completeness

We’ll next study several consequences of the Axiom of Completeness, including:

• Nested Interval Property

• Existence of square roots

• Density of Q in R

Theorem 1.4.1 (Nested Interval Property)

For each n ∈ N, let In = [an, bn] = {x ∈ R : an ≤ x ≤ bn} be a closed interval. Assume theintervals form a nested decreasing sequence. That is,

I1 ⊇ I2 ⊇ I3 ⊇ I4 ⊇ · · · .

Then∞⋂

n=1

In 6= ∅.



Proof.

Because the intervals are nested, we have:

a1 ≤ a2 ≤ a3 ≤ · · · ≤ an ≤ · · · ≤ bn ≤ · · · ≤ b3 ≤ b2 ≤ b1,

as depicted in the figure below.

x[ [ [ [ ] ] ] ]a1 a2 a3 an bn b3 b2 b1︸︷︷︸

A={an : n∈N}

· · · · · · · · ·

Let A := {an : n ∈ N} be the set of all left endpoints of the intervals.

Because of the nesting, every bn is an upper bound on A. So, by the Axiom of Completeness,

x = sup(A)

exists in R.

For each n ∈ N, we havean ≤ x ≤ bn.

Hence,

x ∈∞⋂

n=1

In ⇒∞⋂

n=1

In 6= ∅.



Note

In the Nested Interval Theorem, either

∞⋂n=1

In = {x} (Set containing single point x)

or∞⋂

n=1

In = [a, b] (Set consisting of closed interval where a < b.)

In the last formula, a = sup{a1, a2, a3, . . .} and b = inf{b1, b2, b3, . . .}.



Theorem (Archimedean Property)

(i) Given any x ∈ R, there exists an n ∈ N satisfying n > x.

(ii) Given any real y > 0, there exists an n ∈ N satisfying 0 < 1n< y.

Proof †.

(i) Assume, by way of contradiction, that N is bounded above. By the Axiom of Completeness,there exists a real number α = sup(N). Then α− 1 is not an upper bound on N and so thereexist some n ∈ N such that α− 1 < n. But then α < n+ 1. Since n+ 1 ∈ N, the number αis not an upper bound for N, a contradiction.

(ii) Let y > 0. By part (i), there exists some n ∈ N such that

0 < x =1

y< n.

Then

0 <1

n< y.



Theorem (Density of Q in R)

For every two real numbers a and b with a < b, there exists a rational number r such thata < r < b.

Proof †.

We will find a rational number r = mn

where m ∈ Z and n ∈ N such that

a <m

n< b. (1)

Inequality (??) is equivalent tona < m < nb. (2)

By the Archimedean Property, there exists n ∈ N such that

0 <1

n< b− a. (3)



Proof (continued).

Having chosen n, we’ll find m. Choose m ∈ Z such that

m− 1(4)

≤ na(5)< m.

(The integer m exists by the Archimedean Property.) Inequality (5) implies a < mn

.Also we have

m ≤ na+ 1 by (4)

< n(b− 1n) + 1 since by (3), a < b− 1

n

= nb

which implies mn< b. Thus

a < r =m

n< b.



Existence of Square Roots

Let r ∈ R and r > 0. There exists α ∈ R satisfying α2 = r.

Proof.

Consider the setT = {t ∈ R : t2 < r}.

and setα = sup(T ).

There are three cases:

(1) α2 < r

(2) α2 > r

(3) α2 = r.

By showing that cases (1) and (2) are impossible, we find that the only remaining possibility isα2 = r.

First, α is positive, as follows: By the Archimedean Principle, there exists some k ∈ N such that0 < 1

k< r. Then

0 <1

k< r ⇒ 0 <

1

k2≤

1

k< r ⇒

1

k∈ T ⇒ 0 <

1

k≤ sup(T ) = α.



Proof (continued).

Case (1): Suppose, by way of contradiction, that α2 < r.

For a natural number n ≥ 2, write(α+

1

n

)2

= α2 +2α

n+

1

n2

< α2 +2α

n+

1

n(This merely makes the algebra easier.)

= α2 +2α+ 1

n.

Now, by the Archimedean Principle, there exists a natural number n0 ≥ 2 large enough such that

2α+ 1

n0< r − α2.︸︷︷︸

(> 0 by hypothesis)

Then (α+

1

n0

)2

< α2 +2α+ 1

n0< α2 + (r − α2) = r.

Thus, α+ 1n0∈ T which contradicts the fact that α = sup(T ). So, the case α2 < r is

impossible.



Proof (continued).

Case (2): Suppose, by way of contradiction, that α2 > r.

For a natural number n, write (α−

1

n

)2

= α2 −2α

n+

1

n2

≥ α2 −2α

n.

Now, by the Archimedean Principle there exists a natural number n1 such that

2α

n1< α2 − r.︸︷︷︸

(> 0 by hypothesis)

Then (α−

1

n1

)2

> α2 −2α

n1> α2 + (r − α2) = r.

But α− 1n16∈ T and is an upper bound on T . Since α− 1

n1< α this contradicts the fact that

α = sup(T ). So, the case α2 > r is impossible.

The only remaining possibility is that α2 = r. This completes the proof.



Examples

The Nested Interval Property may hold or may not hold if the intervals are open:

1

∞⋂n=1

(0, 1

n

)= ∅. (Empty so failure of NIP)

2

∞⋂n=1

(− 1

n, 1n

)= {0}.

3

∞⋂n=1

(−1n, 1 + 1

n

)= [0, 1].

Example

The Nested Interval Property can be false if the intervals are closed but not finite:Set Ln = [n,∞). Then

L1 ⊇ L2 ⊇ L3 ⊇ · · ·

but∞⋂

n=1

Ln = ∅.



Theorem

If a and b are any two real numbers with a < b, there exists an irrational number t such that

a < t < b.

Proof †.

By an earlier theorem about the density of Q in R there exists a rational number r such that

a−√2 < r < b−

√2.

Thena < t = r +

√2 < b,

where t is irrational.



End of Section 1.4Next: Section 1.5 – Cardinality


§1.5: Cardinality

Section 1.5 – Cardinality

Recall that last time we studied

• Nested Interval Property: Let I1, I2, I3, . . . be a sequences of closed intervals of finitelength in R such that

I1 ⊇ I2 ⊇ I3 ⊇ · · · .

Then∞⋂

n=1

In 6= ∅.

• Density of Q in R.

Today we’ll discuss the size of sets.

Question: How “large” are sets? Especially Q and R.


§1.5: Cardinality

Cardinality of a Set

Let A be a finite set of the form A = {a1, a2, . . . , an} where a1, . . . , an are distinct. Thecardinality of A, denoted |A| = n, is the number of elements of A.

Examples

∅ = {} A = {a, b, c}|∅| = 0 |A| = 3.

Definitions

Let A and B be sets and let f : A→ B be a function.

1 f is injective (or one-to-one) if a1, a2 ∈ A and a1 6= a2 imply that f(a1) 6= f(a2).

2 f is surjective (or onto) if for every b ∈ B there exists some a ∈ A such that f(a) = b.

3 f is bijective (or a one-to-one correspondence) if f is both injective and surjective.


§1.5: Cardinality

Examples

1 f : N→ Q where f(x) = x is injective but not surjective.

2 g : R→ [0,∞) where g(x) = x2 is surjective but not injective.

3 h : R→ R where h(x) = sin(x) is neither injective nor surjective.

4 k : R→ R where k(x) = −x3 is bijective.

Cardinality

Sets A and B have the same cardinality, written |A| = |B| if there exists a bijective functionf : A→ B.


§1.5: Cardinality

Theorem

|N| = |Z|

Proof.

Consider the function f : N→ Z determined by

f(1) = 0

f(2) = 1

f(3) = −1f(4) = 2

f(5) = −2f(6) = 3

f(7) = −3...

Then f is a bijection from N to Z.


§1.5: Cardinality

Theorem

|N| = |Q|

Proof.

Write every fraction of the form mn

where n ∈ N and m ∈ Z. List these fractions in a semi-infinitearray where fractions of the from m

nare on the nth row from the bottom. Cross out the fractions

that are not in reduced form. (In the diagram below the non-reduced fractions are colored red.)Then wind through the diagram (skipping non-reduced fractions) as indicated in the figure.

−4/5 // −3/5 // −2/5 // −1/5 // 0/5 // 1/5 // 2/5 // 3/5 // 4/5

��−4/4

OO

−3/4

��

−2/4oo −1/4oo 0/4oo 1/4oo 2/4oo 3/4oo 4/4

��−4/3

OO

−3/3

��

−2/3 // −1/3 // 0/3 // 1/3 // 2/3

��

3/3

OO

4/3

��−4/2

OO

−3/2

��

−2/2

OO

−1/2

��

0/2oo 1/2oo 2/2

��

3/2

OO

4/2

��· · · −4/1

OO

−3/1oo −2/1

OO

−1/1oo 0/1 // 1/1

OO

2/1 // 3/1

OO

4/1 // · · ·

This defines a bijection f : N→ Q. Hence, |N| = |Q|.


§1.5: Cardinality

Proof.

The first few values of the bijection f : N→ Q are:

f(1) = 0/1

f(2) = 1/1

f(3) = 1/2

f(4) = −1/2f(5) = −1/1f(6) = −2/1f(7) = −2/3f(8) = −1/3f(9) = 1/3

f(10) = 2/3

f(11) = 2/1

...


§1.5: Cardinality

Example

|(−1, 1)| = |R|

Proof.

Show that the function

f(x) =

x

1− xif 0 ≤ x < 1

x

1 + xif −1 < x < 0,

is strictly increasing (hence injective) and is also surjective.

(The textbook uses f(x) = x1−x2 to show that |(−1, 1)| = |R|. I used an easier function.)

A sketch of this function is given below:

-1.0 -0.5 0.5 1.0

-10

-5

5

10

(Continued on next page.)Chapter 1 – The Real Numbers 40 / 1

§1.5: Cardinality

Proof†.

Proof that f is injective: We will show that if −1 < x1 < x2 < 1, then f(x1) < f(x2) which willprove that f is injective. There are three cases.

(i) 0 ≤ x1 < x2 < 1.

(ii) −1 < x1 < x2 ≤ 0.

(iii) −1 < x1 < 0 < x2 < 1.

Case (i): Suppose0 ≤ x1 < x2 < 1.

Then

1 ≥ 1− x1 > 1− x2 > 0

1 ≤1

1− x1<

1

1− x20 ≤

x1

1− x1<

x2

1− x2.

So,0 ≤ x1 < x2 < 1 ⇒ 0 ≤ f(x1) < f(x2). (1)



§1.5: Cardinality

Proof (continued).

Case (ii): By a very similar argument,

− 1 < x1 < x2 ≤ 0 ⇒ f(x1) < f(x2) ≤ 0. (2)

Case (iii): Combining inequalities (??) and (??) gives

− 1 < x1 < 0 < x2 < 1 ⇒ f(x1) < 0 < f(x2) (3)

From (??), (??), and (??), we have shown in all cases that

−1 < x1 < x2 < 1 ⇒ f(x1) < f(x2).

Thus f is both strictly increasing and injective.



§1.5: Cardinality

Proof (continued).

Proof that f is surjective: We already know that

−1 < x < 0 ⇒ f(x) < 0,

0 < x < 1 ⇒ f(x) > 0,

x = 0 ⇒ f(x) = 0.

Let y ∈ R and y > 0, we wish to solve the equation

y =x

1− x= f(x)

for x in terms of y, if possible.

y =x

1− x⇔ (1− x)y = x

⇔ y = (y + 1)x

⇔ x =y

y + 1.

So, if y > 0, then 0 < x = yy+1 < 1 is a preimage of y.

By a very similar argument, if y < 0, then −1 < x = y1−y < 0 is a preimage of y.

Also, f(0) = 0.

We have shown that for any y ∈ R, there exists x ∈ (−1, 1) such that f(x) = y. Thus f is surjective.

Since f is both injective and surjective, it is bijective. Therefore, |R| = |(−1, 1)|, as desired.Chapter 1 – The Real Numbers 43 / 1

§1.5: Cardinality

Countable and Uncountable

1 If A is an infinite set and |A| = |N|, then A is countable.

2 If A is an infinite set that is not countable, then A is uncountable.

Theorem

R is uncountable.

Proof.

Since N ⊂ R, it is clear that R is an infinite set. Assume, by way of contradiction, that R iscountable. Then we may list the elements of R as

R = {x1, x2, x3, x4, . . .}.

Let I1 = [a1, b1] be a closed interval, with a1 < b1, not containing x1.

Let I2 = [a2, b2] be a closed interval, with a2 < b2, contained in I1 not containing x2.

In general, given a closed interval In = [an, bn], with an < bn, not containing x1, x2, . . . , xn,choose a closed interval In+1 = [an+1, bn+1], with an+1 < bn+1 such that

• In+1 ⊂ In• xn+1 6∈ In+1

(Continued on next page)


§1.5: Cardinality

Proof (continued).

This construction is illustrated below:

xxn

[ [ ] ]xn+1

In+1︷︸︸︷︸︷︷︸

In

By our construction, for any particular xn0 ∈ R, xn0 6∈ In0 . So we have nested closed intervals offinite length such that

∞⋂n=1

In = ∅,

which contradicts the Nested Interval Theorem. Therefore, R is not countable and so R isuncountable.


§1.5: Cardinality

Schroder-Bernstein Theorem

Let X and Y be sets. If there exist injective functions

f : X → Y, and

g : Y → X,

then |X| = |Y |. That is, there exists a bijection from X to Y .

Proof.

The proof is not very long, but it takes a little bit of extra effort to understand. You can find aproof in your Math 290 textbook.


§1.5: Cardinality

Example

Show that |(0, 2]| = |(−1, 2)|.

Proof.

Let f : (0, 2]→ (−1, 2) be defined by

f(x) = x− 1.

Then f((0, 2]

)= (−1, 1] ⊆ (−1, 2), and f is an injection since f is a linear function with nonzero

slope.

Let g : (−1, 2)→ (0, 2] be defined by

g(x) =x+ 1

2.

Then g((−1, 2)

)= (0, 3/2) ⊆ (0, 2], and g is an injection since g is a linear function with nonzero

slope.

By the Schroder-Berstein Theorem, |(0, 2]| = |(−1, 2)|.


§1.5: Cardinality

End of Section 1.5Next: Section 1.6 – Cantor’s Theorem


§1.6: Cantor’s Theorem

Section 1.6 – Cantor’s Theorem

Today we’ll continue to study cardinalities. Recall from last time:

• |A| denotes the cardinality of the set A.

• |A| = |B| means there exists a bijection f : A→ B.

• |A| ≤ |B| means there exists an injection f : A→ B.

• |A| < |B| means there exists an injection f : A→ B but there does not exist a surjectionfrom A to B.

• If |A| = |N| we call A countable.

• |N| = |Z| = |Q|• R is uncountable using a proof based on Nested Interval Theorem

• |R| = |(−1, 1)| by showing that the function g : (−1, 1)→ R

g(x) =

{x

1−xif 0 ≤ x < 1

x1+x

if −1 < x < 0,

is a bijection.

• (From Math 290) Schroder-Bernstein Theorem: If |A| ≤ |B| and |B| ≤ |A|, then |A| = |B|.That is, if there exist injections f : A→ B and g : B → A, then there exists a bijectionh : A→ B.

Today, we’ll apply Cantor’s Diagonalization Method to prove that R is uncountable. (This is oursecond proof that R is uncountable.) We’ll also show that |A| < |P(A)|.



Theorem

The open interval (0, 1) = {x ∈ R : 0 < x < 1} is uncountable.

Proof.

(The proof illustrates Cantor’s diagonalization method.)

First, the function f : N→ (0, 1) give by f(n) = 1n+1

is injective. So, |N| ≤ |(0, 1)|

Now, let f : N→ (0, 1) be any function. We’ll show that f is not a surjection, which will showthat |N| < |(0, 1)|. For each n ∈ N, we’ll write f(n) in decimal format as

f(n) = 0.an1an2an3an4 · · ·

where ank is the kth digit after the decimal point of the number f(n). When we write ourdecimal numbers we will avoid any numbers that terminate in infinitely many 9’s. For example,we would write 0.500000 . . . instead of 0.499999 . . .

f(1) = 0. a11 a12a13a14 . . .

f(2) = 0.a21 a22 a23a24 . . .

f(3) = 0.a31a32 a33 a34 . . .

f(4) = 0.a41a42a43 a44 . . .

...



Proof.

Now, define a real number x ∈ (0, 1)

x = 0.b1b2b3b4 . . .

in decimal format according to the following rule:

bn =

{2 if ann 6= 2,

3 if ann = 2.

Then x 6= f(1) since b1 6= a11, and x 6= f(2) since b2 6= a22, and in general x 6= f(n) sincebn 6= ann. This shows that the function f is not surjective since x does not belong to its image.

Since |N| ≤ |(0, 1)| but there does not exist any surjective function N→ (0, 1), it follows that|N| < |(0, 1)|. Hence, (0, 1) is uncountable.



How large is |R|?We have shown that |N| < |R|. Can we find a more precise relationship between |N| and |R|?

Definition of Power Set

If A is a set, the power set of A, denoted P(A), is the collection of all subsets of A.

Examples

|∅| = 0 P(∅) = {∅} |P(∅)| = 1 = 20

|{a}| = 1 P({a}) = {∅, {a}} |P({a})| = 2 = 21

|{a, b}| = 2 P({a, b}) = {∅, {a}, {b}, {a, b}} |P({a, b})| = 4 = 22

In general, if A is finite and |A| = n, then |A| = n < |P(A)| = 2n.



Cantor’s Theorem

For any set A,|A| < |P(A)|.

In other words, there exists an injection from A into B, but there does not exist a surjection fromA onto B.

Proof.

If A = ∅, then |A| = 0 < |P(A)| = 1. In this case, the theorem holds.

Now, suppose A 6= ∅. The function f : A→P(A) defined by f(x) = {x} is an injection. So,|A| ≤ |P(A)|.

Let g : A→P(A) be any function. We’ll show that g is not surjective which will complete the proof that|A| < |P(A)|.

Consider the “barber” setB := {x ∈ A : x 6∈ g(x)}.

Let y ∈ A. Then

y ∈ g(y) implies y 6∈ B,

y 6∈ g(y) implies y ∈ B.

Either way, y belongs to exactly one of g(y) or B. So,

B 6= g(y)

for any y ∈ A. This shows that the mapping g : A→P(A) is not surjective. Hence, |A| < |P(A)|.



The Cardinality of R|R| = |P(N)|

Proof.

We will exhibit explicit injections f : (0, 1)→P(N) and g : P(N)→ (0, 1). By theSchroder-Bernstein Theorem, it will follow that |R| = |P(N)|.

We will write elements x ∈ (0, 1) in binary (base 2) form and avoid any representation thatterminates with infinitely many repeated 1’s. (For example, 0.1 = 0.0111111 . . .. We will use0.100000 . . . and not 0.011111 . . ..)

For x = 0.x1x2x3 . . . define f(x) by

f(x) = {n ∈ N : xn = 1}.

If x, y ∈ (0, 1), and x 6= y, then they differ in some digit, say the kth digit. Then either k ∈ f(x)and k 6∈ f(y) or k 6∈ f(x) and k ∈ f(y). So f(x) 6= f(y) and this mapping is an injection(0, 1)→P(N).

Next, we’ll define an injective mapping g : P(N)→ (0, 1). If A ⊆ N, then let (a1, a2, a3, . . .) bethe sequence of 0’s and 1’s defined by

an =

{1 if n ∈ A,

0 if n 6∈ A.

(Continued on next page)



Proof.

Then define g(A) by

g(A) = 0 .1 0 a1 0 a2 0 a3 0 a4 0 . . . (base-2)

where the expression on the right is the binary representation of a number in (0, 1). We interlacezero digits with the digits a1, a2, a3, . . .. This ensures that the mapping is injective since noelement in the image eventually terminate in repeated 1’s.

Since f : (0, 1)→P(N) and g : P(N)→ (0, 1) are both injective, the Schroder-BersteinTheorem tells us that there exists a bijective mapping from (0, 1) to P(N). Therefore,|(0, 1)| = |P(N)|, and so |R| = |P(N)|.

Cantor’s Continuum Hypothesis

There does not exist a subset A ⊂ R such that

|N| < |A| < |R| = |P(N)|.

• Kurt Godel (1940) – Proved that it is impossible to disprove the Continuum Hypothesis.

• Paul Cohen (1963) – Proved that it is impossible to prove the Continuum Hypothesis.

Hence, the Continuum Hypothesis is undecidable using the standard axioms of set theory.


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