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Introduction Cardinality Infinity Beyond Infinity

9. To Infinity and Beyond

Cardinality

Terence Sim

October 18, 2012




Introduction

9.1. Introduction

In today’s lesson, you will learn:1 What counting is2 How to count to infinity3 How to compare sizes without counting

4 How some infinities are more infinite than others




Introduction

Quiz

1 ∞ + 1 =?2 ∞ + 1010 =?

3 ∞ + ∞ =?

4 ∞ − ∞ =?

5 ∞ × ∞ =?




Introduction

Infinite hotel videos

http://youtu.be/faQBrAQ87l4http://youtu.be/H2W2vduP22Q

I d i C di li I fi i B d I fi i

http://youtu.be/faQBrAQ87l4

http://youtu.be/H2W2vduP22Q

http://youtu.be/H2W2vduP22Q

http://youtu.be/faQBrAQ87l4




Cardinality

What is counting?

I t d ti C di lit I fi it B d I fi it




Cardinality

Definition 9.2.1

Let Z+

= {1, 2, 3, . . .} denote the set of positive integers.

Let Zn= {1, 2, 3, . . . , n} denote the set of positive integers from 1

to n, inclusive.

Counting is the assignment of consecutive elements in Zn to theobjects being counted.

This assignment is a bijection, i.e. every object is assigned exactlyone integer. Furthermore, the order of assignment does not matter.





Cardinality

Definition 9.2.2 (Finite)

A set S is said to be finite if, and only if,(i) S is the empty set; or

(ii) There exists a bijection from S to Zn for some n ∈ Z+

A set S is said to be infinite if it is not finite.

Definition 9.2.3 (Cardinality)

The cardinality of a finite set S , denoted |S |, equals

(i) 0, if S = ∅; or

(ii) n, if f : S → Zn is a bijection.

We will deal with the cardinality of infinite sets later.





Cardinality

Definition 9.2.4

A set A has the same cardinality as another set B iff there is abijection from A to B .

We write |A| = |B |.

Proposition 9.2.5 (Cardinality relation)

Note that the cardinality relation is an equivalence relation.For all sets A, B , C :

1 (Reflexive) |A| = |A|.

2 (Symmetric) |A| = |B | ⇒ |B | = |A|.3 (Transitive) (|A| = |B |) ∧ (|B | = |C |) ⇒ |A| = |C |.





Cardinality

Proof.

1

(Reflexive) The identity function I A provides a bijection fromA to A.

2 (Symmetric) Let |A| = |B |. This means there exists abijection, f : A → B . Then by Proposition 6.1.12,f −1 : B → A is a bijection.

3 (Transitive) Let f : A → B and g : B → C be bijections.Then g ◦ f : A → C is also a bijection.a

aTry to prove this!





Cardinality

Example 1: Finite set

Let A = {a, b , c }. What is |A|?

It is easy to construct a bijection f : A → Z3. Here’s one, shownas a table:

x f(x)b 1

a 2

c 3

There are a total of 3! = 6 possible bijections. Any one is sufficientto show that |A| = 3.




y y y y

Cardinality

Example 2: Even integers

Define 2Z to be the set of even integers. Prove that 2Z has thesame cardinality as Z.

Proof.

By definition, we need to show that there exists a bijection from

2Z to Z. Since any bijection will do, we simply construct one, i.e.this is a proof by construction.

Define f : 2Z → Z by f (n) = n/2. We need to show that f isbijective.

1 (Injective) ∀m, n ∈ 2Z, f (m) = f (n)⇒ m/2 = n/2 ⇒ m = n.

2 (Surjective) ∀n ∈ Z, clearly, 2n ∈ 2Z and f (2n) = n.

Thus |2Z| = |Z|. QED.




y y y y

Cardinality

Note that 2Z ⊂ Z and 2Z = Z. Yet |2Z| = |Z|. This is a strange

result!

For a finite set A, any proper subset B of A will have |B | < |A|.But this is not true for infinite sets.

Some mathematicians have proposed to use this as the definition

of an infinite set. That is, a set A is infinite iff there exists a set B such that B ⊂ A ∧ B = A ∧ |B | = |A|.




Cardinality

Example 3: Line segments

In the figure below, which line segment, DE or BC , has morepoints?

Answer: They have the same cardinality! Every point on DE canbe paired up with a point on BC , e.g. the points X and Y . Thispairing of points is a bijection.




Infinity

9.3. Infinity

Definition 9.3.1 (Cardinal numbers)

Define ℵ0= |Z+|.

ℵ, pronounced “aleph”, is the first letter of the Hebrew alphabet.

This is the first cardinal number. Others will be defined later.

Definition 9.3.2 (Countably infinite)

A set S is said to be countably infinite (or, S has the cardinality of

integers) iff |S | = |Z+

| = ℵ0.

Definition 9.3.3 (Countable)

A set S is said to be countable iff it is finite or countably infinite.

A set is said to be uncountable if it is not countable.




Infinity

Example 3: Z is countably infinite

Intuitively, Z contains twice as many numbers as Z+

, so we expect|Z| = 2|Z+|. But this is not true. Their cardinalities are equal!

This is how to count them:

That is, define f : Z+ → Z by:

f (n) =

n/2, if n is even

−(n − 1)/2, if n is odd

It is straighforward to show that f is a bijection. So, |Z| = |Z+|.




Infinity

Example 4: What about Z+ × Z+ ?

What if an infinite number of buses, each carrying an infinitenumber of guests, arrive at the Infinite Hotel? Is there room for allof them?




Infinity

Define f : Z+ × Z+ → Z+ by:

f (m, n) = (m + n − 2)(m + n − 1)

2 + m




Infinity

Note that on each diagonal, m + n is constant.

The formula exploits the fact that m, n is the mth element on thediagonal where m + n is constant.

Proving that f is bijective is tricky, but can be done. You try it!

As this example shows, finding bijections can be tricky. Provingthem can be even trickier. It’s time to have some theorems tomake life easier.




Infinity

Proposition 9.3.4 (Listing)

A set A is countably infinite iff its elements can be arranged,without duplication and without omission, in an infinite list a1, a2, a3, . . ..

Proof sketch

Clearly, such as listing is a bijection from Z+ to A.




Infinity

Theorem 9.3.5 (Cartesian product)

If sets A and B are both countably infinite, then so is A × B.

Proof sketch

By Proposition 9.3.4, the elements of A and B can be listed,

respectively, as: a1, a2, a3, . . . and b 1, b 2, b 3, . . ..Then A × B can be arranged in table form, similar to that inZ+ × Z+ (page 17). We can then use the same diagonal countingtechnique to map each element am, b n to some positive integer

k . This establishes a bijection from A × B to Z+

.




Infinity

Corollary 9.3.6 (General Cartesian product)

Given n ≥ 2 countably infinite sets A1, A2, . . . , An, the Cartesianproduct A1 × A2 × . . . × An is also countably infinite.

Proof sketch

This is a proof by induction. The base case was already proven in

Theorem 9.3.5.




Infinity

Theorem 9.3.7 (Unions)

The union of countably many countable sets is countable.

That is, if A1, A2, . . . are all countable sets, then so is

A =∞

i =i

Ai

Proof sketch

By Proposition 9.3.4, the elements in each set Ai may be

enumerated as ai 1, ai 2, ai 3, . . .. Each element in the union A mustcome from some Ai , by the definition of union.

· · ·




Infinity

cont’d

We can therefore enumerate (without duplication or omission) theelements of A in a table:

a11 a12 a13 · · ·a21 a22 a23 · · ·a31 a32 a33 · · ·

... ... ... . . .

Now by the diagonal counting method on page 17, we have ourbijection.




Infinity

Theorem 9.3.8 (Subset)

Any subset of a countable set is countable.

Proof sketch

Let A be a countable set, and B ⊂ A.

Case 1, B is finite: Then, B is countable, by definition.

Case 2, B is infinite:

1 This implies A is countably infinite. By Proposition 9.3.4, listthe elements of A: L = a1, a2, . . .

2 Enumerate the elements of B by sequentially searching list Land selecting only those elements ak that belong to B .

3 You now have a list for B : ak 1, ak 2, . . .. By Proposition 9.3.4,B is countably infinite, and hence countable.




Infinity

Example 5: the set of prime numbers, P

Note: P ⊂ Z

+

1 List the elements of Z+ :L = 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, . . .

2 Mark the elements in L which are prime:

L = 1, 2

∗

, 3

∗

, 4, 5

∗

, 6, 7

∗

, 8, 9, 10, . . .3 Thus, a listing of P is 2, 3, 5, 7, . . .

By Proposition 9.3.4, P is countably infinite.




Infinity

Corollary 9.3.9

The superset of an uncountable set is uncountable.

That is, if B ⊂ A is uncountable, then so is A.

Proof.

This is the contra-positive form of Theorem 9.3.8.




I fi i



Infinity

Step 1: Positive rationals

Define Q+

= { m, n | m, n ∈ Z+ ∧ m⊥n } to be the set of positive rational numbers.

Note that usually, a rational number is written as mn

, but thatis just a notational habit. Here we will use m, n, for reasons

that will become obvious.Also, note that our definition of Q+ requires m and n to beco-prime (see Definition 8.2.6), meaning that each rationalnumber is already “reduced to lowest terms”. For example,the number 0.5 will not be 2, 4 but 1, 2. This ensures that

every rational number has a unique representation.

Clearly, Q+ ⊂ Z+ × Z+.

Therefore by Theorem 9.3.8, Q+ is countable.


I fi it



Infinity

Step 2: Negative rationals

Define Q−

= { −m, n | m, n ∈ Z+ ∧ m⊥n } to be the set of negative rational numbers.

It is easy to define a bijection, f : Q− → Q+, by

f (−m, n) = m, n. And thus by Definition 9.2.4, |Q−

| = |Q+

|,so the set of negative rationals is also countably infinite.


Infinity



Infinity

Step 3: Putting it all together

Define the rational number zero by 0, 1.Now define the set of rational numbers by:

Q = Q+ ∪ {0, 1} ∪ Q−

Then, by Theorem 9.3.7, Q is also countable.

This proves that, contrary to intuition, rational numbers have thesame cardinality as integers; that is, |Q| = |Z|.




Beyond Infinity



Beyond Infinity

Proposition 9.4.1 (The set, (0, 1), is uncountable)

The set of real numbers between 0 and 1, excluding the endpoints:

(0, 1) = { x ∈ R | 0 < x < 1 }

is uncountable.

To prove that a set is uncountable means proving that there is no possibility of a bijection from that set to Z+. Simply saying, “Ican’t find a bijection”, is not enough, since it could mean I didn’tlook hard enough, or I’m not smart enough to find one.

The way to prove is by contradiction. Georg Cantor, who gave usset theory, also gave an ingenius proof that (0, 1) is uncountable.We now refer to it as Cantor’s Diagonalization Argument.


Beyond Infinity



Beyond Infinity

Proof sketch

1 Suppose (0, 1) is countable.

2 Since it is not finite, it is countably infinite.

3 By Proposition 9.3.4, the elements of (0, 1) can be listed in asequence.

x 1= 0.a11a12a13 . . . a1n . . .x 2= 0.a21a22a23 . . . a2n . . .x 3= 0.a31a32a33 . . . a3n . . .... =

...x n= 0.an1an2an3 . . . ann . . .

... = ...

where each aij ∈ {0, 1, . . . , 9} is a single digit.a

aNote that some numbers have two representations, e.g.0.499 . . . = 0.500 . . .. We agree to use only the latter representation.


Beyond Infinity



y y

Proof cont’d

4 Construct a number d = 0.d 1d 2d 3 . . . d n . . . such that

d n =

1, if ann = 1;

2, if ann = 1.

5

Note that ∀n ∈ Z

+

, d n = ann. Thus, d = x n, ∀n ∈ Z

+

.6 But clearly, d ∈ (0, 1), resulting in a contradiction. QED.

In other words, we constructed a new number d ∈ (0, 1) thatdiffers from x 1 in the first digit, from x 2 in the second digit, from

x 3 in the third digit, etc.Clearly, this d is not in the list. Yet d is in the set, resulting in acontradiction.




Beyond Infinity



Proposition 9.4.2 (R is uncountable)

Proof.

Since (0, 1) ⊂ R is uncountable, then by Corollary 9.3.9, R isuncountable. QED.

If |A| = |R|, we say that A has the cardinality of the continuum, aphrase coined by Cantor.


Beyond Infinity



Proposition 9.4.3 (|(0, 1)| = |R|)

Proof.

Define the bijection, h : (0, 1) → R by h(x ) = cot(πx ). ByDefinition 9.2.4, they have the same cardinality. QED.

-0.5 0 0.5 1 1.5

-6

-4

-2

0

2

4

6

x

cot( x)


Beyond Infinity



Definition 9.4.4

Let A, B be sets.If |A| < |B |, then there exists an injection f : A → B , but nosurjection g : A → B .

Theorem 9.4.5 (Cardinality of Power set)

If A is any set, then |A| < |P (A)|.


Beyond Infinity



Proof.

1 Using Definition 9.4.4, we need to show that there exists aninjection f : A → P (A), but no surjection g : A → P (A).

2 Define f (x ) = {x }. This function maps x ∈ A to the singletonset {x } ∈ P (A). To prove it is injective:

(a) Suppose f (x ) = f (y ).

(b) Then, {x } = {y }.(c) By Axiom 4.4.3 of ZFC set theory, x = y .

QED.

3 We prove there is no surjection by contradiction.

(a) Suppose there is a surjection g : A → P (A).(b) This function maps x ∈ A to g (x ) ∈ P (A). That is, g (x ) is asubset of A.

...


Beyond Infinity



Proof cont’d

3 Cont’d.

(c) Define B = {x ∈ A | x /∈ g (x )} ⊂ A.(d) Since B ⊂ A, we have B ∈ P (A).(e) Since g is surjective, there exists a ∈ A such that g (a) = B .(f) Question: a ∈ B or a /∈ B ?

Case 1: If a ∈ B , then the definition of B implies a /∈ g (a), and sinceg (a) = B we get a /∈ B . Contradiction.

Case 2: If a /∈ B , then the definition of B implies a ∈ g (a), and sinceg (a) = B we get a ∈ B . Again, contradiction.

(g) Therefore there is no surjection g : A → P (A).QED.

4 Therefore, |A| < |P (A)|.

QED.


Beyond Infinity



Countably infinite number of infinities

Using Theorem 9.4.5, we can construct the following chain of infinite cardinalities.

ℵ0 = |Z+| < |P (Z+)| < |P (P (Z+))| < |P (P (P (Z+)))| < . . .

The set Z+

is countable; all the others are uncountable.

Furthermore, it is clear that the set containing all the infinite setsproduced in this manner has the cardinality of integers. That is,there are (at least) countably infinitely many infinities.

There could be even more infinities than these. That is, the set of infinite sets could well be uncountable. We just haven’t foundthem yet!


Beyond Infinity



Theorem 9.4.6

|R| = |P (Z+

)|

Proof.

We omit the long and complicated proof.


Beyond Infinity



The Continuum Hypothesis

Cantor himself wondered if there exists a set A such that:

ℵ0 < |A| < |R|

In 1964, Paul Cohen proved that the Continuum Hypothesis is

undecidable ; it is independent of ZFC set theory.In other words, the existence of such a set A cannot be settledwithout adding new axioms to standard mathematics. Youcan assume either way without causing a contradiction!

Assuming the Hypothesis is false (no such A), we can thendefine cardinal numbers as follows:

ℵ1 = 2ℵ0 , ℵ2 = 2ℵ1 , ℵ3 = 2ℵ2 , . . . , ℵk +1 = 2ℵk , . . .

This is called the Generalized Continuum Hypothesis.


Beyond Infinity



Buzz Lightyear’s last word

They even created a new dance for me:http://youtu.be/2okbjYrsupA

http://youtu.be/2okbjYrsupA

http://youtu.be/2okbjYrsupA

12 cardinality

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