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Section 6.5
Cardinality
Copyright © 2013, 2005, 2001 Pearson Education, Inc.
Section 2.4, Slide
1-1 1
How can we compare the sizes of two sets?
If S = {x ∈
: x2 = 9}, then S = {–3, 3} and we say that S has two elements.
If T = {1, 7, 11}, then T has three elements and we think of T as being “larger” than S.
These intuitive ideas are fine for small (finite) sets, but how can we compare the size of
(infinite) sets like
or ?
Certainly, it is reasonable to say that two sets S and T are the same size if there is a
bijective function f : S → T, for this function will set up a one-to-one correspondence
between the elements of each set.
Definition 6.5.1
Two sets S and T are called equinumerous, or have the same cardinality and we write
S ~ T, if there exists a bijective function from S onto T.
It is easy to see that “~” is an equivalence relation on any family of sets, and it partitions
the family into disjoint equivalence classes. With each equivalence class we associate a
cardinal number (the number of elements) that we think of as giving the size of the set.
In dealing with finite sets, it will be convenient to abbreviate the set {1, 2, …, n} by In.
Copyright © 2013, 2005, 2001 Pearson Education, Inc.
Section 2.4, Slide
1-2 2
Lemma 6.5.2
• (“∽” is an equivalence relation)
• Let 𝐴, 𝐵 and 𝐶 be sets.
1. 𝐴 ∽ 𝐴.
2. If 𝐴 ∼ 𝐵, then 𝐵 ∼ 𝐴.
3. If 𝐴 ∼ 𝐵 and 𝐵 ∼ 𝐶, then 𝐴 ∼ 𝐶.
Copyright © 2013, 2005, 2001 Pearson Education, Inc.
Section 2.4, Slide
1-3 3
Definition 6.5.4 (1)
A set S is said to be finite if S = ∅ or if there exists and a bijection f : In → S.
If a set is not finite, it is said to be infinite.
Definition (2)
The cardinal number of In is n, and if S ~ In, we say that S has n elements.
The cardinal number of ∅ is taken to be 0.
If a cardinal number is not finite, it is called transfinite.
Definition (3)
A set S is said to be denumerable if there exists a bijection f :
If a set is finite or denumerable, it is called countable.
If a set is not countable, it is uncountable.
The cardinal number of a denumerable set is denoted by ℵ0.
→ S.
“Aleph” is the first letter of the Hebrew alphabet.
Copyright © 2013, 2005, 2001 Pearson Education, Inc.
Section 2.4, Slide
1-4 4
The relationships between the various “sizes” of sets is shown below.
countable sets
infinite sets
finite
denumerable
uncountable
sets
sets
sets
There are two kinds of countable sets: finite sets and denumerable sets.
The denumerable sets are infinite sets.
The infinite sets that are not denumerable sets are uncountable sets.
We have not yet shown that there exist any infinite sets that are not denumerable.
Before doing so, we look at some properties of countable sets.
Copyright © 2013, 2005, 2001 Pearson Education, Inc.
Section 2.4, Slide
1-5 5
Countable Sets
Example 1
It would seem at first glance that the set
of natural numbers should be “bigger”
and, in fact,
than the set E of even natural numbers. Indeed, E is a proper subset of
it contains only “half” of . But what is “half ” of ℵ0?
Our experience with finite sets is a poor guide here, for
: 1
2
3
4
5
6
and E are actually equinumerous!
7 ...
f (n) = 2n
E: 2
4 6
The function f (n) = 2n is a bijection from
8 10 12 14 . . .
onto E, so E also has cardinality ℵ0.
Copyright © 2013, 2005, 2001 Pearson Education, Inc.
Section 2.4, Slide
1-6 6
P223, Example 2
• The set of natural numbers ℕ and the set of
integers Z have the same cardinality.
Copyright © 2013, 2005, 2001 Pearson Education, Inc.
Section 2.4, Slide
1-7 7
If a nonempty set S is finite, then there exists n ∈
and a bijection f : In → S.
Using the function f , we can count off the members of S as follows:
f (1), f (2), f (3), …, f (n).
Letting f (k) = sk for 1 ≤ k ≤ n, we obtain the more familiar notation S = {s1, s2, …, sn}.
The same kind of counting process is possible for a denumerable set, and this is why
both kinds of sets are called countable.
For example, if T is denumerable, then there exists a bijection g : → T,
and we may write T = {g (1), g (2), g (3),…} or T = {t1, t2, t3, …}, where g (n) = tn.
This ability to list the members of a set as a first, second, third and so on, characterizes
countable sets. If the list terminates, then the set is finite.
If the list does not terminate, then the set is denumerable.
Theorem 6.6.7
Let S be a countable set and let T ⊆ S. Then T is countable.
See Page 235 for proof .
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Section 2.4, Slide
1-8 8
The following theorem follows readily from Theorem 2.4.9. The details of the proof
are in the text.
Theorem 6.6.8 See page 235 for proof
Let S be a nonempty set. The following three conditions are equivalent.
(a) S is countable.
(b) There exists an injection f : S →
(c) There exists a surjection g :
.
→ S.
Example
Suppose S and T are countable sets. Then S ∪ T is countable.
Theorem 6.6.8 implies that there exist surjections f :
Define h:
→ S ∪ T by
  n +1
 f  2 ,
 

h( n) = 
g  n ,
  2 
→ S and g :
→ T.
if n is odd
if n is even.
Then h is surjective, so S ∪ T is countable. We can see this as follows:
Copyright © 2013, 2005, 2001 Pearson Education, Inc.
Section 2.4, Slide
1-9 9
Given surjections f :
→ S and g :
→ T,
let h use the odd integers with f to count S and the even integers with g to count T.
h
f
S
1
2
3
4
5
6
..
.
•
•
•
•
1
2
3
4
f (1) = h(1)
f (2) = h(3)
f (3) = h(5)
..
.
..
.
g
T
•
•
•
•
1
2
3
4
g (1) = h(2)
g (2) = h(4)
g (3) = h(6)
..
.
..
.
h
Copyright © 2013, 2005, 2001 Pearson Education, Inc.
Section 2.4, 1-10
Slide 10
Theorem 6.6.9. and 6.6.10
• Theorem 6.6.9 Let 𝐼 be a non-empty set, and
let {𝐴𝑖 }𝑖∈𝐼 be a family of sets indexed by 𝐼.
Suppose that 𝐴𝑖 is countable for each 𝑖 ∈ 𝐼.
1. ⋂𝑖∈𝐼 𝐴𝑖 is countable.
2. If 𝐼 is countable, then ⋃𝑖∈𝐼 𝐴𝑖 is countable.
• Theorem 6.6.10. Let 𝐴1 , ⋯ 𝐴𝑛 be sets for some
𝑛 ∈ ℕ. Suppose that 𝐴1 , ⋯ 𝐴𝑛 are countable.
Then 𝐴1 × ⋯ × 𝐴𝑛 is countable.
Copyright © 2013, 2005, 2001 Pearson Education, Inc.
Section 2.4, Slide
1-11 11
Theorem 6.6.11. and 6.6.12
• Theorem 6.6.11. Let A be a set. If A is infinite,
then A has a countably infinite subset.
• Theorem 6.6.12. Let A be a set. Then A is finite
if and only if A has no proper subset with the
same cardinality as A.
Copyright © 2013, 2005, 2001 Pearson Education, Inc.
Section 2.4, 1-12
Slide 12
Theorem 6.7.1. the set ℚ of rational numbers is countable.
We begin by constructing a rectangular array of fractions. The first row contains all the
positive integers. The second row contains all the fractions with denominator equal to 2.
The third row contains all the fractions with denominator equal to 3, etc.
1
1
2
2
2
2
3
3
2
4
5
5
2
4
2
...
...
Moving along each diagonal of the array in the
manner indicated, we obtain a listing of all the
elements in the positive rationals + :
1, 2,
1, 1, 2,
2 3 2
3, 4,
3 , 2 , ...
2 3
1
3
2
3
3
3
4
3
5
3
...
This listing defines a surjection f :
so + is countable.
1
4
2
4
3
4
4
4
5
4
...
But,
1
5
2
5
3
5
...
=
+ ∪ {0} ∪
– , so
→
+,
is countable, too.
In the text this is generalized to show that any union
of a countable family of countable sets is countable.
Copyright © 2013, 2005, 2001 Pearson Education, Inc.
Section 2.4, 1-13
Slide 13
Theorem 6.7.3
The set
of real numbers is uncountable.
Proof: Since any subset of a countable set is countable (Theorem *), it suffices to
show that the interval J = (0, 1) is uncountable. If J were countable, we could list its
members and have
J = {x1, x2, x3, …} = {xn : n ∈ }.
We shall show that this leads to a contradiction by constructing a real number that is
in J but is not included in the list of xn’s. Each element of J has an infinite decimal
expansion, so we can write
=
x 0. a a a ⋅⋅⋅,
1
11 12
13
=
x2 0. a21 a22 a23 ⋅⋅⋅,
x3 0. a31 a32 a33 ⋅⋅⋅,
=
⋅⋅
⋅
where each ai j ∈ {0, 1, …, 9}.
2, if ann ≠ 2,
We now construct a real number y = 0. b1 b2 b3 ⋅ ⋅ ⋅ by defining bn = 
3, if ann = 2.
Since each digit in the decimal expansion of y is either 2 or 3, y ∈ J.
But y is not one of the numbers xn, since it differs from xn in the nth decimal place.
This contradicts our assumption that J is countable, so J must be uncountable. ♦
Copyright © 2013, 2005, 2001 Pearson Education, Inc.
Section 2.4, 1-14
Slide 14
Ordering of Cardinal Numbers
Our approach to comparing the size of two sets is built on the definition of equinumerous
and our understanding of functions. Intuitively, if f : S → T is injective, then S can be
no larger than T. Not only is this true for finite sets, but we have observed that it also
holds for countable sets (Theorem 2.4.10). Since we think of cardinal numbers as
representing the size of a set, we shall use them when comparing sizes.
Definition 6.5.9 & 6.6.1
We denote the cardinal number (cardinality) of a set S by | S |, so | S | = | T | iff S and T
have the same cardinality. i.e. | S | = | T | (𝑆 ∼ 𝑇) iff there exists a bijection f : S → T.
We define | S | ≤ | T | (𝑆 ≼ 𝑇) to mean that there exists an injection f : S → T.
As usual, | S | < | T | (𝑆 ≺ 𝑇) means that | S | ≤ | T | and | S | ≠ | T |.
Theorem 6.5.10 (Schroeder–Bernstein Theorem). Let 𝐴 and 𝐵 be sets. Suppose
that 𝐴 ≼ 𝐵 and 𝐵 ≼ 𝐴. Then 𝐴~𝐵.
Theorem 6.5.13 (Trichotomy Law for Sets). Let 𝐴 and 𝐵 be sets. Then 𝐴 ≼ 𝐵
or 𝐵 ≼ 𝐴.
Copyright © 2013, 2005, 2001 Pearson Education, Inc.
Section 2.4, 1-15
Slide 15
Theorem ** (Lemma 6.5.2)
Let S, T, and U be sets.
(a) If S ⊆ T, then | S | ≤ | T | (𝑆 ≼ 𝑇).
(b) | S | ≤ | S |, (𝑆 ≼ 𝑆).
(c) If | S | ≤ | T | (𝑆 ≼ 𝑇 ) and | T | ≤ | U | (𝑇 ≼ 𝑈), then | S | ≤ | U | (𝑆 ≼ 𝑈) .
(d) If m, n ∈ ℕ and m ≤ n, then |{1, 2, …, m}| ≤ |{1, 2, …, n}|.
(e) If S is finite, then | S | < ℵ0.
Notes: (a) This corresponds to our intuitive feeling about the relative sizes of subsets.
(b) This is the reflexive property.
(c) This is the transitive property.
(d) This means that the order of m and n as integers is the same as the order
for the finite cardinals m and n.
(e) Recall that ℵ0 is the transfinite cardinal number of .
Copyright © 2013, 2005, 2001 Pearson Education, Inc.
Section 2.4, 1-16
Slide 16
It is customary to denote the cardinal number of
Since
⊆
In fact, since
by c, for continuum.
, we have ℵ0 ≤ c.
is countable and
is uncountable, we have ℵ0 < c.
Thus Theorem ** (e) implies that ℵ0 and c are unequal transfinite cardinals.
Are there any others?
The answer is an emphatic yes, as we see in our next theorem.
Definition
Given any set S, let P (S) denote the collection of all the subsets of S.
The set P (S) is called the power set of S.
Theorem 6.5.7
For any set S, we have | S | < |P (S) |.
Recall: | S | < | T | means there exists an injection from S to T,
but no bijection from S to T.
Copyright © 2013, 2005, 2001 Pearson Education, Inc.
Section 2.4, 1-17
Slide 17
Theorem 6.5.7
For any set S, we have | S | < |P (S) |.
Proof: The function g : S → P (S) given by g (s) = {s} is clearly injective, so | S | ≤ |P (S) |.
To prove that | S | ≠ |P (S) |, we show that no function from S to P (S) can be surjective.
Suppose that f : S → P (S). Then for each x ∈ S, f (x) is a subset of S.
Now for some x in S it may be that x is in the subset f (x) and for others it may not be.
Let T = {x ∈ S : x ∉ f (x)}.
We have T ⊆ S, so T ∈ P (S). If f were surjective, then T = f ( y) for some y ∈ S.
Now either y ∈ T or y ∉ T, but both possibilities lead to contradictions:
If y ∈ T, then y ∉ f ( y) by the definition of T.
But f ( y) = T, so y ∉ f ( y) implies y ∉ T.
On the other hand, if y ∉ T, then since f ( y) = T, we have y ∉ f ( y).
But then y ∈ T, by the definition of T.
Thus we conclude that no function from S to P (S) can be surjective, so | S | ≠ |P (S) |. ♦
Copyright © 2013, 2005, 2001 Pearson Education, Inc.
Section 2.4, 1-18
Slide 18
By applying Theorem 6.5.7 again and again, we obtain an infinite sequence of
transfinite cardinals each larger than the one preceding:
ℵ0 = |
| < |P (
Does the cardinal c = |
Theorem. |P (ℕ) | = c.
)| < |P (P (
))| < |P (P (P (
)))| < ⋅ ⋅ ⋅
| fit into this sequence?
Theorem 6.6.11. Every infinite set has a denumerable subset.
Since (by Theorem 6.6.7) every infinite subset of a denumerable set is denumerable,
we see that ℵ0 is the smallest transfinite cardinal.
What is the first cardinal greater than ℵ0?
We know that c > ℵ0, but is there any cardinal number λ such that
ℵ0 < λ < c ?
That is, is there any subset of
with size “in between” and ?
The conjecture that there is no such set was first made by Cantor and is known
as the continuum hypothesis. In 1900 it was included as the first of Hilbert’s
famous 23 unsolved problems. Whether it is true or false is still an unanswered—
perhaps unanswerable—question.
Copyright © 2013, 2005, 2001 Pearson Education, Inc.
Section 2.4, 1-19
Slide 19
Exercises
• #6.5.2. Prove that the disk ℝ2 of radius 3
centered at (1,2) has the same cardinality as
the unit disk in ℝ2 centered at the origin.
• #6.5.7. Let 𝐴 and 𝐵 be sets. Prove that 𝐴 ∼
𝐵 implies that 𝑃(𝐴) ∼ 𝑃(𝐵).
• #6.5.10. Let A, B, C and D be sets. Suppose
that 𝐴 ∼ 𝐵 and 𝐶 ∼ 𝐷 . Prove that 𝐴 × 𝐶 ∼
𝐵 × 𝐷.
Copyright © 2013, 2005, 2001 Pearson Education, Inc.
Section 2.4, 1-20
Slide 20
Exercise 6.6.2.
• Let 𝐴 ⊆ ℕ be a subset. Suppose that there is
some 𝑀 ∈ ℕ such that 𝑎 ≤ 𝑀 for all 𝑎 ∈ 𝐴.
Prove that 𝐴 is finite.
Copyright © 2013, 2005, 2001 Pearson Education, Inc.
Section 2.4, 1-21
Slide 21
Recall that
denotes the set of positive integers, also called the natural numbers.
= {1, 2, 3, 4, . . . }
In this text we do not attempt to develop the properties of the natural numbers in a
rigorous way from set theory. Rather, we assume familiarity with the usual arithmetic
operations of addition and multiplication and the relation of “less than.”
There is one additional property of
that we assume as an axiom.
It is based on the intuitive idea that each nonempty subset of
Axiom (Th 6.2.5)
must have a least member.
The Well-Ordering Property of ℕ:
If S is a nonempty subset of
m ≤ k for all k ∈ S.
, then there exists an element m ∈ S such that
In proving theorems about the natural numbers, we need a technique that does not
depend on verifying the validity one number at a time.
This is provided by the Principle of Mathematical Induction.
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Section 2.4, 1-22
Slide 22
HW
•
•
•
•
•
•
•
•
Exercise 6.5.1.
Exercise 6.5.3. Prove Lemma 6.5.2.
Exercise 6.5.4.
Exercise 6.5.5.
Exercise 6.5.9.
Exercise 6.6.1.
Exercise 6.7.1
Due Tue. Dec. 3
Copyright © 2013, 2005, 2001 Pearson Education, Inc.
Section 2.4, 1-23
Slide 23