Download 4. Techniques of Proof: II

Chapter 2
Sets and Functions
Copyright © 2013, 2005, 2001 Pearson Education, Inc.
Section 2.4, Slide
1-1 1
Section 2.4
Cardinality
Copyright © 2013, 2005, 2001 Pearson Education, Inc.
Section 2.4, Slide
1-2 2
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 2.4.1
Two sets S and T are called equinumerous, 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 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
1-3 3
Definition 2.4.3
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 2.4.4
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 2.4.6
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 0.
 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 2.4.7
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 0?
Our experience with finite sets is a poor guide here, for
5
6
and E are actually equinumerous!
: 1
2
3
4
7 ...
E: 2
4
6
8 10 12 14 . . .
f (n) = 2n
The function f (n) = 2n is a bijection from
onto E, so E also has cardinality 0.
Copyright © 2013, 2005, 2001 Pearson Education, Inc.
Section 2.4, Slide
1-6 6
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 2.4.9
Let S be a countable set and let T  S. Then T is countable.
The proof of this “obvious” result is not difficult, but it uses the Well-Ordering Property
of
that we won’t encounter until Chapter 3.
Copyright © 2013, 2005, 2001 Pearson Education, Inc.
Section 2.4, Slide
1-7 7
The following theorem follows readily from Theorem 2.4.9. The details of the proof
are in the text.
Theorem 2.4.10
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 2.4.11(a)
Suppose S and T are countable sets. Then S  T is countable.
Theorem 2.4.10 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
1-8 8
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, Slide
1-9 9
We can adapt this process to show that the set
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
4
2
5
5
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, 1-10
Slide 10
Theorem 2.4.12
The set
of real numbers is uncountable.
Proof: Since any subset of a countable set is countable (Theorem 2.4.9), 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, Slide
1-11 11
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 2.4.14
We denote the cardinal number of a set S by | S |, so that we have | S | = | T | iff S and T
are equinumerous. That is, | S | = | T | iff there exists a bijection f : S  T.
We define | S |  | T | to mean that there exists an injection f : S  T.
As usual, | S | < | T | means that | S |  | T | and | S |  | T |.
The basic properties of our ordering of cardinals are included in Theorem 2.4.15.
The proofs are all straightforward and are left to the exercises.
Copyright © 2013, 2005, 2001 Pearson Education, Inc.
Section 2.4, 1-12
Slide 12
Theorem 2.4.15
Let S, T, and U be sets.
(a) If S  T, then | S |  | T |.
(b) | S |  | S |.
(c) If | S |  | T | and | T |  | U |, then | S |  | U |.
(d) If m, n 
and m  n, then |{1, 2, …, m}|  |{1, 2, …, n}|.
(e) If S is finite, then | S | < 0.
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 0 is the transfinite cardinal number of .
Copyright © 2013, 2005, 2001 Pearson Education, Inc.
Section 2.4, 1-13
Slide 13
It is customary to denote the cardinal number of
Since

In fact, since
by c, for continuum.
, we have 0  c.
is countable and
is uncountable, we have 0 < c.
Thus Theorem 2.4.15(e) implies that 0 and c are unequal transfinite cardinals.
Are there any others?
The answer is an emphatic yes, as we see in our next theorem.
Definition 2.4.16
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 2.4.18
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, 1-14
Slide 14
Theorem 2.4.18
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, 1-15
Slide 15
By applying Theorem 2.4.18 again and again, we obtain an infinite sequence of
transfinite cardinals each larger than the one preceding:
0 = |
| < |P (
)| < |P (P (
))| < |P (P (P (
)))| <   
Does the cardinal c = | | fit into this sequence?
In Exercise 24 we sketch the proof that |P (
) | = c.
In Exercise 11 we show that every infinite set has a denumerable subset.
Since (by Theorem 2.4.9) every infinite subset of a denumerable set is denumerable,
we see that 0 is the smallest transfinite cardinal.
What is the first cardinal greater than 0?
We know that c > 0, but is there any cardinal number  such that
0 <  < 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, 1-16
Slide 16