Class notes Sept. 06
Countable and Uncountable Sets. Cardinality
When discussing finite sets we all know what it means that two sets are of the same size,
it means they have the same number of elements. We also have an idea of what it means
for a set to be infinite, but what does it mean that two infinite sets are of the same size?
How do we measure the various degrees of infinity?
For example the sets A  {0,1, 2} and B  {4,5, 6} are both finite and have three elements.
But how do we compare N and S={2n  1: n  N } . Obviously they are both infinite, and
S is a proper subset of N, but does N have more elements than S?
Definition 1. Let A and B be two sets. We say that A and B have the same cardinality if
there exists a function f : A  B one-to-one and onto. In this case we write A B .
We will also refer to the sets A and B as being equivalent.
It can be easily shown using the properties of functions and compositions of functions
that the following properties hold:
Lemma 1.
Let A, B and C be sets. We have:
(i) A A .
(ii) If A B then B A .
(iii) If A B and B C , then A C .
Example 1.
Let us show that for the set S defined above, S N .
Define f : N  S , f (n)  2n  1 .
If for n, m  N , f (n)  f (m) , we have 2n  1  2m  1, so n  m , proving that f is one-toone. To prove that it is also onto, we need to show that every element in S is the image of
some element in N by the function f. Let m  S . We see from the definition of the
function that m has to be odd. Now if we show that we can find n  N : f (n)  m we are
done, since this would prove that S  f ( N ) , and as f ( N )  S it follows that f ( N )  S .
m 1
 N since m is odd, and we are done.
Now, f (n)  2n  1  m, n 
2
Definition 2. For each positive integer n, let Nn  {1, 2,..., n} . If A is a set we say:
(a) A is finite if A Nn for some n, or A   , the empty set.
(b) A is infinite if A is not finite.
(c) A is countable if A N .
Observation Before we state some important results we clarify a technique that we will
use in the future. Let A be a countable set. According to Definition 2, A N , therefore
there exists a one-to-one and onto function from N to A, f : N  A . Let
f (1)  x1 , f (2)  x2 ,..., f (n)  xn ,... . Then one can arrange the elements of A in a
sequence {x1 , x2 ,..., xn ,...} .
We are now ready to state the first result, which in essence says that there are no
“smaller” infinite sets than the countable sets.
Theorem 1.
Every infinite subset of a countable set is countable.
Proof: Let E  A be a proper subset of A. Arrange the elements of A in a sequence
{x1 , x2 ,..., xn ,...} of distinct elements. Define the sequence {nk , k  1,.2,...} as follows:
Let n1  min{ j  N : x j  E} , that is n1 picks the first element in A that belongs to the
subset E, xn1  E . Now we find the second element from the sequence that belongs to E
by defining n2  min{ j  n1 : x j  E} . Now the construction follows by recurrence: if
n1 , n2 ,..., nk 1 are defined, let nk  min{ j  nk 1 : x j  E} .
By induction then, the sequence {nk } is defined for all k  N .
Define now a function f : N  E , f (k )  xnk . Since the elements in the sequence
defining A are distinct, f is one-to-one. Moreover f is onto: let x  E be arbitrary. Since
E  A, x  xm , for some index m  N . But then x  xm  xnk for some nk , and therefore
x  f (k ) , proving that f is onto.
Therefore, we proved that E N so E is countable.
QED
Theorem 2.
Let {En , n  1, 2,...} be a sequence of countable sets, and let S 

n 1
En . Then S is
countable.
Proof:
We start by arranging every set En in a sequence {xnk , k  1, 2,...} , and arrange all the
elements in an array as below:
x11 x12
x13 x14 ...
x21
x22
x23
x24 ...
x31
x32
x33
x34 ...
x41
x42
x43
The arrows in the array indicate an arrangement of the elements in the union of all En :
{x11 , x21 , x12 , x31 , x22 , x13 ,...} . Notice that some elements may be listed multiple times. Let
us assume that they are all distinct. In order to show that the set S is countable we will
show that it can be put in a one-to-one and onto correspondence with a subset of the set
of positive integers. Define f : S  N , in the following way:
f ( x11 )  1, f ( x21 )  21 , f ( x12 )  21  1  3, f ( x31 )  22  4, f ( x22 )  22  1  5, f ( x13 )  22  2  6,
f ( x41 )  23  8, f ( x32 )  23  1  9, f ( x23 )  23  2  10, f ( x14 )  23  3  11,...
So for each n , f ( xn1 )  2n1 , f ( xn1,2 )  2n1  1,..., f ( x1,n )  2n1  n 1 .
Since f ( S )  N is a subset of the set of positive integers, f ( S ) is at most countable and
therefore S ~ f ( S ) is at most countable. But E1  S is countable, therefore S is
countable. Since S is countable when we assume that all elements are distinct, it will be
countable in the general case as well.
QED
Theorem 3.
Let A be countable, and let Bn  {(a1 , a2 ,..., an ) : ak  A} be the set of all n-tuples formed
with elements of A. In the above description, some elements may be repeated. Then Bn is
countable.
Proof:
We prove the theorem by induction.
For n  1, B1  A which we know is countable.
Suppose that for n  1, Bn1 is countable, for some n  2 . Consider the set Bn . Its
elements are of the form (b, a) : b  Bn1 , a  A . Now for every b  Bn1 ,{(b, a) : a  A} has
{(b, a) : a  A} is a countable
the same cardinality as A, so it is countable. Then Bn 
bBn1
(since Bn1 is countable) union of countable sets, and therefore it is countable by
Theorem 2.
By the principle of mathematical induction, Bn is countable for all n.
QED
Corollary 1.
The set of rational numbers is countable.
Proof:
p

Consider Q   , p, q  Z , q  0  , with the fractions written in lower terms. Now the
q

result follows from Theorem 3 ( n  2 ) and the fact that the set of integers is countable.
QED
Now are preparing to prove the main result of the lecture, that the set of real numbers is
uncountable. In order to do so we need a few auxiliary results.
Theorem 4.
Let A  {(bn , n  1, 2,..), bn  0 or 1} be the set of all binary sequences. Then A is
uncountable.
Proof:
The idea of proof is the following: we prove that any countable subset of A is a proper
subset of A, therefore if A were countable, it would be a proper subset of itself, that is,
A  A , impossible! Therefore A is uncountable.
Let E  A , countable, and let E consists of sequences {s1 , s2 ,...} . Define a sequence s as
follows: if the n-th digit in sn is 1 we let the n-th digit in s be 0, and vice versa. Then
s  A , but s  E since s differs in at least one digit from all elements in E. Therefore
E  A is a proper subset of A.
QED
The idea of the proof which is ingeniously simple, belongs to Georg Cantor, the creator
of set theory. The construction of the sequence s is known as Cantor’s diagonal process.
Theorem 5.

bn
, bn  0 or bn  1 .
n
n 1 2
Every real number x  [0,1] has a binary expansion, i.e. x  
Proof:
Remember
bn  0, then
first

0
2
n 1
n
that
if
bn  1 for all n, then

1
2
n 1
n
 1,
and
of
course
if
all
 0 , so x  0, and x  0 satisfy the Theorem.
1
1
 x  n1 1 , that is
n1
2
2
1
1
n1  min{k : 2k x  1} . If x  n1 then we are done, otherwise x  n1 , so there exists a
2
2
1
1
positive integer n2  min{k : 2k ( x  n1 )  1} , that is we repeat the procedure for x  n1 .
2
2
1
1
1
1
 x  n1 1 , x  n1  n1 , so n2  n1 . Then we consider
Notice that since
n1
2
2
2
2
1
1
x  n1  n2 . If it is zero then we stop, if not we continue the procedure by finding
2
2
1
1
n3  min{k : 2k ( x  n1  n2 )  1} , and so forth. Let the binary sequence be defined in the
2
2
following way: bn  1 if n  nk for some k, and bn  0 otherwise.
Let 0  x  1 . There exists a positive integer n1 such that
m b

Define the set A   nn , m  1, 2,... . Obviously A is bounded above by 1, so by the
 n 1 2

Axiom of Completeness it must have a least upper bound in R. We show that sup A  x .
1
1
1
Notice that if at some moment 2nk ( x  n1  n2  ...  nk 1 )  1, then we find the binary
2
2
2
1
1
1
expansion in a finite number of steps, x  n1  n2  ...  nk , the set A is finite and
2
2
2
x  max A  A .
m
b
Suppose then that x  A , so the set A is infinite. By the construction, x   nn  0 , for
n 1 2
all m  N , so x is an upper bound for the set A.
Suppose x  sup A , so x  sup A . Then there exists   0 , such that x  x    sup A , so
x   is still an upper bound for the set A. For  above there exists a positive integer k
1
1
such that x  k  x    sup A . Since x   is an upper bound for the set A, x  k is an
2
2
m
bn
1
upper bound as well, so  n  x  k for all m. We rewrite this as
2
n 1 2
m
b
1
x   nn  k , for all m . (1)
2
n 1 2
Since A is infinite there exists m  A : bm  1, and m  k . But then by the construction
m
bn
 0 we can find a smallest positive integer l, l  m (as shown in
n
n 1 2
the second step of the construction) such that
m
bn
1
1

x

 l 1 .

l
n
2
2
n 1 2
m
b
1
1
1
1
1
Since l  m  k , l 1  m  k , so x   nn  l 1  k , contradicting (1).
2
2
2
2
2
n 1 2
Therefore our assumption that x  sup A was false and x  sup A .
QED
Corollary 2.
The set of real numbers in the interval (0,1) is uncountable.
Proof:
The proof follows immediately from Theorems 4 and 5. By Theorem 5 every real number
in (0,1) has a binary expansion, and as we saw in Theorem 4, the set of all sequences of
0 and 1 is uncountable.
QED
Corollary 3.
The set of real numbers is uncountable.
Proof:
Since (0,1)  R , R has an uncountable subset, therefore it is uncountable.
QED
Theorem 6.
The set of real numbers has the same cardinality as the interval (0,1) .
Proof:
In order to prove that the two sets have the same cardinality we need to construct a oneto-one and onto function between the two sets. We start by showing that the set of
positive real numbers has the same cardinality as (0,1) .
x
Let f : (0, )  (0,1), f ( x) 
.
x 1
x1
x
 2 . Then a bit of
We prove that f is one-to-one. Suppose f ( x1 )  f ( x2 ), so
x1  1 x2  1
algorithm, if x  
algebra leads to x1  x2 , so f is one-to-one.
To prove that f is onto let y  (0,1) . We want to find x  (0, ) such that
x
y
y  f ( x) 
. Solving for x in the last equation gives x 
 (0, ) , so f is onto.
x 1
1 y
Therefore (0, ) ~ (0,1) .
Now let g : R  (0, ), g ( x)  e x . We already know that the exponential function is oneto-one and onto, so this proves that R ~ (0, ) . Then R ~ (0,1) (Can you find the one-toone and onto function between the two sets?)
QED
EXERCISES
1. Prove that any infinite set contains an infinite set that is countable.
2. Prove that a set is infinite if and only if it can be put in a one-to-one correspondence
with one of its proper subsets.
3. An algebraic number is a number that is a zero for a polynomial with integer
coefficients. (see Ex. 1.4.12. Abbott) A transcendental number is a real number that is
not algebraic. Prove that there exist transcendental numbers. Note that a proof that a
specific number (such as e or  is transcendental is extremely difficult).
4. For a set A   let P( A) represent the collection of all subsets of A. P( A) is known
as the power set of A.
a) Suppose that A has n elements. Prove that P( A) has 2 n elements.
b) Suppose that A is countable. Prove that P( A) is uncountable.
c) Suppose that A is uncountable. Prove that there is no onto correspondence between A
and P( A) .
5. Show that (0,1) (3, 6) .
6. Show that for each a  R : (0,1) (a, a  1) .
7. Show that for each a  R : (0,1) (a, a  2) .
8. Prove that (0,1) (1, ) .
9. Show that if A B, then P( A) ~ P( B) .