Class notes August 21st
Structure of the Real Numbers
There are two well-known ways of representing real numbers:
1. by its decimal expansion;
2. as a point on a line.
However, as it will be pointed out below, there are problems with either representation.
In the decimal expansion the representation is not always unique.
Example 1. We will show that 2  1.9999... .
Denote x  1.9999... . We have 10x  19.999... , and 10 x  x  18, therefore x  2 .
The second representation is geometric in nature, and therefore it depends on the scale on
the line. Also, operations such as multiplication and division of real numbers are hard to
visualize. Moreover, there is another problem philosophical in nature: points have no
dimension but a line, which is made up of points, has length!
Then one may ask: “What is a rigorous mathematical definition of a real number?” A
“natural” construction should be founded on what is previously known, in this case the
rational numbers. We have seen that there are numbers that cannot be represented as
fractions of integers, and moreover, the rational numbers have “gaps”. In Chapter 8.4 you
can read about the construction of the real numbers through the Dedekind cuts. In what
follows we will present an axiomatic approach to the set of real numbers.
Definition 1. A field is a nonempty set F that has two operations defined on it, addition
and multiplication, that have the following properties:
1. x  y  F for all x, y  F
2. x  y  y  x for all x, y  F
3. ( x  y )  z  x  ( y  z ) for all x, y, z  F
4. There exists an element 0  F such that x  0  x for all x  F
5. For each x  F , there exists y  F : x  y  0.
6. xy  F for all x, y  F
7. xy  yx for all x, y  F
8. ( xy) z  x( yz ) for all x, y, z  F
9. There exists an element 1  F such that x  1  x for all x  F
10. For each x  F , x  0, x has an inverse y  F : x  y  1.
11. x( y  z )  xy  xz for all x, y, z  F .
Definition 2. An order “  ” on a set S is a relation that satisfies:
a. x  x for all x  S (the relation is said to be reflexive).
b. If x  y and y  x then x  y (the relation is said to be antisymmetric).
c. If x  y and y  z then x  z (the relation is said to be transitive).
Definition 3. A field F is an ordered field if F is endowed with an order  that satisfies:
a. If y  z , then x  y  x  z for all x  F .
b. If x  0 and y  0, then xy  0.
Example 2. We prove that the element zero of a field F is unique.
Proof:
Let us assume that there are two zero elements: 0 and 0.' Then, using property (4) of
Definiton 1, first with x  0 and 0 as the zero element, and second with x  0' and 0 ' as
the zero element, we get:
0  0'  0, but also 0'  0  0' .
But since the addition is commutative (property 2), it follows that 0  0' .
QED
Example 3. We prove that in a field F we have the following: for any x, y, z  F , if
y  z and x  0 , then xy  xz .
Proof: Since y  z, we have z  y  0 , so by property b of Definition 3, it follows that
x( z  y )  0 . By property (11) of Definition 1, x( z  y )  xz  xy  0 , and using property
(a) of Definition 3, we get: xz  xz  xy  xy  xy, or xy  xz.
QED
EXERCISES
1. Prove that in a field the additive inverse of an element is unique.
2. Prove that in a field the multiplication inverse of an element is unique.
3. Using Definition 3 and the field properties prove the following:
a) If x  0 and y  0 then x  y  0 .
b) x  y if and only if y  x  0 .
All questions that follow pertain to an ordered field F.
4. Prove that if x  0 then x  0 .
5. Prove that 1  0 .
6. Show that for any x  F , x 2  0 , and x 2  0 if and only if x  0 .
Sets
Definition 4. A set is a collection of objects sharing the same property.
We refer to the “objects” of a set as its elements.
Notation: if an element x belongs to a set A we write x  A .
Definition 5. If A and B are sets such that each element of A is also an element of B, then
A is a subset of B or A is contained in B and we write A  B .
If A  B and A  B then A is a proper subset of B, and we denote this by A  B .
Definition 6. Two sets A and B are equal ( A  B ) if and only if every element in A also
belongs to B, and every element in B also belongs to A.
Let us prove the following statement which is very important for two reasons: first is the
nature of the statement itself, a necessary and sufficient condition type of theorem, and
second, it shows the approach to prove equality between sets.
Theorem 1.
Let A and B be sets. Then A  B if and only if A  B and B  A .
Proof: Notice the method of proving an “if and only if” statement. We need to prove two
implications.
First, let us assume A  B . Then by the definition, every element in A belongs to B and
every element in B belongs to A as well. Therefore, if x  A , then x  B , for any x  A ,
which proves that A  B . Similarly, for any x  B , we have x  A , proving that B  A .
So we proved that if A  B , then A  B and B  A .
Now, for the converse statement, let us assume that A  B and B  A . If x  A , then
since A  B , x  B . Also, for any element in B, if y  B , since B  A , then y  A .
Therefore, we proved that any element in A belongs to B, and any element in B belongs to
A, so by definition, A  B .
QED
Basic operations with sets and notation
Intersection: A  B  {x : x  A, and x  B}.
Union: A  B  {x : x  A, or x  B}.
Complement of a set A: Ac  {x : x  A}.
Complement of B relative to A: A \ B  {x : x  A, and x  B}.
The empty set  is the set containing no elements at all.
Theorem 2. Properties of set operations:
Let A, B and C be sets. Then
(1) A  B  B  A
(2) A  B  B  A
(3) ( A  B)  C  A  ( B  C )
(4) ( A  B)  C  A  ( B  C )
(5) A  ( B  C )  ( A  B)  ( A  C )
(6) A  ( B  C )  ( A  B)  ( A  C )
Proof: Parts (1) through (4) are easy and are left to the reader. Let us prove (5). We use
Theorem 1 to show the equality of the two sets. Therefore we need to show
A  ( B  C )  ( A  B)  ( A  C ) (7) and ( A  B)  ( A  C )  A  ( B  C ) (8) .
Let x  A  ( B  C ) . Then x  A and x  B  C . Since x  B  C , then, we either have
x  B or x  C . If x  B then x  A  B so x  ( A  B)  ( A  C ) and (7) follows.
If x  C , then x  A  C and again x  ( A  B)  ( A  C ) , and (7) is proved.
Let us now prove (8). Let x  ( A  B)  ( A  C ) . Then x  A  B or x  A  C .
If x  A  B , then x  A  ( B  C ) , and similarly, if x  A  C , then x  A  ( B  C ) ,
and (8) is proved.
QED
The proofs of the next two theorems are left to the reader.
Theorem 3.
Let A, B and C be sets. Then
(1) A \ ( B  C )  ( A \ B)  ( A \ C )
(2) A \ ( B  C )  ( A \ B)  ( A \ C ) .
Theorem 4. De Morgan’s Laws
For any two sets A, B, we have
(1) ( A  B)c  Ac  B c
(2) ( A  B)c  Ac  B c .
Definition 7. Two sets A, B are said to be disjoint if A  B   .
Definition 8. Let  be a set, and suppose that for each   , a subset A of a given set
S is specified. The collection of sets A is called an indexed family of subsets of S with
 as the index set. We denote this by { A } .
Definition 9. If { A } is an indexed family of sets, define


A  {x : x  A for all   }
A  {x : x  A for some   }={x : there exists    such that x  A }
1
Example 4. Let An  (0, ) , with n  N the set of natural numbers. We show that
n
An  
nN
1
An . Then x0  (0, ) for each n  N . Obviously
n
nN
x0  0 . To continue the proof we need to use the Archimedean Property (Theorem 1.4.2
1
in text and also see below). Consider the number
: by the Archimedean Property there
x0
1
1
1
 n0 . Then
 x0 , and therefore x0  (0, ) , which implies
exists n0 such that
x0
n0
n0
Let us assume there exists x0 
x0 
An . Therefore our assumption lead to a contradiction, so there cannot exist
nN
x0 
An   .
An , so
nN
nN
QED
Theorem 5. Archimedean Property of real numbers
(1) For any real number x  R there exists n  N such that n  x .
1
(2) Given any y  0 there exists n  N such that  y .
n
Theorem 6. De Morgan’s Laws generalized.
c


Ac
 A  
   
c
(1)


Ac (2)
 A  
   
Proof:
We will give the proof of (1) only. Since we need to prove the equality of two sets we
will use Theorem 1.
c


Let x   A  . Then, since x does not belong to the intersection of the family of sets,
  
there must exist an index 0   such that x  A0 . But then x  Ac0 and therefore
c


Ac .
x  A . So we proved  A  


  
c
For the converse, let x  A . Then x  Ac0 for some index 0   . This implies
c

c
x  A0 ,
x
so

A , which means


x   A  . Therefore we proved
  
c


A   A  and (1) is proved.

  
c
QED
EXERCISES
7. If A  B and C is an arbitrary set, then C \ B  C \ A .
8. Under what conditions does A \ ( A \ B)  B ?
9. Prove Theorem 3.
10. Prove Theorem 4.
11. Finish the proof of Theorem 6.
12. Find each of the following sets:
 1 1
(a)
 ,  .
n n
nN 
(b)
 n, n 
nN
1
 1
  ,1  
n
n
nN 
1
 1
(d)
 ,2  .
n
n
nN 
13. Prove that A  B  A if and only if A  B .
14. Prove that if A  B then Ac  Bc .
(c)