1.A     Elementary Concepts from Set Theory

 

Denition: A set is a collection of distinct objects.

 

• Here are some examples of sets:

{1, 2, 3},

{1, 2},

{1},

{book, pen},

N = {1, 2, 3, . . .}, the set of natural (counting) numbers,

= {}, the empty set.

Remark: Not all sets can be enumerated like this, as a (nite or innite) list of

elements. The set of real numbers is one such example.1

Remark: If we       can  write the elements of a set in a list, the order in which we list

them is not important.

 

Denition: We say that a set A is a subset of a set B                  if every element of A is also

an element of B. We write A B.2

 

Denition: We say that a set             A  contains    a set   B  if every element of B       is also an

element of A. We write A B. Note that this denition implies that B A.

1See the excellent article on countability, “How do I love thee? Let me count the ways!” by L.

Marcoux, http://www.pims.math.ca/pi/issue1/page10-14.pdf, 2000.


2Some authors write this as A


B   and reserve the notation A B         for the case where A is a


subset of B but is not identical to B, that is, where A is a proper subset of B. In our notation, if

we want to emphasize that A must be a proper subset of B, we explicitly write A ( B.

 

7


 


 

 

8


 

 

 

CHAPTER 1. REAL NUMBERS


 

Denition: We say that two sets A and B               are equal if A B and B A, that is,


if every element in A is also in B          and vice-versa, so that A and B contain exactly

the same elements. We write A = B.

• {1, 2}={2, 1}.

Denition: The set containing all elements of                    A  and all elements of        B   (but no

additional elements) is called the union of A and B and is denoted A B.

Denition: The set containing exactly those elements common to both A                        and B   is

called the intersection of A and B and is denoted A ∩ B.

These denitions are illustrated in Figure 1.1.

• {1} {2} = {1, 2}.

• {1, 2, 3} ∩ {1, 4}={1}.

• {1, 2} {2}={1, 2}.

 

A ∩ B


 

 

 

A


 

 

 

 

 

 

 

 

 

A B


 

 

 

B


 

Figure 1.1: Venn Diagram

 

 

 

 

1.B     Hierarchy of Sets of Numbers

 

We will nd it useful to consider the following sets ( means is an element of ):

= {} the empty set,

N ={1, 2, 3, . . .}, the set of natural (counting) numbers,

Z = {−n : n N} {0} N, the set of integers,

Q ={pq: p, q Z, q6= 0}, the set of rational numbers,

R, the set of all real numbers.

Notice that N Z Q R.


 


 

 

1.B. HIERARCHY OF SETS OF NUMBERS


 

 

 

9


 

Q.   Why do we need the set R of real numbers to develop calculus? Why can’t we just

use the set Q of rational numbers? One might try to argue, for example, that

every number representable on a (nite-precision) digital computer is rational.

If a subset of Q is good enough for computers, shouldn’t it be good enough for

mathematicians, too?

 

To answer this question, it will be helpful to recall                 Pythagoras’ Theorem, which

states that the square of the length               c  of the hypotenuse of a right-angle triangle

equals the sum of the squares of the lengths a and b of the other two sides. A simple

geometric proof of this important result is illustrated in Figure 1.2. Four identical

 

 

 

 

b

 

 

 

c

a


 

 

a


 

 

b


 

Figure 1.2: Pythagoras’ Theorem

 

 

copies of the triangle, each with area ab/2, are placed around a square of side c, so as

to form a larger square with side a + b. The area c2of the inner square is then just

the area (a + b)2= a2+ 2ab + b2of the large square minus the total area 2ab of the

four triangles. That is, c2= a2+ b2.

Consider now the following problem. Suppose you draw a right-angle triangle

having two sides of length one.

 

 

x


1


 

 

 

1


 

The Greek mathematicians of antiquity noticed that the length of the hypotenuse of

such a triangle cannot possibly be a rational number; that is, it cannot be expressed

as the ratio of two integers. Let us denote the length of the hypotenuse by x. From

Pythagoras’ Theorem, we know that x2= 12+ 12= 2. Suppose that we could indeed

 

write   x  =   P/Q, where     P   and   Q  are integers (with      Q6= 0). By cancelling out any

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