1. Chapter 2
Sets and Functions

2. Section 2.1
Basic Set Operations

3. The idea of a set or collection of things is common in our everyday experience.
We speak of a football team, a flock of geese, or a finance committee.
We do not have a formal definition of the concept “set,” but we use the
informal understanding that
a set is a collection of objects characterized by some defining
property that allows us to think of the objects as a whole.
The objects in a set are called elements or members of the set.
Notation Meaning
a  S Object a is an element of set S.
a  S Object a is not an element of set S.
To define a particular set, we have to indicate the property that characterizes its elements.
For a finite set, this can be done by listing its members.
For example, if set A consists of the elements 1, 2, and 3, then we write A = {1, 2, 3}.
If B consists of just one member, say 5, then we write B = {5}.
That is, we distinguish
between the element 5 and the set that contains 5 as its only member.

4. For an infinite set we cannot list all the members, so a defining rule must be given.
It is customary to set off the rule within braces, as in
C = {x : x is prime}.
Read, “C is the set of all x such that x is prime.”
Definition 2.1.3
Let A and B be sets. We say that A is a subset of B (or A is contained in B) if
every element of A is an element of B, and we denote this by writing A  B.
If A is a subset of B and there exists an element in B that is not in A, then A is called
a proper subset of B.
This definition tells us what we must do if we want to prove A  B.
We must show that “if x  A, then x  B” is a true statement.
That is, we must show that each element of A satisfies the defining condition
that characterizes B.

5. Definition 2.1.4
Let A and B be sets. We say that A is equal to B, written A = B, if A  B and B  A.
When this definition is combined with the definition of subset, we see that
proving A = B is equivalent to proving
x  A  x  B and x  B  x  A.
Note: in describing a set, the order in which the elements appear does not matter,
nor does the number of times they are written.
So the following sets are all equal:
{1, 2, 3, 4} = {2, 4, 1, 3} = {1, 2, 3, 2, 4, 2}.
Although we cannot give a formal definition of them now, it is convenient to name
the following sets:
will denote the set of all positive integers (or natural numbers).
will denote the set of all rational numbers.
will denote the set of all real numbers.

6. the elements are being chosen. Sometimes we abbreviate the notation.
In constructing examples of sets it is often helpful to indicate a larger set from which
the elements are being chosen.
Sometimes we abbreviate the notation.
{x : x  and 0 < x < 1} becomes {x  : 0 < x < 1}
Read, “The set of all x in such that 0 < x < 1.”
There is a standard notation that we use for interval subsets of the real numbers:
[a, b] = {x    : a  x  b}, (a, b) = {x   : a < x < b},
[a, b) = {x    : a  x < b}, (a, b] = {x    : a < x  b}.
We use a square bracket if the endpoint is included
and a round parenthesis if the endpoint is not included.
The set [a, b] is called a closed interval and the set (a, b) is called an open interval.
We also have occasion to refer to the unbounded intervals:
[a, ) = {x    : x  a}, (a, ) = {x    : x > a},
( , b] = {x   : x  b}, ( , b) = {x    : x < b}.
At this time no special significance should be attached to the symbols “  ” and “ –  ”
as in [a, ) and (– , b].
They simply indicate that the interval contains all real numbers
greater than or equal to a, or less than or equal to b, as the case may be.

7. Example 2.1.5*
*Similar to Example in the text.
Let A = {1, 3}, B = {1, 3, 5}, and C = {x  : x2 = – 1}.
Determine whether each statement is true or false.
3  B
True
3 is one of the elements listed in B.
3  C
False
32  – 1.
A  B
True
Every element of A is an element of B.
B  A
False
5  B, but 5  A.
C  A
True
Does C contain any elements that are not in A?
No. So all the elements of C are contained in A.
In fact, set C contains no members. It is an example of the empty set.
We denote the empty set by the symbol  .
Theorem 2.1.7
Let A be a set. Then   A.
Proof:
To prove that   A, we must establish that the implication
if x  , then x  A
is true.
Since  has no members, the antecedent “x  ” is false for all x.
Thus, according to our definition of implies, the implication is always true. 

8. There are three basic ways to form new sets from existing sets.
Definition 2.1.8
Let A and B be sets. The union of A and B (denoted A  B), the intersection of A
and B (denoted A  B), and the complement of B in A (denoted A \ B) are given by
A  B = {x : x  A or x  B}
A  B = {x : x  A and x  B}
A \ B = {x : x  A and x  B}
If A  B = , then A and B are said to be disjoint.
These three set operations given above correspond in a natural way to three of the
basic logical connectives:
x  A  B iff (x  A)  (x  B)
x  A  B iff (x  A)  (x  B)
x  A \ B iff (x  A)  ~ (x  B).

9. Mathematical concepts and proofs always occur within the context of some mathematical
system.
It is customary for the elements of the system to be called the universal set.
Then any set under consideration is a subset of this universal set.
Example
Let A = {1, 2, 3, 4} and B = {2, 4, 6} be subsets of the universal set U = {1, 2, 3, 4, 5, 6}.
Then A  B = {1, 2, 3, 4, 6}.
If you toss the elements of A and B into the same bag, this is what you get.
A  B = {2, 4}
These are the elements that A and B have in common.
A \ B = {1, 3}
If you start with A and throw out anything that’s in B, this is what’s left.
U \B = {1, 3, 5}.
If you start with all of U and throw out anything that’s in B, this is what’s left.
Note: The complement of B in the universal set U, namely U \B, is sometimes called
the complement of B.

10. One way to visualize set operations is by use of Venn diagrams as shown below.
The rectangle represents the universal set U.
Set A is the blue circle on the left.
Set B is the yellow circle on the right.
If we color both circles, the total colored area is the union: A  B.
And the green area where they overlap is the intersection: A  B. U A B

11. Theorem
Let A, B, and C be subsets of a universal set U. Then the following
statements are true.
(a) A  (U \ A) = U
(b) A  (U \ A) = 
(c) U \(U \ A) = A
(d) A  (B  C ) = (A  B)  (A  C )
(e) A  (B  C ) = (A  B)  (A  C )
(f ) A \(B  C ) = (A \ B)  (A \ C)
(g) A \(B  C ) = (A \ B)  (A \ C)
The proofs of most of these are left as exercises, but we will do part (d) to
illustrate the process.

12. Proof: We begin by showing that A  (B  C )  (A  B)  (A  C ).
Theorem (d)
A  (B  C ) = (A  B)  (A  C ).
Proof: We begin by showing that
A  (B  C )  (A  B)  (A  C ).
If x  _____________, then either x  A or x  B  C.
If x  A, then certainly x  A  B and x  A  C. Thus x  __________________. On the other hand, if _____________, then x  B and x  C. But this implies that
x  A  B and _____________, so x  (A  B)  (A  C ). Hence A  (B  C )  (A  B)  (A  C ).
A  (B  C )
(A  B)  (A  C )
x  (B  C)
x  (A  C)

13. Theorem (d)
A  (B  C ) = (A  B)  (A  C ).
Conversely, if y  (A  B)  (A  C ), then ____________ and ___________. There are two cases to consider: when y  A and when y  A. If y  A, then y  A  (B  C ) and this part is done. On the other hand, if ___________ , then since y  A  B, we must have y  B. Similarly, since y  A  C and y  A, we have ___________. Thus ______________, and this implies that y  A  (B  C ). Hence (A  B)  (A  C )  A  (B  C ). 
y  (A  B)
y  (A  C)
y  A
y  C
y  (B  C)

14. Comments on the proof of Theorem 2.1.13 (d).
Notice how the argument divides naturally into parts, the second part being introduced by the word “conversely.” This word is appropriate because the second half of the argument is indeed the converse of the first half.
2. In the first part the point in A  (B  C ) was called x and in the second part the point in (A  B)  (A  C ) was called y. Why is this?
The choice of a name is completely arbitrary, and in fact the same name could have been used in both parts.
It is important to realize that the two parts are separate arguments; we start over from scratch in proving the converse and can use nothing that was derived about the point x in the first part.
By using different names for the points in the two parts we emphasize this separateness.
It is common practice, however, to use the same name (such as x) for the arbitrary point in both parts.

15. Comments on the proof of Theorem 2.1.13 (d).
Notice that each half of the argument also has two parts or cases, the second case being introduced by the phrase “on the other hand.”
This type of division of the argument is necessary when dealing with unions.
If x  S  T, then x  S or x  T. Each of the possibilities must be followed to its logical conclusion, and both “bridges” must lead to the same desired result (or to a contradiction, which would show that only one alternative could occur).
When proving that one set, say S, is a subset of another set, say T, it is common to begin with the phrase “If x  S, then…”
It is also acceptable to begin with “Let x  S ” and then conclude that x  T.
The subtle difference between these phrases is that “Let x  S ” assumes that S is nonempty, so there is an x in S to choose.
This might seem to be an unwarranted assumption, but really it is not. If S is the empty set, then of course S  T, so the only nontrivial case to prove is when S is nonempty.

16. Sometimes we wish to form unions or intersections of more than 2 or 3 sets. To do this
we need to extend our previous definitions.
Definition
If for each element j in a nonempty set J there corresponds a set Aj, then A = {Aj : j  J }
is called an indexed family of sets with J as the index set.
The union of all the sets in A is defined by
= {x: x  Aj for some j  J }
and the intersection is = {x: x  Aj for all j  J }.
If J = , we write and
In general, if B is a nonempty collection of sets, then we let
= {x: x  B for some B  B  }
and = {x: x  B for all B  B  }.

17. Example 2.1.17* For each k  N, let Ak = [1/k, 3 – 1/k]. Find and
*Similar to Example in the text.
For each k  N, let Ak = [1/k, 3 – 1/k].
Find and
We have
etc.
To find the union of all these sets, we note that the left endpoint is getting closer and
closer to 0, and the right endpoint is approaching (but never reaches) 3.
So,
To find the intersection of the sets, we ask, “What numbers are in all of the sets?”
We find