Analysis with an introduction to proof 5th by steven lay ch02e

The objects in a set are called elements or members of the set.. We do not have a formal definition of the concept “set,” but we use the informal understanding that a set is a collection

Chapter 2

Sets and Functions

Section 2.1 Basic Set Operations

That is, we distinguish between the element 5 and the set that contains 5 as its only member

The objects in a set are called elements or members of the set

We speak of a football team, a flock of geese, or a finance committee The idea of a set or collection of things is common in our everyday experience

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

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}.

It is customary to set off the rule within braces, as in

C = {x : x is prime}.

For an infinite set we cannot list all the members, so a defining rule must be given

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

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.

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

In constructing examples of sets it is often helpful to indicate a larger set from which

{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},

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},

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.

To prove that ∅⊆ A, we must establish that the implication

if x ∈∅, then x ∈ A

is true

Example 2.1.5*

Let A = {1, 3}, B = {1, 3, 5}, and C = {x ∈ : x2 = – 1}.

Determine whether each statement is true or false

No So all the elements of C are contained in A.

*Similar to Example 2.1.5 in the text.

In fact, set C contains no members It is an example of the empty set

Theorem 2.1.7 Let A be a set Then ∅⊆ A.

Proof:

Since ∅ has no members, the antecedent “x ∈∅” is false for all x.

Thus, according to our definition of implies, the implication is always true ♦

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)

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 2.1.10

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}.

A ∩ B = {2, 4}

A  \ B = {1, 3}

U \B = {1, 3, 5}.

If you toss the elements of A and B into the same bag, this is what you get.

These are the elements that A and B have in common.

If you start with A and throw out anything that’s in B, this is what’s left.

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.

One way to visualize set operations is by use of Venn diagrams as shown below.

U

The rectangle represents the universal set U.

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.

Theorem 2.1.13

Let A, B, and C be subsets of a universal set U Then the following

statements are true.

(c) U \(U \ A) = A

(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.

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)

Theorem 2.1.13 (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)

Theorem 2.1.13 (d)

A ∪ (B ∩ C ) = (A ∪ B) ∩ (A ∪ C ).

Comments on the proof of Theorem 2.1.13 (d).

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

Comments on the proof of Theorem 2.1.13 (d).

This type of division of the argument is necessary when dealing with unions

the same desired result (or to a contradiction, which would show that only one alternative could occur).

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.

and = {x: x ∈ B for all B ∈B  }.

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 2.1.15

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

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

j

j J ∈ A

=

j

j J ∈ A

=

j ∞ = A

B

B

∈

U

B

B

B

∈ I

B

Example 2.1.17* *Similar to Example 2.1.17 in the text.

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

1 k

k ∞ = A

[ ]

3

U

I