We defined the notation s ∈ S to mean “s belongs to S” or “s is an element of S.” If S and T are two sets, we shall write T ⊆ S to mean that every member of T is also a member of S. In other words, “If s is any element of T then s is a member of S,” or

                                        s ∈ T ⇒ s ∈ S,

where ⇒ is shorthand for implies. When T ⊆ S we say T is subset of S. Sets S and T are equal, S = T, if and only if S ⊆ T and T ⊆ S are both true. If necessary, we can represent the situation where T is a subset of S but S is not equal to T—there is at least one member of S that is not a member of T—by writing S ⊂ T, and we call T a proper subset of S.

Suppose R ⊆ S and S ⊆ T are both true. Any member of R will also be a member of S, which means it is a member of T. So R ⊆ T. This sort of rule is called a transitive law.

It is important not to confuse the two symbols ∈ and ⊆, or their meanings:

Sample Problem 1.1 Suppose S = {0,1}. Which of the following are true:

(i) 0 ∈ S, {0} ∈ S, 0 ⊂ S,

(ii) {0} ⊂ S, 0 ⊆ S, {0} ⊆ S, S ∈ S,

(iii) S ⊂ S, S ⊆ S?

Solution.

(i) 0 is a member of S, but {0} and S are not, so 0 ∈ S is true but {0} ∈ S, and S ∈ S are false.

(ii) As 0 is a member of S, {0} ⊂ S and {0} ⊆ S are true. But 0 is not a set ofelements of S, so 0 ⊂ S and 0 ⊆ S are false.

(iii) S ⊆ S is true, but S ⊂ S would imply S ≠S, so it is false.

Among the standard number sets, many subset relationships exist. Every natural number is an integer, every integer is a rational number, and every rational number is a real number, so N ⊆ Z, Z ⊆ Q, Q ⊆ R. We could write all these relationships down in one expression:

                                                           N ⊆ Z ⊆ Q ⊆ R.

In fact, we know that no two of these sets are equal, so we could write

                                                                N ⊂ Z ⊂ Q ⊂ R.

Given sets S and T, we define two operations: the union of S and T is the set

                                                   S∪T = {x : x ∈ S or x ∈ T (or both)};  

the intersection of S and T is the set

                                              S∩T = {x : x ∈ S and x ∈ T}.

As a kind of opposite to the union, the notation ST denotes the set of all members of S that are not in T.

There is also a special relationship between subsets and the other operations. If S is any subset of T, then S∩T = S and S∪T = T.

Suppose two sets, S and T, have no common element. Then S and T are called disjoint. In that case, S∩T is a set with no elements! There is no problem with the concept of such a set. We shall define the empty set, also called the null set, to be a set that has no elements. This set is denoted 0/. The set 0/ is unique and is a subset of every other set. Then “S and T are disjoint” means S∩T = 0/.

Given sets S and T, the notation ST is used for the set formed by deleting from S all the members that are also in T. Clearly ST is the same as S(S∩T). If S and T are disjoint, then ST = S, while SS = S.

Finally, we can combine two sets S and T to form a new set called the Cartesian product S×T. This consists of all the ordered pairs with the first element a member of S and the second a member of T. For example, If S = {1,3} and T = {2,3,4} then

                                      S×T = {(1,2),(1,3),(1,4),(3,2),(3,3),(3,4)}.

In the following example, remember that a perfect square means a number of the form n², where n is an integer.

Sample Problem 1.2 In each case, are the sets S and T disjoint? If not, what is their intersection?

(i) S is the set of perfect squares, T = RR+.

(ii) S is the set of all multiples of 5, T is the set of all multiples of 7. Solution.

(i) They are not disjoint, because 0 is a perfect square (0 = 02);  S∩T = {0}.

مواضيع ذات صلة

Subset

Set Class

Set

Second Category

Proper Subset

Power Set

Matroid

Inductive Set

Independent Vertex Set

Independent Set

Graphoid

G_delta Set