Introduction to Mathematical Analysis I - Second Edition

9 (e) (Distributive law) A ∪ ( B ∩ C ) = ( A ∪ B ) ∩ ( A ∪ C ) ; (f) (DeMorgan’s law) A \ ( B ∪ C ) = ( A \ B ) ∩ ( A \ C ) ; (g) (DeMorgan’s law) A \ ( B ∩ C ) = ( A \ B ) ∪ ( A \ C ) ; (h) A \ B = A ∩ B c . Proof: We prove some of the results and leave the rest for the exercises. (a) Clearly, A ∪ A c ⊂ X since both A and A c are subsets of X . Now let x ∈ X . Then either x is an element of A or it is not an element of A . In the first case, x ∈ A and, so, x ∈ A ∪ A c . In the second case, x ∈ A c and, so, x ∈ A ∪ A c . Thus, X ⊂ A ∪ A c . (b) No element of x can be simultaneously in A and not in A . Thus, A ∩ A c = /0. (d) Let x ∈ A ∩ ( B ∪ C ) . Then x ∈ A and x ∈ B ∪ C . Therefore, x ∈ B or x ∈ C . In the first case, since x is also in A we get x ∈ A ∩ B and, hence, x ∈ ( A ∩ B ) ∪ ( A ∩ C ) . In the second case, x ∈ A ∩ C and, hence, x ∈ ( A ∩ B ) ∪ ( A ∩ C ) . Thus, in all cases, x ∈ ( A ∩ B ) ∪ ( A ∩ C ) . This shows A ∩ ( B ∪ C ) ⊂ ( A ∩ B ) ∪ ( A ∩ C ) . Now we prove the other inclusion. Let x ∈ ( A ∩ B ) ∪ ( A ∩ C ) . Then x ∈ A ∩ B or x ∈ A ∩ C . In either case, x ∈ A . In the first case, x ∈ B and, hence, x ∈ B ∪ C . It follows in this case that x ∈ A ∩ ( B ∪ C ) . In the second case, x ∈ C and, hence, x ∈ B ∪ C . Again, we conclude x ∈ A ∩ ( B ∪ C ) . Therefore, ( A ∩ B ) ∪ ( A ∩ C ) ⊂ A ∩ ( B ∪ C ) as desired. A set whose elements are sets is often called a collection/family of sets and is often denoted by script letters such as A or B . Let I be a nonempty set such that to each i ∈ I corresponds a set A i . Then the family of all sets A i as i ranges over I is denoted by { A i : i ∈ I } . Such a family of sets is called an indexed family and the set I is called the index set . Consider the indexed family of sets { A i : i ∈ I } . The union and intersection of this family as i ranges over I is defined respectively by [ i ∈ I A i = { x : x ∈ A i for some i ∈ I } and \ i ∈ I A i = { x : x ∈ A i for every i ∈ I } . Example 1.1.1 The following examples illustrate the notation. (a) Let the index set be I = N and for each n ∈ N we have A n = [ − n , n ] . Then [ n ∈ N A n = R \ n ∈ N A n = [ − 1 , 1 ] . (b) Here we let the index set be J = ( 0 , 1 ] and for each α ∈ J we have A α = ( − α , α ) . Then [ α ∈ J A α = ( − 1 , 1 ) \ α ∈ J A α = { 0 } .