9 Theorem 1.1.2 Let A,B, and C be subsets of a universal set X. Then the following hold: (i) A∪ Ac = X. (ii) A∩ Ac = 0/ . (iii) (Ac)c = A. (iv) (Distributive law) A∩ (B ∪C)=(A ∩ B) ∪ (A∩C). (v) (Distributive law) A∪ (B ∩C)=(A ∪ B) ∩ (A∪C). (vi) (DeMorgan’s law) A\ (B ∪C)=(A \ B) ∩ (A\C). (vii) (DeMorgan’s law) A \ (B ∩C)=(A \ B) ∪ (A\C). (viii) A \ B = A ∩ Bc . Proof: We prove some of the results and leave the rest for the exercises. (i) Clearly, A ∪ Ac ⊂ X since both A and Ac 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 frst case, x ∈ A and, so, x ∈ A ∪ Ac . In the second case, x ∈ Ac and, so, x ∈ A ∪ Ac . Thus, X ⊂ A∪ Ac . Applying Theorem (1.1.1) it follows that A ∪ Ac = X. (ii) No element x can be simultaneously in A and not in A. Thus, A∩ Ac = 0/ . (iv) Let x ∈ A ∩ (B ∪C). Then x ∈ A and x ∈ B ∪ C. Therefore, x ∈ B or x ∈ C. In the frst 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 frst 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). Applying Theorem (1.1.1) the equality follows. □ 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 Ai. Then the family of all sets Ai as i ranges over I is denoted by {Ai : 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 {Ai : i ∈ I}. The union and intersection of this family as i ranges over I is defned respectively by [Ai = {x : x ∈ Ai for some i ∈ I} i∈I and \ Ai = {x : x ∈ Ai for every i ∈ I}. 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 An =[−n,n]. Then [ \ An = R and An =[−1,1]. n∈N n∈N
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