Introduction to Mathematical Analysis I - Second Edition

10 1.1 BASIC CONCEPTS OF SET THEORY The proofs of the following properties are similar to those in Theorem 1.1.2 . We include the proof of part (a) and leave the rest as an exercise. Theorem 1.1.3 Let { A i : i ∈ I } be an indexed family of subsets of a universal set X and let B be a subset of X . Then the following hold: (a) B ∪ T i ∈ I A i = T i ∈ I B ∪ A i ; (b) B ∩ S i ∈ I A i = S i ∈ I B ∩ A i ; (c) B \ T i ∈ I A i = S i ∈ I B \ A i ; (d) B \ S i ∈ I A i = T i ∈ I B \ A i ; (e) T i ∈ I A i c = S i ∈ I A c ; (f) S i ∈ I A i c = T i ∈ I A c . Proof of (a) : Let x ∈ B ∪ T i ∈ I A i . Then x ∈ B or x ∈ T i ∈ I A i . If x ∈ B , then x ∈ B ∪ A i for all i ∈ I and, thus, x ∈ T i ∈ I B ∪ A i . If x ∈ T i ∈ I A i , then x ∈ A i for all i ∈ I . Therefore, x ∈ B ∪ A i for all i ∈ I and, hence, x ∈ T i ∈ I B ∪ A i . We have thus showed B ∪ T i ∈ I A i ⊂ T i ∈ I B ∪ A i . Now let x ∈ T i ∈ I B ∪ A i . Then x ∈ B ∪ A i for all i ∈ I . If x ∈ B , then x ∈ B ∪ T i ∈ I A i . If x 6∈ B , then we must have that x ∈ A i for all i ∈ I . Therefore, x ∈ T i ∈ I A i and, hence, x ∈ B ∪ T i ∈ I A i . This proves the other inclusion and, so, the equality. We want to consider pairs of objects in which the order matters. Given objects a and b , we will denote by ( a , b ) the ordered pair where a is the first element and b is the second element. The main characteristic of ordered pairs is that ( a , b ) = ( c , d ) if and only if a = c and b = d . Thus, the ordered pair ( 0 , 1 ) represents a different object than the pair ( 1 , 0 ) (while the set { 0 , 1 } is the same as the set { 1 , 0 } ) 1 . Given two sets A and B , the Cartesian product of A and B is the set defined by A × B : = { ( a , b ) : a ∈ A and b ∈ B } . Example 1.1.2 If A = { 1 , 2 } and B = {− 2 , 0 , 1 } , then A × B = { ( 1 , − 2 ) , ( 1 , 0 ) , ( 1 , 1 ) , ( 2 , − 2 ) , ( 2 , 0 ) , ( 2 , 1 ) } . Example 1.1.3 If A and B are the intervals [ − 1 , 2 ] and [ 0 , 7 ] respectively, then A × B is the rectangle [ − 1 , 2 ] × [ 0 , 7 ] = { ( x , y ) : − 1 ≤ x ≤ 2 , 0 ≤ y ≤ 7 } . We will make use of cartesian products in the next section when we discuss functions. Exercises 1.1.1 Prove the remaining items in Theorem 1.1.2 . 1.1.2 I Let Y and Z be subsets of X . Prove that ( X \ Y ) ∩ Z = Z \ ( Y ∩ Z ) . 1 For a precise definition of ordered pair in terms of sets see [ Lay13 ]