10 1.1 Basic Concepts of Set Theory (b) Here we let the index set be J =(0,1] and for eachs ∈J we have As =(−s,s). Then [s∈J As =(−1,1) and \ s∈J As ={0}. The proofs of the following properties are similar to those in Theorem 1.1.2. We include the proof of part (i) and leave the rest as an exercise. Theorem 1.1.3 Let {Ai : i ∈I} be an indexed family of subsets of a universal set X and let Bbe a subset of X. Then the following hold: (i) B∪ Ti ∈I Ai =Ti ∈I (B∪Ai). (ii) B∩ Si ∈I Ai =Si ∈I (B∩Ai). (iii) B\ Ti ∈I Ai =Si ∈I (B\Ai). (iv) B\ Si ∈I Ai =Ti ∈I (B\Ai). (v) Ti ∈I Ai c =Si ∈I Ac i . (vi) Si ∈I Ai c =Ti ∈I Ac i . Proof of (i): Let x ∈B∪ Ti ∈I Ai . Thenx ∈Bor x ∈Ti ∈I Ai. If x ∈B, thenx ∈B∪Ai for all i ∈I and, thus, x ∈Ti ∈I (B∪Ai). If x ∈Ti ∈I Ai, then x ∈Ai for all i ∈I. Therefore, x ∈B∪Ai for all i ∈I and, hence, x ∈Ti ∈I (B∪Ai). We have thus showedB∪ Ti ∈I Ai ⊂Ti ∈I (B∪Ai). Now let x ∈Ti ∈I (B∪Ai). Then x ∈B∪Ai for all i ∈I. If x ∈B, then x ∈B∪ Ti ∈I Ai . If x̸ ∈B, then we must have that x∈Ai for all i ∈I. Therefore, x∈Ti ∈I Ai and, hence, x∈B∪ Ti ∈I Ai . 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, wewill 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 AandB, the Cartesian product of AandBis the set defined by A×B:={(a,b) : a∈Aandb∈B}. ■ Example 1.1.2 If A={1,2}andB={−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. 1For a precise definition of ordered pair in terms of sets see [Lay13]
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