MATH 220 Homework 7 Solutions Exercise 3.1.15 Let r 2 R with r 6= 1. Prove that n−1 X 1 − rn rk = : 1 − r k=0 P1−1 k 1−r1 Pm−1 k Solution: Fix r 2 R with r 6= 1. Note that k=0 r = 1 = 1−r . Now assume that k=0 r = 1−rm 1−r for some positive integer n. Then we have m m−1 X X 1 − r 1 − rm 1 − rm+1 rk = rm + rk = rm + = : 1 − r 1 − r 1 − r k=0 k=0 Pn−1 k 1−rn Thus, by induction it follows that k=0 r = 1−r is true for all positive integers n. Exercise 4.2.1(g) Let A and B be subsets of some universal set U. Prove that (A [ B) = A \ B. Solution: Suppose x 2 U Let P be the statement \x 2 A", and let Q be the statement \x 2 B". Then x 2 (A [ B) is equivalent to :(P _ Q), and by De Morgan's law this is equivalent to (:P ) ^ (:Q), which is equivalent to x 2 A \ B. Therefore, we have (A [ B) = A \ B. Exercise 4.2.2 Let A and B be sets, where U is the underlying universal set. Prove that A ⊆ B , B ⊆ A: Solution: First assume that A ⊆ B. If B is not a subset of A, then there is some x 2 B such that x2 = A. But then x 2 A and x 2 B, a contradiction since A ⊆ B. Thus B ⊆ A. Conversely, assume that B ⊆ A. Then by the preceding argument, we have A ⊆ B. Since A = A and B = B, we have A ⊆ B. 1 Exercise 4.2.5(e) Let A and B be sets. Prove that (A − B) [ (B − A) = (A [ B) − (A \ B). Solution: Using the identities found in section Section 4.2, this can be done with a few manip- ulations: (A − B) [ (B − A) = (A \ B) [ (B \ A) = (A [ (B \ A)) \ (B [ (B \ A)) = [(A [ B) \ (A [ A)] \ [(B [ B) \ (B [ A)] = (A [ B) \ (B [ A) = (A [ B) \ (A \ B) = (A [ B) − (A \ B): You could also give a a double-inclusion proof. Exercise 4.3.1 For i 2 N, let Ai = (−i; i). S1 T1 (a) Find i=1 Ai and i=1 Ai. (b) Prove your answers to part (a) are correct. Solution: S1 T1 (a) i=1 Ai = R and i=1 Ai = (−1; 1). S1 S1 (b) If x 2 i=1 Ai, then x 2 Ai for some i. Since Ai ⊆ R, x 2 R. Since x 2 i=1 Ai was arbitrary, S1 it follows that i=1 Ai ⊆ R. If x 2 R, then there is some i 2 N such that jxj < i. Thus S1 S1 x 2 (−i; i) ⊆ i=1 Ai, and therefore R = i=1 Ai. T1 If x 2 (−1; 1), then x 2 (−i; i) for all i 2 N and thus (−1; 1) ⊆ i=1 Ai. If x2 = (−1; 1), then T1 x2 = A1, and thus x2 = i=1 Ai. Exercise 4.3.4 1 For i 2 N, let Ai = 0; 1 − i . 2 (a) Find S A and T A . i2N i i2N i (b) Prove your answers to part (a) are correct. Solution: (a) S A = [0; 1) and T A = f0g. i2N i i2N i (b) If x 2 S A , then x 2 A for some j 2 . Thus 0 ≤ x ≤ 1 − 1 < 1, and therefore x 2 [0; 1). i2N i j N j 1 If now x 2 [0; 1), then 0 ≤ x < 1. Thus 0 < 1 − x so there is some j 2 N such that j ≤ 1 − x. Thus 0 ≤ x ≤ 1 − 1 , and therefore x 2 A ⊆ S A . j j i2N i Now suppose x 2 T A . Then in particular x 2 A = [0; 1 − 1] = f0g, and therefore i2N i 1 h T A ⊆ f0g. Conversely, 0 2 0; 1 − 1 = A for all j 2 , and therefore f0g ⊆ T A . i2N i j j N i2N i Exercise 4.3.9(a) Let I be a nonempty set and let fAi : i 2 Ig be an indexed family of sets. Let X be a set. Suppose T that for all i 2 I, X ⊆ Ai. Prove that X ⊆ i2I Ai. Solution: Suppose x 2 X, and let i 2 I be given. Since X ⊆ Ai, x 2 Ai. Since i 2 I was T T arbitrary, x 2 Ai for all i, thus x 2 i2I Ai, and therefore X ⊆ i2I Ai. Exercise 4.3.10(b) Let fAi : i 2 Ng be an indexed family of sets. Assume that for all i 2 N, Ai+1 ⊆ Ai. Prove that S A = A . i2N i 1 Solution: If x 2 A1, then there is some i 2 N such that x 2 Ai (namely, i = 1), and thus x 2 S A . i2N i Next, we show that Ai ⊆ A1 for all i 2 N. Clearly A1 ⊆ A1. If now Am ⊆ A1, then Am+1 ⊆ Am ⊆ A1, so Am+1 ⊆ A1. Thus it follows by induction that An ⊆ A1 for all n 2 N. Now assume x 2 S A . Then there is some n 2 such that x 2 A . Since A ⊆ A , it i2N i N n n 1 follows that x 2 A1. Lecture Notes Exercise 10 1 Consider the sequence (an)n=1 recursively defined as a1 = 2, a2 = 4 and for all n ≥ 3, an = 3an−1 − 2an−2. For all n ≥ 1, find a closed formula for an. 3 n Solution: After performing a few calculations, it should be clear that an = 2 . We prove by induction that this is the correct formula. 1 2 Clearly a1 = 2 = 2 and a2 = 4 = 2 . Now suppose that for some m 2 N with m ≥ 2, we have k ak = 2 whenever 1 ≤ k ≤ m. Then we have m m−1 m m m+1 am+1 = 3am − 2am−2 = 3 · 2 − 2 · 2 = 3 · 2 − 2 = 2 : n Thus, by induction it follows that an = 2 for all n 2 N. Lecture Notes Theorem 13(2) T S Let (Xi)i2I be a collection of sets. Then i2I Xi = i2I Xi. Solution: Suppose each Xi is a subset of the universal set U. Fix x 2 U. For each i 2 I let T P (i) be the statement \x 2 Xi." By definition, the statement x 2 i2I Xi means :[(8i 2 I)P (i)] S which is logically equivalent to (9i 2 I)(:P (i)), which by definition means that x 2 i2I Xi. Thus T S T S x 2 i2I Xi if and only if x 2 i2I Xi, and therefore i2I Xi = i2I Xi. 4.