MATH 220 Homework 7 Solutions

Exercise 3.1.15

Let r ∈ 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 ∈ 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 ∈ U Let P be the statement “x ∈ A”, and let Q be the statement “x ∈ B”. Then x ∈ (A ∪ B) is equivalent to ¬(P ∨ Q), and by De Morgan’s law this is equivalent to (¬P ) ∧ (¬Q), which is equivalent to x ∈ 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 ∈ B such that x∈ / A. But then x ∈ A and x ∈ 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 ∈ N, let Ai = (−i, i).

S∞ T∞ (a) Find i=1 Ai and i=1 Ai.

(b) Prove your answers to part (a) are correct.

Solution:

S∞ T∞ (a) i=1 Ai = R and i=1 Ai = (−1, 1).

S∞ S∞ (b) If x ∈ i=1 Ai, then x ∈ Ai for some i. Since Ai ⊆ R, x ∈ R. Since x ∈ i=1 Ai was arbitrary, S∞ it follows that i=1 Ai ⊆ R. If x ∈ R, then there is some i ∈ N such that |x| < i. Thus S∞ S∞ x ∈ (−i, i) ⊆ i=1 Ai, and therefore R = i=1 Ai. T∞ If x ∈ (−1, 1), then x ∈ (−i, i) for all i ∈ N and thus (−1, 1) ⊆ i=1 Ai. If x∈ / (−1, 1), then T∞ x∈ / A1, and thus x∈ / i=1 Ai.

Exercise 4.3.4

 1  For i ∈ N, let Ai = 0, 1 − i .

2 (a) Find S A and T A . i∈N i i∈N i

(b) Prove your answers to part (a) are correct.

Solution:

(a) S A = [0, 1) and T A = {0}. i∈N i i∈N i

(b) If x ∈ S A , then x ∈ A for some j ∈ . Thus 0 ≤ x ≤ 1 − 1 < 1, and therefore x ∈ [0, 1). i∈N i j N j 1 If now x ∈ [0, 1), then 0 ≤ x < 1. Thus 0 < 1 − x so there is some j ∈ N such that j ≤ 1 − x. Thus 0 ≤ x ≤ 1 − 1 , and therefore x ∈ A ⊆ S A . j j i∈N i Now suppose x ∈ T A . Then in particular x ∈ A = [0, 1 − 1] = {0}, and therefore i∈N i 1 h  T A ⊆ {0}. Conversely, 0 ∈ 0, 1 − 1 = A for all j ∈ , and therefore {0} ⊆ T A . i∈N i j j N i∈N i

Exercise 4.3.9(a)

Let I be a nonempty set and let {Ai : i ∈ I} be an indexed family of sets. Let X be a set. Suppose T that for all i ∈ I, X ⊆ Ai. Prove that X ⊆ i∈I Ai.

Solution: Suppose x ∈ X, and let i ∈ I be given. Since X ⊆ Ai, x ∈ Ai. Since i ∈ I was T T arbitrary, x ∈ Ai for all i, thus x ∈ i∈I Ai, and therefore X ⊆ i∈I Ai.

Exercise 4.3.10(b)

Let {Ai : i ∈ N} be an indexed family of sets. Assume that for all i ∈ N, Ai+1 ⊆ Ai. Prove that S A = A . i∈N i 1 Solution: If x ∈ A1, then there is some i ∈ N such that x ∈ Ai (namely, i = 1), and thus x ∈ S A . i∈N i Next, we show that Ai ⊆ A1 for all i ∈ 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 ∈ N. Now assume x ∈ S A . Then there is some n ∈ such that x ∈ A . Since A ⊆ A , it i∈N i N n n 1 follows that x ∈ A1.

Lecture Notes Exercise 10

∞ 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 ∈ 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 ∈ N.

Lecture Notes Theorem 13(2)

T S Let (Xi)i∈I be a collection of sets. Then i∈I Xi = i∈I Xi.

Solution: Suppose each Xi is a subset of the universal set U. Fix x ∈ U. For each i ∈ I let T P (i) be the statement “x ∈ Xi.” By definition, the statement x ∈ i∈I Xi means ¬[(∀i ∈ I)P (i)] S which is logically equivalent to (∃i ∈ I)(¬P (i)), which by definition means that x ∈ i∈I Xi. Thus T S T S x ∈ i∈I Xi if and only if x ∈ i∈I Xi, and therefore i∈I Xi = i∈I Xi.

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