Formal MethodsKey to Homework Assignment 6, Part 3March 7, 2007• Restate theorems 2.8 and 2.9 (pp. 58 and 59) for the special case in which Λ = {1, 2}.You don’t need to prove anything.1. Theorem 2.8. Let A = {Aλ: λ ∈ Λ} be an indexed family of sets and let β ∈ Λ.Then(a) Aβ⊆ ∪λ∈ΛAλ, and(b) ∩λ∈ΛAλ⊆ Aβ.If Λ = {1, 2}, then β = 1 or β = 2, and(a) Aβ⊆ A1∪ A2, and(b) A1∩ A2⊆ Aβ.2. Theorem 2.9. Let A = {Aλ: λ ∈ Λ} be an indexed family of sets and let B be aset. Then(a) B ∩ (∪λ∈ΛAλ) = ∪λ∈Λ(B ∩ Aλ) .(b) B ∪ (∩λ∈ΛAλ) = ∩λ∈Λ(B ∪ Aλ) .(c) (∩λ∈ΛAλ)0= ∪λ∈ΛA0λ(d) (∪λ∈ΛAλ)0= ∩λ∈ΛA0λIf Λ = {1, 2}, then(a) B ∩ (A1∪ A2) = (B ∩ A1) ∪ (B ∩ A2) .(b) B ∪ (A1∩ A2) = (B ∪ A1) ∩ (B ∪ A2) .(c) (A1∩ A2)0= A01∪ A02.(d) (A1∪ A2)0= A01∩ A02.• What are the left- and right-hand sides of each part of theorem 2.9 in the case in whichΛ = ∅? You can assume that each Aλis a subset of some universal set U. You don’tneed to prove anything.Recall that if Λ = ∅, we defined∪λ∈ΛAλ= ∅, and ∩λ∈ΛAλ= U.So in Theorem 2.9 we have11. B ∩ (∪λ∈ΛAλ) = B ∩ ∅ = ∅. Furthermore, the family {B ∩ Aλ: λ ∈ Λ} is alsoempty. So ∪λ∈Λ(B ∩ Aλ) = ∅.2. B ∪ (∩λ∈ΛAλ) = B ∪ U = U. We also have that the family {B ∪ Aλ: λ ∈ Λ} isempty. So ∩λ∈Λ(B ∪ Aλ) = U.3. (∩λ∈ΛAλ)0= U0= ∅. Since the family {A0λ: λ ∈ Λ} is empty, we also have∪λ∈ΛA0λ= ∅.4. (∪λ∈ΛAλ)0= ∅0= U, and ∩λ∈ΛA0λ= U.33.(d) Prove Theorem 2.6(d), which says that if A and B are subsets of some universal set U,then A − B = A ∩ B0.Proof.x ∈ A − B ⇔ (x ∈ A) ∧ (x 6∈ B)⇔ (x ∈ A) ∧ (x ∈ B0)⇔ x ∈ A ∩ B038. Use Theorem 2.6(a) and (b) to prove Theorem 2.6(c).Parts (a), (b), and (c) of Theorem 2.6 say: Let A and B be subsets of some universalset U. Then(a) (A ∪ B)0= A0∩ B0(b) (A0)0= A(c) (A ∩ B)0= A0∪ B0Starting with the left-hand side of part (c) we haveA0∪ B0= [(A0∪ B0)0]0by part (b)= [(A0)0∩ (B0)0]0by part (a)= (A ∩ B)0by part (b), again.50. We give a “proof” and a “counterexample” to the following conjecture: “If A ⊆ B andC ⊆ D, then A ∩ C ⊆ B ∩ D.Which, if either, of these is correct?1. “Counterexample:” Let A = {1, 2}, B = {1, 2, 3, 4}, C = {{1}, {2}}, and D ={1, 2, 3, 4}.This is definitely not a counterexample. In the first place, C 6⊆ D. For example,{1} 6∈ D. In the second place, A∩C = ∅ and B∩D = {1, 2, 3, 4}. So A∩C ⊆ B∩D.2. “Proof:” Let x ∈ A ∩ C. Then x ∈ A and x ∈ C. Since A ⊆ B and C ⊆ D, x ∈ Band x ∈ D. Therefore x ∈ B ∩ D, so A ∩ C ⊆ B ∩ D.This is fine.257. (a) For each natural number n, let An= (n, n + 1). Find ∪n∈NAnand ∩n∈NAn.(a) The interval (n, n + 1) contains all of the real numbers strictly between thepositive integers n and n + 1. So any real number greater than or equal to 1that is not an integer will belong to ∪n∈NAn. Furthermore, no positive integerwill be in ∪n∈NAn, since no integer belongs to any An. So ∪n∈NAn= [1, ∞)−N.(b) If m and n are positive integers and m 6= n, then (m, m + 1) ∩ (n, n + 1) = ∅.So ∩n∈NAn= ∅.(b) For each real number x, let Ax= (x, x + 1). Find ∪x∈RAxand ∩x∈RAx.(a) Every real number y belongs to the interval (y−1/2, y+1/2). So ∪x∈RAx= R.(b) If y is any real number y 6∈ (y, y + 1). So y 6∈ ∩x∈RAxand ∩x∈RAx= ∅.63. Give an example of a collection of sets indexed by a set consisting of 5 members.There are a lot of possibilities here. T he only requirements are that Λ have 5 elementsand A have elements that are indexed by Λ.One possibility is Λ = {1, 2, 3, 4, 5} and Aλ= {x ∈ R : x > λ.}64. Give an example of a collection of sets indexed by the set of prime numbers.Once again there are a lot of possibilities. We must have Λ equal to the set of primenumbers. One possibility isAλ= {n ∈ N : λ|n}.72. Give an example of a family C = {Cn: n ∈ N} such that ∩C = ∅, but for each n ∈ N,∩{Ck: 1 ≤ k ≤ n} 6= ∅.We had such an example in the last homework set: number 54. Define Cn= {k ∈ N :k ≥ n}. Then ∩∞n=1Cn= ∅, because if m is any positive integer, m 6∈ Cm+1. However,if n is any positive integer ∩nm=1Cm6= ∅ because n ∈ Cmfor all m ≤ n, and hencen ∈ ∩nm=1Cm. (Note that if Λ = {1, 2, . . . , n}, then we often write ∩nm=1Amfor ∩λ∈ΛAλand ∪nm=1Amfor