Download Formal Methods Key to Homework Assignment 6, Part 3

Formal Methods
Key to Homework Assignment 6, Part 3
March 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∈Λ Aλ , and
(b) ∩λ∈Λ Aλ ⊆ 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 a
set. Then
(a)
(b)
(c)
(d)
B ∩ (∪λ∈Λ Aλ ) = ∪λ∈Λ (B ∩ Aλ ) .
B ∪ (∩λ∈Λ Aλ ) = ∩λ∈Λ (B ∪ Aλ ) .
(∩λ∈Λ Aλ )0 = ∪λ∈Λ A0λ
(∪λ∈Λ Aλ )0 = ∩λ∈Λ A0λ
If Λ = {1, 2}, then
(a)
(b)
(c)
(d)
B ∩ (A1 ∪ A2 ) = (B ∩ A1 ) ∪ (B ∩ A2 ) .
B ∪ (A1 ∩ A2 ) = (B ∪ A1 ) ∩ (B ∪ A2 ) .
(A1 ∩ A2 )0 = A01 ∪ A02 .
(A1 ∪ A2 )0 = A01 ∩ A02 .
• What are the left- and right-hand sides of each part of theorem 2.9 in the case in which
Λ = ∅? You can assume that each Aλ is a subset of some universal set U. You don’t
need to prove anything.
Recall that if Λ = ∅, we defined
∪λ∈Λ Aλ = ∅, and ∩λ∈Λ Aλ = U.
So in Theorem 2.9 we have
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1. B ∩ (∪λ∈Λ Aλ ) = B ∩ ∅ = ∅. Furthermore, the family {B ∩ Aλ : λ ∈ Λ} is also
empty. So ∪λ∈Λ (B ∩ Aλ ) = ∅.
2. B ∪ (∩λ∈Λ Aλ ) = B ∪ U = U. We also have that the family {B ∪ Aλ : λ ∈ Λ} is
empty. So ∩λ∈Λ (B ∪ Aλ ) = U.
3. (∩λ∈Λ Aλ )0 = U 0 = ∅. Since the family {A0λ : λ ∈ Λ} is empty, we also have
∪λ∈Λ A0λ = ∅.
4. (∪λ∈Λ Aλ )0 = ∅0 = U, and ∩λ∈Λ A0λ = 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 ∩ B 0 .
Proof.
x ∈ A − B ⇔ (x ∈ A) ∧ (x 6∈ B)
⇔ (x ∈ A) ∧ (x ∈ B 0 )
⇔ x ∈ A ∩ B0
38. 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 universal
set U. Then
(a) (A ∪ B)0 = A0 ∩ B 0
(b) (A0 )0 = A
(c) (A ∩ B)0 = A0 ∪ B 0
Starting with the left-hand side of part (c) we have
A0 ∪ B 0 = [(A0 ∪ B 0 )0 ]0
by part (b)
0 0
0 0 0
= [(A ) ∩ (B ) ] by part (a)
= (A ∩ B)0
by part (b), again.
50. We give a “proof” and a “counterexample” to the following conjecture: “If A ⊆ B and
C ⊆ 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 ∈ B
and x ∈ D. Therefore x ∈ B ∩ D, so A ∩ C ⊆ B ∩ D.
This is fine.
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57. (a) For each natural number n, let An = (n, n + 1). Find ∪n∈N An and ∩n∈N An .
(a) The interval (n, n + 1) contains all of the real numbers strictly between the
positive integers n and n + 1. So any real number greater than or equal to 1
that is not an integer will belong to ∪n∈N An . Furthermore, no positive integer
will be in ∪n∈N An , since no integer belongs to any An . So ∪n∈N An = [1, ∞)−N.
(b) If m and n are positive integers and m 6= n, then (m, m + 1) ∩ (n, n + 1) = ∅.
So ∩n∈N An = ∅.
(b) For each real number x, let Ax = (x, x + 1). Find ∪x∈R Ax and ∩x∈R Ax .
(a) Every real number y belongs to the interval (y−1/2, y+1/2). So ∪x∈R Ax = R.
(b) If y is any real number y 6∈ (y, y + 1). So y 6∈ ∩x∈R Ax and ∩x∈R Ax = ∅.
63. Give an example of a collection of sets indexed by a set consisting of 5 members.
There are a lot of possibilities here. The only requirements are that Λ have 5 elements
and 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 prime
numbers. One possibility is
Aλ = {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=1 Cn = ∅, because if m is any positive integer, m 6∈ Cm+1 . However,
if n is any positive integer ∩nm=1 Cm 6= ∅ because n ∈ Cm for all m ≤ n, and hence
n ∈ ∩nm=1 Cm . (Note that if Λ = {1, 2, . . . , n}, then we often write ∩nm=1 Am for ∩λ∈Λ Aλ
and ∪nm=1 Am for ∪λ∈Λ Aλ .)
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