discrete_mathematics_with_applications/chapter_6/exercises.md

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Exercise Set 6.1

1. In each of (a) -(f), answer the following questions: Is A \subseteq B? Is B \subseteq A? Is either A or B a proper subset of the other?

a. A = \{2, \{2\}, (\sqrt{2})^2\}, B = \{2, \{2\}, \{\{2\}\}\}

A \subseteq B ?:

 A = \{2, \{2\}, (\sqrt{2})^2\} = \{2, \{2\}, 2\} = \{2, \{2\}\} 

Yes, every element in A is in B.

B \subseteq A ?:

No, because \{\{2\}\} is an element of B, but is not an element of A, so \B \nsubseteq A.

Is either A or B a proper subset of the other?

Yes, A is a proper subset of B, because every element in A is in B, but not every element in B is in A.

b. A = \{3, \sqrt{5^2 - 4^2}, 24 \mod 7\}, B = \{8 \mod 5\}

A \subseteq B ?:

 A = \{3, \sqrt{5^2 - 4^2}, 24 \mod 7\} = \{3, 3, 3\} = \{3\} 

 B = \{8 \mod 5\} = \{3\} 

Yes, A is a subset of B since every element of A is in B.

B \subseteq A ?:

Yes, B is a subset of A since every element of B is in A.

Is either A or B a proper subset of the other?

Yes, both A and B are proper subsets of the other since A = B.

c. A = \{\{1, 2\}, \{2, 3\}\}, B = \{1, 2, 3\}

A \subseteq B ?:

No, because there are no elements in A that are in B, A \nsubseteq B

B \subseteq A ?:

No, because there are no elements in B that are in A, B \nsubseteq A

Is either A or B a proper subset of the other?

No, since neither set share any elements, neither is a proper subset of the other.

d. A = \{a, b, c\}, B = \{\{a\}, \{b\}, \{c\}\}

A \subseteq B ?:

No, because there are no elements in A that are in B, A \nsubseteq B

B \subseteq A ?:

No, because there are no elements in B that are in A, B \nsubseteq A

Is either A or B a proper subset of the other?

No, since neither set share any elements, neither is a proper subset of the other.

e. A = \{\sqrt{16}, \{4\}\}, B = \{4\}

A \subseteq B ?:

 A = \{\sqrt{16}, \{4\}\} = \{4, \{4\}\} 

No, because every element of A is not an element in B (4 is not in B), A \nsubseteq B.

B \subseteq A ?:

Yes, because every element in B is an element in A, B \subseteq A.

Is either A or B a proper subset of the other?

Yes, B is a proper subset of A since B \subseteq A and A \nsubseteq B.

f. A = \{x \in \mathbb{R} | \cos x \in \mathbb{Z}\}, B = \{x \in \mathbb{R} | \sin x \in \mathbb{Z}\}

From trigonometry, we know that \cos x = -1 \text{ or } 0 \text{ or } 1 and \sin x = -1 \text{ or } 0 \text{ or } 1 . When we evaluate for x in these cases we find:

 A = \{\dots, -\frac{5\pi}{2}, -\frac{3\pi}{2}, \pi, -\frac{\pi}{2}, 0, \frac{\pi}{2}, \frac{3\pi}{2}, \frac{5\pi}{2}, \dots\} 

 B = \{\dots, -\frac{5\pi}{2}, -\frac{3\pi}{2}, \pi, -\frac{\pi}{2}, 0, \frac{\pi}{2}, \frac{3\pi}{2}, \frac{5\pi}{2}, \dots\} 

A \subseteq B ?: Yes.

B \subseteq A ?: Yes.

Yes, B is a proper subset of A since B \subseteq A and A \nsubseteq B.

Yes, since A = B.

2. Complete the proof from Example 6.1.3: Prove that B \subseteq A where

 A = \{m \in \mathbb{Z} | m, = 2a \text{ for some integer } a\} 

and

 B = \{n \in \mathbb{Z} | n = 2b - 2 \text{ for some integer } b\} 

Part 2, Proof that B \subseteq A:

Suppose x is a particular but arbitrarily chosen element of B.

By definition of B, there is an integer, say b, such that x = 2b - 2.

To prove that B \subseteq A, we must show that there is some x that can equal both 2a, for some integer a, and that same x can also equal 2b - 2.

 2b - 2 = 2a 

 a = b - 1 

By the difference integers, a is an integer. Then, by substitution:

 2a = 2(b - 1) 

 = 2b - 2 

 = x 

Thus, by definition of A, x is an element of A.

Q.E.D.

3. Let sets R, S, and T be defined as follows:

 R = \{x \in \mathbb{Z} | x \text{ is divisible by } 2\}  

 S = \{y \in \mathbb{Z} | y \text{ is divisible by } 3\}  

 T = \{z \in \mathbb{Z} | z \text{ is divisible by } 6\}  

Prove or disprove each of the following statements.

a. R \subseteq T

R \nsubseteq T since 2 \in R since 2 \mid 2, but 2 \notin T since 6 \cancel{\mid} 2.

b. T \subseteq R

Proof:

Suppose n is any integer such that 6 \mid n, therefore n \in T.

By the definition of divisibility:

 n = 6m 

for some integer m.

 n = 2(3m) 

By the product of integers, 3m is an integer. It follows that n = 2 \cdot (\text{some integer}). Thus 2 \mid n, so n \in R. This is what was to be shown.

Q.E.D.

c. T \subseteq S

Proof:

Suppose n is any integer such that 6 \mid n, therefore n \in T.

By the definition of divisibility:

 n = 6m 

for some integer m.

 n = 3(2m) 

By the product of integers, 2m is an integer. It follows that n = 3 \cdot (\text{some integer}). Thus 3 \mid n, so n \in S. This is what was to be shown.

Q.E.D.

4. Let A = \{n \in \mathbb{Z} | n = 5r \text{ for some integer } r\} and B = \{m \in \mathbb{Z} | m = 20s \text{ for some integer } s\}. Prove or disprove each of the following statements.

a. A \subseteq B

A \nsubseteq B since 5 \in A since 5 \mid 5, but 5 \notin B since 5 \cancel{\mid} 20.

b. B \subseteq A

Proof:

Suppose n is any integer such that 20 \mid n, therefore n \in B.

By the definition of divisibility:

 n = 20m 

for some integer m.

 n = 5(4m) 

By the product of integers, 4m is an integer. It follows that n = 5 \cdot (\text{some integer}). Thus 5 \mid n, so n \in A. This is what was to be shown.

Q.E.D.

5. Let C = \{n \in \mathbb{Z} | n = 6r - 5 \text{ for some integer } r\} and D = \{m \in \mathbb{Z} | m = 3s + 1 \text{ for some integer } s\}. Prove or disprove each of the following statements.

a. C \subseteq D

Proof:

Suppose n is any integer such that n = 6r - 5 for some integer r, which means that n \in C.

Also suppose that m is any integer such that m = 3s + 1 for some integer s, which means that m \in S.

We must show that there exists some r that when substituted for s will satisfy the definition of n.

Let s = 2r - 2. Then, by substitution:

 m = 3(2r - 2) + 1 

 = 6r - 6 + 1 

 = 6r - 5 

 = n 

By the product and difference of integers, 6r - 5 is an integer, therefore n \in D. This is what was to be shown.

Q.E.D.

b. D \subseteq C

Disproof:

D \nsubseteq C because there are elements in D that are not in C. For example 4 is in D because 4 = 3(1) + 1, but 4 is not in C. If 4 were in C, this would mean:

 4 = 6r - 5 

for some integer r.

 4 + 5 = 6r 

 9 = 6r 

 \frac{9}{6} = r 

 \frac{3}{2} = r 

But \dfrac{3}{2} is not an integer. This is a contradiction, therefore D \nsubseteq C.

Q.E.D.

6. Let A = \{x \in \mathbb{Z} | x = 5a + 2 \text{ for some integer } a\}, B = \{y \in \mathbb{Z} | y = 10b - 3 \text{ for some integer } b\}, and C = \{z \in \mathbb{Z} | z = 10c + 7 \text{ for some integer } c\}.

Prove or disprove each of the following statements.

a. A \subseteq B

Disproof (by counterexample):

Suppose n is any integer such that n = 5a + 2 for some integer a. This means that n \in A.

Suppose also that there is some integer m such that m = 10b - 3 for some integer b. This means that m \in B.

To show that there is some integer b that will satisfy n, we must relate it to a:

 5a + 2 = 10b - 3 

 5a + 5 = 10b 

 5a + 5 = 10b 

 \frac{1}{2}a + \frac{1}{2} = b 

 \frac{a + 1}{2} = b 

In order for A \subseteq B, every element of A must be in B. If a = 0, then n = 2, so n \in A. If a = 0, then b = \dfrac{1}{2}, which is not an integer, thus 2 \notin B. Therefore A \nsubseteq B.

Q.E.D.

b. B \subseteq A

Proof:

Suppose y is any integer such that y = 10b - 3 for some integer b. This means that y \in B.

Let's first find a as it relates to y.

 y = 5a + 2 

 10b - 3 = 5a + 2 

 10b - 5 = 5a 

 2b - 1 = a 

So let a = 2b - 1.

Then substitute in for the condition for A:

 x = 5a + 2 

 = 5(2b - 1) + 2 

 = 10b - 5 + 2 

 = 10b - 3 

 = y 

Therefore y \in A.

Q.E.D.

c. B = C

To prove B = C, we must prove both that B \subseteq C and C \subseteq B.

_Prove B \subseteq C:

Suppose y is any integer such that y = 10b - 3 for some integer b. This means that y \in B.

Let's first find some integer c as it relates to y:

 y = 10c + 7 

 10b - 3 = 10c + 7 

 10b - 10 = 10c 

 b - 1 = c 

So, let c = b - 1.

Then substitute in for the condition for C:

 z = 10c + 7 

 = 10(b - 1) + 7 

 = 10b - 10 + 7 

 = 10b - 3 

 = y 

Thus y \in C, and therefore B \subseteq C.

_Prove C \subseteq B:

Suppose z is any integer such that z = 10c + 7 for some integer c. This means that z \in C.

Let's first find some integer b as it relates to z:

 z = 10b - 3 

 10c + 7 = 10b - 3 

 10c + 10 = 10b 

 c + 1 = b 

So, let b = c + 1.

Then substitute in for the condition for B:

 y = 10b - 3 

 = 10(c + 1) - 3 

 = 10c + 10 - 3 

 = 10c + 7 

 = z 

Therefore z \in B.

Thus z \in B, and therefore C \subseteq B.

Since B \subseteq C and C \subseteq B, it follows that B = C. This is what was to be shown.

Q.E.D.

7. Let A = \{x \in \mathbb{Z} | x = 6a + 4 \text{ for some integer } a\}, B = \{y \in \mathbb{Z} | y = 18b - 2 \text{ for some integer } b\}, and C = \{z \in \mathbb{Z} | z = 18c + 16 \text{ for some integer } c\}.

Prove or disprove each of the following statements.

a. A \subseteq B

Disproof (by counterexample):

Suppose x is any integer such that x = 6a + 4 for some integer a. This means that x \in A.

Let's first find some integer b as it relates to a.

 x = 18b - 2 

 6a + 4 = 18b - 2 

 6a + 6 = 18b 

 \frac{6}{18}a + \frac{6}{18} = b 

 \frac{1}{3}a + \frac{1}{3} = b 

 \frac{a + 1}{3} = b 

By definition of b, b must always be an integer for all a.

Suppose a = 0, then:

 x = 6(0) + 4 = 4 

so 4 \in A, but:

 b = \frac{0 + 1}{3} = \frac{1}{3} 

so 4 \notin B. We can see this as 4 = 18b - 2 results in b = \dfrac{1}{3}, but b must be an integer.

b. B \subseteq A

Proof:

Suppose y is any integer such that y = 18b - 2 for some integer b. This means that y \in B.

Let's first find some integer a as it relates to b.

 y = 6a + 4 

 18b - 2 = 6a + 4 

 18b - 6 = 6a 

 3b - 1 = a 

Now, substitute a in for the condition for A:

 x = 6a + 4 

 = 6(3b - 1) + 4 

 = 18b - 2 

 = y 

Therefore B \subseteq A.

c. B = C

To prove B = C, we must prove both that B \subseteq C and C \subseteq B.

_Prove B \subseteq C:

Suppose y is any integer such that y = 18b - 2 for some integer b. This means that y \in B.

Let's first find some integer c as it relates to b.

 y = 18c + 16 

 18b - 2 = 18c + 16 

 18b - 18 = 18c 

 b - 1 = c 

So, let c = b - 1. Now substitute c in for the condition of C:

 z = 18c + 16 

 = 18(b - 1) + 16 

 = 18b - 18 + 16 

 = 18b - 2 

 = y 

Therefore B \subseteq C.

_Prove C \subseteq B:

Suppose z is any integer such that z = 18c + 16 for some integer c.

Let's first find some b as it relates to c.

 z = 18b - 2 

 18c + 16 = 18b - 2 

 18c + 18 = 18b 

 c + 1 = b 

So, let b = c + 1. Now, let's substitute b in for the condition for B.

 y = 18b - 2 

 = 18(c + 1) - 2 

 = 18c + 18 - 2 

 = 18c + 16 

 = z 

Therefore C \subseteq B.

Since B \subseteq C and C \subseteq B, we conclude that B = C. This is what was to be shown.

Q.E.D.

8. Write in words to read each of the following out loud. Then write each set using the symbols for union, intersection, set difference, or set complement.

a. \{x \in U | x \in A \text{ and } x \in B\}

In words:

The set of all x in U such that x is in A and x is in B.

In symbolic notation:

 A \cap B 

b. \{x \in U | x \in A \text{ or } x \in B\}

In words:

The set of all x in U such that x is in A or x is in B.

In symbolic notation:

 A \cup B 

c. \{x \in U | x \in A \text{ and } x \notin B\}

In words:

The set of all x in U such that x is in A and x is not in B.

In symbolic notation:

 A - B 

d. \{x \in U | x \notin A\}

In words:

The set of all x in U such that x is not in A.

In symbolic notation:

 A^c 

9. Complete the following sentences without using the symbols \cup, \cap, or -.

a. x \notin A \cup B if, and only if, _____.

x is not in A and x is not in B.

b. x \notin A \cap B if, and only if, _____.

x is not in A or x is not in B.

c. x \notin A - B if, and only if, _____.

x is not in A, or x is in B, or both.

Note: recall that the negation of an "and", which is A - B, is an "or", thus:

 x \in (A - B) \to x \in A \wedge x \notin B 

so:

 \neg(x \in (A - B)) \to \neg(x \in A \wedge x \notin B) \to x \notin A \vee x \in B 

10. Let A = \{1, 3, 5, 7, 9\}, b = \{3, 6, 9\}, and C = \{2, 4, 6, 8\}. Find each of the following:

a. A \cup B

 A \cup B = \{1, 3, 5, 6, 7, 9\} 

b. A \cap B

 A \cap B = \{3, 9\} 

c. A \cup C

 A \cup C = \{1, 2, 3, 4, 5, 6, 7, 8, 9\} 

d. A \cap C

 A \cap C = \emptyset 

e. A - B

 A - B = \{1, 5, 7\} 

f. B - A

 B - A = \{6\} 

g. B \cup C

 B \cup C = \{2, 3, 4, 6, 8, 9\} 

h. B \cap C

 B \cap C = \{6\} 

11. Let the universal set \mathbb{R}, the set of all real numbers, and let A = \{x \in \mathbb{R} | 0 < x \leq 2\}, B = \{x \in \mathbb{R} | 1 \leq x < 4\}, and C = \{x \in \mathbb{R} | 3 \leq x < 9\}. Find each of the following:

a. A \cup B

 A \cup B = \{x \in \mathbb{R} | 0 < x < 4\} 

b. A \cap B

 A \cap B = \{x \in \mathbb{R} | 1 \leq x \leq 2\} 

c. A^c

 A^c = \{x \in \mathbb{R} | x \leq 0 \text{ or } x > 2\} 

d. A \cup C

 A \cup C = \{x \in \mathbb{R} | 0 < x \leq 2 \text{ or } 3 \leq x < 9 \} 

e. A \cap C

 A \cap C = \emptyset 

f. B^c

 B^c = \{x \in \mathbb{R} | x < 1 \text{ or } x \geq 4\} 

g. A^c \cap B^c

 A^c = \{x \in \mathbb{R} | x \leq 0 \text{ or } x > 2\} 

 B^c = \{x \in \mathbb{R} | x < 1 \text{ or } x \geq 4\} 

 A^c \cap B^c = \{x \in \mathbb{R} | x \leq 0 \text{ or } x \geq 4\}  

h. A^c \cup B^c

 A^c = \{x \in \mathbb{R} | x \leq 0 \text{ or } x > 2\} 

 B^c = \{x \in \mathbb{R} | x < 1 \text{ or } x \geq 4\} 

 A^c \cup B^c = \{x \in \mathbb{R} | x < 1 \text{ or } x > 2\} 

i. (A \cap B)^c

 A \cap B = \{x \in \mathbb{R} | 1 \leq x \leq 2\} 

 (A \cap B)^c = \{x \in \mathbb{R} | x < 1 \text{ or } x > 2 \} 

j. (A \cup B)^c

 A \cup B = \{x \in \mathbb{R} | 0 < x < 4\} 

 (A \cup B)^c = \{x \in \mathbb{R} | x \leq 0 \text{ or } x \geq 4\} 

12. Let the universal set be \mathbb{R}, the set of all real numbers, and let A = \{x \in \mathbb{R} | -3 \leq x \leq 0\}, B = \{x \in \mathbb{R} | -1 < x < 2\}, and C = \{x \in \mathbb{R} | 6 < x \leq 8\}. Find each of the following:

a. A \cup B

 A \cup B = \{x \in \mathbb{R} | -3 \leq x < 2\} 

b. A \cap B

 A \cap B = \{x \in \mathbb{R} | -1 < x \leq 0\} 

c. A^c

 A^c = \{x \in \mathbb{R} | x < -3 \text{ or } x > 0 \} 

d. A \cup C

 A \cup C = \{x \in \mathbb{R} | -3 \leq x \leq 0 \text{ or } 6 < x \leq 8\} 

e. A \cap C

 A \cap C = \emptyset 

f. B^c

 B^c = \{x \in \mathbb{R} | x \leq -1 \text{ or } x \geq 2 \} 

g. A^c \cap B^c

 A^c = \{x \in \mathbb{R} | x < -3 \text{ or } x > 0 \} 

 B^c = \{x \in \mathbb{R} | x \leq -1 \text{ or } x \geq 2 \} 

 A^c \cap B^c = \{x \in \mathbb{R} | x < -3 \text{ or } x \geq 2 \} 

h. A^c \cup B^c

 A^c = \{x \in \mathbb{R} | x < -3 \text{ or } x > 0 \} 

 B^c = \{x \in \mathbb{R} | x \leq -1 \text{ or } x \geq 2 \} 

 A^c \cup B^c = \{x \in \mathbb{R} | x \leq -1 \text{ or } x > 0 \} 

i. (A \cap B)^c

 A \cap B = \{x \in \mathbb{R} | -1 < x \leq 0\} 

 (A \cap B)^c = \{x \in \mathbb{R} | x \leq -1 \text{ or } x > 0\} 

j. (A \cup B)^c

 A \cup B = \{x \in \mathbb{R} | -3 \leq x < 2\} 

 (A \cup B)^c = \{x \in \mathbb{R} | x < -3 \text{ or } x \geq 2 \}

13. Let S be the set of all strings of $0$'s and $1$'s of length 4, and let A and B be the following subsets of S: A = \{1110, 1111, 1000, 1001\} and B = \{1100, 0100, 1111, 0111\}. Find each of the following:

a. A \cap B

 A \cap B = \{1111\} 

b. A \cup B

 A \cup B = \{1100, 0100, 1110, 1111, 0111, 1000, 1001\} 

c. A - B

 A - B = \{1110, 1000, 1001\} 

d. B - A

 B - A = \{1100, 0100, 0111\} 

14. In each of the following, draw a Venn diagram for sets A, B, and C that satisfy the given conditions.

a. A \subseteq B, C \subseteq B, A \cap C = \emptyset

Done physically.

b. C \subseteq A, B \cap C = \emptyset

Done physically.

15. In each of the following, draw a Venn diagram for sets A, B, and C that satisfy the given conditions.

a. A \cap B = \emptyset, A \subseteq C, C \cap B \neq \emptyset

Done physically.

b. A \subseteq B, C \subseteq B, A \cap C \neq \emptyset

Done physically.

c. A \cap B \neq \emptyset, B \cap C \neq \emptyset, A \cap C = \emptyset, A \nsubseteq B, C \nsubseteq B

Done physically.

16. Let A = \{a, b, c\}, B = \{b, c, d\}, and C = \{b, c, e\}.

a. Find A \cup (B \cap C), (A \cup B) \cap C, and (A \cup B) \cap (A \cup C). Which of these sets are equal?

 B \cap C = \{b, c\} 

 A \cup (B \cap C) = \{a, b, c\} 

 A \cup B = \{a, b, c, d\} 

 (A \cup B) \cap C = \{b, c\} 

 A \cup C = \{a, b, c, e\} 

 (A \cup B) \cap (A \cup C) = \{a, b, c\} 

 A \cup (B \cap C) = (A \cup B) \cap (A \cup C) 

b. Find A \cap (B \cup C), (A \cap B) \cup C, and (A \cap B) \cup (A \cap C). Which of these sets are equal?

 B \cup C = \{b, c, d, e\} 

 A \cap (B \cup C) = \{b, c\} 

 A \cap B = \{b, c\} 

 (A \cap B) \cup C = \{b, c, e\} 

 A \cap C = \{b, c\} 

 (A \cap B) \cup (A \cap C) = \{b, c\} 

 A \cap (B \cup C) = A \cap C = (A \cap B) \cup (A \cap C) 

c. Find (A - B) - C and A - (B - C). Are these sets equal?

 A - B = \{a\} 

 (A - B) - C = \{a\} 

 B - C = \{d\} 

 A - (B - C) = \{a, b, c\} 

 (A - B) - C \neq A - (B - C) 

17. Consider the following Venn diagram. For each of (a)-(f), copy the diagram and shade the region corresponding to the indicated set.

a. A \cap B

Omitted.

b. B \cup C

Omitted.

c. A^c

Omitted.

d. A - (B \cup C)

Omitted.

e. (A \cup B)^c

Omitted.

f. A^c \cap B^c

Omitted.

(See page 412 for image)

a. Is the number 0 in \emptyset? Why?

No, by the definition of \emptyset, there are no elements in \emptyset. In other words \emptyset \neq \{0\}.

b. Is \emptyset = \{\emptyset\}? Why?

No, by the definition of \emptyset, there are no elements in \emptyset. In other words \emptyset \neq \{\emptyset\}.

c. Is \emptyset \in \{\emptyset\} Why?

Yes, because \emptyset itself can be an element in a set, it is true that \emptyset \in \{\emptyset\}.

d. Is \emptyset \in \emptyset? Why?

No, by the definition of \emptyset, it is empty, it has no elements, therefore \emptyset cannot contain itself. \emptyset \notin \emptyset.

19. Let A_i = \{i, i^2\} for each integer i = 1, 2, 3, 4.

a. A_1 \cup A_2 \cup A_3 \cup A_4 = \text{ ?}

A_1 = {1, 1^2} = {1, 1} = {1} \ A_2 = {2, 2^2} = {2, 4} \ A_3 = {3, 3^2} = {3, 9} \ A_4 = {4, 4^2} = {4, 16} \

 A_1 \cup A_2 \cup A_3 \cup A_4 = \{1, 2, 3, 4, 9, 16\} 

b. A_1 \cap A_2 \cap A_3 \cap A_4 = \text{ ?}

 A_1 \cap A_2 \cap A_3 \cap A_4 = \emptyset 

c. Are A_1, A_2, A_3, and A_4 mutually disjoint? Explain.

No, since A_2 and A_4 both contain the element 4, they are not mutually disjoint.

20. Let B_i = \{x \in \mathbb{R} | 0 \leq x\leq i\} for each integer i = 1, 2, 3, 4.

a. B_1 \cup B_2 \cup B_3 \cup B_4 = \text{ ?}

 B_1 \cup B_2 \cup B_3 \cup B_4 = \{x \in \mathbb{R} | 0 \leq x \leq 4\} 

b. B_1 \cap B_2 \cap B_3 \cap B_4 = \text{ ?}

 B_1 \cap B_2 \cap B_3 \cap B_4 = \{x \in \mathbb{R} | 0 \leq x \leq 1\} 

c. Are B_1, B_2, B_3, and B_4 mutually disjoint? Explain.

No, since all sets include all real numbers within the range 0 \leq x \leq 1, they are not mutually disjoint.

21. Let C_i = \{i, -i\} for each nonnegative integer i.

C_0 = {0, -0} = {0} \ C_1 = {1, -1} \ C_2 = {2, -2} \ C_3 = {3, -3} \ C_4 = {4, -4} \

a. \bigcup_{i = 0}^{4}C_i = \text{ ?}

 \bigcup_{i = 0}^{4}C_i = C_0 \cup C_1 \cup C_2 \cup C_3 \cup C_4 

 \bigcup_{i = 0}^{4}C_i = \{-4, -3, -2, -1, 0, 1, 2, 3, 4\} 

b. \bigcap_{i = 0}^{4}C_i = \text{ ?}

 \bigcap_{i = 0}^{4}C_i = \emptyset 

c. Are C_0, C_1, C_2, \dots mutually disjoint? Explain.

Yes, since none of the sets have any elements in common, they are mutually disjoint.

d. \bigcup_{i = 0}^{n}C_i = \text{ ?}

 \bigcup_{i = 0}^{n}C_i = \{-n, -(n - 1), \dots -2, -1, 0, 1, 2, \dots (n - 1), n\} 

e. \bigcap_{i = 0}^{n}C_i = \text{ ?}

 \bigcap_{i = 0}^{n}C_i = \emptyset 

f. \bigcup_{i = 0}^{\infty}C_i = \text{ ?}

 \bigcup_{i = 0}^{\infty}C_i = \{-\infty, \dots, -2, -1, 0, 1, 2, \dots, \infty\} = \mathbb{Z} 

g. \bigcap_{i = 0}^{\infty}C_i = \text{ ?}

 \bigcap_{i = 0}^{\infty}C_i = \emptyset 

22. Let D_i = \{x \in \mathbb{R} | -i \leq x \leq i\} = [-i, i] for each nonnegative integer i.

D_0 = [-0, 0] = {0} \ D_1 = [-1, 1] \ D_2 = [-2, 2] \ D_3 = [-3, 3] \ D_4 = [-4, 4] \

a. \bigcup_{i = 0}^{4}D_i = \text{ ?}

 \bigcup_{i = 0}^{4}D_i = \{x \in \mathbb{R} | -4 \leq x \leq 4\} = [-4, 4] 

b. \bigcap_{i = 0}^{4}D_i = \text{ ?}

 \bigcap_{i = 0}^{4}D_i = \{0\} 

c. Are D_0, D_1, D_2, \dots mutually disjoint? Explain.

No, in fact all sets have at least \{0} in common , as i increases, so does the amount of elements all sets have in common, or D_k \subseteq D_{k + 1}.

d. \bigcup_{i = 0}^{n}D_i = \text{ ?}

 \bigcup_{i = 0}^{n}D_i = \{x \in \mathbb{R} | -n \leq x \leq n\} = [-n, n] 

e. \bigcap_{i = 0}^{n}D_i = \text{ ?}

 \bigcap_{i = 0}^{n}D_i = \{0\} 

f. \bigcup_{i = 0}^{\infty}D_i = \text{ ?}

 \bigcup_{i = 0}^{\infty}D_i = (-\infty, \infty) = \mathbb{R} 

g. \bigcap_{i = 0}^{\infty}D_i = \text{ ?}

 \bigcap_{i = 0}^{\infty}D_i = \{0\} 

23. Let V_i = \{x \in \mathbb{R} | -\dfrac{1}{i} \leq x \leq \dfrac{1}{i}\} = \left[-\dfrac{1}{i}, \dfrac{1}{i}\right] for each positive integer i.

V_1 = \left[-\frac{1}{1}, \frac{1}{1}\right] = [-1, 1] \ V_2 = \left[-\frac{1}{2}, \frac{1}{2}\right] \ V_3 = \left[-\frac{1}{3}, \frac{1}{3}\right] \ V_4 = \left[-\frac{1}{4}, \frac{1}{4}\right] \

a. \bigcup_{i = 1}^{4}V_i = \text{ ?}

 \bigcup_{i = 1}^{4}V_i = [-1, 1] 

b. \bigcap_{i = 1}^{4}V_i = \text{ ?}

 \bigcap_{i = 1}^{4}V_i = \left[-\frac{1}{4}, \frac{1}{4}\right] 

c. Are V_1, V_2, V_3, \dots mutually disjoint? Explain.

No, every set includes 0.

d. \bigcup_{i = 1}^{n}V_i = \text{ ?}

 \bigcup_{i = 1}^{n}V_i = [-1, 1] 

e. \bigcap_{i = 1}^{n}V_i = \text{ ?}

 \bigcap_{i = 1}^{n}V_i = \left[-\frac{1}{n}, \frac{1}{n}\right] 

f. \bigcup_{i = 1}^{\infty} = \text{ ?}

 \bigcup_{i = 1}^{\infty} = [-1, 1] 

g. \bigcap_{i = 1}^{\infty} = \text{ ?}

 \bigcap_{i = 1}^{\infty} = \{0\} \text{ because as } i \to \infty \text{ then } \frac{1}{i} \to 0 

24. Let W_i = \{x \in \mathbb{R} | x > i\} = (i, \infty) for each nonnegative integer i.

W_0 = (0, \infty) \ W_1 = (1, \infty) \ W_2 = (2, \infty) \ W_3 = (3, \infty) \ W_4 = (4, \infty) \

a. \bigcup_{i = 0}^{4}W_i = \text{ ?}

 \bigcup_{i = 0}^{4}W_i = (0, \infty) 

b. \bigcap_{i = 0}^{4}W_i = \text{ ?}

 \bigcap_{i = 0}^{4}W_i = (4, \infty) 

c. Are W_0, W_1, W_2, \dots mutually disjoint? Explain.

No, because they all have (i, \infty) in common, or W_{i + 1} \subseteq W_i.

d. \bigcup_{i = 0}^{n}W_i = \text{ ?}

 \bigcup_{i = 0}^{n}W_i = (0, \infty) 

e. \bigcap_{i = 0}^{n}W_i = \text{ ?}

 \bigcap_{i = 0}^{n}W_i = (n, \infty) 

f. \bigcup_{i = 0}^{\infty}W_i = \text{ ?}

 \bigcup_{i = 0}^{\infty}W_i = (0, \infty) 

g. \bigcap_{i = 0}^{\infty}W_i = \text{ ?}

 \bigcap_{i = 0}^{\infty}W_i = \emptyset 

There is no real number greater than every positive integer, so no element belongs to all W_i.

25. Let R_i = \{x \in \mathbb{R} | 1 \leq x \leq 1 + \dfrac{1}{i}\} = \left[1, 1 + \dfrac{1}{i}\right] for each positive integer i.

R_1 = \left[1, 1 + \frac{1}{1}\right] = [1, 2] \ R_2 = \left[1, 1 + \frac{1}{2}\right] = \left[1, \frac{3}{2}\right] \ R_3 = \left[1, 1 + \frac{1}{3}\right] = \left[1, \frac{4}{3}\right] \ R_4 = \left[1, 1 + \frac{1}{4}\right] = \left[1, \frac{5}{4}\right] \

a. \bigcup_{i = 1}^{4}R_i = \text{ ?}

 \bigcup_{i = 1}^{4}R_i = [1, 2] 

b. \bigcap_{i = 1}^{4}R_i = \text{ ?}

 \bigcap_{i = 1}^{4}R_i = \left[1, \frac{5}{4}\right] 

c. Are R_1, R_2, R_3, \dots mutually disjoint? Explain.

No, they all include the element 1.

d. \bigcup_{i = 1}^{n}R_i = \text{ ?}

 \bigcup_{i = 1}^{n}R_i = [1, 2] 

e. \bigcap_{i = 1}^{n}R_i = \text{ ?}

 \bigcap_{i = 1}^{n}R_i = \left[1, 1 + \frac{1}{n}\right]

f. \bigcup_{i = 1}^{\infty}R_i = \text{ ?}

 \bigcup_{i = 1}^{\infty}R_i = [1, 2] 

g. \bigcap_{i = 1}^{\infty}R_i = \text{ ?}

 \bigcap_{i = 1}^{\infty}R_i = \{1\} 

Because \dfrac{1}{\infty} \to 0 and \left(1 + \dfrac{1}{\infty}\right) \to 1.

26. Let S_i = \{x \in \mathbb{R} | 1 < x < 1 + \dfrac{1}{i}\} = \left(1, 1 + \dfrac{1}{i}\right) for each positive integer i.

S_1 = \left(1, 1 + \frac{1}{1}\right) = (1, 2) \ S_2 = \left(1, 1 + \frac{1}{2}\right) = \left(1, \frac{3}{2}\right) \ S_3 = \left(1, 1 + \frac{1}{3}\right) = \left(1, \frac{4}{3}\right) \ S_4 = \left(1, 1 + \frac{1}{4}\right) = \left(1, \frac{5}{4}\right) \

a. \bigcup_{i = 1}^{4}S_i = \text{ ?}

 \bigcup_{i = 1}^{4}S_i = (1, 2) 

b. \bigcap_{i = 1}^{4}S_i = \text{ ?}

 \bigcap_{i = 1}^{4}S_i = \left(1, \frac{5}{4}\right) 

c. Are S_1, S_2, S_3, \dots mutually disjoint? Explain.

No, any element sufficiently close to 1 are in all the sets.

d. \bigcup_{i = 1}^{n}S_i = \text{ ?}

 \bigcup_{i = 1}^{n}S_i = (1, 2) 

e. \bigcap_{i = 1}^{n}S_i = \text{ ?}

 \bigcap_{i = 1}^{n}S_i = \left(1, 1 + \frac{1}{n}\right) 

f. \bigcup_{i = 1}^{\infty}S_i = \text{ ?}

 \bigcup_{i = 1}^{\infty}S_i = (1, 2) 

g. \bigcap_{i = 1}^{\infty}S_i = \text{ ?}

 \bigcap_{i = 1}^{\infty}S_i = \emptyset 

Because the range converges on 1, but cannot include 1, the set is empty.

a. Is \{\{a, d, e\}, \{b, c\}, \{d, f\}\} a partition of \{a, b, c, d, e, f\}?

No, since d is an element in two sets, the sets are not mutually disjoint, and so therefore is not a partition.

b. Is \{\{w, x, v\}, \{u, y, q\}, \{p, z\}\} a partition of \{p, q, u, v, w, x, y, z\}?

 \{w, x, v\} \cup \{u, y, q\} \cup \{p, z\} = \{p, q, u, v, w, x, y, z\} 

and:

 \{w, x, v\} \cap \{u, y, q\} \cap \{p, z\} = \emptyset 

So yes, the given sets are a partition of the overall set.

c. Is \{\{5, 4\}, \{7, 2\}, \{1, 3, 4\}, \{6, 8\}\} a partition of \{1, 2, 3, 4, 5, 6, 7, 8\}?

No, as 4 is an element in two of the given sets, and so the given sets are not a partition of the overall set.

d. Is \{\{3, 7, 8\}, \{2, 9\}, \{1, 4, 5\}\} a partition of \{1, 2, 3, 4, 5, 6, 7, 8, 9\}?

No, since none of the sets contain 6.

e. Is \{\{1, 5\}, \{4, 7\}, \{2, 8, 6, 3\}\} a partition of \{1, 2, 3, 4, 5, 6, 7, 8\}?

Yes, since none of the elements in each of the given sets are in any other of the given sets and all of the elements make up the overall set.

28. Let E be the set of all even integers and O the set of all odd integers. Is \{E, O\} a partition of \mathbb{Z}, the set of all integers? Explain your answer.

Yes, since no integer is both even and odd, and all integers are either even or odd, \{E, O\} is a partition of \mathbb{Z}.

29. Let \mathbb{R} be the set of all real numbers. Is \{\mathbb{R}^+, \mathbb{R}^-, \{0\}\} a partition of \mathbb{R}? Explain your answer.

Yes, since all real numbers are either positive, negative, or 0, and \mathbb{R}^+, \mathbb{R}^- and \{0\} do not have any elements in common, these subsets all form a partition of \mathbb{R}.

30. Let \mathbb{Z} be the set of all integers and let

 A_0 = \{n \in \mathbb{Z} | n = 4k, \text{ for some integer } k\} 

 A_1 = \{n \in \mathbb{Z} | n = 4k + 1, \text{ for some integer } k\} 

 A_2 = \{n \in \mathbb{Z} | n = 4k + 2, \text{ for some integer } k\} 

and

 A_3 = \{n \in \mathbb{Z} | n = 4k + 3, \text{ for some integer } k\} 

Is \{A_0, A_1, A_2, A_3\} a partition of \mathbb{Z}? Explain your answer.

Yes. These sets are mutually disjoint, and by the quotient-remainder theorem, every integer has exactly one of the forms n = 4k, n = 4k + 1, n = 4k + 2, n = 4k + 3.

31. Suppose A = \{1, 2\} and B = \{2, 3\}. Find each of the following:

a. \mathscr{P}(A \cap B)

 A \cap B = \{2\} 

 \mathscr{P}(A \cap B) = \{\emptyset, \{2\}\} 

b. \mathscr{P}(A)

 \mathscr{P}(A) = \{\emptyset, \{1\}, \{2\}, \{1, 2\}} 

c. \mathscr{P}(A \cup B)

 A \cup B = \{1, 2, 3\} 

 \mathscr{P}(A \cup B) = \{\emptyset, \{1\}, \{2\}, \{3\}, \{1, 2\}, \{1, 3\}, \{2, 3\}, \{1, 2, 3\}\} 

d. \mathscr{P}(A \times B)

 A \times B = \{(1, 2), (1, 3), (2, 2), (2, 3)\} 

 \mathscr{P}(A \times B) = \{\emptyset, \{(1, 2)\}, \{(1, 3)\}, \{(2, 2)\}, \{(2, 3)\}, \{(1, 2), (1, 3)\}, \{(1, 2), (2, 2)\}, \{(1, 2,), (2, 3)\}, \{(1, 3), (2, 2)\}, \{(1, 3), (2, 3)\}, \{(2, 2), (2, 3)\}, \{(1, 2), (1, 3), (2, 2)\}, \{(1, 2), (1, 3), (2, 3)\}, \{(1, 2), (2, 2), (2, 3)\}, \{(1, 3), (2, 2), (2, 3)\}, \{(1, 2), (1, 3), (2, 2), (2, 3)\}\} 

a. Suppose A = \{1\} and B = \{u, v\}. Find \mathscr{P}(A \times B).

 A \times B = \{(1, u), (1, v)\} 

 \mathscr{P}(A \times B) = \{\emptyset, \{(1, u)\}, \{(1, v)\}, \{(1, u), (1, v)\}\} 

b. Suppose X = \{a, b\} and Y = \{x, y\}. Find \mathscr{P}(X \times Y).

 X \times Y = \{(a, x), (a, y), (b, x), (b, y)\} 

 \mathscr{P}(X \times Y) = \{\emptyset, \{(a, x)\}, \{(a, y)\}, \{(b, x)\}, \{(b, y)\}, \{(a, x), (a, y)\}, \{(a, x), (b, x)\}, \{(a, x), (b, y)\}, \{(a, y), (b, x)\}, \{(a, y), (b, y)\}, \{(b, x), (b, y)\}, \{(a, x), (a, y), (b, x)\}, \{(a, x), (a, y), (b, y)\}, \{(a, x), (b, x), (b, y)\}, \{(a, y), (b, x), (b, y)\} \{(a, x), (a, y), (b, x), (b, y)\}\} 

a. Find \mathscr{P}(\emptyset).

 \mathscr{P}(\emptyset) = \{\emptyset\} 

b. Find \mathscr{P}(\mathscr{P}(\emptyset)).

 \mathscr{P}(\mathscr{P}(\emptyset)) = \{\emptyset, \{\emptyset\}\} 

b. Find \mathscr{P}(\mathscr{P}(\mathscr{P}(\emptyset))).

 \mathscr{P}(\mathscr{P}(\mathscr{P}(\emptyset))) = \{\emptyset, \{\emptyset\}, \{\emptyset, \{\emptyset\}\}, \{\{\emptyset\}\}\} 

34. let A_1 = \{1\}, A_2 = \{u, v\}, and A_3 = \{m, n\}. Find each of the following sets:

a. A_1 \cup (A_2 \times A_3)

 A_2 \times A_3 = \{(u, m), (u, n), (v, m), (v, n)\} 

 A_1 \cup (A_2 \times A_3) = \{1, (u, m), (u, n), (v, m), (v, n)\} 

b. (A_1 \cup A_2) \times A_3

 A_1 \cup A_2 = \{1, u, v\} 

 (A_1 \cup A_2) \times A_3 = \{(1, m), (1, n), (u, m), (u, n), (v, m), (v, n)\} 

35. let A = \{a, b\}, B = \{1, 2\}, and C = \{2, 3\}. Find each of the following sets:

a. A \times (B \cup C)

 B \cup C = \{1, 2, 3\} 

 A \times (B \cup C) = \{(a, 1), (a, 2), (a, 3), (b, 1), (b, 2), (b, 3)\} 

b. (A \times B) \cup (A \times C)

 A \times B = \{(a, 1), (a, 2), (b, 1), (b, 2)\} 

 A \times C = \{(a, 2), (a, 3), (b, 2), (b, 3)\} 

 (A \times B) \cup (A \times C) = \{(a, 1), (a, 2), (a, 3), (b, 1), (b, 2), (b, 3)\} 

c. A \times (B \cap C)

 B \cap C = \{2\} 

 A \times (B \cap C) = \{(a, 2), (b, 2)\} 

d. (A \times B) \cap (A \times C)

 A \times B = \{(a, 1), (a, 2), (b, 1), (b, 2)\} 

 A \times C = \{(a, 2), (a, 3), (b, 2), (b, 3)\} 

 (A \times B) \cap (A \times C) = \{(a, 2), (b, 2)\} 

36. Trace the action of Algorithm 6.1.1 on the variables i, j, \text{found}, and \text{answer} for m = 3, n = 3, and sets A and B represented as the arrays a[1] = u, a[2] = v, a[3] = w, b[1] = w, b[2] = u, and b[3] = v.

Omitted.

37. Trace the action of Algorithm 6.1.1 on the variables i, j, \text{found}, and \text{answer} for m = 4, n = 4 and sets A and B represented as the arrays a[1] = u, a[2] = v, a[3] = w, a[4] = x, b[1] = r, b[2] = u, b[3] = y, b[4] = z.

Omitted.

38. Write an algorithm to determine whether a given element x belongs to a given set that is represented as the array a[1], a[2], \dots, a[n].

Omitted.

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Page 427

Exercise Set 6.2

a. To say that an element is in A \cap (B \cup C) means that it is in __ (1) __ and in __ (2) __.

b. To say that an element is in (A \cap B) \cup C means that it is in __ (1) __ or in __ (2) __.

c. To say that an element is in A - (B \cap C) means that it is in __ (1) __ and not in __ (2)__.

d. To prove that (A \cup B) \cap C \subseteq A \cup (B \cap C), we suppose that x is any element in __ (1) __. Then we must show that __ (2) __.

e. If A, B, and C are any sets such that B \subseteq C, to prove that A \cap B \subseteq A \cap C, we suppose that x is any element in __ (1) __. Then we must show that __ (2) __.

2. The following are two proofs that for all sets A and B, A - B \subseteq A. The first is less formal, and the second is more formal. Fill in the blanks.

a. Proof: Suppose A and B are any sets. To show that A - B \subseteq A, we must show that every element in __ (1) __ is in __ (2) __. But any element in A - B is in __ (3) __ and not in __ (4) __ (by definition of A - B). In particular, such an element is in A.

b. Proof: Suppose A and B are any sets and x \in A - B. _[We must show that __ (1) _.] By definition of set difference, x \in __ ( 2 ) __ and x \notin __ (3) __. In particular, x \in __ (4) __ [which is what was to be shown].

In 3 and 4, supply explanations of the stesp in the given proofs.

3. Theorem: For all sets A, B, and C, if A \subseteq C, B \subseteq C, then A \subseteq C.

Proof:

Statement	Explanation
Suppose A, B, and C are any sets such that A \subseteq B and B \subseteq C	starting point
We must show that A \subseteq C.	conclusion to be shown
Let x be any element in A.	start of an element proof
Then x is in B.	__ (a) __
It follows that x is in C.	__ (b) __
Thus every element in A is in C	since x could be any element of A
Therefore, A \subseteq C [as was to be shown].	__ c __

4. Theorem: For all sets A and B, if A \subseteq B, then A \cup B \subseteq B.

Proof:

Statement	Explanation
Suppose A, B, and C are any sets such that A \subseteq B.	starting point
We must show that A \cup B \subseteq B	conclusion to be shown
Let x be any element in A \cup B.	start of an element proof
Then x is in A or x is in B.	__ (a) __
In case x is in A, then x is in B	__ (b) __
In case x is in B, then x is in B.	tautology (p \to p)
So in either case x is in B.	proof by division into cases
Thus every element in A \cup B is in B	since x could be any element of A \cup B
Therefore, A \cup B \subseteq B [as was to be shown].	__ c __

5. Prove that for all sets A and B, (B - A) = B \cap A^c.

6. Let \cap and \cup stand for the words "intersection" and "union", respectively. Fill in the blanks in the following proof that for all sets A, B, and C, A \cap (B \cup C) = (A \cap C) \cup (A \cap C).

Proof: Suppose A, B, and C are any sets.

(1) Proof that A \cap (B \cup C) \subseteq (A \cap B) \cup (A \cap C):

Let x \in A \cap (B \cup C). [We must show that x \in __ (a) __ ].

By definition of \cap, x \in __ (b) __ and x \in B \cup C.

Thus x \in A and, by definition of \cup, x \in B or __ c __.

Case 1 (x \in A \text{ and } x \in B): In this case, x \in A \cap B by definition of \cap.

Case 2 (x \in A \text{ and } x \in C): IN this case, x \in A \cap C by definition of \cap.

By cases 1 and 2, x \in A \cap B or x \in A \cap C, and so, by definition of \cup, __ (d) __.

[So A \cap (B \cup C) \subseteq (A \cap B) \cup (A \cap C) by definition of subset.]

(2) Proof that (A \cap B) \cup (A \cap C) \subseteq A \cap (B \cup C):

Let x \in (A \cap B) \cup (A \cap C). [We must show that x \in A \cap (B \cup C).]

By definition of \cup, x \in A \cap B __ (a) __ x \in A \cap C.

Case 1 (x \in A \cap B): In this case, by definition of \cap, x \in A and $x \in B$$.

Since x \in B, then x \in B \cup C by definition of \cup.

Case 2 (x \in A \cap C): In this case, by definition of \cap, x \in A __ (b) __ x \in C.

Since x \in C, then x \in B \cup C by definition of \cup.

In both cases x \in A and $$ix \in B \cup C, and so, by definition of \cap, __ c __.

[So (A \cap B) \cup (A \cap C) \subseteq A \cap (B \cup C) by definition of __ (d) __ .]

(3) Conclusion: _[Since both subset relations have been proved, it follows, by definition of set equality, that __ (a) _.]

Use an element argument to prove each statement in 7-22. Assume that all sets are subsets of a universal set U.

7. For all sets A and B, (A \cap B)^c = A^c \cup B^c.

8. For all sets A and B, (A \cap B) \cup (A \cap B^c) = A.

(This property is used in Section 9.9.)

9. For all sets A, B, and C,

 (A - B) \cup (C - B) = (A \cup C) - B 

10. For all sets A, B, and C,

 (A \cup B) \cap C \subseteq A \cup (B \cap C) 

11. For all sets A, B, and C,

 A \cap (B - C) \subseteq (A \cap B) - (A \cap C) 

12. For all sets A, B, and C,

 (A \cup B) - C \subseteq (A - C) \cup (B - C) 

13. For all sets A, B, and C,

 (A - B) \cap (C - B) = (A \cap C) - B 

14. For all sets A and B, A \cup (A \cap B) = A.

15. For every set A, A \cup \emptyset = A.

16. For all sets A, B, and C, if A \subseteq B then A \cap C \subseteq B \cap C.

17. For all sets A, B, and C, if A \subseteq B then A \cup C \subseteq B \cup C.

18. For all sets A and B, if A \subseteq B then B^c \subseteq A^c.

19. For all sets A, B, and C, if A \subseteq B and A \subseteq C then A \subseteq B \cap C.

20. For all sets A, B, and C, if A \subseteq C and B \subseteq C then A \cup B \subseteq C.

21. For all sets A, B, and C,

 A \times (B \cup C) = (A \times B) \cup (A \times C) 

22. For all sets A, B, and C,

 A \times (B \cap C) = (A \times B) \cap (A \times C) 

23. Find the mistake in the following "proof" that for all sets A, B, and C, if A \subseteq B and B \subseteq C then A \subseteq C.

Proof: Suppose A, B, and C are any sets such that A \subseteq B and B \subseteq C. Since A \subseteq B, there is an element x such that x \in A and x \in B, and since B \subseteq C, there is an element x such that x \in B and x \in C. Hence there is an element x such that x \in A and x \in C and so A \subseteq C.

24. Find the mistake in the following "proof."

Theorem: For all sets A and B, A^c \cup B^c \subseteq (A \cup B)^c^c

Proof: Suppose A and B are any sets, and x \in A^c \cup B^c. Then x \in A^c or x \in B^c by definition of union. It follows that x \notin A or x \notin B by definition of complement, and so x \notin A \cup B by definition of union. Thus x \in (A \cup B)^c by definition of complement, and hence A^c \cup B^c \subseteq (A \cup B)^c.

25. Find the mistake in the following "proof" that for all sets A and B, (A - B) \cup (A \cap B) \subseteq A.

Proof: Suppose A and B are any sets, and suppose x \in (A - B) \cup (A \cap B). If x \in A then x \in A - B, and so, by definition of difference, x \in A and x \notin B. In particular, x \in A, and, therefore, (A - B) \cup (A \cap B) \subseteq A by definition of subset.

26. Consider the Venn diagram below.

(See page 429 for image.)

a. Illustrate one of the distributive laws by shading in the region corresponding to A \cup (B \cap C) on one copy of the diagram and (A \cup B) \cap (A \cup C) on another.

b. Illustrate the other distributive law by shading in the region corresponding to A \cap (B \cup C) on one copy of the diagram and (A \cap B) \cup (A \cap C) on another.

c. Illustrate one of De Morgan's laws by shading in the region corresponding to (A \cup B)^c on one copy of the diagram and A^c \cap B^c on the other. (Leave the set C out of your diagrams.)

d. Illustrate the other De Morgan's law by shading in the region corresponding to (A \cap B)^c on one copy of the diagram and A^c \cup B^c on the other. (Leave the set C out of your diagrams.)

27. Fill in the blanks in the following proof that for all sets A and B, (A - B) \cap (B - A) = \emptyset.

Proof:

Let A and B be any sets and suppose (A - B) \cap (B - A) \neq \emptyset. That is, suppose there is an element x in __ (a) __. BY definition of __ (b) __, x \in A - B and x \in __ c __. Then by definition of set difference, x \in A and x \notin B and x \in __ (d) __ and x \notin __ (e) __. IN particular x \in A and x \notin __ (f) __, which is a contradiction. Hence [the supposition that (A - B) \cap (B - A) \neq \emptyset is false, and so] __ (g) __.

Use the element method for proving a set equals the empty set to prove each statement in 28-38. Assume that all sets are subsets of a universal set U.

28. For all sets A and B, (A \cap B) \cap (A \cap B^c) = \emptyset. (This property is used in Section 9.9.)

29. For all sets A, B, and C,

 (A - C) \cap (B - C) \cap (A - B) = \emptyset 

30. For every subset A of a universal set U, A \cap A^c = \emptyset.

31. If U denotes a universal set, then U^c = \emptyset.

32. For every set A, A \times \emptyset = \emptyset.

33. For all sets A and B, if A \subseteq B then A \cap B^c = \emptyset.

34. For all sets A and B, if B \subseteq A^c then A \cap B = \emptyset.

35. For all sets A, B, and C, if A \subseteq B and B \cap C = \emptyset then A \cap C = \emptyset.

36. For all sets A, B, and C, if C \subseteq B - A, then A \cap C = \emptyset.

37. For all sets A, B, and C, if B \cap C \subseteq A, then (C - A) \cap (B - A) = \emptyset.

38. For all sets A, B, C, and D, if A \cap C = \emptyset then (A \times B) \cap (C \times D) = \emptyset.

Prove each statement in 39-44.

39. For all sets A and B,

a. (A - B) \cup (B - A) \cup (A \cap B) = A \cup B

b. The sets (A - B), (B - A), and (A \cap B) are mutually disjoint.

40. For every positive integer n, if A and B_1, B_2, B_3, \dots are any sets, then

 A \cap \left(\bigcup_{i = 1}^{n}B_i\right) = \bigcup_{i = 1}^{n}(A \cap B_i) 

41. For every positive integer n, if A_1, A_2, A_3, \dots and B are any sets, then

 \bigcap_{i = 1}^{n}(A_i - B) = \left(\bigcup_{i = 1}^{n}A_i\right) - B 

42. For every positive integer n, if A_1, A_2, A_3, \dots and B are any sets, then

 \bigcap_{i = 1}^{n}(A_i - B) = \left(\bigcap_{i = 1}^{n}A_i\right) - B 

43. For every positive integer n, if A and B_1, B_2, B_3, \dots are any sets, then

 \bigcup_{i = 1}^{n}(A \times B_i) = A \times \left(\bigcup_{i = 1}^{n}B_i\right) 

44. For every positive integer n, if A and B_1, B_2, B_3, \dots are any sets, then

 \bigcap_{i = 1}^{n}(A \times B_i) = A \times \left(\bigcap_{i = 1}^{n}B_i\right)