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) __.

(1) A

(2) B \cup C

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

(1) A \cap B

(2) C

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

(1) A

(2) B \cap C

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) __.

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

(2) x \in A \cup (B \cap C)

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) __.

(1) A \cap B

(2) A \cap C

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.

(1) A - B

(2) A

(3) A

(4) B

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].

(1) x \in A

(2) A

(3) B

(4) A

In 3 and 4, supply explanations of the steps 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 __

a. by definition of a subset (because A is a subset of B)

b. by definition of a subset (because B is a subset of C)

c. by definition of a subset

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 __

a. by the definition of a union (because A \cup B)

b. by definition of a subset (because A \subseteq B)

c. by definition of a subset

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

Proof:

Let A and B be any sets.

To prove that (B - A) = B \cap A^c, we must first prove (B - A) \subseteq B \cap A^c and then prove B \cap A^c \subseteq (B - A).

Proof ((B - A) \subseteq B \cap A^c):

Suppose x is some element such that x \in (B - A).

By the definition of the difference of sets, this means that x \in B and x \notin A. It then follows by the definition of the complement of sets that x \in B and x \in A^c.

By definition of an intersection, it then follows further that x \in B \cap A^c.

Therefore every element that is in (B - A) is also in B \cap A^c. This is what was to be shown.

Proof (B \cap A^c \subseteq (B - A)):

Suppose x is some element such that x \in B \cap A^c.

By definition of the intersection of sets, this means that x \in B and x \in A^c. By definition of the complement of sets, this means that x \in B and x \notin A.

It follows that if x \in B and x \notin A, then by the definition of the difference of sets x \in (B - A).

Therefore every element that is in B \cap A^c is in (B - A). This is what was to be shown.

Since both relations have been proved, it is concluded that (B - A) = B \cap A^c, by definition of set equality.

Q.E.D.

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.

(1a) x \in (A \cap B) \cup (A \cap C)

(1b) A

(1c) x \in C

(1d) x \in (A \cap B) \cup (A \cap C)

(2a) or

(2b) and

(2c) x \in A \cap (B \cup C)

(2d) subset

(3a) for all sets A, B, and C, A \cap (B \cup C) = (A \cap B) \cup (A \cap C)

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

Proof:

Let A and B be any sets.

To prove that (A \cap B)^c = A^c \cup B^c, we must show that (A \cap B)^c \subseteq A^c \cup B^c and also show that A^c \cup B^c \subseteq (A \cap B)^c.

Proof ((A \cap B)^c \subseteq A^c \cup B^c):

Suppose x is some element such that x \in (A \cap B)^c.

By the definition of complement, this means that x \notin (A \cap B).

By the definition of intersection of sets (and by De Morgan's Laws of negation of sets), this means that x \notin A or x \notin B. It follows by the definition of complement of sets that x \in A^c or x \in B^c.

Thus, by definition of the union of sets, x \in A^c \cup B^c.

Therefore all elements in (A \cap B)^c are in A^c \cup B^c. Specifically, (A \cap B)^c \subseteq A^c \cup B^c.

This is what was to be shown.

Proof (A^c \cup B^c \subseteq (A \cap B)^c):

Suppose x is some element such that x \in A^c \cup B^c. By the definition of the union of sets, x \in A^c or x \in B^c. By the definition of complement of sets, this means that x \notin A or x \notin B.

By definition of the union of sets, it follows that x \in A^c \cup B^c. Then by De Morgan's Laws of Sets, x \in (A \cap B)^c.

Therefore all elements in A^c \cup B^c are in (A \cap B)^c. Specifically, A^c \cup B^c \subseteq (A \cap B)^c.

This is what was to be shown.

Conclusion:

Since both sets have been shown to be subsets of the other, it is concluded that A^c \cup B^c = (A \cap B)^c, by definition of set equality.

Q.E.D.

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

(This property is used in Section 9.9.)

Proof:

Let A and B be any sets.

To prove that (A \cap B) \cup (A \cap B^c) = A, it must be shown that (A \cap B) \cup (A \cap B^c) \subseteq A and also that A \subseteq (A \cap B) \cup (A \cap B^c).

Proof ((A \cap B) \cup (A \cap B^c) \subseteq A):

Suppose x is any element such that x \in (A \cap B) \cup (A \cap B^c).

By definition of \cup, this means that x \in A \cap B or x \in A \cap B^c

Case x \in A \cap B:

By definition of \cap, this means that x \in A and x \in B. By definition of \subseteq, if x \in A and x \in B, then A \cap B \subseteq A.

Case x \in A \cap B^c:

By definition of \cap, this means that x \in A and x \in B^c. By definition of complement, this means that x \in A and x \notin B. By tautology and by definition of \subseteq, if x \in A and x \notin B, then x \in A \cap B^c \subseteq A.

Thus in both cases, it has been shown that any element in A \cap B or A \cap B^c is in A. Specifically (A \cap B) \cup (A \cap B^c) \subseteq A.

This is what was to be shown.

Proof (A \subseteq (A \cap B) \cup (A \cap B^c)):

Suppose x is any element such that x \in A.

To prove x \in A \subseteq (A \cap B) \cup (A \cap B^c), we must prove that either x \in B or x \in B^c (by definition of \cup and the complement of sets.)

Case x \in B:

By the supposition, it is known that x \in A and x \in B. By the definition of \cap, this means that x \in A \cap B.

Case x \notin B:

By the supposition, it is known that x \in A and x \notin B. By the definition of \cap and the complement of sets, this means that x \in A \cap B^c.

In the case that x \in B, it has been shown that then x \in A \cap B. In the case that x \notin B, it has been shown that x \in A \cap B^c. Thus it can be stated that x \in A \cap B or x \in A \cap B^c. By the definition of \cup, it follows that x \in (A \cap B) \cup (A \cap B^c).

Therefore it can be said that every element in A is in (A \cap B) \cup (A \cap B^c). Specifically, A \subseteq (A \cap B) \cup (A \cap B^c).

This is what was to be shown.

Conclusion:

It has been shown that (A \cap B) \cup (A \cap B^c) \subseteq A and A \subseteq (A \cap B) \cup (A \cap B^c). By the definition of the equality of sets, this means that (A \cap B) \cup (A \cap B^c) = A.

Q.E.D.

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

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

Proof:

Let A, B, and C be any sets.

To prove (A - B) \cup (C - B) = (A \cup C) - B, it must be shown that (A - B) \cup (C - B) \subseteq (A \cup C) - B and that (A \cup C) - B \subseteq (A - B) \cup (C - B).

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

Suppose x is some element such that x \in (A - B) \cup (C - B).

By the definition of \cup, this means that x \in (A - B) or x \in (C - B).

In the case that x \in (A - B), x \in A and x \notin B. In the case that x \in (C - B), x \in C and x \notin B. In both cases x \notin B.

It follows that x \in A or x \in C. Specifically x \in (A \cup C). In either case, x \notin B. Hence, by the definition of difference of sets, x \in (A \cup C) - B.

Therefore every element in (A - B) \cup (C - B) is in (A \cup C) - B. Specifically, (A - B) \cup (C - B) \subseteq (A \cup C) - B.

This is what was to be shown.

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

Suppose x is some element such that x \in (A \cup C) - B.

By the definition of difference of sets, this means that x \in (A \cup C) and x \notin B. By the definition of \cup, this means that x \in A or x \in C.

In the case that x \in A, then x \in A and x \notin B. By the definition of complements, this means that x \in A \cap B^c. It follows by the set difference law, that x \in A - B.

In the case that x \in C, then x \in A and x \notin B. By the definition of complements, this means that x \in C \cap B^c. It follows by the set difference law, that x \in C - B.

Thus it can be said that x \in A - B or x \in C - B. By the definition of \cup, it follows that x \in (A - B) \cup (C - B).

Therefore every element in (A \cup C) - B is in (A - B) \cup (C - B). Specifically, A \cup C - B \subseteq (A - B) \cup (C - B).

This is what was to be shown.

Conclusion:

Since both subset relations have been proved, it is concluded that (A - B) \cup (C - B) = (A \cup C) - B by definition of set equality.

Q.E.D.

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

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

Proof:

Let A, B, and C be any sets.

Suppose x is some element such that x \in (A \cup B) \cap C.

By the definition of \cap, this means that x \in (A \cup B) and x \in C. By the definition of \cup, this means that x \in A or x \in B.

Case x \in A:

Since x \in A and x \in C, it follows that x \in A \cup (B \cap C), since x \in A.

Case x \in B:

Since x \in A and x \in C, this means that x \in B \cap C. It follows that x \in A \cup (B \cap C), since x \in B \cap C.

Thus in both cases x \in A \cup (B \cap C).

Therefore every element in (A \cup B) \cap C is in A \cup (B \cap C). By the definition of a subset, this means that (A \cup B) \cap C \subseteq A \cup (B \cap C).

This is what was to be shown.

Q.E.D.

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

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

Proof:

Let A, B, and C be any sets.

Suppose x is some element such that x \in A \cap (B - C).

By definition of \cap, this means that x \in A and x \in (B - C). By the definition of difference of sets, this means that x \in A and x \in B and x \notin C.

Since x \in A and x \in B, it follows that x \in A \cap B.

Since x \in A and x \notin C, by the definition of complement, it can be said that x \in A \cap C^c, or x \notin A \cap C.

Thus x \in A \cap B and x \notin A \cap C. Hence, by the difference of sets, x \in (A \cap B) - (A \cap C).

Therefore it can be said that every element in A \cap (B - C) is in (A \cap B) - (A \cap C). Specifically A \cap (B - C) \subseteq (A \cap B) - (A \cap C).

This is what was to be shown.

Q.E.D.

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

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

Proof:

Let A, B, and C be any sets.

Suppose x is some element such that x \in (A \cup B) - C.

By the definition of difference, this means that x \in A \cup B and x \notin C.

By the definition of \cup, it follows that x \in A or x \in B.

Case x \in A:

Since x \in A and x \notin C, by the definition of difference, it can be said that x \in A - C.

Case x \in B:

Since x \in B and x \notin C, by the definition of difference, it can be said that x \in B - C.

Hence it can be said that x \in A - C or x \in B - C. By the definition of \cup, it follows that x \in (A - C) \cup (B - C).

Therefore it can said that any element in (A \cup B) - C is also in (A - C) \cup (B - C). Specifically, by definition of a subset, (A \cup B) - C \subseteq (A - C) \cup (B - C).

This is what was to be shown.

Q.E.D.

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

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

Let A, B, and C be any sets.

To prove (A - B) \cap (C - B) = (A \cap C) - B, it must be shown that (A - B) \cap (C - B) \subseteq (A \cap C) - B and that (A \cap C) - B \subseteq (A - B) \cap (C - B).

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

Suppose x is some element such that x \in (A - B) \cap (C - B).

By the definition of \cap, this means that x \in A - B and x \in C - B.

By the definition of difference, this means that x \in A and x \notin B and x \in C and x \notin B.

Thus x is in A and C, or (by definition of \cap), x \in A \cap C. Since x \notin B, it follows then that x \in (A \cap C) \cap B^c, by the definition of complement.

By the set difference law, it follows that x \in (A \cap C) - B.

Thus every element in (A - B) \cap (C - B) is in (A \cap C) - B. By the definition of subset, it follows that (A - B) \cap (C - B) \subseteq (A \cap C) - B.

This is what was to be shown.

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

Suppose x is some element such that x \in (A \cap C) - B.

By the definition of difference this means that x \in A \cap C and x \notin B. By the definition of \cap, this means that x \in A or x \in C and x \notin B.

Case x \in A:

Since x \in A and x \notin B, this means that x \in A \cap B^c. By the set difference law, this means that x \in A - B.

Case x \in C:

Since x \in C and x \notin B, this means that x \in C \cap B^c. By the set difference law, this means that x \in C - B.

It follows that x \in A - B or x \in C - B. By the definition of \cap, this means that x \in (A - B) \cap (C - B).

Thus every element in (A \cap C) - B is in (A - B) \cap (C - B). By the definition of subset, this means that (A \cap C) - B \subseteq (A - B) \cap (C - B).

This is what was to be shown.

Q.E.D.

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

Proof:

Let A and B be any sets.

To prove A \cup (A \cap B) = A, it must be shown that A \cup (A \cap B) \subseteq A and A \subseteq A \cup (A \cap B).

Proof (A \cup (A \cap B) \subseteq A):

Suppose x is some element such that x \in A \cup (A \cap B).

By the definition of \cup, this means that x \in A or x \in A \cap B.

Case x \in A:

Since x \in A, by tautology, x \in A.

Case x \in A \cap B:

By the definition of \cap, x \in A and x \in B.

In either case x \in A. By the definition of subset, this means that x \subseteq A.

Thus every element in A \cup (A \cap B) is in A. By the definition of subset, this means that A \cup (A \cap B) \subseteq A.

This is what was to be shown.

Proof (A \subseteq A \cup (A \cap B)):

Suppose x is some element such that x \in A.

By tautology, x \in A \to x \in A.

Since x \in A, x \in A \cap B, by virtue of x \in A.

It follows that x \in A or x \in A \cap B.

Thus it can be said that every element in A is in A \cup (A \cap B). By definition of subset, this means that A \subseteq A \cup (A \cap B).

This is what was to be shown.

Conclusion:

Since both subset relations have been proved, it has been shown that A \cup (A \cap B) = A.

Q.E.D.

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

Proof:

Let A be any set.

To prove that A \cup \emptyset = A, it must be shown that A \cup \emptyset \subseteq A, and that A \subseteq A \cup \emptyset.

Proof (A \cup \emptyset \subseteq A):

Suppose x is some element such that x \in A \cup \emptyset.

By the definition of \cup, this means that x \in A or x \in emptyset. But x \notin \emptyset, as \emptyset can have no elements.

Hence x \in A, and therefore A \cup \emptyset \subseteq A.

Proof (A \subseteq A \cup \emptyset):

Suppose x is some element such that x \in A. It follows that x \in A or x \in \emptyset. By definition of \cup, this means that x \in A \cup \emptyset.

Therefore A \subseteq A \cup \emptyset.

Since both subset relations have been proved, it can be said that A \cup \emptyset = A by the definition of set equality.

This is what was to be proved.

Q.E.D.

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

Proof:

Let A, B, and C be any sets such that A \subseteq B.

Suppose x is some element such that x \in A \cap C.

By the definition of \cap, this means that x \in A and x \in C.

Since x \in A and A \subseteq B, then x \in B by definition of subset.

Since x \in B and x \in C, by the definition of \cap, it can be said that x \in B \cap C.

Thus it has been shown that any element in A \cap C is in B \cap C. Specifically A \cap C \subseteq B \cap C by the definition of subset.

This is what was to be shown.

Q.E.D.

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

Proof:

Let A, B, and C be any sets such that A \subseteq B.

Suppose x is some element such that x \in A \cup C.

By definition of \cup, this means that x \in A or x \in C.

Case x \in A:

Since x \in A and since A \subseteq B, this means that x \in B.

Case x \in C:

By tautology, x \in C.

It follows that x \in B or x \in C. By the definition of \cup, this is x \in B \cup C.

Thus it can be said that any element in A \cup C is in B \cup C, or A \cup C \subseteq B \cup C by the definition of subset.

This is what was to be shown.

Q.E.D.

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

Proof:

Let A and B be any sets such that A \subseteq B.

Suppose x is some element such that x \in B^c.

By the definition of complement, this means that x \notin B.

Since A \subseteq B, it follows that any element not in B is not in A, thus x \notin A. By the definition of complement, this means that x \in A^c

Hence it can said that any element in B^c is in A^c, or B^c \subseteq A^c by the definition of subset.

This is what was to be shown.

Q.E.D.

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

Proof:

Let A, B, and C be any sets such that A \subseteq B and A \subseteq C.

Suppose x is some element such that x \in A.

Since x \in A and A \subseteq B, it follows by the definition of subset that x \in B.

Since x \in A and A \subseteq C, it follows by the definition of subset that x \in C.

Thus it can be said that x \in B and $x \in C. By the definition of \cap, this is x \in B \cap C.

Therefore it has been shown that any element in A is in B \cap C, or A \subseteq B \cap C, by the definition of subset.

This is what was to be shown.

Q.E.D.

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

Proof:

Let A, B, and C be any sets such that A \subseteq C and B \subseteq C.

Suppose x is some element such that x \in A \cup B.

By the definition of \cup, this means that x \in A or x \in B.

Case x \in A:

Since x \in A and A \subseteq C, this means, by definition of subset, that x \in C.

Case x \in B:

Since x \in B and B \subseteq C, this means, by definition of subset, that x \in C.

In either case, x \in C.

Therefore it can be said that any element in A \cup B is in C, or A \cup B \subseteq C, by definition of subset.

This is what was to be shown.

Q.E.D.

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

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

Proof:

Let A, B, and C be any set.

To prove A \times (B \cup C) = (A \times B) \cup (A \times C), it must be shown that A \times (B \cup C) \subseteq (A \times B) \cup (A \times C) and (A \times B) \cup (A \times C) \subseteq A \times (B \cup C).

Proof (A \times (B \cup C) \subseteq (A \times B) \cup (A \times C)):

Suppose (x, y) are any Cartesian pair such that (x, y) \in A \times (B \cup C).

By the definition of a Cartesian pair, this means that x \in A and y \in B \cup C.

By definition of \cup, this means that y \in B or y \in C.

Case y \in B:

Since x \in A and y \in B, by definition of Cartesian product, (x, y) \in A \times B.

Case y \in C:

Since x \in A and y \in C, by definition of Cartesian product, (x, y) \in A \times C.

Thus it can be said that (x, y) \in A \times B or (x, y) \in A \times C. By definition of \cup, this is (x, y) \in (A \times B) \cup (A \times C).

Thus it can be said that any Cartesian pair of elements in A \times (B \cup C) are in (A \times B) \cup (A \times C), or A \times (B \cup C) \subseteq (A \times B) \cup (A \times C).

This is what was to be shown.

Proof ((A \times B) \cup (A \times C) \subseteq A \times (B \cup C)):

Suppose (x, y) are some Cartesian pair such that (x, y) \in (A \times B) \cup (A \times C).

By the definition of \cup, this means that (x, y) \in (A \times B) or (x, y) \in (A \times C).

Case (x, y) \in (A \times B):

This means that x \in A and y \in B. Since y \in B, it follows that y \in B \cup C, by virtue of y \in B.

Thus it can be said, by the definition of Cartesian product, that (x, y) \in A \times (B \cup C).

Case (x, y) \in (A \times C):

This means that x \in A and y \in C. Since y \in C, it follows that y \in B \cup C, by virtue of y \in C.

Thus it can be said, by the definition of Cartesian product, that (x, y) \in A \times (B \cup C).

Hence in both cases (x, y) \in A \times (B \cup C).

Thus it has been shown that every Cartesian pair in (A \times B) \cup (A \times C) is in A \times (B \cup C), or (A \times B) \cup (A \times C) \subseteq A \times (B \cup C).

This is what was to be shown.

Conclusion:

Since both subset relations have been proven, it can be concluded that A \times (B \cup C) = (A \times B) \cup (A \times C) by the definition of set equality.

This is what was to be shown.

Q.E.D.

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

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

Proof:

Let A, B, and C be any sets.

To prove A \times (B \cap C) = (A \times B) \cap (A \times C), it must be shown that A \times (B \cap C) \subseteq (A \times B) \cap (A \times C) and that (A \times B) \cap (A \times C) \subseteq A \times (B \cap C).

Proof (A \times (B \cap C) \subseteq (A \times B) \cap (A \times C)):

Suppose (x, y) be some elements such that (x, y) \in A \times (B \cap C).

This means that x \in A and y \in B \cap C.

By the definition of \cap, this means that y \in B and y \in C.

Since x \in A and y \in B, this means that (x, y) \in A \times B (by the definition of Cartesian product).

Furthermore, since x \in A and y \in C, this means that (x, y) \in A \times C (by the definition of Cartesian product).

Thus (x, y) \in A \times B and (x, y) \in A \times C or (x, y) \in (A \times B) \cap (A \times C) (by the definition of \cap).

Hence it has been shown that A \times (B \cap C) \subseteq (A \times B) \cap (A \times C).

Proof ((A \times B) \cap (A \times C) \subseteq A \times (B \cap C)):

Suppose (x, y) be some elements such that (x, y) \in (A \times B) \cap (A \times C).

By the definition of \cap, this means that (x, y) \in A \times B and (x, y) \in A \times C.

Since (x, y) \in A \times B, x \in A and y \in B.

Since (x, y) \in A \times C, this means that x \in A and y \in C.

Since y \in B and y \in C, y \in B \cap C (by the definition of \cap).

Since x \in A and y \in B \cap C, by the definition of Cartesian product, (x, y) \in A \times (B \cap C).

Hence it has been shown that (A \times B) \cap (A \times C) \subseteq A \times (B \cap C).

Conclusion:

Since both subset relations have been proven, it is concluded that A \times (B \cap C) = (A \times B) \cap (A \times C) by the definition of set equality.

This is what was to be shown.

Q.E.D.

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.

There is more than one error in this "proof." The most serious is the misuse of the definition of subset. To say that A is a subset of B means that for every x, if x \in A then x \in B. It does not mean that there exists an element of A that is also an element of B. The second error in the proof occurs in the last sentence. Even if there is an element in A that is in B and an element in B that is in C, it does not follow that there is an element in A that is in C.

For instance, suppose A = \{1, 2\}, B = \{2, 3\}, and C = \{3, 4\}. Then there is an element in A that is in B (namely 2) and there is an element in B that is in C (namely, 3), but there is no element in A that is in 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.

The mistake in this "proof" occurs when the author misuses the definition of union in the sentence "and so x \notin A \cup B by definition of union."

For example, take x = 1, A = \{1\} and B = \{2\}, then A^c \cup B^c is true since B^c is true, but x \notin A \cup B (x \in (A \cup B)^c) is false since 1 \in \{1, 2\}.

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.

The author of this "proof" makes a mistake when they assume the conclusion, "If x \in A." The supposition should be "Suppose x is some element such that $x \in (A - B) \cup (A \cap B)$" and follow from there.

Furthermore it does not follow that if x \in A, then x \in A - B. Suppose A = B = \{x\}, then x \in A, but A - B = \emptyset, and by definition of \emptyset, x \notin \emptyset, so x \notin A - B.

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.

Omitted.

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.

Omitted.

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.)

Omitted.

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.)

Omitted.

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) __.

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

b. intersection

c. B - A

d. B

e. A

f. A

g. (A - B) \cap (B - A) = \emptyset

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.)

Proof (by contradiction):

Let A and B be any sets and suppose (A \cap B) \cap (A \cap B^c) \neq \emptyset.

Suppose x is some element such that x \in (A \cap B) \cap (A \cap B^c).

By the definition of \cap, this means that x \in (A \cap B) and x \in (A \cap B^c).

Since x \in (A \cap B), this means that x \in A and x \in B.

Since x \in (A \cap B^c), this means that x \in A and x \notin B.

So x \in B and x \notin B, which is a contradiction.

Hence the supposition is false, and therefore (A \cap B) \cap (A \cap B^c) = \emptyset.

Q.E.D.

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

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

Proof (by contradiction):

Let A, B, and C be any sets, and suppose (A - C) \cap (B - C) \cap (A - B) \neq \emptyset.

Suppose x is some element such that x \in (A - C) \cap (B - C) \cap (A - B).

By the definition of \cap, this means that x \in (A - C) and x \in (B - C) and x \in (A - B).

By the definition of difference, this means that x \in A and x \notin C and x \in B and x \notin C and x \in A and x \notin B.

So x \in B and x \notin B, which is a contradiction.

Hence the supposition is false, and therefore (A - C) \cap (B - C) \cap (A - B) = \emptyset.

Q.E.D.

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

Proof (by contradiction):

Let A be any set and suppose A \cap A^c \neq \emptyset.

Suppose x is some element such that x \in A \cap A^c.

By the definition of \cap, this means that x \in A and x \in A^c.

By the definition of complement, this means that x \in A and x \notin A, which is a contradiction.

Hence the supposition is false, and therefore A \cap A^c = \emptyset.

Q.E.D.

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

Proof (by contradiction):

Let U be the universal set of all elements, and suppose U^c \neq \emptyset.

Suppose x is some element such that x \in U^c.

By definition of complement, this means that x \notin U.

Since U is the universal set of all elements, it follows that x \in U.

So x \notin U and x \in U, which is a contradiction.

Hence the supposition is false, and therefore U^c = \emptyset.

Q.E.D.

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

Proof (by contradiction):

Let A be any set and suppose A \times \emptyset \neq \emptyset.

Suppose (x, y) are any element pair such that (x, y) \in A \times \emptyset.

By the definition of Cartesian product, this means that x \in A and y \in emptyset. By the definition of \emptyset, y \notin \emptyset.

So y \in \emptyset and y \notin \emptyset, which is a contradiction.

Hence the supposition is false, and therefore A \times \emptyset = \emptyset.

Q.E.D.

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

Proof (by contradiction):

Let A and B be any sets such that A \subseteq B.

Suppose A \cap B^c \neq \emptyset. Then let x be some element such that x \in A \cap B^c.

By the definition of \cap, this means that x \in A and x \in B^c. By the definition of complement, this means that x \in A and x \notin B.

Since x \in A and A \subseteq B, it follows that x \in B by definition of subset.

So x \notin B and x \in B, which is a contradiction.

Hence the supposition is false, therefore A \cap B^c = \emptyset.

Q.E.D.

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

Proof (by contradiction):

Let A and B be any sets such that B \subseteq A^c.

Suppose A \cap B \neq \emptyset. Then let x be some element such that x \in A \cap B.

By the definition of \cap, this means that x \in A and x \in B.

Since x \in B and B \subseteq A^c, it follows that x \notin A.

So x \in A and x \notin A, which is a contradiction.

Hence the supposition is false, and therefore A \cap B = \emptyset.

Q.E.D.

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

Proof (by contradiction):

Let A, B, and C be any sets such that A \subseteq B and B \cap C = \emptyset.

Suppose A \cap C \neq \emptyset, then let x be some element such that x \in A \cap C.

By the definition of \cap, this means that x \in A and x \in C.

Since x \in A and A \subseteq B, then x \in B by definition of subset.

Thus x \in B and x \in C, which is, by definition of \cap, x \in B \cap C.

B \cap C = \emptyset, so x \in emptyset.

But by the definition of \emptyset, x \notin \emptyset.

So x \in \emptyset and x \notin \emptyset, which is a contradiction.

Hence the supposition is false, and therefore A \cap C = \emptyset.

Q.E.D.

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

Proof (by contradiction):

Let A, B, and C be any sets such that C \subseteq B - A.

Suppose A \cap C \neq \emptyset, then let x be some element such that x \in A \cap C.

By the definition of \cap, this means that x \in A and x \in C.

Since x \in C, and C \subseteq B - A, this means that x \in B - A. Furthermore, by the definition of difference, this means that x \in B and x \notin A.

So x \in A and x \notin A, which is a contradiction.

Hence the supposition is false, and therefore A \cap C = \emptyset.

Q.E.D.

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

Proof (by contradiction):

Let A, B, and C be any sets such that B \cap C \subseteq A.

Suppose (C - A) \cap (B - A) \neq \emptyset, then let x be some element such that x \in (C - A) \cap (B - A).

By the definition of \cap, this means that x \in (C - A) and x \in (B - A).

By the definition of difference, this means that x \in C and x \notin A and x \in B and x \notin A.

Since x \in B and x \in C, this means that x \in B \cap C.

B \cap C \subseteq A, so x \in A, by definition of subset.

So x \notin A and x \in A, which is a contradiction.

Hence the supposition is false, and therefore (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.

Omitted.

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

Omitted.

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

Omitted.

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) 

Omitted.

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 

Omitted.

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 

Omitted.

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) 

Omitted.

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) 

Omitted.

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Page 435

Exercise Set 6.3

For each of 1-4 find a counterexample to show that the statement is false. Assume all sets are subsets of a universal set U.

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

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

Disproof (by counterexample):

Let A, B, and C be any sets where A has an element not in C. For example:

 A = \{1, 2\}, B = \{2\}, C = \{2\} 

Then, the left-hand side of the given equation becomes:

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

 (A \cup B) \cap C = \{\2\} 

Then, the right-hand side of the given equation becomes:

 B \cap C = \{2\} 

 A \cup (B \cap C) = \{1, 2\} 

Thus we can see that:

 (A \cup B) \cap C = \{2\} \neq \{1, 2\} = A \cup (B \cap C) 

as $1 \in A \cup (B \cap C), but 1 \notin (A \cup B) \cap C.

Hence it has been shown that (A \cup B) \cap C \neq A \cup (B \cap C) by definition of subset.

Q.E.D.

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

Disproof (by counterexample):

Let U = \{1, 2\}, A = \{1\}, B = \{2\}.

Then, left-hand side:

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

 (A \cup B)^c = U - \{1, 2\} 

 = \emptyset 

Then, right-hand side:

 A^c = U - \{1\} 

 B^c = U - \{2\} 

 A^c \cup B^c = (U - \{1\}) \cup (U - \{2\}) 

 = \{2\} \cup \{1\} 

 = \{1, 2\} 

Thus the given equality does not hold.

 (A \cup B)^c = \emptyset \neq \{1, 2\} = A^c \cup B^c 

Therefore it has been shown that (A \cup B)^c \neq A^c \cup B^c by the definition of subset.

Q.E.D.

3. For all sets A, B, and C, if A \nsubseteq B and B \nsubseteq C then A \nsubseteq C.

Disproof (by counterexample):

Suppose A, B, and C are any sets such that A \nsubseteq B and B \nsubseteq C, but A has an element that is in C. For example:

Let A = \{1\}, B = \{2\}, C = \{1, 3\}.

It is trivially true that A \nsubsteq B and B \nsubseteq C (since B does not have any elements that are in A), but notice that A \subseteq C since \{1\} \in \{1, 3\}.

Therefore the supposition is false by the definition of subset.

Q.E.D.

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

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

Disproof (by counterexample):

Suppose A, B, and C are any sets such that B \cup C \subseteq A. For example:

Let A, = \{1, 2, 3\}, B = \{2\}, C =\{3\}.

It is trivially true that B \cup C \subseteq A, since \{2\} \cup \{3\} = \{2, 3\} \subseteq \{1, 2, 3\}.

Then evaluating the left-hand side of the given equality:

 (A - B) \cap (A - C) = (\{1, 2, 3\} - \{2\} ) \cap (\{1, 2, 3\} - \{3\}) 

 = \{1, 3\} \cap \{1, 2\} 

 = \{1\} 

And note by the definition of \emptyset that:

 \{1\} \neq \emptyset 

Therefore the supposition is false since (A - B) \cap (A - C) \neq \emptyset.

Q.E.D.

For each of 5-21 prove each statement that is true and find a counterexample for each statement that is false. Assume all sets are subsets of a universal set U.

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

 A - (B - C) = (A - B) - C 

Disproof (by counterexample):

Let A = \{1, 2, 3\}, B = \{2, 3\}, C = \{3\}.

Then, evaluating the left-hand side of the given equality:

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

 = \{1, 2, 3\} - \{2\} 

 = \{1, 3\} 

Then, evaluating the right-hand side of the given equality:

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

= {1} - {3}

= {1}

Notice that:

 A - (B - C) = \{1, 3\} \neq \{1\} = (A - B) - C 

Therefore the supposition is false since A - (B - C) \neq (A - B) - C.

Q.E.D.

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

Proof:

Let A and B be any sets.

To prove A \cap (A \cup B) = A, it must be shown that A \cap (A \cup B) \subseteq A and also A \subseteq A \cap (A \cup B).

Proof A \cap (A \cup B) \subseteq A:

Suppose x is any element such that x \in A \cap (A \cup B).

By the definition of \cap, this means that x \in A and x \in A \cup B. Furthermore, by the definition of \cup, this means that x \in A and x \in A or x \in B. In particular, it is known in either case that x \in A.

Therefore, by definition of a subset, A \cap (A \cup B) \subseteq A.

This is what was to be shown.

Proof A \subseteq A \cap (A \cup B):

Suppose x is any element such that x \in A. By definition of \cup, this means that x \in A \cup B. Furthermore, by definition of \cap, it follows that x \in A \cap (A \cup B).

Therefore, by definition of subset, A \subseteq A \cap (A \cup B).

This is what was to be shown.

Therefore, since both subset relations have been proved, it can be concluded that A \cap (A \cup B) = A.

Q.E.D.

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

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

Disproof (by counterexample):

Suppose A, B, and C are any sets such that they are disjoint. For example:

Let A = \{1\}, B = \{2\}, C = \{3\}.

Then evaluating the left-hand side of the given equality:

 (A - B) \cap (C - B) = (\{1\} - \{2\}) \cap (\{3\} - \{2\}) 

 = \{1\} \cap \{3\} 

 = \emptyset 

Then evaluating the right-hand side of the given equality:

 A - (B \cup C) = \{1\} - (\{2\} \cup \{3\}) 

 = \{1\} - {2, 3\} 

 = \{1\} 

Notice that:

 (A - B) \cap (C - B) = \emptyset \neq \{1\} = A - (B \cup C) 

Therefore, the supposition is false since (A - B) \cap (C - B) \neq A - (B \cup C).

Q.E.D.

8. For all sets A and B, if A^c \subseteq B then A \cup B = U.

Proof:

Let A and B be any sets such that A^c \subseteq B.

In order to prove that A \cup B = U, it must be shown that A \cup B \subseteq U and U \subseteq A \cup B.

Proof A \cup B \substeq U:

Suppose x is any element such that x \in A \cup B.

By definition of \cup, this means that x \in A or x \in B. Regardless of whether x \in A or x \in B, x \in U, by definition of the universal set.

Therefore it has been shown that A \cup B \subseteq U.

Proof U \subseteq A \cup B:

Suppose x is any element. Since x is any element, by definition of the universal set U, x \in U.

It follows then that either x \in A or x \in A^c.

Case x \in A:

Since x \in A, by the definition of \cup, x \in A \cup B.

Case x \in A^c:

Since x \in A^c, by the supposition, this means that x \in B since A^c \subseteq B.

Since x \in B, by definition of \cup, x \in A \cup B.

In either case x \in A \cup B. This is what was to be shown.

Conclusion:

Since both subset relations have been proved, it can be concluded that A \cup B = U.

Q.E.D.

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

Proof:

Suppose A, B, and C are any sets such that A \subseteq C and B \subseteq C.

Let x be any element such that x \in A \cup B.

By definition of \cup, this means that x \in A or x \in B.

Case x \in A:

Since x \in A, x \in C since A \subseteq C (by the supposition and the definition of subset).

Case x \in B:

Since x \in B, x \in C since B \subseteq C (by the supposition and the definition of subset).

Conclusion:

In either case, x \in C, so by definition of subset, it can be concluded that A \cup B \subseteq C.

Q.E.D.

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

Proof (by contradiction):

Let A and B be any sets such that A \subseteq B and A \cap B^c \neq \emptyset.

Let x be any element such that x \in A \cap B^c

By definition of \cap, this means that x \in A and x \in B^c. By definition of complement, it follows that this means that x \in A and x \notin B.

Since x \in A, x \in B since A \subseteq B (by the supposition and by the definition of subset).

Thus x \in B and x \notin B. This is a contradiction.

Hence the supposition is false, and A \cap B^c = \emptyset.

Q.E.D.

11. For all sets A, B, and C, if A \subseteq B then A \cap (B \cap C)^c = \emptyset.

Disproof (by counterexample):

Suppose A, B, C, are any sets and U is the universal set such that A \subseteq B and C \subseteq A and C \subseteq B. For example:

Let U = \{1, 2, 3, 4\}, A = \{1, 2\}, B = \{1, 2, 3\}, C = \{2\}.

Then, the left-hand side of the given equality becomes:

 A \cap (B \cap C)^c = \{1, 2\} \cap (\{1, 2, 3\} \cap \{2\})^c 

 = \{1, 2\} \cap (\{2\})^c 

 = \{1, 2\} \cap (U - \{2\}) 

 = \{1, 2\} \cap (\{1, 2, 3, 4\} - \{2\}) 

 = \{1, 2\} \cap \{1, 3, 4\} 

 = \{1\} 

Thus A \cap (B \cap C)^c \neq \emptyset since \{1\} \neq \emptyset.

Q.E.D.

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

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

Proof:

Suppose A, B, and C are any sets.

To prove A \cap (B - C) = (A \cap B) - (A \cap C), it must be shown that A \cap (B - C) \subseteq (A \cap B) - (A \cap C) and that (A \cap B) - (A \cap C) \subseteq A \cap (B - C).

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

Let x be any element such that x \in A \cap (B - C).

By the definition of \cap, this means that x \in A and x \in (B - C). By the definition of difference, this means x \in A and x \in B and x \notin C.

Since x \in A and x \in B, it follows that x \in A \cap B (by the definition of \cap).

Since x \in A and x \notin C, it follows that x \notin A \cap C (by the definition of \cap.)

Hence x \in (A \cap B) - (A \cap C), by the definition of difference.

Therefore it has been shown that A \cap (B - C) \subseteq (A \cap B) - (A \cap C), which is what was to be shown.

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

Let x is any element such that x \in (A \cap B) - (A \cap C).

This can be rewritten as x \in (A \cap B) \cap (A \cap C)^c, by the set difference law.

Furthermore, this can be written as x \in (A \cap B) \cap (A^c \cup C^c) by De Morgan's Laws of sets.

This means that x \in A and x \in B and x \notin A or x \notin C.

Since we know that x \in A, then x \notin C must be true.

Hence x \in A and x \in B and x \notin C, or:

 x \in A \cap (B - C) 

Therefore it has been shown that (A \cap B) - (A \cap C) \subseteq A \cap (B - C).

This is what was to be shown.

Conclusion:

Since both subset relations have been proved, it can be concluded that A \cap (B - C) = (A \cap B) - (A \cap C).

Q.E.D.

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

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

Disproof (by counterexample):

Suppose A, B, and C are any sets such that they are disjoint. For example:

Let A = \{1}, B = \{2\}, C = \{3\}.

The left-hand side of the equality is:

 A \cup (B - C) = \{1\} \cup (\{2\} - \{3\}) 

 = \{1\} \cup \{2\} 

 = \{1, 2\} 

The right-hand side of the equality is:

 (A \cup B) - (A \cup C) = (\{1\} \cup \{2\}) - (\{1\} \cup \{3\}) 

 = \{1, 2\} - \{1, 3\} 

 = \{2\} 

Note that:

 A \cup (B - C) = \{1, 2\} \neq \{2\} = (A \cup B) - (A \cup C) 

Therefore the supposition has been disproven.

Q.E.D.

14. For all sets A, B, and C, if A \cap C = B \cap C and A \cup C = B \cup C, then A = B.

Proof:

Let A, B, and C be any sets such that A \cap C = B \cap C and A \cup C = B \cup C.

To prove A = B, it must be shown that A \subseteq B, and B \subseteq A.

Proof A \buseteq B:

Suppose $x is any element such that x \in A.

By the supposition, we know that A \cap C \subseteq B \cap C since A \cap C = B \cap C.

Therefore since x \in A, we must determine if x \in C to determine if A \cap C is true.

Case x \in C:

If x \in C, then, since x \in A also, it follows that A \cap C is true, and thus by the supposition A \cap C = B \cap C is true.

It follows that x \in B \cap C, and thus x \in B.

Case x \notin C:

If x \notin C, then A \cup C is true (by virtue of x \in A). Then, by supposition, A \cup C = B \cup C. It follows that x \in B \cup C. Since x\notin C, this means that x \in B.

In either case x \in B. Therefore it has been shown that A \subseteq B.

Proof B \buseteq A:

Suppose $x is any element such that x \in B.

By the supposition, we know that B \cap C \subseteq A \cap C since A \cap C = B \cap C.

Therefore since x \in B, we must determine if x \in C to determine if B \cap C is true.

Case x \in C:

If x \in C, then, since x \in B also, it follows that B \cap C is true, and thus by the supposition B \cap C = A \cap C is true.

It follows that x \in A \cap C, and thus x \in A.

Case x \notin C:

If x \notin C, then B \cup C is true (by virtue of x \in B). Then, by supposition, B \cup C = A \cup C. It follows that x \in A \cup C. Since x\notin C, this means that x \in A.

In either case x \in A. Therefore it has been shown that B \subseteq A.

Conclusion:

Since both subset relations have been proved, it can be concluded that A = B.

Q.E.D.

15. For all sets A, B, and C, (A - B) \cup C \subseteq A \cup (C - B).

Disproof (by counterexample):

Suppose A, B, and C are any sets such that A \cap B \neq \emptyset, and B \cap C \neq \emptyset, but A \cap C = \emptyset. For example:

Let A = \{1, 2\}, B = \{2, 3\}, C = \{3, 4\}.

Then, evaluating the left-hand side of the given equality:

 (A - B) \cup C = (\{1, 2\} - \{2, 3\}) \cup \{3, 4\} 

 = \{1\} \cup \{3, 4\} 

 = \{1, 3, 4\} 

Then, evaluating the right-hand side of the given equality:

 A \cup (C - B) = \{1, 2\} \cup (\{3, 4\} - \{2, 3\}) 

 = \{1, 2\} \cup \{4\} 

 = \{1, 2, 4\} 

Note that:

(A - B) \cup C = \{1, 3, 4\} \nsubseteq \{1, 2, 4\} = A \cup (C - B)

Hence the given supposition is false since 3 \in (A - B) \cup C and 3 \notin A \cup (C - B).

Q.E.D.

16. For all sets A and B, if A \cap B = \emptyset then A \times B = \emptyset.

Disproof (by counterexample):

Suppose A and B are any sets such that A \cap B = \emptyset. For example:

Let A = \{1\}, B = \{2\}.

Note that the supposition is true given the example since:

 A \cap B = \{1\} \cap \{2\} = \emptyset 

By the definition of \cap.

Note, though, that:

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

Hence the given supposition is false since \{1\} \cap \{2\} = \emptyset, but (1, 2) \neq \emptyset.

Q.E.D.

17. For all sets A and B, if A \subseteq B then \mathscr{P}(A) \subseteq \mathscr{P}(B).

Proof:

Let A and B be any sets such that A \subseteq B.

Suppose X is any set such that X \in \mathscr{P}(A). By the definition of power set, it follows that X \subseteq A.

Since X \subseteq A, X \subseteq B by the supposition/definition of subset (A \subseteq B).

Since X \subseteq B, this means that X \in \mathscr{P}(B), by definition of power set.

Hence \mathscr{P}(A) \subseteq \mathscr{P}(B).

Q.E.D.

18. For all sets A and B, \mathscr{P}(A \cup B) \subseteq \mathscr{P}(A) \cup \mathscr{P}(B).

Disproof (by counterexample):

Let A and B be any two disjoint sets. For example:

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

Then, the left-hand side of the given subset relation is:

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

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

 = \{\emptyset, \{1\}, \{2\}, \{1, 2\}\} 

Then, the right-hand side of the given subset relation is:

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

 = \{\emptyset, \{1\}\} \cup \{\emptyset, \{2\}\} 

 = \{\emptyset, \{1\}, \{2\}\} 

Now, note that:

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

Hence it has been shown that the supposition is false since \{1, 2\} is not in \{\emptyset, \{1\}, \{2\}\}.

Q.E.D.

19. For all sets A and B, \mathscr{P}(A) \cup \mathscr{P}(B) \subseteq \mathscr{P}(A \cup B).

Proof:

Suppose A and B are any sets.

Let X be some set such that X \in \mathscr{P}(A) \cup \mathscr{P}(B).

By the definition of \cup, this means that X \in \mathscr{P}(A) or X \in \mathscr{P}(B).

Case X \in \mathscr{P}(A):

By the definition of power set, X \subseteq A.

Since X \subseteq A, it follows that X \subseteq A \cup B, by definition of \cup.

Case X \in \mathscr{P}(B):

By the definition of power set, X \subseteq B.

Since X \subseteq B, it follows that X \subseteq A \cup B, by definition of \cup.

In either case X \subseteq A \cup B. By the definition of power set, this means that X \in \mathscr{P}(A \cup B).

This is what was to be shown.

Q.E.D.

20. For all sets A and B, \mathscr{P}(A \cap B) = \mathscr{P}(A) \cap \mathscr{P}(B).

Proof:

Let A and B be any sets.

To prove that \mathscr{P}(A \cap B) = \mathscr{P}(A) \cap \mathscr{P}(B), it must be shown that \mathscr{P}(A \cap B) \subseteq \mathscr{P}(A) \cap \mathscr{P}(B) and that \mathscr{P}(A) \cap \mathscr{P}(B) \subseteq \mathscr{P}(A \cap B).

Proof \mathscr{P}(A \cap B) \subseteq \mathscr{P}(A) \cap \mathscr{P}(B):

Suppose X is some set such that X \in \mathscr{P}(A \cap B).

By the definition of power set, this means that X \subseteq A \cap B.

By definition of \cap, this means that X \subseteq A and X \subseteq B.

By the definition of power set, since X \subseteq A, this means that X \in \mathscr{P}(A).

By the definition of power set, since X \subseteq B, this means that X \in \mathscr{P}(B).

Hence X \in \mathscr{P}(A) \cap \mathscr{P}(B).

Thus it has been shown that \mathscr{P}(A \cap B) \subseteq \mathscr{P}(A) \cap \mathscr{P}(B).

Proof \mathscr{P}(A) \cap \mathscr{P}(B) \subseteq \mathscr{P}(A \cap B):

Suppose X is some set such that X \in \mathscr{P}(A) \cap \mathscr{P}(B).

By the definition of \cap, this means that X \in \mathscr{P}(A) and X \in \mathscr{P}(B).

Since X \in \mathscr{P}(A), by the definition of power set, X \subseteq A.

Since X \in \mathscr{P}(B), by the definition of power set, X \subseteq B.

Since X \subseteq A and X \subseteq B, it follows that X \subseteq A \cap B, by the definition of subset and \cap.

By the definition of power set, since X \subseteq A \cap B, this means that X \in \mathscr{P}(A \cap B).

Thus it has been shown that \mathscr{P}(A) \cap \mathscr{P}(B) \subseteq \mathscr{P}(A \cap B).

Conclusion:

Since both subset relations have been proved, it can be concluded that \mathscr{P}(A \cap B) = \mathscr{P}(A) \cap \mathscr{P}(B).

Q.E.D.

21. For all sets A and B, \mathscr{P}(A \times B) = \mathscr{P}(A) \times \mathscr{P}(B).

Disproof (by counterexample):

Suppose A and B are any sets such that they are disjoint. For example:

Let A = \{1\} and B = \{2\}.

Then, the left-hand side of the given equality is:

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

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

 = \{\emptyset, \{(1, 2)\}\} 

Then, the right-hand side of the given equality is:

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

 = \{\emptyset, \{1\}\} \times \{\emptyset, \{2\}\} 

 = \{(\emptyset, \emptyset), (\emptyset, \{2\}), (\{1\}, \emptyset), (\{1\}, \{2\})\} 

Note that:

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

Hence, the supposition is false since \{\emptyset, \{(1, 2)\}\} \neq \{(\emptyset, \emptyset), (\emptyset, \{2\}), (\{1\}, \emptyset), (\{1\}, \{2\})\}.

Q.E.D.

22. Write a negation for each of the following statements. Indicate which is true, the statement or its negation. Justify your answers.

a. \forall sets S, \exists a set T such that S \cap T = \emptyset.

Negation:

\exists a set S such that \forall sets T, S \cap T \neq \emptyset.

The original statement is true, consider if T = S^c, then S \cap T = S \cap S^c = \emptyset.

b. \exists a set S such that \forall sets T, S \cup T = \emptyset.

Negation:

\forall sets S, \exists a set T such that S \cup T \neq \emptyset.

The negation is true. Consider T \neq \emptyset, then no matter whether S = \emptyset or S \neq \emptyset, S \cup T \neq \emptyset will always be true since T \neq \emptyset.

23. Let S =\{a, b, c\}, and for each integer i = 0, 1, 2, 3, let S_i be the set of all subsets of S that have i elements. List the elements in S_0, S_1, S_2, and S_3. Is \{S_0, S_1, S_2, S_3\} a partition of \mathscr{P}(S)?

 S_0 = \{\emptyset\} 

 S_1 = \{\{a\}, \{b\}, \{c\}\} 

 S_2 = \{\{a, b\}, \{a, c\}, \{b, c\}\} 

 S_3 = \{\{a, b, c\}\} 

Since all S_i are mutually disjoint, nonempty, and their union is all of \mathscr{P}(S), the elements in all S_i are a partition of \mathscr{P}(S).

24. Let A = \{t, u, v, w\}, and let S_1 be the set of all subsets of A that do not contain w and S_2 the set of all subsets of A that contain w.

a. Find S_1.

 S_1 = \{\emptyset, \{t\}, \{u\}, \{v\}, \{t, u\}, \{t, v\}, \{u, v\}, \{t, u, v\}\} 

b. Find S_2.

 S_2 = \{\{w\}, \{t, w\}, \{u, w\}, \{v, w\}, \{t, u, w\}, \{t, v, w\}, \{u, v, w\}, \{t, u, v, w\}\} 

c. Are S_1 and S_2 disjoint?

Yes.

d. Compare the sizes of S_1 and S_2.

Their sizes are equal (8).

e. How many elements are in S_1 \cup S_2?

f. What is the relation between S_1 \cup S_2 and \mathscr{P}(A)?

 S_1 \cup S_2 = \mathscr{P}(A) 

25. Use mathematical induction to prove that for every integer n \geq 2, if a set S has n elements, then the number of subsets of S with an even number of elements equals the number of subsets of S with an odd number of elements.

Proof (by mathematical induction):

Let P(n) be the sentence:

If a set S has n elements, then the number of subsets of S with an even number of elements equals the number of subsets of S with an odd number of elements.

Basis Step:

Prove P(2), that is:

If a set S has 2 elements, then the number of subsets of S with an even number of elements equals the number of subsets of S with an odd number of elements.

Consider S = \{1, 2\}. Then the subsets of S, or the power set of S, would be: \mathscr{P}(S) = \{\emptyset, \{1\}, \{2\}, \{1, 2\}\}.

The number of subsets with an even number of elements is 2 (\emptyset, \{1, 2\}).

The number of subsets with an odd number of elements is 2 (\{1\}, \{2\}).

The number of subsets with an even number of elements is equal to the number of subsets with an odd number of elements. Therefore P(2) is true.

Inductive Step:

Let k be any integer such that k \geq 2.

Suppose P(k), that is:

If a set S has k elements, then the number of subsets of S with an even number of elements equals the number of subsets of S with an odd number of elements.

This is the inductive hypothesis.

Prove P(k + 1), that is:

If a set S has k + 1 elements, then the number of subsets of S with an even number of elements equals the number of subsets of S with an odd number of elements.

Let X be some set such that X has k + 1 elements, and let x be some element such that x \in X.

Then, let Y be some set such that Y = X - \{x\}. This means that Y has k elements.

Every subset of X either contains x or doesn't. The subsets of X that do not contain x are the subsets of Y, and, by the inductive hypothesis, have an equal number of subsets containing even and odd amounts of elements.

The subsets containing x are each of the form Z \cup \{x\}, where Z \subseteq Y. Adding x flips the parity of each subset (i.e. even number of subsets now becomes odd and odd number of subsets become even.) Note, though, that the number of subsets with even number of elements and the number of subsets with odd number of elements remain equal.

Therefore, P(k + 1) is true.

Q.E.D.

26. The following problem, devised by Ginger Bolton, appeared in the January 1989 issue of the College Mathematics Journal (Vol. 20, No. 1, p. 68): Given a positive integer n \geq 2, let S be the set of all nonempty subsets of \{2, 3, \dots, n\}. For each S_i \in S, let P_i be the product of the elements of S_i. Prove or disprove that

 \sum_{i = 1}^{]2^{n - 1} - 1}{P_i} = \frac{(n + 1)!}{2} - 1 

Omitted.

In 27 and 28 supply a reason for each step in the derivationl.l

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

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

Proof:

Suppose A, B, and C are any sets. Then

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

by __ (a) __

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

by __ (b) __

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

by __ c __

a. by commutative law for \cap

b. by distributive law

c. by commutative law for \cap

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

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

Proof:

Suppose A, B, and C are any sets. Then

 (A \cup B) - (C - A) = (A \cup B) \cap (C - A)^c 

by __ (a) __

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

by __ (b) __

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

by __ c __

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

by __ (d) __

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

by __ (e) __

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

by __ (f) __

 = A \cup (B - C) 

by __ (g) __

a. by the set difference law

b. by the set difference law

c. by the commutative law for \cap

d. by De Morgan's Law

e. by the double complement law

f. by the distributive law

g. by the set difference law

29. Some steps are missing from the following proof that for all sets A and B, (A \cup B^c) - B = (A - B) \cup B^c. Indicate what they are, and then write the proof correctly.

Proof:

Let any sets A and B be given. Then

 (A \cup C^c) - B = (A \cup B^c) \cap B^c 

by the set difference law

 = (B^c \cap A) \cup (B^c \cap B^c) 

by the distributive law

 = (B^c \cap A) \cup B^c 

by the idempotent law for \cup

 (A - B) \cup B^c 

by the set difference law.

Proof:

Let any sets A and B be given. Then

 (A \cup B^c) - B = (A \cup B^c) \cap B^c 

by the set difference law

 = B^c \cap (A \cup B^c) 

by the commutative law

 = (B^c \cap A) \cup (B^c \cap B^c) 

by the distributive law

 = (B^c \cap A) \cup B^c 

by the idempotent law for \cap.r

 = (A \cap B^c) \cup B^c 

by the commutative law

 = (A - B) \cup B^c 

Q.E.D.

In 30-40, construct an algebraic proof for the given statement. Cite a property from Theorem 6.2.2 for every step.

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

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

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

by commutative law for \cup

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

by distributive laws

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

by commutative laws for \cup

31. For all sets A and B, A \cup (B - A) = A \cup B.

 A \cup (B - A) = A \cup (B \cap A^c) 

by set difference law

 = (A \cup B) \cap (A \cup A^c) 

by distributive laws

 = (A \cup B) \cap U 

by complement laws for \cup

 = A \cup B 

by identity laws for \cap

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

 (A - B) \cup (A \cap B) = (A \cap B^c) \cup (A \cap B) 

by set difference law

 = A \cap (B^c \cup B) 

by distributive laws

 = A \cap U 

by complement laws

 = A 

by identity laws

33. For all sets A and B, (A - B) \cap (A \cap B) = \emptyset.

 (A - B) \cap (A \cap B) = (A \cap B^c) \cap (A \cap B) 

by set difference law

 = A \cap A \cap B^c \cap B 

by associative laws for \cap

 = A \cap B^c \cap B 

by idempotent laws for \cap

 = A \cap \emptyset 

by complement laws for \cap.

 = \emptyset 

by universal bound laws for \cap

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

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

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

by set difference law

 = (A \cap B^c) \cap C^c 

by set difference law

 = A \cap (B^c \cap C^c) 

by associative laws for \cap

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

by De Morgan's laws

 = A - (B \cup C) 

by set difference law

35. For all sets A and B, A - (A - B) = A \cap B.

 A - (A - B) = A - (A \cap B^c) 

by set difference law

 = A \cap (A \cap B^c)^c 

by set difference law

 = A \cap (A^c \cup (B^c)^c) 

by De Morgan's laws

 = A \cap (A^c \cup B) 

by double complement law

 = (A \cap A^c) \cup (A \cap B) 

by the distributive laws for \cap

 = \emptyset \cup (A \cap B) 

by complement laws

 = (A \cap B) \cup \emptyset 

by commutative laws for \cup

 = A \cap B 

by identity laws

36. For all sets A and B, ((A^c \cup B^c) - A)^c = A.

 ((A^c \cup B^c) - A)^c 

 = ((A^c \cup B^c) \cap A^c)^c   

by set difference law

 = (A^c \cup B^c)^c \cup (A^c)^c 

by De Morgan's laws

 = (A^c \cup B^c)^c \cup A 

by double complement law

 = ((A^c)^c \cap (B^c)^c) \cup A 

by De Morgan's laws

 = (A \cap B) \cup A 

by double complement law

 = A \cup (A \cap B) 

by commutative laws for \cup

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

by distributive laws

 = A \cap (A \cup B) 

by idempotent laws

 = A 

by absorption laws for \cap

37. For all sets A and B, (B^c \cup (B^c - A))^c = B.

(B^c \cup (B^c - A))^c $

= (B^c \cup (B^c \cap A^c))^c $

by set difference law

= ((B^c)^c \cap (B^c \cap A^c)^c) $

by De Morgan's laws

= (B \cap (B^c \cap A^c)^c) $

by double complement law

= B \cap ((B^c)^c \cup (A^c)^c) $

by De Morgan's laws

= B \cap (B \cup A) $

by double complement law

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

by distributive laws

 = B \cup (B \cap A) 

by idempotent laws

 = B 

by absorption laws for \cup

38. For all sets A and B, (A \cap B)^c \cap A = A - B.

 (A \cap B)^c \cap A 

 = (A^c \cup B^c) \cap A 

by De Morgan's laws

 = A \cap (A^c \cup B^c) 

by commutative laws for \cap

 = (A \cap A^c) \cup (A \cap B^c) 

by distributive laws

 = \emptyset \cup (A \cap B^c) 

by complement laws

 = (A \cap B^c) \cup \emptyset 

by commutative laws

 = A \cap B^c 

by identity laws

 = A - B 

by set difference law

39. For all sets A and B,

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

 (A - B) \cup (B - A) 

 = (A \cap B^c) \cup (B \cap A^c) 

by set difference law

 = [(A \cap B^c) \cup B] \cap [(A \cap B^c) \cup A^c] 

by distributive laws

 = [B \cup (A \cap B^c)] \cap [A^c \cup (A \cap B^c)] 

by commutative laws

 = [(B \cup A) \cap (B \cup B^c)] \cap [(A^c \cup A) \cap (A^c \cup B^c)] 

by distributive laws

 = [(A \cup B) \cap (B \cup B^c)] \cap [(A \cup A^c) \cap (A^c \cup B^c)] 

by commutative laws

 = [(A \cup B) \cap U] \cap [U \cap (A^c \cup B^c)] 

by complement laws

 = [(A \cup B) \cap U] \cap [(A^c \cup B^c) \cap U] 

by commutative laws

 = (A \cup B) \cap (A^c \cup B^c) 

by identity laws

 = (A \cup B) \cap (A \cap B)^c 

by De Morgan's laws

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

by set difference law

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

 (A - B) - (B - C) = A - B 

 (A - B) - (B - C) 

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

by set difference law

 = (A \cap B^c) \cap (B \cap C^c)^c 

by set difference law

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

by De Morgan's laws

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

by double complement law

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

by associative laws

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

by distributive laws

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

By idempotent laws

 = A \cap B^c 

by absorption laws

 = A - B 

by set difference law

In 41-43 simplify the given expression. Cite a property from Theorem 6.2.2 for every step.

41. A \cap ((B \cup A^c) \cap B^c)

Omitted.

42. (A - (A \cap B)) \cap (B - (A \cap B))

Omitted.

43. ((A \cap (B \cup C)) \cap (A - B)) \cap (B \cup C^c)

Omitted.

44. Consider the following set property: For all sets A and B, A - B and B are disjoint.

a. Use an element argument to derive the property.

Omitted.

b. Use an algebraic argument to derive the property (by applying properties from Theorem 6.2.2).

Omitted.

c. Comment on which method you found easier.

Omitted.

35. Consider the following set property: For all sets A, B, and C,

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

a. Use an element argument to derive the property.

Omitted.

b. Use an algebraic argument to derive the property (by applying properties from Theorem 6.2.2).

Omitted.

c. Comment on which method you found easier.

Omitted.

Definition:

Given sets A and B, the symmetric difference of A and $B$, denoted A \Delta B, is, is

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

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

a. A \Delta B

Omitted.

b. B \Delta C

Omitted.

c. A \Delta C

Omitted.

d. (A \Delta B) \Delta C

Omitted.

Refer to the definition of symmetric difference given above. Prove each of 47-52, assuming that A, B, and C are all subsets of a universal set U.

47. A \Delta B = B \Delta A

Omitted.

48. A \Delta \emptyset = A

Omitted.

49. A \Delta A^c = U

Omitted.

50. A \Delta A = \emptyset

Omitted.

51. If A \Delta C = B \Dcelta C, then A = B.

Omitted.

52. (A \Delta B) \Delta C = A \Delta (B \Delta C)

Omitted.

53. Derive the set identity A \cup (A \cap B) = A from the properties listed 8n Theorem 6.2.2(1)-(9). Start by showing that for every subset B of a universal set U, U \cup B = U. Then intersect both sides with A and deduce the identity.

Omitted.

54. Derive the set identity A \cap (A \cup B) = A from the properties listed in Theorem 6.2.2(1)-(9). Start by showing that for every subset B of a universal set U, \emptyset = \emptyset \cap B. Then take the union of both sides with A and deduce the identity.

Omitted.