4929 lines
126 KiB
Markdown
4929 lines
126 KiB
Markdown
Page 411
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**Exercise Set 6.1**
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1. In each of (a) -(f), answer the following questions: Is $A \subseteq B$? Is
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$B \subseteq A$? Is either $A$ or $B$ a proper subset of the other?
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a. $A = \{2, \{2\}, (\sqrt{2})^2\}$, $B = \{2, \{2\}, \{\{2\}\}\}$
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$A \subseteq B$ ?:
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$$ A = \{2, \{2\}, (\sqrt{2})^2\} = \{2, \{2\}, 2\} = \{2, \{2\}\} $$
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Yes, every element in $A$ is in $B$.
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$B \subseteq A$ ?:
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No, because $\{\{2\}\}$ is an element of $B$, but is not an element of $A$, so
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$\B \nsubseteq A$.
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Is either $A$ or $B$ a proper subset of the other?
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Yes, $A$ is a proper subset of $B$, because every element in $A$ is in $B$, but
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not every element in $B$ is in $A$.
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b. $A = \{3, \sqrt{5^2 - 4^2}, 24 \mod 7\}$, $B = \{8 \mod 5\}$
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$A \subseteq B$ ?:
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$$ A = \{3, \sqrt{5^2 - 4^2}, 24 \mod 7\} = \{3, 3, 3\} = \{3\} $$
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$$ B = \{8 \mod 5\} = \{3\} $$
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Yes, $A$ is a subset of $B$ since every element of $A$ is in $B$.
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$B \subseteq A$ ?:
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Yes, $B$ is a subset of $A$ since every element of $B$ is in $A$.
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Is either $A$ or $B$ a proper subset of the other?
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Yes, both $A$ and $B$ are proper subsets of the other since $A = B$.
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c. $A = \{\{1, 2\}, \{2, 3\}\}$, $B = \{1, 2, 3\}$
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$A \subseteq B$ ?:
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No, because there are no elements in $A$ that are in $B$, $A \nsubseteq B$
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$B \subseteq A$ ?:
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No, because there are no elements in $B$ that are in $A$, $B \nsubseteq A$
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Is either $A$ or $B$ a proper subset of the other?
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No, since neither set share any elements, neither is a proper subset of the
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other.
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d. $A = \{a, b, c\}$, $B = \{\{a\}, \{b\}, \{c\}\}$
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$A \subseteq B$ ?:
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No, because there are no elements in $A$ that are in $B$, $A \nsubseteq B$
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$B \subseteq A$ ?:
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No, because there are no elements in $B$ that are in $A$, $B \nsubseteq A$
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Is either $A$ or $B$ a proper subset of the other?
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No, since neither set share any elements, neither is a proper subset of the
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other.
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e. $A = \{\sqrt{16}, \{4\}\}$, $B = \{4\}$
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$A \subseteq B$ ?:
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$$ A = \{\sqrt{16}, \{4\}\} = \{4, \{4\}\} $$
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No, because every element of $A$ is not an element in $B$ ($4$ is not in $B$),
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$A \nsubseteq B$.
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$B \subseteq A$ ?:
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Yes, because every element in $B$ is an element in $A$, $B \subseteq A$.
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Is either $A$ or $B$ a proper subset of the other?
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Yes, $B$ is a proper subset of $A$ since $B \subseteq A$ and $A \nsubseteq B$.
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f. $A = \{x \in \mathbb{R} | \cos x \in \mathbb{Z}\}$,
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$B = \{x \in \mathbb{R} | \sin x \in \mathbb{Z}\}$
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From trigonometry, we know that $\cos x = -1 \text{ or } 0 \text{ or } 1$ and
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$\sin x = -1 \text{ or } 0 \text{ or } 1 $. When we evaluate for $x$ in these
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cases we find:
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$$ A = \{\dots, -\frac{5\pi}{2}, -\frac{3\pi}{2}, \pi, -\frac{\pi}{2}, 0, \frac{\pi}{2}, \frac{3\pi}{2}, \frac{5\pi}{2}, \dots\} $$
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$$ B = \{\dots, -\frac{5\pi}{2}, -\frac{3\pi}{2}, \pi, -\frac{\pi}{2}, 0, \frac{\pi}{2}, \frac{3\pi}{2}, \frac{5\pi}{2}, \dots\} $$
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$A \subseteq B$ ?: Yes.
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$B \subseteq A$ ?: Yes.
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Yes, $B$ is a proper subset of $A$ since $B \subseteq A$ and $A \nsubseteq B$.
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Yes, since $A = B$.
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2. Complete the proof from Example 6.1.3: Prove that $B \subseteq A$ where
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$$ A = \{m \in \mathbb{Z} | m, = 2a \text{ for some integer } a\} $$
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and
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$$ B = \{n \in \mathbb{Z} | n = 2b - 2 \text{ for some integer } b\} $$
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_Part 2, Proof that $B \subseteq A$:_
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Suppose $x$ is a particular but arbitrarily chosen element of $B$.
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By definition of $B$, there is an integer, say $b$, such that $x = 2b - 2$.
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To prove that $B \subseteq A$, we must show that there is some $x$ that can
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equal both $2a$, for some integer $a$, and that same $x$ can also equal
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$2b - 2$.
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$$ 2b - 2 = 2a $$
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$$ a = b - 1 $$
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By the difference integers, $a$ is an integer. Then, by substitution:
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$$ 2a = 2(b - 1) $$
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$$ = 2b - 2 $$
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$$ = x $$
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Thus, by definition of $A$, $x$ is an element of $A$.
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Q.E.D.
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3. Let sets $R$, $S$, and $T$ be defined as follows:
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$$ R = \{x \in \mathbb{Z} | x \text{ is divisible by } 2\} $$
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$$ S = \{y \in \mathbb{Z} | y \text{ is divisible by } 3\} $$
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$$ T = \{z \in \mathbb{Z} | z \text{ is divisible by } 6\} $$
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Prove or disprove each of the following statements.
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a. $R \subseteq T$
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$R \nsubseteq T$ since $2 \in R$ since $2 \mid 2$, but $2 \notin T$ since
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$6 \cancel{\mid} 2$.
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b. $T \subseteq R$
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**Proof:**
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Suppose $n$ is any integer such that $6 \mid n$, therefore $n \in T$.
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By the definition of divisibility:
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$$ n = 6m $$
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for some integer $m$.
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$$ n = 2(3m) $$
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By the product of integers, $3m$ is an integer. It follows that
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$n = 2 \cdot (\text{some integer})$. Thus $2 \mid n$, so $n \in R$. This is what
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was to be shown.
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Q.E.D.
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c. $T \subseteq S$
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**Proof:**
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Suppose $n$ is any integer such that $6 \mid n$, therefore $n \in T$.
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By the definition of divisibility:
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$$ n = 6m $$
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for some integer $m$.
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$$ n = 3(2m) $$
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By the product of integers, $2m$ is an integer. It follows that
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$n = 3 \cdot (\text{some integer})$. Thus $3 \mid n$, so $n \in S$. This is what
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was to be shown.
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Q.E.D.
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4. Let $A = \{n \in \mathbb{Z} | n = 5r \text{ for some integer } r\}$ and
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$B = \{m \in \mathbb{Z} | m = 20s \text{ for some integer } s\}$. Prove or
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disprove each of the following statements.
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a. $A \subseteq B$
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$A \nsubseteq B$ since $5 \in A$ since $5 \mid 5$, but $5 \notin B$ since
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$5 \cancel{\mid} 20$.
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b. $B \subseteq A$
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**Proof:**
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Suppose $n$ is any integer such that $20 \mid n$, therefore $n \in B$.
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By the definition of divisibility:
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$$ n = 20m $$
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for some integer $m$.
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$$ n = 5(4m) $$
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By the product of integers, $4m$ is an integer. It follows that
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$n = 5 \cdot (\text{some integer})$. Thus $5 \mid n$, so $n \in A$. This is what
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was to be shown.
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Q.E.D.
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5. Let $C = \{n \in \mathbb{Z} | n = 6r - 5 \text{ for some integer } r\}$ and
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$D = \{m \in \mathbb{Z} | m = 3s + 1 \text{ for some integer } s\}$. Prove or
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disprove each of the following statements.
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a. $C \subseteq D$
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**Proof:**
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Suppose $n$ is any integer such that $n = 6r - 5$ for some integer $r$, which
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means that $n \in C$.
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Also suppose that $m$ is any integer such that $m = 3s + 1$ for some integer
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$s$, which means that $m \in S$.
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We must show that there exists some $r$ that when substituted for $s$ will
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satisfy the definition of $n$.
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Let $s = 2r - 2$. Then, by substitution:
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$$ m = 3(2r - 2) + 1 $$
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$$ = 6r - 6 + 1 $$
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$$ = 6r - 5 $$
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$$ = n $$
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By the product and difference of integers, $6r - 5$ is an integer, therefore
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$n \in D$. This is what was to be shown.
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Q.E.D.
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b. $D \subseteq C$
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**Disproof:**
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$D \nsubseteq C$ because there are elements in $D$ that are not in $C$. For
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example $4$ is in $D$ because $4 = 3(1) + 1$, but $4$ is not in $C$. If $4$ were
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in $C$, this would mean:
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$$ 4 = 6r - 5 $$
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for some integer $r$.
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$$ 4 + 5 = 6r $$
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$$ 9 = 6r $$
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$$ \frac{9}{6} = r $$
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$$ \frac{3}{2} = r $$
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But $\dfrac{3}{2}$ is not an integer. This is a contradiction, therefore
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$D \nsubseteq C$.
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Q.E.D.
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6. Let $A = \{x \in \mathbb{Z} | x = 5a + 2 \text{ for some integer } a\}$,
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$B = \{y \in \mathbb{Z} | y = 10b - 3 \text{ for some integer } b\}$, and
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$C = \{z \in \mathbb{Z} | z = 10c + 7 \text{ for some integer } c\}$.
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Prove or disprove each of the following statements.
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a. $A \subseteq B$
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**Disproof (by counterexample):**
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Suppose $n$ is any integer such that $n = 5a + 2$ for some integer $a$. This
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means that $n \in A$.
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Suppose also that there is some integer $m$ such that $m = 10b - 3$ for some
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integer $b$. This means that $m \in B$.
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To show that there is some integer $b$ that will satisfy $n$, we must relate it
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to $a$:
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$$ 5a + 2 = 10b - 3 $$
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$$ 5a + 5 = 10b $$
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$$ 5a + 5 = 10b $$
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$$ \frac{1}{2}a + \frac{1}{2} = b $$
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$$ \frac{a + 1}{2} = b $$
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In order for $A \subseteq B$, every element of $A$ must be in $B$. If $a = 0$,
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then $n = 2$, so $n \in A$. If $a = 0$, then $b = \dfrac{1}{2}$, which is not an
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integer, thus $2 \notin B$. Therefore $A \nsubseteq B$.
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Q.E.D.
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b. $B \subseteq A$
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**Proof:**
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Suppose $y$ is any integer such that $y = 10b - 3$ for some integer $b$. This
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means that $y \in B$.
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Let's first find $a$ as it relates to $y$.
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$$ y = 5a + 2 $$
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$$ 10b - 3 = 5a + 2 $$
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$$ 10b - 5 = 5a $$
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$$ 2b - 1 = a $$
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So let $a = 2b - 1$.
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Then substitute in for the condition for $A$:
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$$ x = 5a + 2 $$
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$$ = 5(2b - 1) + 2 $$
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$$ = 10b - 5 + 2 $$
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$$ = 10b - 3 $$
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$$ = y $$
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Therefore $y \in A$.
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Q.E.D.
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c. $B = C$
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To prove $B = C$, we must prove both that $B \subseteq C$ and $C \subseteq B$.
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_Prove $B \subseteq C$:
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Suppose $y$ is any integer such that $y = 10b - 3$ for some integer $b$. This
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means that $y \in B$.
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Let's first find some integer $c$ as it relates to $y$:
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$$ y = 10c + 7 $$
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$$ 10b - 3 = 10c + 7 $$
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$$ 10b - 10 = 10c $$
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$$ b - 1 = c $$
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So, let $c = b - 1$.
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Then substitute in for the condition for $C$:
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$$ z = 10c + 7 $$
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$$ = 10(b - 1) + 7 $$
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$$ = 10b - 10 + 7 $$
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$$ = 10b - 3 $$
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$$ = y $$
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Thus $y \in C$, and therefore $B \subseteq C$.
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_Prove $C \subseteq B$:
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Suppose $z$ is any integer such that $z = 10c + 7$ for some integer $c$. This
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means that $z \in C$.
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Let's first find some integer $b$ as it relates to $z$:
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$$ z = 10b - 3 $$
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$$ 10c + 7 = 10b - 3 $$
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$$ 10c + 10 = 10b $$
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$$ c + 1 = b $$
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So, let $b = c + 1$.
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Then substitute in for the condition for $B$:
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$$ y = 10b - 3 $$
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$$ = 10(c + 1) - 3 $$
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$$ = 10c + 10 - 3 $$
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$$ = 10c + 7 $$
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$$ = z $$
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Therefore $z \in B$.
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Thus $z \in B$, and therefore $C \subseteq B$.
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Since $B \subseteq C$ and $C \subseteq B$, it follows that $B = C$. This is what
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was to be shown.
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Q.E.D.
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7. Let $A = \{x \in \mathbb{Z} | x = 6a + 4 \text{ for some integer } a\}$,
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$B = \{y \in \mathbb{Z} | y = 18b - 2 \text{ for some integer } b\}$, and
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$C = \{z \in \mathbb{Z} | z = 18c + 16 \text{ for some integer } c\}$.
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Prove or disprove each of the following statements.
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a. $A \subseteq B$
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**Disproof (by counterexample):**
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Suppose $x$ is any integer such that $x = 6a + 4$ for some integer $a$. This
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means that $x \in A$.
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Let's first find some integer $b$ as it relates to $a$.
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$$ x = 18b - 2 $$
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$$ 6a + 4 = 18b - 2 $$
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$$ 6a + 6 = 18b $$
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$$ \frac{6}{18}a + \frac{6}{18} = b $$
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$$ \frac{1}{3}a + \frac{1}{3} = b $$
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$$ \frac{a + 1}{3} = b $$
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By definition of $b$, $b$ must always be an integer for all $a$.
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Suppose $a = 0$, then:
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$$ x = 6(0) + 4 = 4 $$
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so $4 \in A$, but:
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$$ b = \frac{0 + 1}{3} = \frac{1}{3} $$
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so $4 \notin B$. We can see this as $4 = 18b - 2$ results in $b = \dfrac{1}{3}$,
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but $b$ must be an integer.
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b. $B \subseteq A$
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**Proof:**
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Suppose $y$ is any integer such that $y = 18b - 2$ for some integer $b$. This
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means that $y \in B$.
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Let's first find some integer $a$ as it relates to $b$.
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$$ y = 6a + 4 $$
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$$ 18b - 2 = 6a + 4 $$
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$$ 18b - 6 = 6a $$
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$$ 3b - 1 = a $$
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Now, substitute $a$ in for the condition for $A$:
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$$ x = 6a + 4 $$
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$$ = 6(3b - 1) + 4 $$
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$$ = 18b - 2 $$
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$$ = y $$
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Therefore $B \subseteq A$.
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c. $B = C$
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To prove $B = C$, we must prove both that $B \subseteq C$ and $C \subseteq B$.
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_Prove $B \subseteq C$:
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Suppose $y$ is any integer such that $y = 18b - 2$ for some integer $b$. This
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means that $y \in B$.
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Let's first find some integer $c$ as it relates to $b$.
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$$ y = 18c + 16 $$
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$$ 18b - 2 = 18c + 16 $$
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$$ 18b - 18 = 18c $$
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$$ b - 1 = c $$
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So, let $c = b - 1$. Now substitute $c$ in for the condition of $C$:
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$$ z = 18c + 16 $$
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$$ = 18(b - 1) + 16 $$
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$$ = 18b - 18 + 16 $$
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$$ = 18b - 2 $$
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$$ = y $$
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Therefore $B \subseteq C$.
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_Prove $C \subseteq B$:
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Suppose $z$ is any integer such that $z = 18c + 16$ for some integer $c$.
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Let's first find some $b$ as it relates to $c$.
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$$ z = 18b - 2 $$
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$$ 18c + 16 = 18b - 2 $$
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$$ 18c + 18 = 18b $$
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$$ c + 1 = b $$
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So, let $b = c + 1$. Now, let's substitute $b$ in for the condition for $B$.
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$$ y = 18b - 2 $$
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$$ = 18(c + 1) - 2 $$
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$$ = 18c + 18 - 2 $$
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$$ = 18c + 16 $$
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$$ = z $$
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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)
|
|
|
|
18.
|
|
|
|
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.
|
|
|
|
27.
|
|
|
|
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. $\mathscr{P}(A \times B)$
|
|
|
|
$$ A \times B = \{(1, 2), (1, 3), (2, 2), (2, 3)\} $$
|
|
|
|
$$ \mathscr{P}(A \times B) = \{\emptyset, \{(1, 2)\}, \{(1, 3)\}, \{(2, 2)\}, \{(2, 3)\}, \{(1, 2), (1, 3)\}, \{(1, 2), (2, 2)\}, \{(1, 2,), (2, 3)\}, \{(1, 3), (2, 2)\}, \{(1, 3), (2, 3)\}, \{(2, 2), (2, 3)\}, \{(1, 2), (1, 3), (2, 2)\}, \{(1, 2), (1, 3), (2, 3)\}, \{(1, 2), (2, 2), (2, 3)\}, \{(1, 3), (2, 2), (2, 3)\}, \{(1, 2), (1, 3), (2, 2), (2, 3)\}\} $$
|
|
|
|
32.
|
|
|
|
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 $\mathscr{P}(X \times Y)$.
|
|
|
|
$$ X \times Y = \{(a, x), (a, y), (b, x), (b, y)\} $$
|
|
|
|
$$ \mathscr{P}(X \times Y) = \{\emptyset, \{(a, x)\}, \{(a, y)\}, \{(b, x)\}, \{(b, y)\}, \{(a, x), (a, y)\}, \{(a, x), (b, x)\}, \{(a, x), (b, y)\}, \{(a, y), (b, x)\}, \{(a, y), (b, y)\}, \{(b, x), (b, y)\}, \{(a, x), (a, y), (b, x)\}, \{(a, x), (a, y), (b, y)\}, \{(a, x), (b, x), (b, y)\}, \{(a, y), (b, x), (b, y)\} \{(a, x), (a, y), (b, x), (b, y)\}\} $$
|
|
|
|
33.
|
|
|
|
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.
|
|
|
|
---
|
|
|
|
Page 427
|
|
|
|
**Exercise Set 6.2**
|
|
|
|
1.
|
|
|
|
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 $B$)
|
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|
|
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]._ | __ \(c\) __ |
|
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|
|
a. by the definition of a union (because $A \cup B$)
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|
|
|
b. by definition of a subset (because $A \subseteq B$)
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|
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c. by definition of a subset
|
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|
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 $x \in B \cup C$.
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|
|
|
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.
|
|
|
|
---
|
|
|
|
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 $S_1 \cup S_2$?
|
|
|
|
16.
|
|
|
|
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.
|
|
|
|
---
|
|
|
|
Page 445
|
|
|
|
**Exercise Set 6.4**
|
|
|
|
In 1-3 assume that $B$ is a Boolean algebra with operations $+$ and $\cdot$.
|
|
Give the reasons needed to fill in the blanks in the proofs using only the
|
|
axioms for a Boolean algebra.
|
|
|
|
1. _Idempotent law for $\cdot$:_ For every $a$ in $B$, $a \cdot a = a$.
|
|
|
|
**Proof:**
|
|
|
|
Let $a$ be any element of $B$. Then
|
|
|
|
$$ a = a \cdot 1 $$
|
|
|
|
__ (a) __
|
|
|
|
$$ = a \cdot (a + \overline{a}) $$
|
|
|
|
__ (b) __
|
|
|
|
$$ = (a \cdot a) + (a \cdot \overline{a}) $$
|
|
|
|
__ \(c\) __
|
|
|
|
$$ = (a \cdot a) + 0 $$
|
|
|
|
__ (d) __
|
|
|
|
$$ = a \cdot a $$
|
|
|
|
__ (e) __
|
|
|
|
a. by the identity law for $\cdot$
|
|
|
|
b. by the complement law for $+$
|
|
|
|
c. by the distributive law for $+$ over $\cdot$
|
|
|
|
d. by the complement law for $\cdot$
|
|
|
|
e. by the identity law for $+$
|
|
|
|
2. _Universal bound law for $+$:_ For every $a$ in $B$, $a + 1 = 1$.
|
|
|
|
**Proof:**
|
|
|
|
Let $a$ be any element in $B$. Then
|
|
|
|
$$ a + 1 = a + (a + \overline{a}) $$
|
|
|
|
__ (a) __
|
|
|
|
$$ = (a + a) + \overline{a} $$
|
|
|
|
__ (b) __
|
|
|
|
$$ = a + \overline{a} $$
|
|
|
|
by Example 6.4.2
|
|
|
|
$$ = 1 $$
|
|
|
|
__ \(c\) __
|
|
|
|
a. by the complement law for $+$
|
|
|
|
b. by the associative law for $+$
|
|
|
|
c. by the complement law for $+$
|
|
|
|
3. _Absorption law for $\cdot$ over $+$:_ For all $a$ and $b$ in $B$,
|
|
$(a + b) \cdot a = a$.
|
|
|
|
**Proof:** Let $a$ be any element of $B$. Then
|
|
|
|
$$ (a + b) \cdot a = a \cdot (a + b) $$
|
|
|
|
__ (a) __
|
|
|
|
$$ = a \cdot a + a \cdot b $$
|
|
|
|
__ (b) __
|
|
|
|
$$ = a + a \cdot b $$
|
|
|
|
by exercise 1
|
|
|
|
$$ = a \cdot 1 + a \cdot b $$
|
|
|
|
__ \(c\) __
|
|
|
|
$$ = a \cdot (1 + b) $$
|
|
|
|
__ (d) __
|
|
|
|
$$ = a \cdot (b + 1) $$
|
|
|
|
__ (e) __
|
|
|
|
$$ = a \cdot 1 $$
|
|
|
|
by exercise 2
|
|
|
|
$$ = a $$
|
|
|
|
__ (f) __
|
|
|
|
a. by the commutative law for $\cdot$
|
|
|
|
b. by the distributive law of $\cdot$ over $+$
|
|
|
|
c. because $1$ is an identity for $\cdot$
|
|
|
|
d. by the distributive law of $\cdot$ over $+$
|
|
|
|
e. by the commutative law for $+$
|
|
|
|
f. because $1$ is an identity for $\cdot$
|
|
|
|
In 4-10 assume that $B$ is a Boolean algebra with operations $+$ and $\cdot$.
|
|
Prove each statement using only the axioms for a Boolean algebra and statements
|
|
proved in the text or in lower-numbered exercises.
|
|
|
|
4. _Universal bound for $0$:_ For every $a$ in $B$, $a \cdot 0 = 0$.
|
|
|
|
**Proof:**
|
|
|
|
$$ a \cdot 0 = a \cdot (a \cdot \overline{a}) $$
|
|
|
|
by the complement law for $\cdot$
|
|
|
|
$$ = (a \cdot a) \cdot \overline{a} $$
|
|
|
|
by the associative law for $\cdot$
|
|
|
|
$$ = a \cdot \overline{a} $$
|
|
|
|
by exercise 1
|
|
|
|
$$ = 0 $$
|
|
|
|
by the complement law for $\cdot$
|
|
|
|
5. _Complements of $0$ and $1$:_
|
|
|
|
a. $\overline{0} = 1$
|
|
|
|
**Proof:**
|
|
|
|
$$ 0 = 0 \cdot 1 $$
|
|
|
|
because $1$ is an identity for $\cdot$, and
|
|
|
|
$$ 0 + 1 = 1 + 0 $$
|
|
|
|
because $+$ is commutative and $0$ is an identity for $+$.
|
|
|
|
Since $0 = 0 \cdot 1$ and $0 + 1 = 1 + 0$, $1 = \overline{0}$ by the uniqueness
|
|
of the complement laws.
|
|
|
|
b. $\overline{1} = 0$
|
|
|
|
$$ 1 = 1 + 0 $$
|
|
|
|
$$ 1 = 1 + \overline{1} $$
|
|
|
|
by the complement law for $+$
|
|
|
|
$$ 0 = \overline{1} $$
|
|
|
|
by the uniquness of $0$ law.
|
|
|
|
6. _Uniqueness of $0$:_ There is only one element of $B$ that is an identity for
|
|
$+$.
|
|
|
|
**Proof:**
|
|
|
|
Suppose $0$ and $0'$ are elements of $B$ both of which are identities for $+$.
|
|
Then both $0$ and $0'$ satisfy the identity, complement, and universal bound
|
|
laws.
|
|
|
|
_[We will show that $0 = 0'$.]_
|
|
|
|
By the identity law for $+$, for every $a \in B$,
|
|
|
|
$$ a + 0 = a(*) \quad \text{ and } \quad a + 0' = a(**) $$
|
|
|
|
It follows that
|
|
|
|
$$ 0' = 0' + 0 $$
|
|
|
|
by (*) with $a = 0'$
|
|
|
|
$$ = 0 + 0' $$
|
|
|
|
by the commutative law for $+$
|
|
|
|
$$ = 0 $$
|
|
|
|
by (**) with $a = 0$.
|
|
|
|
_[This is what was to be shown.]_
|
|
|
|
7. _Uniqueness of $1$:_ There is only one element of $B$ that 8s an identity for
|
|
$\cdot$.
|
|
|
|
**Proof:**
|
|
|
|
Suppose $1$ and $1'$ are elements of $B$ both of which are identities for
|
|
$\cdot$. Then both $1$ and $1'$ satisfy the identity, complement, and universal
|
|
bound laws.
|
|
|
|
_[We will show that $1 = 1'$.]_
|
|
|
|
By the identity law for $\cdot$, for every $a \in B$,
|
|
|
|
$$ a \cdot 1 = a(*) \quad \text{ and } \quad a \cdot 1' = a(**) $$
|
|
|
|
It follows that
|
|
|
|
$$ 1' = 1' \cdot 1 $$
|
|
|
|
by (*) with $a = 1'$
|
|
|
|
$$ = 1 \cdot 1' $$
|
|
|
|
by the commutative law for $\cdot$
|
|
|
|
$$ = 1 $$
|
|
|
|
by (**) with $a = 1$.
|
|
|
|
_[This is what was to be shown.]_
|
|
|
|
8. _De Morgan's law for $\cdot$:_ For all $a$ and $b$ in $B$,
|
|
$\overline{a \cdot b} = \overline{a} + \overline{b}$. (_Hint:_ Prove that
|
|
$(a \cdot b) + (\overline{a} + \overline{b}) = 1$ and that
|
|
$(a \cdot b) \cdot (\overline{a} + \overline{b}) = 0$, and use the fact that
|
|
$a \cdot b$ has a unique complement.)
|
|
|
|
**Proof:**
|
|
|
|
Suppose $B$ is a Boolean algebra, and that $a$ and $b$ are elements of $B$.
|
|
|
|
Prove that $(a \cdot b) + (\overline{a} + \overline{b}) = 1$:
|
|
|
|
$$ (a \cdot b) + (\overline{a} + \overline{b}) $$
|
|
|
|
$$ = ((a \cdot b) + \overline{a}) + \overline{b} $$
|
|
|
|
by the associative law for $+$
|
|
|
|
$$ = ((b \cdot a) + \overline{a}) + \overline{b} $$
|
|
|
|
by the commutative law for $+$
|
|
|
|
$$ = ((b + \overline{a}) \cdot (a + \overline{a})) + \overline{b} $$
|
|
|
|
by the distributive law for $+$ over $\cdot$
|
|
|
|
$$ = ((b + \overline{a}) \cdot 1) + \overline{b} $$
|
|
|
|
by the complement law for $+$
|
|
|
|
$$ = (b + \overline{a}) + \overline{b} $$
|
|
|
|
by the identity law for $\cdot$
|
|
|
|
$$ = b + (\overline{b} + \overline{a}) $$
|
|
|
|
by the commutative law for $+$
|
|
|
|
$$ = (b + \overline{b}) + \overline{a} $$
|
|
|
|
by the associative law for $+$
|
|
|
|
$$ = 1 + \overline{a} $$
|
|
|
|
by the complement law for $+$
|
|
|
|
$$ = 1 $$
|
|
|
|
by the universal bound law for $+$
|
|
|
|
Prove that $(a \cdot b) \cdot (\overline{a} + \overline{b}) = 0$:
|
|
|
|
$$ (a \cdot b) \cdot (\overline{a} + \overline{b}) $$
|
|
|
|
$$ = ((a \cdot b) \cdot \overline{a}) + ((a \cdot b) \cdot \overline{b}) $$
|
|
|
|
by the distributive law of $\cdot$ over $+$
|
|
|
|
$$ = ((a \cdot \overline{a}) \cdot b) + (a \cdot (b \cdot \overline{b})) $$
|
|
|
|
by the commutative and associative laws
|
|
|
|
$$ = (0 \cdot b) + (a \cdot 0) $$
|
|
|
|
by the complement laws
|
|
|
|
$$ = 0 + 0 $$
|
|
|
|
by the universal bound laws
|
|
|
|
$$ = 0 $$
|
|
|
|
by the identity laws
|
|
|
|
_Conclusion:_
|
|
|
|
Since $(a \cdot b) + (\overline{a} + \overline{b}) = 1$ and
|
|
$(a \cdot b) \cdot (\overline{a} + \overline{b}) = 0$, it can be concluded, by
|
|
the uniqueness of complement laws, that
|
|
$\overline{a \cdot b} = \overline{a} + \overline{b}$. This is what was to be
|
|
shown.
|
|
|
|
Q.E.D.
|
|
|
|
9. _De Morgan's law for $+$:_ For all $a$ and $b$ in $B$,
|
|
$\overline{a + b} = \overline{a} \cdot \overline{b}$.
|
|
|
|
**Proof:**
|
|
|
|
Suppose $B$ is a Boolean algebra, and that $a$ and $b$ are elements of $B$.
|
|
|
|
Prove that $(a + b) + (\overline{a} \cdot \overline{b}) = 1$:
|
|
|
|
$$ (a + b) + (\overline{a} \cdot \overline{b}) $$
|
|
|
|
$$ = ((a + b) + \overline{a}) \cdot ((a + b) + \overline{b}) $$
|
|
|
|
by the distributive laws for $+$ over $\cdot$
|
|
|
|
$$ = ((a + \overline{a}) + b) \cdot (a + (b + \overline{b})) $$
|
|
|
|
by the associative and commutative laws
|
|
|
|
$$ = (1 + b) \cdot (a + 1) $$
|
|
|
|
by the complement laws for $+$
|
|
|
|
$$ = 1 \cdot 1 $$
|
|
|
|
by the universal bound laws for $+$
|
|
|
|
$$ = 1 $$
|
|
|
|
by the identity laws for $\cdot$
|
|
|
|
Prove that $(a + b) \cdot (\overline{a} \cdot \overline{b}) = 0$:
|
|
|
|
$$ (a + b) \cdot (\overline{a} \cdot \overline{b}) $$
|
|
|
|
$$ = (a \cdot \overline{a}) \cdot (b \cdot \overline{b}) $$
|
|
|
|
by the commutative and associative laws for $\cdot$
|
|
|
|
$$ = 0 \cdot 0 $$
|
|
|
|
by the complement laws for $\cdot$
|
|
|
|
$$ = 0 $$
|
|
|
|
by the universal bound laws for $\cdot$
|
|
|
|
_Conclusion:
|
|
|
|
Since $(a + b) + (\overline{a} \cdot \overline{b}) = 1$ and
|
|
$(a + b) \cdot (\overline{a} \cdot \overline{b}) = 0$, it can be concluded, by
|
|
the uniqueness of complement laws, that
|
|
$\overline{a + b} = \overline{a} \cdot \overline{b}$. This is what was to be
|
|
shown.
|
|
|
|
Q.E.D.
|
|
|
|
10. _Cancellation law:_ For all $x$, $y$, and $z$ in $B$, if $x + y = x + z$ and
|
|
$x \cdot y = x \cdot z$, then $y = z$.
|
|
|
|
**Proof:**
|
|
|
|
Suppose $B$ is a Boolean Algebra, and that $x$, $y$, and $z$ are elements in $B$
|
|
such that $x + y = x + z$ and $x \cdot y = x \cdot z$.
|
|
|
|
$$ y = (y + x) \cdot y $$
|
|
|
|
by exercise 3
|
|
|
|
$$ = y \cdot (y + x) $$
|
|
|
|
by the commutative laws
|
|
|
|
$$ = y \cdot (x + y) $$
|
|
|
|
by the commutative laws
|
|
|
|
$$ = y \cdot (x + z) $$
|
|
|
|
by the supposition
|
|
|
|
$$ = (y \cdot x) + (y \cdot z) $$
|
|
|
|
by the distributive laws for $\cdot$ over $+$
|
|
|
|
$$ = (x \cdot y) + (y \cdot z) $$
|
|
|
|
by the commutative laws
|
|
|
|
$$ = (x \cdot z) + (y \cdot z) $$
|
|
|
|
by the supposition
|
|
|
|
$$ = (z \cdot x) + (z \cdot y) $$
|
|
|
|
by the commutative laws
|
|
|
|
$$ = z \cdot (x + y) $$
|
|
|
|
by the distributive laws of $\cdot$ over $+$
|
|
|
|
$$ = z \cdot (x + z) $$
|
|
|
|
by the supposition
|
|
|
|
$$ = (z \cdot x) + (z \cdot z) $$
|
|
|
|
by the distributive laws of $\cdot$ over $+$
|
|
|
|
$$ = (z \cdot x) + z $$
|
|
|
|
by the idempotent laws
|
|
|
|
$$ = (z \cdot x) + (z \cdot 1) $$
|
|
|
|
by the identity laws
|
|
|
|
$$ = z \cdot (x + 1) $$
|
|
|
|
by the distributive laws of $\cdot$ over $+$
|
|
|
|
$$ = z \cdot 1 $$
|
|
|
|
by the universal bound laws
|
|
|
|
$$ = z $$
|
|
|
|
by the identity laws
|
|
|
|
Q.E.D.
|
|
|
|
11. Let $S = \{0, 1\}$, and define operations $+$ and $\cdot$ on $S$ by the
|
|
following tables:
|
|
|
|
| $+$ | $0$ | $1$ |
|
|
| --- | --- | --- |
|
|
| $0$ | $0$ | $1$ |
|
|
| $1$ | $1$ | $1$ |
|
|
|
|
| $\cdot$ | $0$ | $1$ |
|
|
| ------- | --- | --- |
|
|
| $0$ | $0$ | $0$ |
|
|
| $1$ | $0$ | $1$ |
|
|
|
|
a. Show that the elements of $S$ satisfy the following properties:
|
|
|
|
i. the commutative law for $+$.
|
|
ii. the commutative law for $\cdot$.
|
|
iii. the associative law for $+$.
|
|
iv. the associative law for $\cdot$.
|
|
v. the distributive law for $+$ over $\cdot$.
|
|
vi. the distributive law for $\cdot$ over $+$.
|
|
|
|
i.
|
|
|
|
$$ 0 + 1 = 1 + 0 $$
|
|
|
|
$$ 1 = 1 $$
|
|
|
|
ii.
|
|
|
|
$$ 0 \cdot 1 = 1 \cdot 0 $$
|
|
|
|
$$ 0 = 0 $$
|
|
|
|
iii.
|
|
|
|
$$ (0 + 0) + 1 = 0 + (0 + 1) $$
|
|
|
|
$$ 1 = 1 $$
|
|
|
|
iv.
|
|
|
|
$$ (0 \cdot 0) \cdot 1 = 0 \cdot (0 \cdot 1) $$
|
|
|
|
$$ 0 \cdot 1 = 0 \cdot 0 $$
|
|
|
|
$$ 0 = 0 $$
|
|
|
|
v.
|
|
|
|
$$ 0 + (0 \cdot 1) = (0 + 0) \cdot (0 + 1) $$
|
|
|
|
$$ 0 + 0 = 0 \cdot 1 $$
|
|
|
|
$$ 0 = 0 $$
|
|
|
|
vi.
|
|
|
|
$$ 0 \cdot (0 + 1) = (0 \cdot 0) + (0 \cdot 1) $$
|
|
|
|
$$ 0 \cdot 1 = 0 + 0 $$
|
|
|
|
$$ 0 = 0 $$
|
|
|
|
NOTE: part a many cases are omitted as you have to explore each case (2 for both
|
|
commutative and associative, 8 for distributive).
|
|
|
|
b. Show that $0$ is an identity element for $+$ and that $1$ is an identity
|
|
element for $\cdot$.
|
|
|
|
_Hint:_ Verify that $0 + x = x$ and that $1 \cdot x = x$ for every $x \in S$.
|
|
|
|
$0 + 0 = 0$ and $0 + 1 = 1$, so $0$ is an identity for $+$.
|
|
|
|
$1 \cdot 0 k 0$ and $1 \cdot 1 = 1$, so $1$ is an identity for $\cdot$.
|
|
|
|
c. Define $\overline{0} = 1$ and $\overline{1} = 0$. Show that for every $a$ in
|
|
$S$, $a + \overline{a} = 1$ and $a \cdot \overline{a} = 0$. It follows from
|
|
parts (a)-\(c\) that $S$ is a Boolean algebra with the operations $+$ and
|
|
$\cdot$.
|
|
|
|
$a = 0$:
|
|
|
|
$$ 0 + \overline{0} = 0 + 1 = 1 $$
|
|
|
|
$$ 0 \cdot \overline{0} = 0 \cdot 1 = 0 $$
|
|
|
|
$a = 1$:
|
|
|
|
$$ 1 + \overline{1} = 1 + 0 = 1 $$
|
|
|
|
$$ 1 \cdot \overline{1} = 1 \cdot 0 = 0 $$
|
|
|
|
Exercises 12-15 provide an outline for a proof that the associative laws, which
|
|
were included as an axiom for a Boolean algebra, can be derived from the other
|
|
four axioms. The outline is from _Introduction to Boolean Algebra_ by S. Givant
|
|
and P. Halmos, Springer, 2009. In order to avoid unneeded parentheses, assume
|
|
that $\cdot$ takes precedence over $+$.
|
|
|
|
12. The universal bound law for $+$ states that for every element $a$ in a
|
|
Boolean algebra, $a + 1 = 1$. The proof shown in exercise 2 used the
|
|
associative law for $+$. Rederive the law without using the associative law
|
|
and using only the other four axioms for a Boolean algebra.
|
|
|
|
Omitted.
|
|
|
|
13. The absorption law for $+$ states that for all elements $a$ and $b$ in a
|
|
Boolean algebra, $a \cdot b + a = a$. Prove this law without using the
|
|
associative law and using only the other four axioms for a Boolean algebra
|
|
plus the result of exercise 12.
|
|
|
|
Omitted.
|
|
|
|
14. _Test for equality law:_ For all elements $a$, $b$, and $c$ in a Boolean
|
|
algebra,
|
|
|
|
If $b \cdot a = c \cdot a$ and $b \cdot \overline{a} = c \cdot \overline{a}$,
|
|
then $b = c$.
|
|
|
|
Without using the associative law, derive this law from the other four laws in
|
|
the axioms for a Boolean algebra plus the result of exercise 12.
|
|
|
|
Omitted.
|
|
|
|
15. The associative law for $+$ states that for all elements $a$, $b$, and $c$
|
|
in a Boolean algebra, $a + (b + c) = (a + b) + c$. Show that this law, as
|
|
well as the associative law for $\cdot$, can be derived from the other four
|
|
axioms in the definition and axioms for a Boolean algebra. Then explain how
|
|
to use your work to obtain a derivation for the associative law for $\cdot$.
|
|
|
|
_Hints:_ To prove this theorem, suppose $a$, $b$, and $c$ are any elements in a
|
|
Boolean algebra $B$, and divide the proof into three parts. _Part 1:_ Prove that
|
|
$(a + (b + c)) \cdot a = ((a + b) + c) \cdot a$. _Part 2:_ Prove that
|
|
$(a + (b + c)) \cdot \overline{a} = ((a + b) + c) \cdot \overline{a}$. _Part 3:_
|
|
Use the results of parts 1 and 2 to prove that $a + (b + c) = (a + b) + c$. You
|
|
may use the universal bound law for $+$, the absorption law for $+$, and the
|
|
test for equality law from exercises 12, 13, and 14 because the associative laws
|
|
were not used to derive these properties.
|
|
|
|
Omitted.
|
|
|
|
In 16-21 determine whether each sentence is a statement. Explain your answers.
|
|
|
|
16. This sentence is false.
|
|
|
|
In order for a sentence to be a statement, it must be either true or false.
|
|
|
|
The sentence, "This sentence is false.", is not a statement. If the sentence is
|
|
false, then "This sentence is false", is false and therefore the sentence is
|
|
true. On the other hand, if the sentence is true, then "This sentence is false."
|
|
is true and therefore the sentence is false. Consequently, the sentence is both
|
|
true and false and not a statement.
|
|
|
|
17. If $1 + 1 = 3$, then $1 = 0$.
|
|
|
|
This sentence is a statement. By logical deduction, if $1 + 1 = 3$, which is a
|
|
false hypothesis, then $1 = 0$, which is a false conclusion. Thus the sentence
|
|
is vacuously true.
|
|
|
|
18. $\boxed{\text{The sentence in this box is a lie.}}$
|
|
|
|
This sentence is not a statement. Since the sentence is in the box, the
|
|
hypothesis is true. The conclusion however can be either true or false for much
|
|
the same reasons as given in problem 16.
|
|
|
|
19. All positive integers with negative squares are prime.
|
|
|
|
This sentence is a statement. The hypothesis is that for all positive integers
|
|
with negative squares, but there are no such integers. This hypothesis is false,
|
|
therefore the conclusion that they are all prime is vacuously true.
|
|
|
|
20. This sentence is false or $1 + 1 = 3$.
|
|
|
|
This is not a statement. Since the conditional starts with the paradoxical
|
|
statement from problem 16, the addition of an "or" conditional does not change
|
|
the fact that this is not a statement.
|
|
|
|
21. This sentence is false and $1 + 1 = 2$.
|
|
|
|
This is not a statement. For reasons similar to 20. Think on the wording "true
|
|
and false" and "true." This is what this sentence is saying. It is not a
|
|
statement.
|
|
|
|
22.
|
|
|
|
a. Assuming that the following sentence is a statement, prove that $1 + 1 = 3$:
|
|
|
|
If this sentence is true, then $1 + 1 = 3$.
|
|
|
|
**Proof (by contradiction):**
|
|
|
|
Suppose that the sentence "If this sentence is true, then $1 + 1 = 3$" is false.
|
|
|
|
Since the sentence is false, then the hypothesis "If this sentence is true,"
|
|
must be true, and the conclusion, "$1 + 1 = 3$", must be false.
|
|
|
|
So the sentence is true (by the hypothesis), and false (by the conclusion). This
|
|
is a contradiction.
|
|
|
|
Therefore the supposition that the sentence is false is false. Therefore, the
|
|
supposition must be true, and its conclusion that $1 + 1 = 3$ must follow.
|
|
|
|
Q.E.D.
|
|
|
|
b. What can you deduce from part (a) about the status of "This sentence is
|
|
true"? Why? (This example is known as Lob's paradox.)
|
|
|
|
We can deduce that "This sentence is true" is paradoxical, _i.e._ it is both
|
|
true and not true. As such, "This sentence is true" is a sentence, but not a
|
|
statement.
|
|
|
|
It is worth noting that any conclusion that follows it is true by logical
|
|
deduction. This makes any conclusion following the hypothesis true, and thereby
|
|
any sentence true.
|
|
|
|
23. The following two sentences were devised by the logician Saul Kripke. While
|
|
not intrinsically paradoxical, they could be paradoxical under certain
|
|
circumstances. Describe such circumstances.
|
|
|
|
i. Most of Nixon's assertions about Watergate are false.
|
|
|
|
ii. Everything Jones says about Watergate is true.
|
|
|
|
(_Hint:_ Suppose Nixon says (ii) and the only utterance Jones makes about
|
|
Watergate is (i).)
|
|
|
|
Omitted.
|
|
|
|
24. Can there exist a computer program that has as output a list of all the
|
|
computer programs that do not list themselves in their output? Explain your
|
|
answer.
|
|
|
|
No. Suppose there exists a computer program $P$ that has as output a list of all
|
|
computer programs that do not list themselves in their output. If $P$ lists
|
|
itself as output, then it would be on the output list of $P$, which consists of
|
|
all computer programs that do not list themselves in their output. Hence $P$
|
|
would not list itself as output. But if $P$ does not list itself as output, then
|
|
$P$ would be a member of the list of all computer programs that do not list
|
|
themselves in their output, and this list is exactly the output of $P$. Hence
|
|
$P$ would list itself as output. This analysis shows that the assumption of the
|
|
existence of such a program $P$ is contradictory, and so no such program exists.
|
|
|
|
25. Can there exist a book that refers to all those books and only those books
|
|
that do not refer to themselves? Explain your answer.
|
|
|
|
This is the same as number 24.
|
|
|
|
Say there is a book $B$ that refers to all books that do not refer to
|
|
themselves. If $B$ refers to itself, then by the definition of $B$, $B$ would
|
|
not refer to itself. On the other hand, if $B$ does not refer to itself, then by
|
|
definition of $B$, $B$ would refer to itself. This is a paradox and therefore no
|
|
such book can exist.
|
|
|
|
26. Some English adjectives are descriptive of themselves (for instance, the
|
|
word _polysyllabic_ is polysyllabic) whereas others are not (for instance,
|
|
the word _monosyllabic_ is not monosyllabic). The word _heterological_
|
|
refers to an adjective that does not describe itself. Is _heterological_
|
|
heterological? Explain your answer.
|
|
|
|
If _heterological_ is heterological, then _heterological_ does not describe
|
|
itself. But since _heterological_ is heterological, it is describing itself by
|
|
the supposition. This is a contradiction.
|
|
|
|
If _heterological_ is not heterological, then _heterological_ does describe
|
|
itself, which contradicts its own definition.
|
|
|
|
It is paradoxical, _heterological_ is both heterological and not heterological.
|
|
|
|
27. As strange as it may seem, it is possible to give a precise-looking verbal
|
|
definition of an integer that, in fact, is not a definition at all. The
|
|
following was devised by an English librarian, G.G. Berry, and reported by
|
|
Bertrand Russell. Explain how it leads to a contradiction. Let $n$ be "the
|
|
smallest integer not describable in fewer than 12 English words." (Note that
|
|
the total number of strings consisting of 11 or fewer English words is
|
|
finite.)
|
|
|
|
Omitted.
|
|
|
|
28. Is there an algorithm which, for a fixed quantity $a$ and any input
|
|
algorithm $X$ and data set $D$, can determine whether $X$ prints $a$ when
|
|
run with data set $D$? Explain. (This problem is called the **printing
|
|
problem**.)
|
|
|
|
Omitted.
|
|
|
|
29. Use a technique similar to that used to derive Russell's paradox to prove
|
|
that for any set $A$, $\mathscr{P}(A) \nsubseteq A$.
|
|
|
|
Omitted.
|