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@ -7,17 +7,106 @@ Page 411
a. $A = \{2, \{2\}, (\sqrt{2})^2\}$, $B = \{2, \{2\}, \{\{2\}\}\}$
$A \subseteq B$ ?:
$$ A = \{2, \{2\}, (\sqrt{2})^2\} = \{2, \{2\}, 2\} = \{2, \{2\}\} $$
Yes, every element in $A$ is in $B$.
$B \subseteq A$ ?:
No, because $\{\{2\}\}$ is an element of $B$, but is not an element of $A$, so
$\B \nsubseteq A$.
Is either $A$ or $B$ a proper subset of the other?
Yes, $A$ is a proper subset of $B$, because every element in $A$ is in $B$, but
not every element in $B$ is in $A$.
b. $A = \{3, \sqrt{5^2 - 4^2}, 24 \mod 7\}$, $B = \{8 \mod 5\}$
$A \subseteq B$ ?:
$$ A = \{3, \sqrt{5^2 - 4^2}, 24 \mod 7\} = \{3, 3, 3\} = \{3\} $$
$$ B = \{8 \mod 5\} = \{3\} $$
Yes, $A$ is a subset of $B$ since every element of $A$ is in $B$.
$B \subseteq A$ ?:
Yes, $B$ is a subset of $A$ since every element of $B$ is in $A$.
Is either $A$ or $B$ a proper subset of the other?
Yes, both $A$ and $B$ are proper subsets of the other since $A = B$.
c. $A = \{\{1, 2\}, \{2, 3\}\}$, $B = \{1, 2, 3\}$
$A \subseteq B$ ?:
No, because there are no elements in $A$ that are in $B$, $A \nsubseteq B$
$B \subseteq A$ ?:
No, because there are no elements in $B$ that are in $A$, $B \nsubseteq A$
Is either $A$ or $B$ a proper subset of the other?
No, since neither set share any elements, neither is a proper subset of the
other.
d. $A = \{a, b, c\}$, $B = \{\{a\}, \{b\}, \{c\}\}$
$A \subseteq B$ ?:
No, because there are no elements in $A$ that are in $B$, $A \nsubseteq B$
$B \subseteq A$ ?:
No, because there are no elements in $B$ that are in $A$, $B \nsubseteq A$
Is either $A$ or $B$ a proper subset of the other?
No, since neither set share any elements, neither is a proper subset of the
other.
e. $A = \{\sqrt{16}, \{4\}\}$, $B = \{4\}$
$A \subseteq B$ ?:
$$ A = \{\sqrt{16}, \{4\}\} = \{4, \{4\}\} $$
No, because every element of $A$ is not an element in $B$ ($4$ is not in $B$),
$A \nsubseteq B$.
$B \subseteq A$ ?:
Yes, because every element in $B$ is an element in $A$, $B \subseteq A$.
Is either $A$ or $B$ a proper subset of the other?
Yes, $B$ is a proper subset of $A$ since $B \subseteq A$ and $A \nsubseteq B$.
f. $A = \{x \in \mathbb{R} | \cos x \in \mathbb{Z}\}$,
$B = \{x \in \mathbb{R} | \sin x \in \mathbb{Z}\}$
From trigonometry, we know that $\cos x = -1 \text{ or } 0 \text{ or } 1$ and
$\sin x = -1 \text{ or } 0 \text{ or } 1 $. When we evaluate for $x$ in these
cases we find:
$$ A = \{\dots, -\frac{5\pi}{2}, -\frac{3\pi}{2}, \pi, -\frac{\pi}{2}, 0, \frac{\pi}{2}, \frac{3\pi}{2}, \frac{5\pi}{2}, \dots\} $$
$$ B = \{\dots, -\frac{5\pi}{2}, -\frac{3\pi}{2}, \pi, -\frac{\pi}{2}, 0, \frac{\pi}{2}, \frac{3\pi}{2}, \frac{5\pi}{2}, \dots\} $$
$A \subseteq B$ ?: Yes.
$B \subseteq A$ ?: Yes.
Yes, $B$ is a proper subset of $A$ since $B \subseteq A$ and $A \nsubseteq B$.
Yes, since $A = B$.
2. Complete the proof from Example 6.1.3: Prove that $B \subseteq A$ where
$$ A = \{m \in \mathbb{Z} | m, = 2a \text{ for some integer } a\} $$
@ -26,6 +115,32 @@ and
$$ B = \{n \in \mathbb{Z} | n = 2b - 2 \text{ for some integer } b\} $$
_Part 2, Proof that $B \subseteq A$:_
Suppose $x$ is a particular but arbitrarily chosen element of $B$.
By definition of $B$, there is an integer, say $b$, such that $x = 2b - 2$.
To prove that $B \subseteq A$, we must show that there is some $x$ that can
equal both $2a$, for some integer $a$, and that same $x$ can also equal
$2b - 2$.
$$ 2b - 2 = 2a $$
$$ a = b - 1 $$
By the difference integers, $a$ is an integer. Then, by substitution:
$$ 2a = 2(b - 1) $$
$$ = 2b - 2 $$
$$ = x $$
Thus, by definition of $A$, $x$ is an element of $A$.
Q.E.D.
3. Let sets $R$, $S$, and $T$ be defined as follows:
$$ R = \{x \in \mathbb{Z} | x \text{ is divisible by } 2\} $$
@ -38,26 +153,135 @@ Prove or disprove each of the following statements.
a. $R \subseteq T$
$R \nsubseteq T$ since $2 \in R$ since $2 \mid 2$, but $2 \notin T$ since
$6 \cancel{\mid} 2$.
b. $T \subseteq R$
**Proof:**
Suppose $n$ is any integer such that $6 \mid n$, therefore $n \in T$.
By the definition of divisibility:
$$ n = 6m $$
for some integer $m$.
$$ n = 2(3m) $$
By the product of integers, $3m$ is an integer. It follows that
$n = 2 \cdot (\text{some integer})$. Thus $2 \mid n$, so $n \in R$. This is what
was to be shown.
Q.E.D.
c. $T \subseteq S$
**Proof:**
Suppose $n$ is any integer such that $6 \mid n$, therefore $n \in T$.
By the definition of divisibility:
$$ n = 6m $$
for some integer $m$.
$$ n = 3(2m) $$
By the product of integers, $2m$ is an integer. It follows that
$n = 3 \cdot (\text{some integer})$. Thus $3 \mid n$, so $n \in S$. This is what
was to be shown.
Q.E.D.
4. Let $A = \{n \in \mathbb{Z} | n = 5r \text{ for some integer } r\}$ and
$B = \{m \in \mathbb{Z} | m = 20s \text{ for some integer } s\}$. Prove or
disprove each of the following statements.
a. $A \subseteq B$
$A \nsubseteq B$ since $5 \in A$ since $5 \mid 5$, but $5 \notin B$ since
$5 \cancel{\mid} 20$.
b. $B \subseteq A$
**Proof:**
Suppose $n$ is any integer such that $20 \mid n$, therefore $n \in B$.
By the definition of divisibility:
$$ n = 20m $$
for some integer $m$.
$$ n = 5(4m) $$
By the product of integers, $4m$ is an integer. It follows that
$n = 5 \cdot (\text{some integer})$. Thus $5 \mid n$, so $n \in A$. This is what
was to be shown.
Q.E.D.
5. Let $C = \{n \in \mathbb{Z} | n = 6r - 5 \text{ for some integer } r\}$ and
$D = \{m \in \mathbb{Z} | m = 3s + 1 \text{ for some integer } s\}$. Prove or
disprove each of the following statements.
a. $C \subseteq D$
**Proof:**
Suppose $n$ is any integer such that $n = 6r - 5$ for some integer $r$, which
means that $n \in C$.
Also suppose that $m$ is any integer such that $m = 3s + 1$ for some integer
$s$, which means that $m \in S$.
We must show that there exists some $r$ that when substituted for $s$ will
satisfy the definition of $n$.
Let $s = 2r - 2$. Then, by substitution:
$$ m = 3(2r - 2) + 1 $$
$$ = 6r - 6 + 1 $$
$$ = 6r - 5 $$
$$ = n $$
By the product and difference of integers, $6r - 5$ is an integer, therefore
$n \in D$. This is what was to be shown.
Q.E.D.
b. $D \subseteq C$
**Disproof:**
$D \nsubseteq C$ because there are elements in $D$ that are not in $C$. For
example $4$ is in $D$ because $4 = 3(1) + 1$, but $4$ is not in $C$. If $4$ were
in $C$, this would mean:
$$ 4 = 6r - 5 $$
for some integer $r$.
$$ 4 + 5 = 6r $$
$$ 9 = 6r $$
$$ \frac{9}{6} = r $$
$$ \frac{3}{2} = r $$
But $\dfrac{3}{2}$ is not an integer. This is a contradiction, therefore
$D \nsubseteq C$.
Q.E.D.
6. Let $A = \{x \in \mathbb{Z} | x = 5a + 2 \text{ for some integer } a\}$,
$B = \{y \in \mathbb{Z} | y = 10b - 3 \text{ for some integer } b\}$, and
$C = \{z \in \mathbb{Z} | z = 10c + 7 \text{ for some integer } c\}$.
@ -66,10 +290,141 @@ Prove or disprove each of the following statements.
a. $A \subseteq B$
**Disproof (by counterexample):**
Suppose $n$ is any integer such that $n = 5a + 2$ for some integer $a$. This
means that $n \in A$.
Suppose also that there is some integer $m$ such that $m = 10b - 3$ for some
integer $b$. This means that $m \in B$.
To show that there is some integer $b$ that will satisfy $n$, we must relate it
to $a$:
$$ 5a + 2 = 10b - 3 $$
$$ 5a + 5 = 10b $$
$$ 5a + 5 = 10b $$
$$ \frac{1}{2}a + \frac{1}{2} = b $$
$$ \frac{a + 1}{2} = b $$
In order for $A \subseteq B$, every element of $A$ must be in $B$. If $a = 0$,
then $n = 2$, so $n \in A$. If $a = 0$, then $b = \dfrac{1}{2}$, which is not an
integer, thus $2 \notin B$. Therefore $A \nsubseteq B$.
Q.E.D.
b. $B \subseteq A$
**Proof:**
Suppose $y$ is any integer such that $y = 10b - 3$ for some integer $b$. This
means that $y \in B$.
Let's first find $a$ as it relates to $y$.
$$ y = 5a + 2 $$
$$ 10b - 3 = 5a + 2 $$
$$ 10b - 5 = 5a $$
$$ 2b - 1 = a $$
So let $a = 2b - 1$.
Then substitute in for the condition for $A$:
$$ x = 5a + 2 $$
$$ = 5(2b - 1) + 2 $$
$$ = 10b - 5 + 2 $$
$$ = 10b - 3 $$
$$ = y $$
Therefore $y \in A$.
Q.E.D.
c. $B = C$
To prove $B = C$, we must prove both that $B \subseteq C$ and $C \subseteq B$.
_Prove $B \subseteq C$:
Suppose $y$ is any integer such that $y = 10b - 3$ for some integer $b$. This
means that $y \in B$.
Let's first find some integer $c$ as it relates to $y$:
$$ y = 10c + 7 $$
$$ 10b - 3 = 10c + 7 $$
$$ 10b - 10 = 10c $$
$$ b - 1 = c $$
So, let $c = b - 1$.
Then substitute in for the condition for $C$:
$$ z = 10c + 7 $$
$$ = 10(b - 1) + 7 $$
$$ = 10b - 10 + 7 $$
$$ = 10b - 3 $$
$$ = y $$
Thus $y \in C$, and therefore $B \subseteq C$.
_Prove $C \subseteq B$:
Suppose $z$ is any integer such that $z = 10c + 7$ for some integer $c$. This
means that $z \in C$.
Let's first find some integer $b$ as it relates to $z$:
$$ z = 10b - 3 $$
$$ 10c + 7 = 10b - 3 $$
$$ 10c + 10 = 10b $$
$$ c + 1 = b $$
So, let $b = c + 1$.
Then substitute in for the condition for $B$:
$$ y = 10b - 3 $$
$$ = 10(c + 1) - 3 $$
$$ = 10c + 10 - 3 $$
$$ = 10c + 7 $$
$$ = z $$
Therefore $z \in B$.
Thus $z \in B$, and therefore $C \subseteq B$.
Since $B \subseteq C$ and $C \subseteq B$, it follows that $B = C$. This is what
was to be shown.
Q.E.D.
7. Let $A = \{x \in \mathbb{Z} | x = 6a + 4 \text{ for some integer } a\}$,
$B = \{y \in \mathbb{Z} | y = 18b - 2 \text{ for some integer } b\}$, and
$C = \{z \in \mathbb{Z} | z = 18c + 16 \text{ for some integer } c\}$.
@ -78,49 +433,234 @@ Prove or disprove each of the following statements.
a. $A \subseteq B$
**Disproof (by counterexample):**
Suppose $x$ is any integer such that $x = 6a + 4$ for some integer $a$. This
means that $x \in A$.
Let's first find some integer $b$ as it relates to $a$.
$$ x = 18b - 2 $$
$$ 6a + 4 = 18b - 2 $$
$$ 6a + 6 = 18b $$
$$ \frac{6}{18}a + \frac{6}{18} = b $$
$$ \frac{1}{3}a + \frac{1}{3} = b $$
$$ \frac{a + 1}{3} = b $$
By definition of $b$, $b$ must always be an integer for all $a$.
Suppose $a = 0$, then:
$$ x = 6(0) + 4 = 4 $$
so $4 \in A$, but:
$$ b = \frac{0 + 1}{3} = \frac{1}{3} $$
so $4 \notin B$. We can see this as $4 = 18b - 2$ results in $b = \dfrac{1}{3}$,
but $b$ must be an integer.
b. $B \subseteq A$
**Proof:**
Suppose $y$ is any integer such that $y = 18b - 2$ for some integer $b$. This
means that $y \in B$.
Let's first find some integer $a$ as it relates to $b$.
$$ y = 6a + 4 $$
$$ 18b - 2 = 6a + 4 $$
$$ 18b - 6 = 6a $$
$$ 3b - 1 = a $$
Now, substitute $a$ in for the condition for $A$:
$$ x = 6a + 4 $$
$$ = 6(3b - 1) + 4 $$
$$ = 18b - 2 $$
$$ = y $$
Therefore $B \subseteq A$.
c. $B = C$
To prove $B = C$, we must prove both that $B \subseteq C$ and $C \subseteq B$.
_Prove $B \subseteq C$:
Suppose $y$ is any integer such that $y = 18b - 2$ for some integer $b$. This
means that $y \in B$.
Let's first find some integer $c$ as it relates to $b$.
$$ y = 18c + 16 $$
$$ 18b - 2 = 18c + 16 $$
$$ 18b - 18 = 18c $$
$$ b - 1 = c $$
So, let $c = b - 1$. Now substitute $c$ in for the condition of $C$:
$$ z = 18c + 16 $$
$$ = 18(b - 1) + 16 $$
$$ = 18b - 18 + 16 $$
$$ = 18b - 2 $$
$$ = y $$
Therefore $B \subseteq C$.
_Prove $C \subseteq B$:
Suppose $z$ is any integer such that $z = 18c + 16$ for some integer $c$.
Let's first find some $b$ as it relates to $c$.
$$ z = 18b - 2 $$
$$ 18c + 16 = 18b - 2 $$
$$ 18c + 18 = 18b $$
$$ c + 1 = b $$
So, let $b = c + 1$. Now, let's substitute $b$ in for the condition for $B$.
$$ y = 18b - 2 $$
$$ = 18(c + 1) - 2 $$
$$ = 18c + 18 - 2 $$
$$ = 18c + 16 $$
$$ = z $$
Therefore $C \subseteq B$.
Since $B \subseteq C$ and $C \subseteq B$, we conclude that $B = C$. This is
what was to be shown.
Q.E.D.
8. Write in words to read each of the following out loud. Then write each set
using the symbols for union, intersection, set difference, or set complement.
a. $\{x \in U | x \in A \text{ and } x \in B\}$
_In words:_
The set of all $x$ in $U$ such that $x$ is in $A$ and $x$ is in $B$.
_In symbolic notation:_
$$ A \cap B $$
b. $\{x \in U | x \in A \text{ or } x \in B\}$
_In words:_
The set of all $x$ in $U$ such that $x$ is in $A$ or $x$ is in $B$.
_In symbolic notation:_
$$ A \cup B $$
c. $\{x \in U | x \in A \text{ and } x \notin B\}$
_In words:_
The set of all $x$ in $U$ such that $x$ is in $A$ and $x$ is not in $B$.
_In symbolic notation:_
$$ A - B $$
d. $\{x \in U | x \notin A\}$
_In words:_
The set of all $x$ in $U$ such that $x$ is not in $A$.
_In symbolic notation:_
$$ A^c $$
9. Complete the following sentences without using the symbols $\cup$, $\cap$, or
$-$.
a. $x \notin A \cup B$ if, and only if, _____.
$x$ is not in $A$ and $x$ is not in $B$.
b. $x \notin A \cap B$ if, and only if, _____.
$x$ is not in $A$ or $x$ is not in $B$.
c. $x \notin A - B$ if, and only if, _____.
$x$ is not in $A$, or $x$ is in $B$, or both.
Note: recall that the negation of an "and", which is $A - B$, is an "or", thus:
$$ x \in (A - B) \to x \in A \wedge x \notin B $$
so:
$$ \neg(x \in (A - B)) \to \neg(x \in A \wedge x \notin B) \to x \notin A \vee x \in B $$
10. Let $A = \{1, 3, 5, 7, 9\}$, $b = \{3, 6, 9\}$, and $C = \{2, 4, 6, 8\}$.
Find each of the following:
a. $A \cup B$
$$ A \cup B = \{1, 3, 5, 6, 7, 9\} $$
b. $A \cap B$
$$ A \cap B = \{3, 9\} $$
c. $A \cup C$
$$ A \cup C = \{1, 2, 3, 4, 5, 6, 7, 8, 9\} $$
d. $A \cap C$
$$ A \cap C = \emptyset $$
e. $A - B$
$$ A - B = \{1, 5, 7\} $$
f. $B - A$
$$ B - A = \{6\} $$
g. $B \cup C$
$$ B \cup C = \{2, 3, 4, 6, 8, 9\} $$
h. $B \cap C$
$$ B \cap C = \{6\} $$
11. Let the universal set $\mathbb{R}$, the set of all real numbers, and let
$A = \{x \in \mathbb{R} | 0 < x \leq 2\}$,
$B = \{x \in \mathbb{R} | 1 \leq x < 4\}$, and
@ -128,24 +668,56 @@ h. $B \cap C$
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
@ -153,24 +725,56 @@ j. $(A \cup B)^c$
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
@ -178,66 +782,146 @@ j. $(A \cup B)^c$
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. $A_1 \cup A_2 \cup A_3 \cup A_4 = \text{ ?}$

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chapter_6/test_yourself.md
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@ -4,25 +4,51 @@ Page 411$a
1. The notation $A \subseteq B$ is read "_____" and means that _____.
The set $A$ is a subset of the set $B$; if $x \in A$ then $x \in B$
2. To use an element argument for proving that a set $X$ is a subset of a set
$Y$, you suppose that _____ and show that _____.
$x$ is a particular but arbitrarily chosen element of $X$; $x$ is an element of
$Y$.
3. To disprove that a set $X$ is a subset of a set $Y$, you show that there is
_____.
an element in $X$ that is not in $Y$.
4. An element $x$ is in $A \cup B$ if, and only if, _____.
$x$ is in either $A$ or $B$.
5. An element $x$ is in $A \cap B$ if, and only if, _____.
$x$ is in both $A$ and $B$.
6. An element $x$ is in $B - A$ if, and only if, _____.
$x$ is in $B$ but not in $A$.
7. An element $x$ is in $A^c$ if, and only if, _____.
$x$ is in the universal set and is not in $A$.
8. The empty set is a set with _____.
no elements.
9. The power set of a set $A$ is _____.
the set of all subsets of $A$.
10. Sets $A$ and $B$ are disjoint if, and only if, _____.
they have no elements in common, or $A \cap B = \emptyset$.
11. A collection of nonempty sets $A_1, A_2, A_3, \dots$ is a partition of a set
$A$ if, and only if, _____.
all $A_i$ are a subset of $A$, but are also disjoint.
$A$ is the union of all the sets $A_1, A_2, A_3, \dots$ and
$A_i \cap A_j = \emptyset$ whenever $i \neq j$.