Page 401 **Element Argument: The Basic Method for Proving That One Set is a Subset of Another** Let sets $X$ and $Y$ be given. To prove that $X \subseteq Y$, 1. **suppose** that $x$ is a particular but arbitrarily chosen element of $X$, 2. **show** that $x$ is an element of $Y$. --- Page 402 **Definition** Given sets $A$ and $B$, $A$ **equals** $B$, written $A = B$, if, and only if, every element of $A$ is in $B$ and every element of $B$ is in $A$. Symbolically: $$ A = B \Leftrightarrow A \subseteq B \text{ and } B \subseteq A $$ --- Page 404 Let $A$ and $B$ be the subsets of a universal set $U$. 1. The **union** of $A$ and $B$, denoted $A \cup B$, is the set of all elements that are in at least one of $A$ or $B$. 2. The **intersection of $A$ and $B$, denoted $A \cap B$, is the set of all elements that are common to both $A$ and $B$. 3. The **difference** of $B$ minus $A$ (or **relative complement** of $A$ in $B$), denoted $B - A$, is the set of all elements that are in $B$ and not $A$. 4. The **complement** of $A$, denoted $A^c$, is the set of all elements in $U$ that are not in $A$. Symbolically: $$ A \cup B = \{x \in U | x \in A \text{ or } x \in B\} $$ $$ A \cap B = \{x \in U | x \in A \text{ and } x \in B\} $$ $$ B - A = \{x \in U | x \in B \text{ and } x \notin A\} $$ $$ A^c = \{x \in U | x \notin A\} $$ --- Page 405: **Interval Notation:** Given real numbers $a$ and $b$ with $a \leq b$: $$ (a, b) = \{x \in \mathbb{R} | a < x < b\} $$ $$ [a, b] = \{x \in \mathbb{R} | a \leq x \leq b\} $$ $$ (a, b] = \{x \in \mathbb{R} | a < x \leq b\} $$ $$ [a, b) = \{x \in \mathbb{R} | a \leq x < b\} $$ The symbols $\infty$ and $-\infty$ are used to indicate intervals that are unbounded either on the right or on the left: $$ (a, \infty) = \{x \in \mathbb{R} | x > a\} $$ $$ [a, \infty) = \{x \in \mathbb{R} | x \geq a\} $$ $$ (-\infty, b) = \{x \in \mathbb{R} | x < b\} $$ $$ (-\infty, b] = \{x \in \mathbb{R} | x \leq b\} $$ --- Page 406 **Definition** **Unions and Intersections of an Indexed Collection of Sets** Given sets $A_0, A_1, A_2, \dots$ that are subsets of a universal set $U$, and given a nonnegative integer $n$, $$ \bigcup_{i = 0}^{n}A_i = \{x \in U | x \in A_i \text{ for at least one } i = 0, 1, 2, \dots, n\} $$ $$ \bigcup_{i = 0}^{\infty}A_i = \{x \in U | x \in A_i \text{ for at least one nonnegative integer } i\} $$ $$ \bigcap_{i = 0}^{n}A_i = \{x \in U | x \in A_i \text{ for every } i = 0, 1, 2, \dots, n\} $$ $$ \bigcap_{i = 0}^{\infty}A_i = \{x \in U | x \in A_i \text{ for every nonnegative integer } i\} $$ --- Page 408 **Definition** Two sets are called **disjoint** if, and only if, they have no elements in common. Symbolically: $$ A \text{ and } B \text{ are disjoint } \Leftrightarrow A \cap B = \emptyset $$ --- Page 408 **Definition** Sets $A_1, A_2, A_3, \dots$ are **mutually disjoint** (or **pairwise disjoint** or **nonoverlapping**) if, and only if, no two sets $A_i$ and $A_j$ with distinct subscripts have any elements in common. More precisely, for all integers $i$ and $j = 1, 2, 3, \dots$ $$ A_i \cap A_j = \emptyset \text{ whenever } i \neq j $$ --- Page 408 **Definition** A finite or infinite collection of nonempty sets $\{A_1, A_2, A_3, \dots\}$ is a **partition** of a set $A$ if, and only if, 1. $A$ is the union of all the $A_i$. 2. the sets $A_1, A_2, A_3, \dots$ are mutually disjoint. --- Page 409 **Definition** Given a set $A$, the **power** set of $A$, denoted $\mathscr{P}(A)$, is the set of all subsets of $A$. --- Page 410 **Algorithm 6.1.1 Testing whether $A \subseteq B$** _[The input sets $A$ and $B$ are represented as one-dimensional arrays $a[1], a[2], \dots, a[m]$ and $b[1], b[2], \dots, b[n]$, respectively. Starting with $a[1]$ and for each successive $a[i]$ in $A$, a check is made to see whether $a[i]$ is in $B$. To do this, $a[i]$ is compared to successive elements of $B$. If $a[i]$ is not equal to any element of $B$, then the output string, called answer, is given the value "$A \nsubseteq B$." If $a[i]$ equals some element of $B$, the next successive element in $A$ is checked to see whether it is in $B$. If every successive element of $A$ is found to be in $B$, then the answer never changes from its initial value "$A \subseteq B$."]_ **Input:** _$m$ [a positive integer], $a[1], a[2], \dots, a[m]$ [a one-dimensional array representing the set $A$], $n$ [a positive integer], $b[1], b[2], \dots, b[n]$ [a one-dimensional array representing the set $B$]_ **Algorithm Body:** $i := 1, \text{answer} := A \subseteq B\\ \text{\textbf{while}} (i \leq m \text{ and answer } = A \subseteq B )\\ \ \ j := 1, \text{found} := \text{"no"}\\ \ \ \text{\textbf{while }} (j \neq n \text{ and } \text{found}= \text{"no"})\\ \ \ \ \ \text{\textbf{if }} a[i] = b[j] \text{\textbf{ then }} \text{found} := \text{"yes"}\\ \ \ \ \ j := j + 1\\ \ \ \text{\textbf{end while}}\\ \ \ \text{[If found has not been given the value "yes" when execution reaches this point, then } a[i] \neq B\text{ .]}\\ \ \ \text{\textbf{if }} \text{found} = \text{"no"} \text{\textbf{ then }} \text{answer} := A \nsubseteq B\\ \ \ i := i + 1\\ \text{\textbf{end while}}$ **Output:** _answer [a string]_ --- Page 414 **Theorem 6.2.1 Some Subset Relations** 1. _Inclusion of Intersection:_ For all sets $A$ and $B$, $$ \text{(a) } A \cap B \subseteq A \quad \text{ and } \quad \text{ (b) } A \cap B \subseteq B $$ 2. _Inclusion in Union:_ For all sets $A$ and $B$, $$ \text{(a) } A \subseteq A \cup B \quad \text{ and } \quad \text{ (b) } B \subseteq A \cup B $$ 3. _Transitive Property of Subsets:_ For all sets $A$, $B$, $C$, $$ \text{if } A \subseteq B \text{ and } B \subseteq C \text{, then } A \subseteq C $$ --- Page 415 **Procedural Versions of Set Definitions** Let $X$ and $Y$ be subsets of a universal set $U$ and suppose $x$ and $y$ are elements of $U$. 1. $x \in X \cup Y \Leftrightarrow x \in X \text{ or } x \in Y$ 2. $x \in X \cap Y \Leftrightarrow x \in X \text{ and } x \in Y$ 3. $x \in X - Y \Leftrightarrow x \in X \text{ and } x \notin Y$ 4. $x \in X^c \Leftrightarrow x \notin X$ 5. $(x, y) \in X \times Y \Leftrightarrow x \in X \text{ and } y \in Y$ --- Page 417 **Theorem 6.2.2 Set Identities** Let all sets referred to below be subsets of a universal set $U$. 1. _Commutative Laws:_ For all sets $A$ and $B$, $$ \text{(a) } A \cup B = B \cup A \quad \text{ and } \quad \text{ (b) } A \cap B = B \cap A $$ 2. _Associative Laws:_ For all sets $A$, $B$, and $C$, $$ \text{(a) } (A \cup B) \cup C = A \cup (B \cup C) \quad \text{ and } \quad \text{ (b) } (A \cap B) \cap C = A \cap (B \cap C) $$ 3. _Distributive Laws:_ For all sets $A$, $B$, and $C$, $$ \text{(a) } A \cup (B \cap C) = (A \cup B) \cap (A \cup C) \quad \text{ and } \quad \text{ (b) } A \cap (B \cup C) = (A \cap B) \cup (A \cap C) $$ 4. _Identity Laws:_ For every set $A$, $$ \text{(a) } A \cup \emptyset = A \quad \text{ and } \quad \text{ (b) } A \cap U = A $$ 5. _Complement Laws:_ For every set $A$, $$ \text{(a) } A \cup A^c = U \quad \text{ and } \quad A \cap A^c = \emptyset $$ 6. _Double Complement Law:_ For every set $A$, $$ (A^c)^c = A $$ 7. _Idempotent Laws:_ For every set $A$, $$ \text{(a) } A \cup A = A \quad \text{ and } \quad \text{ (b) } A \cap A = A $$ 8. _Universal Bound Laws:_ For every set $A$, $$ \text{(a) } A \cup U = U \quad \text{ and } \quad \text{ (b) } A \cap \emptyset = \emptyset $$ 9. _De Morgan's Laws:_ For all sets $A$ and $B$, $$ \text{(a) } (A \cup B)^c = A^c \cap B^c \quad \text{ and } \quad \text{ (b) } (A \cap B)^c = A^c \cup B^c $$ 10. _Absorption Laws:_ For all sets $A$ and $B$, $$ \text{(a) } A \cup (A \cap B) = A \quad \text{ and } \quad \text{ (b) } A \cap (A \cup B) = A $$ 11. _Complements of $U$ and $\emptyset$:_ $$ \text{(a) } U^c = \emptyset \quad \text{ and } \quad \text{ (b) } \emptyset^c = U $$ 12. _Set Difference Law:_ For all sets $A$ and $B$, $$ A - B = A \cap B^c $$ --- Page 418 **Basic Method for Proving That Sets Are Equal** Let sets $X$ and $Y$ be given. To prove that $X = Y$: 1. Prove that $X \subseteq Y$. 2. Prove that $Y \subseteq X$. --- Page 420 **Theorem 6.2.2(3)(a) A Distributive Law for Sets** (Too lengthy, see page 420) --- Page 422 **Theorem 6.2.2(9)(a) A De Morgan's Law for Sets** For all sets $A$ and $B$, $(A \cup B)^c = A^c \cap B^c$. **Proof:** Suppose $A$ and $B$ are sets. _Proof that $(A \cup B)^c \subseteq A^c \cap B^c$:_ _[We must show that $\forall x, \text{ if } x \in (A \cup B)^c \text{ then } x \in A^c \cap B^c$.]_ Suppose $x \in (A \cup B)^c$. _[We must show that $x \in A^c \cap B^c$.]_ By definition of complement, $$ x \notin A \cup B $$ Now to say that $x \notin A \cup B$ means that it is false that ($x$ is in $A$ or $x$ is in $B$). By De Morgan's laws of logic, this implies that $x$ is not in $A$ and $x$ is not in $B$, which can be written $$ x \notin A \quad \text{ and } \quad x \notin B $$ Hence $x \in A^c$ and $x \in B^c$ by definition of complement. It follows, by definition of intersection, that $x \in A^c \cap B^c$ _[as was to be shown]._ So $(A \cup B)^c \subseteq A^c \cap B^c$ by definition of subset. _Proof that $A^c \cap B^c \subseteq (A \cup B)^c$:_ _[We must show that $\forall x, \text{ if } x \in A^c \cap B^c \text{ then } x \in (A \cup B)^c$.]_ Suppose $x \in A^c \cap B^c$. _[We must show that $x \in (A \cup B)^c$.]_ By definition of intersection, $x \in A^c$ and $x \in B^c$, and by definition of complement, $$ x \notin A \quad \text{ and } \quad x \notin B $$ In other words, $x$ is not in $A$ and $x$ is not in $B$. By De Morgan's laws of logic this implies that it is false that ($x$ is in $A$ or $x$ is in $B$), which can be written $$ x \notin A \cup B $$ by definition of union. Hence, by definition of complement, $x \in (A \cup B)^c$ _[as was to be shown]._ It follows that $A^c \cap B^c \subseteq (A \cup B)^c$ by definition of subset. _Conclusion:_ Since both set containments have been proved, $(A \cup B)^c = A^c \cap B^c$ by definition of set equality. --- Page 423 **Theorem 6.2.3 Intersection and Union with a Subset** For any sets $A$ and $B$, if $A \subseteq B$, then $$ \text{(a) } A \cap B = A \quad \text{ and } \quad \text{ (b) } A \cup B = B $$ **Proof:** _Part (a):_ Suppose $A$ and $B$ are sets with $A \subseteq B$. To show part (a) we must show both that $A \cap B \subseteq A$ and that $A \subseteq A \cap B$. We already know that $A \cap B \subseteq A$ by the inclusion of intersection property. To show that $A \subseteq A \cap B$, let $x$ be any element in $A$. _[We must show that $x$ is in $A \cap B$.]_ But, because of the hypothesis that $A \subseteq B$, we can conclude that $x$ is also in $B$ by definition of subset. Hence $$ x \in A \quad \text{ and } x \in B $$ and thus $$ x \in A \cap B $$ by definition of intersection _[as was to be shown]._ **Proof:** _Part (b):_ The proof of part (b) is left as an exercise. --- Page 424 **Theorem 6.2.4 A Set with No Elements Is a Subset of Every Set** If $E$ is a set with no elements and $A$ is any set, then $E \subseteq A$. **Proof (by contradiction):** Suppose not. _[We take the negation of the theorem and suppose it to be true.]_ Suppose there exists a set $E$ with no elements and a set $A$ such that $E \nsubseteq A$. _[We must deduce a contradiction.]_ Then there would be an element of $E$ that is not an element of $A$ _[by definition of subset]_. But there can be no such element since $E$ has no elements. This is a contradiction. _[Hence the supposition that there are sets $E$ and $A$, where $E$ has no elements and $E \nsubseteq A$, is false, and so the theorem is true.]_ --- Page 424 **Corollary 6.2.5 Uniqueness of the Empty Set** There is only one set with no elements. **Proof:** Suppose $E_1$ and $E_2$ are both sets with no elements. By Theorem 6.2.4, $E_1 \subseteq E_2$ since $E_1$ has no elements. Also $E_2 \subseteq E_1$ since $E_2$ has no elements. Thus $E_1 = E_2$ by definition of set equality. --- Page 425 **Proposition 6.2.6** For all sets $A$, $B$, and $C$, if $A \subseteq B$ and $B \subseteq C^c$, then $A \cap C = \emptyset$. **Proof:** Suppose $A$, $B$, and $C$ are sets such that $A \subseteq B$ and $B \subseteq C^c$. We must show that $A \cap C = \emptyset$. Suppose not. That is, suppose there is an element $x$ in $A \cap C$. By definition of intersection, $x \in A$ and $x \in C$. Then, since $A \subseteq B$, $x \in B$ by definition of subset. Also, since $B \subseteq C^c$, then $x \in C^c$ by definition of subset again. It follows by definition of complement that $x \notin C$. Thus $x \in C$ and $x \notin C$, which is a contradiction. So the supposition that there is an element $x$ in $A \cap C$ is false, and thus $A \cap C = \emptyset$ _[as was to be shown]_. --- Page 433 **Theorem 6.3.1** For every integer $n \geq 0$, if a set $X$ has $n$ elements, then $\mathscr{P}(X)$ has $2^n$ elements. **Proof (by mathematical induction):** Let the property $P(n)$ be the sentence Any set with $n$ elements has $2^n$ subsets. _Show that $P(0)$ is true:_ To establish $P(0)$, we must show that Any set with $0$ elements has $2^0$ subsets. Now the only set with zero elements is the empty set, and the only subset of the empty set is itself. Thus a set with zero elements has one subset. Since $1 = 2^0$, we have that $P(0)$ is true. _Show that for every integer $k \geq 0$, if $P(k)$ is true then $P(k + 1)$ is also true:_ _[Suppose that $P(k)$ is true for a particular but arbitrarily chosen integer $k \geq 0$. That is:]_ Suppose that $k$ is any integer with $k \geq 0$ such that Any set with $k$ elements has $2^k$ subsets. _[We must show that $P(k + 1)$ is true. That is:]_ We must show that Any set with $k + 1$ elements has $2^{k + 1}$ subsets. Let $X$ be a set with $k + 1$ elements. Since $k + 1 \geq 1$, we may pick an element $z$ in $X$. Observe that any subset of $X$ either contains $z$ or does not. Furthermore, any subset of $X$ that does not contain $z$ is a subset of $X - \{z\}$. And any subset $A$ of $X - \{z\}$ can be matched up with a subset $B$, equal to $A \cup \{z\}$, of $X$ that contains $z$. Consequently, there are as many subsets of $X$ that contain $z$ as do not, and thus there are twice as many subsets of $X$ as there are subsets of $X - \{z\}$. It follows that since $X - \{z\}$ has $k$ elements, then, by inductive hypothesis, the number of subsets of $X - \{z\} = 2^k$ Therefore, the number of subsets $X = 2 \cdot (\text{the number of subsets of } X - \{z\})$ $$ = 2 \cdot (2^k) $$ $$ = 2^{k + 1} $$ _[This is what was to be shown.]_ _[Since we have proved both the basis step and the inductive step, we conclude that the theorem is true.]_