172 lines
5.1 KiB
Markdown
172 lines
5.1 KiB
Markdown
Page 401
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**Element Argument: The Basic Method for Proving That One Set is a Subset of
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Another**
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Let sets $X$ and $Y$ be given. To prove that $X \subseteq Y$,
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1. **suppose** that $x$ is a particular but arbitrarily chosen element of $X$,
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2. **show** that $x$ is an element of $Y$.
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---
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Page 402
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**Definition**
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Given sets $A$ and $B$, $A$ **equals** $B$, written $A = B$, if, and only if,
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every element of $A$ is in $B$ and every element of $B$ is in $A$.
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Symbolically:
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$$ A = B \Leftrightarrow A \subseteq B \text{ and } B \subseteq A $$
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---
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Page 404
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Let $A$ and $B$ be the subsets of a universal set $U$.
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1. The **union** of $A$ and $B$, denoted $A \cup B$, is the set of all elements
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that are in at least one of $A$ or $B$.
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2. The **intersection of $A$ and $B$, denoted $A \cap B$, is the set of all
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elements that are common to both $A$ and $B$.
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3. The **difference** of $B$ minus $A$ (or **relative complement** of $A$ in
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$B$), denoted $B - A$, is the set of all elements that are in $B$ and not
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$A$.
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4. The **complement** of $A$, denoted $A^c$, is the set of all elements in $U$
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that are not in $A$.
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Symbolically:
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$$ A \cup B = \{x \in U | x \in A \text{ or } x \in B\} $$
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$$ A \cap B = \{x \in U | x \in A \text{ and } x \in B\} $$
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$$ B - A = \{x \in U | x \in B \text{ and } x \notin A\} $$
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$$ A^c = \{x \in U | x \notin A\} $$
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---
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Page 405:
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**Interval Notation:**
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Given real numbers $a$ and $b$ with $a \leq b$:
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$$ (a, b) = \{x \in \mathbb{R} | a < x < b\} $$
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$$ [a, b] = \{x \in \mathbb{R} | a \leq x \leq b\} $$
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$$ (a, b] = \{x \in \mathbb{R} | a < x \leq b\} $$
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$$ [a, b) = \{x \in \mathbb{R} | a \leq x < b\} $$
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The symbols $\infty$ and $-\infty$ are used to indicate intervals that are
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unbounded either on the right or on the left:
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$$ (a, \infty) = \{x \in \mathbb{R} | x > a\} $$
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$$ [a, \infty) = \{x \in \mathbb{R} | x \geq a\} $$
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$$ (-\infty, b) = \{x \in \mathbb{R} | x < b\} $$
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$$ (-\infty, b] = \{x \in \mathbb{R} | x \leq b\} $$
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---
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Page 406
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**Definition**
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**Unions and Intersections of an Indexed Collection of Sets**
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Given sets $A_0, A_1, A_2, \dots$ that are subsets of a universal set $U$, and
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given a nonnegative integer $n$,
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$$ \bigcup_{i = 0}^{n}A_i = \{x \in U | x \in A_i \text{ for at least one } i = 0, 1, 2, \dots, n\} $$
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$$ \bigcup_{i = 0}^{\infty}A_i = \{x \in U | x \in A_i \text{ for at least one nonnegative integer } i\} $$
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$$ \bigcap_{i = 0}^{n}A_i = \{x \in U | x \in A_i \text{ for every } i = 0, 1, 2, \dots, n\} $$
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$$ \bigcap_{i = 0}^{\infty}A_i = \{x \in U | x \in A_i \text{ for every nonnegative integer } i\} $$
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---
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Page 408
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**Definition**
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Two sets are called **disjoint** if, and only if, they have no elements in
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common.
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Symbolically:
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$$ A \text{ and } B \text{ are disjoint } \Leftrightarrow A \cap B = \emptyset $$
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---
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Page 408
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**Definition**
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Sets $A_1, A_2, A_3, \dots$ are **mutually disjoint** (or **pairwise disjoint**
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or **nonoverlapping**) if, and only if, no two sets $A_i$ and $A_j$ with
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distinct subscripts have any elements in common. More precisely, for all
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integers $i$ and $j = 1, 2, 3, \dots$
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$$ A_i \cap A_j = \emptyset \text{ whenever } i \neq j $$
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---
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Page 408
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**Definition**
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A finite or infinite collection of nonempty sets $\{A_1, A_2, A_3, \dots\}$ is a
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**partition** of a set $A$ if, and only if,
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1. $A$ is the union of all the $A_i$.
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2. the sets $A_1, A_2, A_3, \dots$ are mutually disjoint.
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---
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Page 409
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**Definition**
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Given a set $A$, the **power** set of $A$, denoted $\mathscr{P}(A)$, is the set
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of all subsets of $A$.
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---
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Page 410
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**Algorithm 6.1.1 Testing whether $A \subseteq B$**
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_[The input sets $A$ and $B$ are represented as one-dimensional arrays
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$a[1], a[2], \dots, a[m]$ and $b[1], b[2], \dots, b[n]$, respectively. Starting
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with $a[1]$ and for each successive $a[i]$ in $A$, a check is made to see
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whether $a[i]$ is in $B$. To do this, $a[i]$ is compared to successive elements
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of $B$. If $a[i]$ is not equal to any element of $B$, then the output string,
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called answer, is given the value "$A \nsubseteq B$." If $a[i]$ equals some
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element of $B$, the next successive element in $A$ is checked to see whether it
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is in $B$. If every successive element of $A$ is found to be in $B$, then the
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answer never changes from its initial value "$A \subseteq B$."]_
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**Input:** _$m$ [a positive integer], $a[1], a[2], \dots, a[m]$ [a
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one-dimensional array representing the set $A$], $n$ [a positive integer],
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$b[1], b[2], \dots, b[n]$ [a one-dimensional array representing the set $B$]_
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**Algorithm Body:**
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$i := 1, \text{answer} := A \subseteq B\\ \text{\textbf{while}} (i \leq m \text{ and answer } = A \subseteq B )\\ \ \ j := 1, \text{found} := \text{"no"}\\ \ \ \text{\textbf{while }} (j \neq n \text{ and } \text{found}= \text{"no"})\\ \ \ \ \ \text{\textbf{if }} a[i] = b[j] \text{\textbf{ then }} \text{found} := \text{"yes"}\\ \ \ \ \ j := j + 1\\ \ \ \text{\textbf{end while}}\\ \ \ \text{[If found has not been given the value "yes" when execution reaches this point, then } a[i] \neq B\text{ .]}\\ \ \ \text{\textbf{if }} \text{found} = \text{"no"} \text{\textbf{ then }} \text{answer} := A \nsubseteq B\\ \ \ i := i + 1\\ \text{\textbf{end while}}$
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**Output:** _answer [a string]_
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