discrete_mathematics_with_applications/chapter_6/notes.md

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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.

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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 

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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\} 

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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\} 

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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\} 

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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 

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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 

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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.

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Page 409

Definition

Given a set A, the power set of A, denoted \mathscr{P}(A), is the set of all subsets of A.

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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]