discrete_mathematics_with_applications/chapter_6/notes.md

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.

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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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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 

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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]

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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 

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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

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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 

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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.

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Page 420

Theorem 6.2.2(3)(a) A Distributive Law for Sets

(Too lengthy, see page 420)

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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.

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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.

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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.]

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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.

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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].

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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.]

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Page 439

Definition and Axioms for a Boolean Algebra

A Boolean algebra is a set B together with two operations, generally denoted + and \cdot, such that for all a and b in B both a + b and a \cdot b are in B and the following axioms are assumed to hold:

1. Commutative Laws: For all a and b in B,

 \text{(a) } a + b = b + a \quad \text{ and } \quad \text{(b) } a \cdot b = b \cdot a 

2. Associative Laws: For all a, b, and c in B,

 \text{(a) } (a + b) + c = a + (b + c) \quad \text{ and } \quad \text{(b) } (a \cdot b) \cdot c = a \cdot (b \cdot c) 

3. Distributive Laws: For all a, b, and c in B,

 \text{(a) } a + (b \cdot c) = (a + b) \cdot (a + c) \quad \text{ and } \quad \text{(b) } a \cdot (b + c) = (a \cdot b) + (a \cdot c) 

4. Identity Laws: There exist distinct elements 0 and 1 in B such that for each a in B,

 \text{(a) } a + 0 = a \quad \text{ and } \quad \text{(b) } a \cdot 1 = a 

5. Complement Laws: For each a in B, there exists an element in B, denoted \overline{a} and called the complement or negation of a, such that

 \text{(a) } a + \overline{a} = 1 \quad \text{ and } \quad \text{(b) } a \cdot \overline{a} = 0 

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Page 439

Theorem 6.4.1 Properties of a Boolean Algebra

Let B be any Boolean algebra.

1. Uniqueness of the Complement Laws: For all a and x in B, if a + x = 1 and a \cdot x = 0 then x = \overline{a}.

2. Uniqueness of 0 and 1: If there exists x in B such that a + x = a for every a in B, then x = 0, and if there exists y in B such that a \cdot y = a for every a in B, then y = 1.

3. Double Complement Law: For every a \in B, \overline{(\overline{a})} = a.

4. Idempotent Laws: For every a \in B ,

 \text{(a) } a + a = a \quad \text{ and } \quad \text{(b) } a \cdot a = a 

5. Universal Bound Laws: For every a \in B,

 \text{(a) } a + 1 = 1 \quad \text{ and } \quad \text{(b) } a \cdot 0 = 0 

6. De Morgan's Laws: For all a and b \in B,

 \text{(a) } \overline{a + b} = \oveline{a} \cdot \overline{b} \quad \text{ and } \quad \text{(b) } \overline{a \cdot b} = \overline{a} + \overline{b}  

7. Absorption Laws: For all a and b \in B,

 \text{(a) } (a + b) \cdot a = a \quad \text{ and } \quad \text{(b) } (a \cdot b) + a = a 

8. Complements of 0 and 1:

 \text{(a) } \overline{0} = 1 \quad \text{ and } \quad \text{(b) } \overline{1} = 0 

Proof:

Part 1: Uniqueness of the Complement Law

Suppose a and x are particular, but arbitrarily chosen, elements of B that satisfy the following hypothesis: a + x = 1 and a \cdot x = 0. Then

 x = x \cdot 1 

because 1 is an identity for \cdot

 = x \cdot (a + \overline{a}) 

by the complement law for +

 = x \cdot a + x \cdot \overline{a} 

by the distributive law for \cdot over +

 =  a \cdot x + x \cdot \overline{a} 

by the commutative law for \cdot

 = 0 + x \cdot \overline{a} 

by hypothesis

 = a \cdot \overline{a} + x \cdot \overline{a} 

by the complement law for \cdot

 = (\overline{a} \cdot a) + (\overline{a} \cdot x) 

by the commutative law for \cdot

 = \overline{a} \cdot (a + x) 

by the distributive law for \cdot over +

 = \overline{a} \cdot 1 

by hypothesis

 = \overline{a} 

because 1 is an identity for \cdot.

Proofs of the other parts of the theorem are discussed in the examples that follow and in the exercises.

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Page 441

Theorem 6.4.1(3) Double Complement Law

For every element a in a Boolean algebra B, \overline{(\overline{a})} = a.

Proof:

Suppose B is a Boolean algebra and a is any element of B. Then

 \overline{a} + a = a + \overline{a} 

by the commutative law for +

 = 1 

by the complement law for 1

and

 \overline{a} \cdot a = a \cdot \overline{a} 

by the commutative law for \cdot

 = 0 

by the complement law for 0

Thus a satisfies the two equations with respect to \overline{a} that are satisfied by the complement of \overline{a}. From the fact that the complement of a is unique, we conclude that \overline{(\overline{a})} = a.

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Page 444

Theorem 6.4.2

There is no computer algorithm that will accept any algorithm X and data set D as input and then will output "halts" or "loops forever" to indicate whether or not X terminates in a finite number of steps when X is run with data set D.

Proof (by contradiction):

Suppose there is an algorithm, CheckHalt, such that if an algorithm X and a data set D are input, then

\text{CheckHalt}(X, D) prints

"halts" if X terminates in a finite number of steps when run with data set D

or

"loops forever" if X does not terminate in a finite number of steps when run with data set D.

[To show that no algorithm such as CheckHalt can exist, we will deduce a contradiction.]

Observe that the sequence of characters making up an algorithm X can be regarded as a data set itself. Thus it is possible to consider running CheckHalt with input (X, X). Define a new algorithm, Test, as follows: For any input algorithm X,

\text{Test}(X)

loops forever if \text{CheckHalt}(X, X) prints "halts"

or

stops if \text{CheckHalt}(X, X) prints "loops forever".

Now run algorithm Test with input Test. If \text{Test}(\text{Test}) terminates after a finite number of steps, then the value of \text{Checkhalt}(\text{Test}, \text{Test}) is "halts" and so \text{Test}(\text{Test}) loops forever.

On the other hand, if \text{Test}(\text{Test}) does not terminate after a finite number of steps, then \text{CheckHalt}(\text{Test}, \text{Test}) prints "loops forever" and so \text{Test}(\text{Test}) terminates.

The two paragraphs above show that \text{Test}(\text{Test}) loops forever and also that it terminates. This is a contradiction. But the existence of Test follows logically from the supposition of the existence of an algorithm CheckHalt that can check any algorithm and data set for termination. [Hence the supposition must be false, and there is no such algorithm.]