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Discrete Mathematics I Lectures Chapter 6. Some material adapted from lecture notes provided by Dr. Chungsim Han and Dr. Sam Lomonaco. Dr. Adam Anthony Spring 2011. … and now for something completely different…. Set Theory.
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Discrete Mathematics ILectures Chapter 6 Some material adapted from lecture notes provided by Dr. Chungsim Han and Dr. Sam Lomonaco Dr. Adam Anthony Spring 2011
… and now for something completely different… Set Theory Actually, you will see that logic and set theory are very closely related.
Set Theory • Set: Collection of objects (“elements”) • aA“a is an element of A” “a is a member of A” • aA“a is not an element of A” • A = {a1, a2, …, an} “A contains…” • Order of elements is meaningless • It does not matter how often the same element is listed – repeats are OK!.
Set Equality • Sets A and B are equal if and only if they contain exactly the same elements. • Examples: • A = {9, 2, 7, -3}, B = {7, 9, -3, 2} : A = B • A = {dog, cat, horse}, B = {cat, horse, squirrel, dog} : A B • A = {dog, cat, horse}, B = {cat, horse, dog, dog} : A = B
Examples for Sets • “Standard” Sets: • Natural numbers N = {0, 1, 2, 3, …} • Integers Z = {…, -2, -1, 0, 1, 2, …} • Positive Integers Z+ = {1, 2, 3, 4, …} • Real Numbers R = {47.3, -12, , …} • Rational Numbers Q = {1.5, 2.6, -3.8, 15, …}(correct definition will follow)
Examples for Sets • A = “empty set/null set” • A = {z} Note: zA, but z {z} • A = {{b, c}, {c, x, d}} • A = {{x, y}} Note: {x, y} A, but {x, y} {{x, y}} • A = {x | P(x)}“set of all x such that P(x)” • A = {x | xN x > 7} = {8, 9, 10, …}“set builder notation”
Examples for Sets • We are now able to define the set of rational numbers Q: • Q = {a/b | aZ bZ+} or • Q = {a/b | aZ bZ b0} • And how about the set of real numbers R? • R = {r | r is a real number}That is the best we can do.
Exercise 1 • What are the members of the following sets? • A = {x| x R x2 = 5} • B = {n | n Z n 2} • Use set-builder notation to describe the following sets: • The set of all positive integers that are divisible by 3 • The set of all animals with black and white stripes
Why Sets? • Sets add another layer of notational convenience • Call it Logic 3.0! • We can use sets to simplify logic expressions • Let BW = {x | x is an animal and x has black and white stripes} • All animals with black and white stripes are easy to see • a BW, easy-to-see(a) • There exists an animal without black and white stripes that is easy to see • aBW, easy-to-see(a)
Subsets • A B“A is a subset of B” • A B if and only if every element of A is also an element of B. • We can completely formalize this: • A B x (xA xB) • Examples: A = {3, 9}, B = {5, 9, 1, 3}, A B ? true A = {3, 3, 3, 9}, B = {5, 9, 1, 3}, A B ? true A = {1, 2, 3}, B = {2, 3, 4}, A B ? false
U B C A Subsets • Useful rules: • A = B (A B) (B A) • (A B) (B C) A C (see Venn Diagram)
Subsets • Useful rules: • A for any set A • A A for any set A • Proper subsets: • A B “A is a proper subset of B” • A B x (xA xB) x (xB xA) • or • A B x (xA xB) x (xB xA)
Exercise 2 Let A = {1,3,4}, B = {1,2,4}, C = {4,3,1,1}, D = {1,2,3,4} • Is 3A? • Is {3} A? • Is {3} A? • Is 3 A? • Is 3 B? • Is 3 B? • Is A B? • Is A D? • Is C A? • Is D A? • Is A = C? • Is A = D?
Exercise 3 • Let A = {a | a Z a -3}, B = {b | b Z bk = 5 for some k Z+}, C = {c | c Z c is prime 7 < c < 10} • Describe A, B and C by listing their elements • Is B A? • Is A B? • Is C A?
Cardinality of Sets • If a set S contains n distinct elements, nN,we call S a finite set with cardinality n. • Examples: • A = {Mercedes, BMW, Porsche}, |A| = 3 B = {1, {2, 3}, {4, 5}, 6} |B| = 4 C = |C| = 0 D = { xN | x 7000 } |D| = 7001 E = { xN | x 7000 } E is infinite!
The Power Set • (A) “power set of A” • (A) = {B | B A} (contains all subsets of A) • Examples: • A = {x, y, z} • (A)= {, {x}, {y}, {z}, {x, y}, {x, z}, {y, z}, {x, y, z}} • A = • (A) = {} • Note: |A| = 0, | (A)| = 1
A 1 2 3 4 5 6 7 8 x x x x x x x x x y y y y y y y y y z z z z z z z z z The Power Set • Cardinality of power sets: • | (A) | = 2|A| • Imagine each element in A has an “on/off” switch • Each possible switch configuration in A corresponds to one element in 2A • For 3 elements in A, there are 222 = 8 elements in P(A)
Exercise 4 • Find the power sets ({1,2}) and ({1,2,3})
Ordered Pairs • We are familiar with ordered pairs in algebra when we plot on a Cartesian coordinate system: • The pair (x,y) is orderedso that we know how to plot it on the axes • Important: (a,b) = (c,d) a = c and b = d • The order matters! (x,y)
Cartesian Product • The ordered n-tuple(a1, a2, a3, …, an) is an ordered collection of objects. • Just a general version of the ordered pair: • Two ordered n-tuples (a1, a2, a3, …, an) and (b1, b2, b3, …, bn) are equal if and only if they contain exactly the same elements in the same order, i.e. ai = bi for 1 i n. • The Cartesian product of two sets is defined as: • AB = {(a, b) | aA bB} • Example: A = {x, y}, B = {a, b, c}AB = {(x, a), (x, b), (x, c), (y, a), (y, b), (y, c)}
Cartesian Product • The Cartesian product of two sets is defined as: AB = {(a, b) | aA bB} • Example: • A = {good, bad}, B = {student, prof} • AB = { (bad, prof)} (good, student), (good, prof), (bad, student), (prof, bad)} BA = { (student, good), (prof, good), (student, bad),
Cartesian Product • Note that: • A = • A = • For non-empty sets A and B: AB AB BA • |AB| = |A||B| • The Cartesian product of two or more sets is defined as: • A1A2…An = {(a1, a2, …, an) | aiA for 1 i n}
Set Operations • Union: AB = {x | xA xB} • Example: A = {a, b}, B = {b, c, d} AB = {a, b, c, d} • Intersection: AB = {x | xA xB} • Example: A = {a, b}, B = {b, c, d} AB = {b}
Set Operations • Two sets are called disjoint if their intersection is empty, that is, they share no elements: • AB = • If a set C = A B and A and B are disjoint, then the set P = {A,B} is referred to as a partition of C • The difference between two sets A and B contains exactly those elements of A that are not in B: • A-B = {x | xA xB}Example:A = {a, b}, B = {b, c, d}, A-B = {a}
Set Operations • Given a Universe identified by the set U, the complement of a set A contains exactly those elements under consideration that are not in A: • Ac = U-A • Example:U = N, B = {250, 251, 252, …} Bc = {0, 1, 2, …, 248, 249}
Exercise 5 • Let the Universe U = {0,1,2,3,…,9} and: A = {0,2,4,6,8}B = {1, 2, 4, 7}C = {1,5,7} • Find the following: • A B • A B • Bc • A - B • (A B)c • A (B C) • (A B) C • Ac Bc
Exercise 6 • Let A = {1,2,3,4,5,6,7} Which collections form a partition of A? • P1 = {1,3}, {2,5}, {4,6,7} • P2 ={1}, {2,4,6}, {3,5} • P3 ={1,3,5,7}, {2,4,6}, {3,6} • Let B = {x | x R x -2 }, C = {x | x R -2 <x 4}, D = {x | x R x > 4} • Is P4 = {B,C,D} a partition of R?
Venn Diagrams of Operations • NOTE: The amount that A and B overlap is not always guaranteed A B
Venn Diagrams of Operations • A A B
Venn Diagrams of Operations • B A B
Venn Diagrams of Operations • A B A B
Venn Diagrams of Operations • A B
Venn Diagrams of Operations • A - B
Exercise 7 • Draw a Venn Diagram to represent the following expressions over the sets A,B and C: • A B = • A B = A and C B • C A B • Ac – C • A B and A C and B C BUT A B C
Section 6.2,6.3: Properties of Sets • Unifying Logic and Sets • Proofs with sets • Set identities
Logic and Sets • We can interpret our set operations logically: • A B x A x B (Set-based implication) • A B x A x B (Set-based OR) • A B x A x B (Set-based AND) • x Ac x A (Set-based Negation) • A – B x A x B (NEW expressive power!) • (x,y) A x B x A y B (NEW expressive power!)
Element Arguments • Prove that for all sets A and B, A B A • Normal starting Point: “Suppose A and B are arbitrarily chosen sets” • Where does this get us? • We make more progress using an ‘element argument’: • Note that A B A translates to “If x A B then x A” • From this, we get the statement: “Suppose A and B are arbitrarily chosen sets and x A B” • Idea: focusing on a single arbitrary element in a set lets us prove something about every single element in the set!
Transitivity Revisited • Prove, using logical interpretations, the transitive property for subsets: • For all sets A, B and C, if A B and B C, then A C.
Proving Equality • How can we Prove A(BC) = (AB)(AC)? • Recall: A = B A B B A • To prove A = B, you must show both subset claims are true • So an equality proof really requires 2 element proofs!
Prove A(BC) = (AB)(AC). • Prove that A(BC) (AB)(AC): Suppose that x is an arbitrarily chosen element of A(BC). There are 2 cases for x: x A or x (BC)Case 1: x ASince x A, by definition of union, x (A B) and x (A C). Therefore, by definition, x (AB)(AC). Case 2: x (BC). By definition of intersection, x B and x C. By definition of union, x (AB) and x (AC). Therefore, by definition of intersection, x (AB)(AC)
Prove A(BC) = (AB)(AC). • Prove that (AB)(AC) A(BC) • Drawing a Venn diagram might help us think this through… Suppose that x is an arbitrarily chosen element of (AB)(AC). We’ll analyze this one element in two cases: x A and x A. Case 1: x A It immediately follows that x A(BC) by definition of union. Case 2: x A Since x (AB)(AC), it must be true that x (AB) and x (AC). Since x A in this case, it must follow that x B and x C or else x would not be a member of (AB)(AC). Therefore, by definition of intersection, x (BC) and by definition of Union, x A(BC)
OR…Use logic instead! • How can we prove A(BC) = (AB)(AC)? • Suppose x is an arbitrary element of A(BC). Then by definition, the following statements are true: x(A(BC)) • (xA) (x(BC)) • (xA) (xB xC) • (xA xB) (xA xC)(distributive law for logical expressions) • (x(AB)) (x(AC)) • x((AB)(AC)) Since each statement is a biconditional with the one before it, x(A(BC)) x((AB)(AC)) and x((AB)(AC)) x(A(BC)) . So by definition, (A(BC)) ((AB)(AC)) and ((AB)(AC)) (A(BC)). Therefore, A(BC) = (AB)(AC)
Which method to use? • You may use an element argument or logic conversions to prove set identities on assignments and exams • Which one you choose depends on your strengths • Neither approach is ‘easier’ than the other • Element arguments provide more direction about where you’re headed • Logic arguments have less direction, but tend to be more concise
Exercise 1 • Use Venn Diagrams to verify the following set-theoretic DeMorgan’s law: • (A B)c = Ac Bc
Exercise 2 • Recall that two sets are disjoint if their intersection is the empty set (A B = ). • Draw venn diagrams for the sets A – B and B – A • Use Theorem 6.2.2 to construct a proof that A – B and B – A are disjoint
Exercise 3 • Use Theorem 6.2.2 to prove the following: (A (B C) ((A B) – C) = A B
A Technology Tie-In • Don’t forget your set theory—You’ll use it in Databases! • Each ‘table’ in a database is called a relation, filled with tuples • Database theory says that a relation is actually a subset of the universe of all possible tuples Sailors
A Technology Tie-In • SQL (structured query language): Select SID, SNAME, RATING, AGE FROM SAILORS WHERE AGE > 40 • S = {X SAILORS | AGE > 40} • See the similarity? Remembering set-theory will really help you work with databases Sailors