Student Responsibilities — Week 4

I Reading: Textbook, Section 2.1–2.3

Mat 2345 I Attendance: Strongly Encouraged

Week 4

I Week 4 Overview

Fall 2013 I 2.1 Sets

I 2.2 Set Operations

I 2.3 Functions

2.1 Sets Notation used to specify sets

I list the elements between braces; listing an object more than once does not change the set—ordering means nothing. I Set: an unordered collection or group of objects, which are said to be elements, or members of the set S = a, b, c, d = b, c, a, d, d { } { }

I Set builder notation – specify by predicate; here, S contains all I A set is said to contain its elements elements from U which make the predicate P true S = x P(x) { | } I There must be an underlying Universal Set, U, either specifically stated or understood I brace notation with ellipses; here, the negative integers: S = ..., 3, 2, 1 { − − − }

Common Universal Sets Notation

I R — Real Numbers I x S — x is a member of S, or x is an element of S ∈

I x / S — x is not an element of S I N — Natural Numbers: 0, 1, 2, 3,... ∈ { }

I Z — Integers: ..., 3, 2, 1, 0, 1, 2, 3,... Set Equality — Definition #1 { − − − } I Two sets are equal if and only if they have the same elements. + I Z — Positive Integers I That is, if A and B are sets, then A and B are equal if and only if x[x A x B]. ∀ ∈ ↔ ∈ p I Q — Rational Numbers: p, q Z q = 0 { q | ∈ ∧ 6 } I We write A = B if A and B are equal sets. Subsets Power Set

I Subset: Let A and B be sets. Then A B x [ x A x B ] I Power Set: P(A) is the set of all possible subsets of the set A ⊆ ⇔ ∀ ∈ → ∈

I If A = a, b , then I Empty, void, or Null Set: ∅ is the set with no members { } P(A) = ∅, a , b , a, b { { } { } { }} I the assertion x ∅ is always false, thus: ∈ I What is the power set of the set B = 0, 1, 2 ? x [ x ∅ x B] is always (vacuously) true, ∀ ∈ → ∈ { } and therefore ∅ is a subset of every set

I Note: a set B is always a subset of itself: B B ⊆ I How many elements would P( a, b, c, d ) have? { } I Proper subset: A B if A B, but A = B ⊂ ⊆ 6

Cardinality Example

I Cardinality: A is the number of distinct elements in A | | I Let A = ∅, ∅ I If the cardinality is a natural number (in N), then the set is { { }} I A has two elements and hence four subsets: called finite; otherwise, it’s called infinite ∅, ∅ , ∅ , ∅, ∅ { } {{ }} { { }} I Example: Let A = a, b I Note that ∅ is both a member of A and a subset of A { } I A = a, b = 2 | | |{ }| I P(A) = P( a, b ) = 4 I Russell’s Paradox: Let S be the set of all sets which are not | | | { } | I A is finite, and so is P(A) members of themselves. Is S a member of itself or not? n I Note : A = n P(A) = 2 1 | | → | | I Note2: N is infinite since N is not a natural number — I The Paradox of the Barber of Seville: The (male) barber of | | it is called a transfinite cardinal number Seville shaves all and only men who do not shave themselves. Who shaves the barber of Seville? I Note3: Sets can be both members and subsets of other sets

Cartesian Product

I Cartesian Product of A with B: A B is the set of ordered I The Cartesian product of the sets A , A ,..., A , denoted by × 1 2 n pairs: < a, b > a A b B A A A is the set of ordered n–tuples { | ∈ ∧ ∈ } 1 × 2 × · · · × n < a1, a2,..., an >, where ai Ai for 1 i n. I Notation: Πn A = < a , a ,..., a > a A , ∈ ≤ ≤ i=1 i { 1 2 n | i ∈ i } an n–tuple I A A A = 1 × 2 × · · · × n I The Cartesian product of any set with is — why? < a1, a2,..., an > ai Ai , i = 1, 2,..., n ∅ ∅ { | ∈ }

I Example 1. Let A = a, b and B = 1, 2, 3 { } { } I If A = a, b and B = 1, 2, 3 , what is A B A? A B = < a, 1 >, < a, 2 >, < a, 3 >, < b, 1 >, < b, 2 >, < b, 3 > { } { } × × × { } What is B A? ×

If A = m and B = n, what is A B ? | | | | | × | Quantifiers Truth Sets The Universe of Discourse, also known as the Domain of Discourse, is often referred to simply as the domain. I Let P be a predicate and D a domain. The Truth Set of P is the set of elements x D P(x) is true. The domain specifies the possible values of our variables. ∈ 3 I The truth set of P(x) is denoted: x D P(x) We can use quantifiers to restrict the domain { ∈ | } I Assume the domain is the set of integers. What are the truth sets: I x S[P(x)] denotes x[x S P(x)] ∀ ∈ ∀ ∈ → I P = x Z x = 1 Truth Set: Ex: x R[x2 0] means: { ∈ | | | } ∀ ∈ ≥ 2 2 I Q = x Z x = 2 Truth Set: for every real number x, x is non-negative { ∈ | } I R = x Z x = x Truth Set: Note: The meaning of the universal quantifier changes when we { ∈ | | | } change the domain. I Note: xP(x) is true over the domain U IFF the truth set of P ∀ is U. I x S[P(x)] denotes x[x S P(x)] ∃ ∈ ∃ ∈ ∧ I Note: xP(x) is true over the domain U IFF the truth set of Ex: x Z[x2 = 1] means: ∃ ∃ ∈ P = ∅. there exists an integer x such that x2 = 1 6

2.2 Set Operations Equality of Sets

By a previous logical equivalence, we have: I Boolean Algebra: an algebraic system, instances of which are propositional calculus and set theory. A = B IFF I The operators in set theory are defined in terms of the x [(x A x B) (x B x A)] corresponding operator in propositional calculus. ∀ ∈ → ∈ ∧ ∈ → ∈

— or another definition — I As before, there must be a universe, U, and all sets are assumed to be subsets of U. A = B IFF A B and B A ⊆ ⊆

Set Operations More Set Operations

I Union of A and B, denoted A B, is the set ∪ x x A x B { | ∈ ∨ ∈ } I Difference of A and B, or the complement of B relative to A, denoted A B, is the set − A B I Intersection of A and B, denoted A B, is the set ∩ ∩ Note: The absolute complement of A is U A x x A x B − { | ∈ ∧ ∈ } If the intersection is void, A and B are said to be disjoint I Symmetric Difference of A and B, denoted A B, is the set ⊕ (A B) (B A) − ∪ − I Complement of A, denoted A, is the set x (x A) = x x A { | ¬ ∈ } { | 6∈ } Examples Venn Diagrams

U = 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 { } A = 1, 2, 3, 4, 5 and B = 4, 5, 6, 7, 8 Venn Diagrams are a useful geometric visualization tool for 3 or { } { } fewer sets. I A B = 1, 2, 3, 4, 5, 6, 7, 8 ∪ { } I A B = 4, 5 I The Universal set U is a rectangular box ∩ { }

I A = 0, 6, 7, 8, 9, 10 I Each set is represented by a circle and its interior { } I B = 0, 1, 2, 3, 9, 10 { } I All possible combinations of the sets must be represented I A B = 1, 2, 3 − { } I Shade the appropriate region to represent the given set operation I B A = 6, 7, 8 − { } I A B = 1, 2, 3, 6, 7, 8 ⊕ { }

Examples U U A B AB

For 2 sets A B ∩ U A B U A B

C C For 3 sets A (C B) ∪ ∩

Set Identities Universal Instantiation

Set identities correspond to the logical equivalences. We now apply an important rule of inference called Example The complement of the union is the intersection of the Universal Instantiation complements: In a proof, we can eliminate the universal quantifier A B = A B ∪ ∩ which binds a variable if we do not assume anything To prove this statement, we would need to show: about the variable other than it is an arbitrary mem- x [x A B x A B] ber of the Universe. We can then treat the resulting ∀ ∈ ∪ ↔ ∈ ∩ predicate as a proposition.

To show two sets are equal:

I we can show for all x that x is a member of one set IFF it is a We say, “Let x be arbitrary.” Then we can treat the predicates member of the other, or as propositions. I show that each is a subset of the other Universal Generalization Assertion Reason x A B x / (A B) Defn of complement We can apply another rule of inference ∈ ∪ ⇔ ∈ ∪ [x (A B)] Defn of / ⇔ ¬ ∈ ∪ ∈ [(x A) (x B)] Defn of union ⇔ ¬ ∈ ∨ ∈ (x A) (x B) DeMorgan’s Laws Universal Generalization ⇔ ¬ ∈ ∧ ¬ ∈ (x / A) (x / B) Defn of / ⇔ ∈ ∧ ∈ ∈ We can apply a universal quantifier to bind a variable (x A) (x B) Defn of complement ⇔ ∈ ∧ ∈ if we have shown the predicate to be true for all values x (A B) Defn of intersection of the variable in the Universe ⇔ ∈ ∩

Hence, x A B x A B is a tautology since and claim the assertion is true for all x, i.e., ∈ ∪ ↔ ∈ ∩ I x was arbitrary, and x[x A B x A B] I we have used only logically equivalent assertions and definitions ∀ ∈ ∪ ↔ ∈ ∩ Q.E.D. — an abbreviation for the Latin phrase “Quod Erat Demonstrandum” – “which was to be demonstrated” – used to signal the end of a proof.

Alternative Identity

Note: as an alternative which might be easier in some cases, use the identity: A = B [A B and B A] Now we need to show ⇔ ⊆ ⊆ x A (B A) x Example ∈ ∩ − → ∈ is a tautology. Show A (B A) = ∩ − The empty set is a subset of every set. Hence, But the consequent is always false. A (B A) ∩ − ⊇ Therefore, the antecedent (or premise) must also be false. Therefore, it suffices to show A (B A) ∩ − ⊆ or x[x A (B A) x ] ∀ ∈ ∩ − → ∈ So, as before, we say “let x be arbitrary”

Indexed Collections

We proceed by applying the definitions: Let A1, A2,..., An be an indexed collection of sets. Union and intersection are associative (because and and or are), Assertion Reason we have: x A (B A) (x A) [x (B A)] Defn of intersection n A = A A A ∈ ∩ − ⇔ ∈ ∧ ∈ − i=1 i 1 ∪ 2 ∪ · · · ∪ n (x A) [(x B) (x / A)] Defn of difference ⇔ ∈ ∧ ∈ ∧ ∈ S and [(x A) (x / A)] (x B) Comm Prop of and ⇔ ∈ ∧ ∈ ∧ ∈ n F (x B) Table 6 in textbook Ai = A1 A2 An ⇔ ∧ ∈ i=1 ∩ ∩ · · · ∩ F Domination ⇔ T Examples Let A = [ i, ), 1 i < Hence, because P P is always false, the implication is a i ∞ ≤ ∞ ∧ ¬ tautology. The result follows by Universal Generalization. n A = [1, ) Q.E.D. i=1 i ∞ S n A = [n, ) i=1 i ∞ T 2.3 Functions

I If F (x) = y: I Function: Let A and B be sets. Then a function (mapping, map) f from A to B, denoted f : A B, is a subset of A B → × I y is called the image of x under f such that I x is called a preimage of y x [x A y [ y B < x, y > f ]] ∀ ∈ → ∃ ∈ ∧ ∈ and I Note: there may be more than one preimage of y, but there is [ < x, y1 > f < x, y2 > f ] y1 = y2 only one image of x. ∈ ∧ ∈ →

I Note: f associates with each x A one and only one y B. ∈ ∈ I The range of f is the set of all images of points in A under f ; it is denoted by f (A) I A is called the domain of f

I B is called the codomain of f

Injections, Surjections, and Bijections Example I

A B Let f be a function from A to B f I Injection: f is one–to–one (denoted 1–1) or Injective if a preimages are unique X

Note: this means that if x1 = x2 then f (x1) = f (x2) b 6 6 Y c I Surjection: f is onto or surjective if every y in B has a preimage Z d Note: this means that for every y in B there must be an x in A such that f (x) = y 1. Is this an injection? I Bijection: f is bijective if it is surjective and injective, in other 2. Is this a surjection? words, 1–1 and onto. 3. Is this a bijection? 4. How do we determine the answers?

Example II If S is a subset of A, then f (S) = f (s) s S A B { | ∈ } f a A B X f a b X Y b c Y Z c d Z d

I the codomain is B = X , Y , Z { } I f (a) = Z; the image of d is Z I the preimage of Y is b

I f ( a, d ) = Z I the preimages of Z are a, c, and d { } { } I the domain of f is A = a, b, c, d I f ( a, b ) = Y , Z { } { } { } I The range of f is f (A) = Y , Z { } Example III Example IV

A B g A B h a V a V b W b W c X c X d Y d Y Z

1. Is this an injection? 1. Is this an injection? 2. Is this a surjection? 2. Is this a surjection? 3. Is this a bijection? 3. Is this a bijection?

Cardinality

I Note: whenever there is a bijection from A to B, the two sets Let E be the set of even nonnegative integers, 0, 2, 4, 6,... { } must have the same number of elements or the same cardinality Then there is a bijection f from N to E, the even nonnegative integers, defined by: I That will become our definition, especially for infinite sets. f (x) = 2x Let A = B = R, the reals Determine which are injections, surjections, bijections: Hence, the set of even nonnegative integers has the same Function Injection? Surjection? Bijection? cardinality as the set of natural numbers. . . 1—1 onto

f (x) = x OH, NOES! IT CAN’T BE. . . E is only half as big!! f (x) = x2 (But it’s TRUE.) f (x) = x3 f (x) = x | |

Inverse Functions Example Let f be defined by the diagram: A f B a V

b W I Inverse Function: Let f be a bijection from A to B. Then the 1 inverse of f , denoted f − , is the function from B to A defined c X as: d Y 1 f − (y) = x iff f (x) = y

1 −1 Then f − is defined as: A f B a V

I Note: no inverse exists unless f is a bijection. b W

c X

d Y Inverse Applied to a Subset Composition

I Composition: Let f : B C, g : A B. I Inverse Function over Subsets: Let S be a subset of B. Then → → 1 The composition of f with g, denoted f g, is the function f − (S) = x f (x) S ◦ { | ∈ } from A to C defined by I Example: Let f be the following function – f g(x) = f (g(x)) ◦ A f B I Example: a X A g f C A f g C b Y B a V a c Z H H b W b d I I c X c J J f 1( X , Y ) = a, b d Y d − { } { }

Other Examples Discussion

Let f (x) = x2 and g(x) = 2x + 1 I Suppose f : B C, g : A B, and f g is injective → → ◦ I What can we say about f and g? f g(x)= f (g(x)) ◦ I Using the definition of injective, we know that if a = b, then = f (2x + 1) 6 f (g(a)) = f (g(b)), since the composition is injective = (2x + 1)2 6 = 4x2 + 4x + 1 I Since f is a function, it cannot be the case that g(a) = g(b), since f would have two different images for the same point.

I Hence, g(a) = g(b) 6 g f (x)= g(f (x)) ◦ I It follows that g must be an injection = g(x2) I However, f need not be an injection. . . how could you show this? 2 = 2x + 1 (counterexample)

floor and ceiling Functions

I Floor: The floor function, denoted f (x) = x or f (x) = floor(x) b c is the largest integer less than or equal to x.

I Ceiling: The ceiling function, denoted f (x) = x or f (x) = ceiling(x) d e is the smallest integer greater than or equal to x.

I Examples: 3.5 = 3, 3.5 = 4 b c d e

I Note: The floor function is equivalent to truncation for positive numbers