Sets

Definitions (I)

Collection of elements such that:There are no duplicatesThere is no order

Special setsUniverse (U or E): all elements under considerationEmpty set ({ } or ): set with no elements Others: N, Z, R, etc.

Notation { }Enumeration: {1, 2, 3}, {1, 2, …}, {1, 2, … , 100}, etc.Set building: { x | P(x) } all elements in E that satisfy

property P (e.g., { x in N | x>5 x<10 } = {6, 7, 8, 9}

Definitions (II)

Element of:x A

Cardinality:|A| = size or number of elements in A

Set EqualityA = B iff A and B have the same elements

A = B xA xB

Subset/SupersetA B xA xB (subset or equal)

A B A B x(xB xA) (proper subset)

Set Operations: Intersection

A B {x | xA xB}

Example: {1, 2, 3} {2, 3, 4} = {2, 3}

Prove: A B ABy definition, A B A xAB xA

1. xA negate conclusion2. xAB premise3. xA xB def of 4. xA 3, simplification5. xA xA 1&4, conjunction6. F 5, contradiction

Proof by contradiction

A B

Set Operations: Union

A B {x | xA xB}

Example: {1, 2, 3} {2, 3, 4} = {1, 2, 3, 4}No duplicates!

Prove: A A BBy definition, A AB xA xA xB

1. xA premise2. xA xB 1, law of addition

A B

Set Operations: Difference

A – B {x | xA xB}

Example: {1, 2, 3} – {2, 3, 4} = {1}Remove elements of B from A

Prove: A – B ABy definition, A – B A xA–B xA

1. x A – B premise2. x A x B definition3. x A simplification

A B

Set Operations: Complement

~ A E – A {x | xE xA}

Example: ~{1, 2, 3} = {4} if E = {1, 2, 3, 4}

Prove: A ~A = A ~A =

A ~A A ~A set equality A ~A T is subset

of every set A ~A identity x A x ~A x def of and x A x E x A x def of ~ F x comm.,

contradict., dominat. T

A

Basic Set LawsSet Algebra Name

A ~A = EA ~A =

Complementation lawExclusion law

A E = AA = A

Identity laws

A E = EA =

Domination laws

A A = AA A = A

Idempotent laws

Duals: and E

Basic Set Identities (continued…)Set Algebra Name

~(~A) = A Double Compl.

A B = B A A B = B A

Commutative laws

(A B) C = A (B C) (A B) C = A (B C)

Associative laws

A (B C) = (A B) (A C) A (B C) = (A B) (A C)

Distributive laws

~ (A B) = ~A ~B~ (A B) = ~A ~B

De Morgan’s laws

Example: Set LawsAbsorption

A (A B) = AA (A B) = A

Venn Diagram “Proof”Direct Proof

A (A B)= (A ) (A B) ident.= A ( B) distrib.= A

dominat.= A

ident.

A B

Practice ExercisesPE1

Prove that if R is a subset of S and S is a subset of T, then R is a subset of T.

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PE2

Prove De Morgan's law for sets (do not use a Venn diagram): ~ (A B) = ~A ~B

Tuples

Collection of elements, such that:All elements are ordered

Notation: ( )(x1, x2, …, xn)

Tuples of 2 elements are known as pairsTypically, elements are taken from known sets

x females, y males(Mary, Jim) – might mean: Mary and Jim are a married couple

x people, y cars(Mary, red sports car17) – might mean: Mary owns red sports car17

x, y, z integers(3, 4, 7) – might mean: 3 + 4 = 7

Cartesian/Cross Product

A1 … An = {(x1, …, xn) | xA1 … xnAn}

Example: A = {1, 2}, B = {a, b, c}A B = {(1, a), (1, b), (1, c), (2, a), (2, b),

(2, c)}

A = {1, 2}, B = {a, b, c}, C = {, }A B C = {(1,a,), (1,a,), (1,b,), (1,b,),

(1,c,), (1,c,), (2,a,), (2,a,), (2,b,), (2,b,), (2,c,), (2,c,)} |A1 … An| = |A1| … |An|

Can get large:A = set of students at BYU (30,000)B = set of BYU student addresses (10,000)C = set of BYU student phone#’s (60,000)|A| |B| |C| = 1.8 1013

RelationsRelation

Subset of the cross product

Examples:A = {1, 2} & B = {a, b, c}

R = {(1, a), (2, b), (2, c)}A = {1, 2} & B = {a, b, c} & C = {, }

R = {(1, a, ), (2, c, )}Marriage: subset of the cross product of

males and females

If 2 sets, the relation is binary

Functions

A function is a special kind of binary relation

A binary relation f A B is a function if for each a A there is a unique b B

Function Definition

1

2

3

α

β

γ

x

y

NOT Functions

1

2

3

α

β

γ

f = {(1, α), (2, β)} “For each” violated

Some x’s do not have corresponding y’s

x

y

NOT Functions

Uniqueness violated for some x’s

x

y

1

2

3

α

β

γ

f = {(1, α), (2, β), (3, β), (3, γ)} uniqueness violated for 3 appears twice

Functionswith N-Dimensional

Domains An (n+1)-ary relation f A1 A2 …

An B is a function if for each < a1, a2, …, an> A1 A2 … An there is a unique b B.

α

β

γ

<1,1>

<1,2>

<1,3>

We can use various notation for functions: for f = {(1, α),(2, β),(3, β)}

Notation for Functions

Notation (x, y) f f : x→y y = f(x)

Example (2, β) f f : 2→β β = f(2)

• In the notation, x is the argument or preimage and y is the image.

• For functions with n-ary domains, use <x0, x1, …, xn> in place of x.

Function Domain and Range

f : A → BA is the domain space

same as the domain (since all elements participate)dom f, dom(f), or domain(f)

B is the range spacemay or may not be the same as the range, which is:

{y | x(y=f(x))}All rhs values in pairs (all that get “hit”)

Bf

ran f, ran(f), range(f)

f : D1 D2 … Dn → Z

f : Dn → Z (when all domains are the same)

Remove the requirement that each a A must participate. Retain the uniqueness requirement.

Partial Functions

Partial Function:

α

β

γ

f = {(<1,2>, β),(<1,3>, β),(<1,3>, γ)} <1,3> not unique

<1,1>

<1,2>

<1,3>

α

β

γ

<1,1>

<1,2>

<1,3>

NOT a Partial Function:

α

β

γ

<1,1>

<1,2>

<1,3>

Partial Function: (A Total Function is also a Partial Function.)

Identity FunctionIA : A → A

IA = {(x, x) | x A}

Constant FunctionC : A → BC = {(x, c) | x A c B }Often A and B are the same

C : A → AC= {(x, c) | x A c A}

Special Functions

Composition of Functions

Composition is written “°”

Range space of f = domain space of g

a

c

1

2

4

fg

b

α

β3

f(a) = 2 g(2) = α g(f(a)) = α

g°f(a) = α

f(b) = 2 g(2) = α g(f(b)) = α

g°f(b) = α

f(c) = 4 g(4) = β g(f(c)) = β

g°f(c) = β

Injection: “one-to-one” or “1-1”xy(f(x) = f(y) x = y)For f : A → B, the elements in B are “hit” at most once

Injection

a

b

d

1

2

3

c

Injective

a

b

d

1

2

3

c

NOT Injective

x

y

x

y

Surjection: “onto”yx(y = f(x))For f : A → B, the elements in B are all “hit” at least once

Surjection

1

2

4

a

b

c3

Surjective NOT Surjective

x

y

x

y

1

2

4

a

b

c3

{ not “hit”

Bijection: “one-to-one and onto” or “1-1 correspondence”xy(f(x) = f(y) x = y) yx(y = f(x))For f : A → B, every B element is “hit” once and only once

Bijection

1

2

a

b

c3

Bijective NOT Bijective

x

y

x

y

1

2

4

a

b

c3

NOT Surjective

NOT injective

Notes on Bijection

1. |A| = |B|An “extra” B cannot be “hit” (not a surjection)

An “extra” A requires that at least one B must be “hit” twice (not an injection)

2. If f is a bijection, swapping the elements of the ordered pairs is a function

Called the inverse

Denoted f-1

Is also a bijection

f-1(f(x)) is the identity function, i.e. f-1(f(x)) = x.

Practice ExercisesPE1

If A={a,b,c,d}, are the following functions from A to A injective, surjective or bijective?

{(d, a), (d, c), (b, b), (b, d)}

{(a, b), (b, b), (c, d), (d, d)}

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PE2

If f(x)=2x+3 and g(x)=x-3, what is g°f?

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PE3

Which is the larger set?

E (even numbers) vs. O (odd numbers)

N (natural numbers) vs. Z (integers)

N (natural numbers) vs. [0,1] (real numbers between 0 and 1)

Which is bigger? N or Z?

f(x) =x odd: (x+1)/−2

x even: x/2

y negative: −2x−1

y positive: 2xg(y) =

{{

x y 0 0 1 −1 2 1 3 −2 4 2

Since g = f−1, there is a bijection from N to Z and thus |N| = |Z|

Which is bigger? N or [0,1]?

Assume |N| = |[0,1]|, then there exists a bijection, e.g.,

1 0.34234…2 0.34987… diagonalization3 0.00040…

But now, there exists a number in [0,1] such that d1 = not 3, d2 = not 4, d3 = not 0, etc. Hence, not surjective and thus not bijective

ALL THAT THE FATHER HATH

A Functional “Proof”

Language

Power Set

Set of all subsets of a set AA = {1,2}P(A) = 2A = { {}, {1}, {2}, {1,2} }

Each element of A is either present (1) or not present (0)Treat the elements of A as a sequence (e.g., A={a,b,c,d})

Use bit-string representation to say which elements are present (e.g., 0110 means {b,c})

Can represent all subsets of A, from = 0000 to A = 1111

Number of subsets in power set|2A| = 2 · 2 · … · 2 (|A| times) = 2|A|

Motivates the notation 2A for the power set

Bit-String Operations

With bit string representationsSet intersection: = pairwise Set union: = pairwise Set complement: ~ = bit complementSet minus: – = mask out using 1’s = complement 2nd operand

and do pairwise

E.g. using {a,b,c,d}1011 1101 = 1001 i.e. {a,c,d} {a,b,d} = {a,d}1011 1101 = 1111 i.e. {a,c,d} {a,b,d} = {a,b,c,d}~1011 = 0100 i.e. ~{a,c,d} = {b}1010 – 1100 = 0010 i.e. {a,c} – {a,b} = {c}

Practice ExercisesPE1

What is the power set of {a, 1, x, 2}?

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PE2

What is the power set of (i.e., the empty set)?

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PE3

What is the power set of { } (i.e., the set containing the empty set)?

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PE4

What is the power set of { a, { } }?

Language

Let V be a set of symbols, known as an alphabet or a vocabulary

A string is any finite sequence of symbols from V

Strings have length

Vn denotes the set of all strings of length n

V* denotes the set of all strings, or sentences, over V

A language L is a subset of V*, i.e., L V*

Programming languageSet of all possible programs (valid, very long string)

Language Representation

• Finite– Enumerate all sentences

• Infinite language– Cannot be specified by enumeration– Use a generative device, i.e., a grammar

• Specifies the set of all legal sentences• Defined recursively (or inductively)