Snick snack CPSC 121: Models of Computation 2011 Winter Term 1 Sets Steve Wolfman, based on notes by Patrice Belleville and others 1

snick

snack

CPSC 121: Models of Computation2011 Winter Term 1

Sets

Steve Wolfman, based on notes by Patrice Belleville and others

1

Outline

• Prereqs, Learning Goals, and Quiz Notes• Out-of-class notes on set definitions• What’s the Use of Sets (history & DFAs)• More set operations

– Cardinality (size)– Power set (and an induction proof)– Cartesian products (and application to DFAs)

• Set proofs

• Next Lecture Notes2

Learning Goals: Pre-Class

By the start of class, you should be able to:– Define the set operations union, intersection, complement,

and set difference and the logical operations subset and set equality in terms of predicate logic and set membership ().

– Translate between sets represented explicitly (possibly using ellipses “…”, e.g., {4, 6, 8, …}) and using “set builder” notation (e.g., {x Z+ | x2 > 2 x is even}).

– Execute the union, intersection, complement, set difference, subset, and set equality operations on sets expressed explicitly, using set builder notation, or a combination of these and set operators.

– Interpret the empty set symbol , including the fact that the empty set has no members and that it is a subset of any set.

3

Learning Goals: In-Class

By the end of this unit, you should be able to:– Define the power set and cartesian product

operations in terms of predicate logic and set membership/subset relations.

– Execute the power set, cartesian product, and cardinality operations on sets expressed through any of the notations discussed so far.

– Apply your proof skills to proofs involving sets.– Relate DFAs to sets.

4

Outline



• Set proofs

• Next Lecture Notes

CORRESPONDS TO TEXTBOOK READING(NOT COVERED IN CLASS)

5

Sets

A set is a collection of elements:– the set of students in this class– the set of lowercase letters in English– the set of natural numbers (N)– the set of all left-handed students in this class

An element is either in the set (x S) or not (x S).

Is there a set of everything?


6

Quantifier Example

Someone in this class is left-handed (where C is the set of people in this class and L(p) means p is left-handed):

x C, L(x)


7

What is a Set?

A set is an unordered collection of objects.

The objects in a set are called members.(a S indicates a is a member of S; a S indicates a is not a member of S)

A set contains its members.


8

Describing Sets (1/4)

Some sets…A = {1, 3, 9}

B = {1, 3, 9, 27, snow}

C = {1, 1, 3, 3, 9, 9}

D = {A, B}

D' = { {1, 3, 9}, {1, 3, 9, 27, snow} }

E = { }


9


Some sets…

A = {1, 5, 25, 125, …}

B = {…, -2, -1, 0, 1, 2, …}

C = {1, 2, 3, …, 98, 99, 100}

(The set of powers of 5, the set of integers, and the set of integers between 1 and 100.)

“…” is an ellipsis


10


Some sets, using set builder notation:

A = {x N | y N, x = 5y}

B = {2i - 1 | i is a prime}

C = {n Z | 0 < n 100}

To read, start with “the set of all”. Read “|” as “such that”.

A: “the set of all natural numbers x such that x is a power of 5”B: “the set of all numbers of the form 2i-1 such that i is a prime”C: “the set of all integers n such that 0 < n 100”


11


Graphical depiction of sets: Venn diagrams.

Draw the set of all five-letter things.

All red things? All red, five-letter things?

fire truck

snows

happiness

Texas

heart

books

seven

U


12

U is the universal

set of everything.


Graphical depiction of sets: Venn diagrams.

Draw the set of all five-letter things.

All red things? All red, five-letter things?

fire truck

snows

happiness

Texas

heart

books

seven

U


13

Containment

A set A is a subset of a set B iffx U, x A x B.

We write A is a subset of B as A B.

If A B, can B have elements that are not elements of A?


14

Containment

A set A is a subset of a set B iffx U, x A x B.

We write A is a subset of B as A B.

If A B, can B have elements that are not elements of A? Yes, but A can’t have elements that are not elements of B.


15

Membership and Containment

A = {1, {2}}

Is 1 A?

Is {1} A?

Is 1 A?

Is {1} A?

Is 2 A?

Is {2} A?

Is 2 A?

Is {2} A?


16

Membership and Containment

A = {1, {2}}

Is 1 A? Yes

Is {1} A? Yes

Is 1 A? Not meaningful since 1 is not a set.

Is {1} A? No

Is 2 A? No

Is {2} A? No

Is 2 A? Not meaningful since 2 is not a set.

Is {2} A? Yes


17

Thought Question

What if A B and B A?


18

Set Equality

Sets A and B are equal — denoted A = B — if and only if x U, x A x B.

Can we prove that that’s equivalent to A B and B A?


19

Set Equality

Sets A and B are equal — denoted A = B — if and only if x U, x A x B.

Can we prove that that’s equivalent to A B and B A? Yes, using a standard predicate logic proof in which we note that p q is logically equivalent to p q p q.


20

Set Union

The union of A and B — denoted A B — is {x U | x A x B}.

A B is the blue region...

A B

U


21

Set Intersection

The intersection of A and B — denoted A B — is {x U | x A x B}.

A B is the dark blue region...

A B

U


22

Set Difference

The difference of A and B — denoted A - B — is {x U | x A x B}.

A – B is the pure blue region.

A B

U


23

Set Complement

The complement of A — denoted A — is {x U | x A}.

A is everything but the blue region.

A

U

Can we express this as a set difference?


24

Set Operation Identities

Many logical equivalences have analogous set operation identities. Here are a few… read more in the text!

A B = B A Commutative Law

(A B) C = (A C) (B C) Distributive Law

(A B) = A B DeMorgan’s Law

A U = A U as identity for ...


25

Outline



• Set proofs


What Are Sets Good For?

Historically:

Mathematicians attempted to formalize set theory to create a foundation for all of mathematics. Essentially all mathematical constructs can be defined in terms of sets.

Good news: this means sets are a powerful way to communicate many types of ideas.

27

Bad News: Russell’s Paradox(and other problems)

Does the “set of all sets that contain themselves” contain itself?

a.Yes, definitely.

b.Maybe, either way is fine.

c.No, definitely not.

d.None of these.

28

Bad News: Russell’s Paradox(and other problems)

Does the “set of all sets that do not contain themselves” contain itself?

a.Yes, definitely.

b.Maybe, either way is fine.

c.No, definitely not.

d.None of these.

Same question, different form: “Imagine a barber that shaves every man in town who

does not shave himself. Does the barber shave himself?”29

What Are Sets Good For?

Applications for us:

Codifying and communicating ideas. For example, formalizing DFAs...

30

What is a DFA?

The input language is: ________________

The states are: ______

The start state is: ____

The accepting states are: _____________

What’s left?

a

b

b

a,b

a,ba

31

The Transition Function

The arrows in our DFA have the following properties:

• An arrow starts from every state, for every letter in the input alphabet.

• The arrows lead to other states.

We’ll formalize this as the function N soon, which will complete our DFA!

a

b

b

a,b

a,ba

32

What is a DFA?(Deterministic Finite Automaton)I the (finite) set of letters

in the input language.

S the (finite) set of states.

s0 the initial state; s0 S.

F the set of accepting (“final”) states; F S.

N the next-state function, to be described.

These are just standard names we use,but the constraints make sense regardless of names.

a

b

b

a,b

a,ba

33

Problem: Formalizing an Example DFA

For this DFA:

I = { }

S = { }

s0 =

F = { }

N: ( , )

( , )

( , )

( , )

( , )

( , )

( , )

( , )

a

b

b

a,b

a,ba

34

Problem: Testing the DFA Formalism

I the (finite) set of letters in the input language.

S the (finite) set of states.

s0 the initial state; s0 S.

F the set of accepting (“final”) states; F S.

N the next-state function, to be described.

Must a DFA have an initial state?

Can a DFA have more than one initial state?

Must a DFA have an accepting state?

Can all states in a DFA be accepting?

Can the initial state be accepting?

a. Yes b. No c. Not enough information.36

Testing the DFA Formalism

Must a DFA have an initial state?

Can a DFA have more than one initial state?

Must a DFA have an accepting state?

Can all states in a DFA be accepting?

Can the initial state be accepting?

s0 is the initial state; so, yes.

s0 is the only initial state; so, no.

F S. Is S?

F S. Is S S?

s0 S and F S. Can s0 F?

37

Must a DFA Have an Accepting State?

What does a DFA with no accepting states look like? What language does it accept?

38

Can All States in a DFA Be Accepting?

What does a DFA with no rejecting states look like? What language does it accept?

39

Outline



• Set proofs


Cardinality

The cardinality of a set A — denoted |A| — is the set’s size.

For finite sets, the cardinality of the set is the number of elements it contains…

|A| = |{1, 3}| = 2|B| = |{1, 3, 27, snow}| = 4|C| = |{ {1, 3}, {1, 3, 27, snow} }| = 2|D| = || = 0

41

How Big Is a Set?

How big are these sets?

A = {1, 3}B = {1, 3, 27, snow}C = { {1, 3}, {1, 3, 27, snow} }D = E = NF = { …, -4, -2, 0, 2, 4, … }

a. 0 b. 1 c. 2 d. 3 e. None of these42

Cardinality of Infinite Sets?

For now, we won’t worry about the cardinality of infinite sets. Why?

Well, we can’t count the members of such sets.

Side note: Depending on our definition, however, the results can be surprising. For example, under the standard definition of cardinality, these have the same cardinality:E = N = {1, 2, 3, … }F = { …, -4, -2, 0, 2, 4, … }

Why? Try “folding” the set F over creatively.43

Problem: Cardinality Exercises

Given the definitions:

A = {1, 2, 3}

B = {2, 4, 6, 8}

What are:

|A| = _________

|B| = _________

|A B| = _________

|A B| = _________

|A – B| = _________

|B – A| = _________

|{{}}| = _________

|{}| = _________

|{{}}| = _________

a. 0 b. 1 c. 2 d. 3 e. None of these44

Outline



• Set proofs


A Touch of Logic

Consider the four propositional logic variables p, q, r, and s.

What are all the different ways we can assign truth values to each of these?

Now, imagine a set S that contains exactly the true variables among p, q, r, and s.

What are the possible “Ss” (depending on p, q, r, and s’s truth values)?

47

A Touch of Logic: Rephrased

Consider the set {p, q, r, s}.

What are all the subsets of this set?

Equivalently, what is: {S U | S {p,q,r,s}}?

48

Power Sets

The power set of a set T — denoted P(T) — is the set of all subsets of T:{ S U | S T }

49

Using Logic to Build Power Sets

p q r s Set from P(…)

F F F F {}

F F F T {s}

F F T F {r}

F F T T {r,s}

F T F F {q}

F T F T {q,s}

F T T F {q,r}

F T T T {q,r,s}

p q r s Set from P(…)

T F F F {p}

T F F T {p,s}

T F T F {p,r}

T F T T {p,r,s}

T T F F {p,q}

T T F T {p,q,s}

T T T F {p,q,r}

T T T T {p,q,r,s}

So, how big is the power set of a set?

P({p,q,r,s}) = …

50

Problem: Power Set Exercises

Given the definitions: A = {1, 2} B = {2, 4, 6}

What are: P(A) = _____________________________ P(B) = _____________________________ |P(A)| = _____________________________ |P(B)| = _____________________________ |P(P(A))|= _____________________________ P(P(A)) = _____________________________ P(A B) = _____________________________

51

Cardinality of a Finite Power Set

Theorem: |P(S)| = 2|S|

Base case: When |S| = 0, what is S?

What is P(S)? How do we know?

53

Cardinality of a Finite Power Set

Inductive Hypothesis: Assume for all sets S of cardinality k 0, |P(S)| = 2|S|.

Inductive step:

To prove: Given the induction hypothesis, for all sets T, |P(T)| = 2|T| when |T| = k+1.

The key is to drop an element from a set of size k+1.Then, take the power set.

Then, consider how to put the element back in.54

Outline



• Set proofs


Tuples

A tuple is an ordered collection of elements. (An n-tuple is a tuple with n elements.)

Two tuples are equal when each pair of corresponding elements is the same:

(a, 1, ) = (a, 5 – 4, A A)(a, b, c) (a, c, b)

Actual fact: database people love the word “tuple”.And… why shouldn’t they? Say “two-tuple”. Isn’t that great?

59

Formalizing DFAs: Where the Arrows Come From




We’ll formalize this as a function soon.

What is the set of “things” for which there is an arrow?

a

b

b

a,b

a,ba

t

m

b

g

60

Cartesian Products

The Cartesian product of the sets A and B — denoted A B — is the set of all tuples whose first element is drawn from A and whose second element is drawn from B.

In other words, A B = {(a,b) | a A b B}

61

Cartesian Products andthe Cartesian Plane

Let’s visualize N N:

1

2

3

4

5

1 2 3 4 5 6 7 8

62

Calculating Cartesian Products

What is {a,b} {1,2,3}:

1

2

3

a b

( , )

( , )

( , )

( , )

( , )

( , )

63

Formalizing DFAs: Where the Arrows Come From




We’ll formalize this as a function soon.

Each tuple of a state and a letter has an arrow.There’s an arrow for every member of S I

a

b

b

a,b

a,ba

t

m

b

g

65

Example DFA, Revisited

For this DFA:

I = { a, b }

S = { t, m, b, g}

s0 = t

F = { t, b }

N: (t, a) m

(t, b) b

(m, a) g

(m, b) b

(b, a) b

(b, b) b

(g, a) g

(g, b) g

a

b

b

a,b

a,ba

t

m

b

g

66

Outline



• Set proofs


Problem: Proofs with Sets (1/2)

Prove by contradiction that for all sets A and B, if A B then A B = { }.

(Note: B contains all elements in the universe that are not in B.)

68

Problem: Proofs with Sets (2/2)

Prove for all sets A and B, if A B then P(A) P(B).

(Note: if A B and B C, then A C.)

69

Outline



• Set proofs


Learning Goals: In-Class

By the end of this unit, you should be able to:– Define the power set and cartesian product

operations in terms of predicate logic and set membership/subset relations.

– Execute the power set, cartesian product, and cardinality operations on sets expressed through any of the notations discussed so far.

– Apply your proof skills to proofs involving sets.– Relate DFAs to sets.

71

Next Lecture Learning Goals: Pre-Class

By the start of class, you should be able to:– Define the terms domain, co-domain, range,

image, and pre-image– Use appropriate function syntax to relate

these terms (e.g., f : A B indicates that f is a function mapping domain A to co-domain B).

– Determine whether f : A B is a function given a definition for f as an equation or arrow diagram.

72

Next Lecture Prerequisites

See the “Functions” readings on the “Textbook and References” section of the website.

Complete the quiz on Vista before class.

73

snick

snack

Extra Slides: Worked Proofs

(Thanks to Meghan Allen)

74

Proofs with Sets #1 version 1

Prove that for all sets A and B:A B = A B

Proof #1:A B = {x U | x (A B)}

= {x U | ~(x A x B)} = {x U | x A x B}

= {x U | x A x B} = A B

Def’n of

De Morgan’s

Def’n of

Def’n of

Def’n of

75


Remember that for any two sets C and D, C = D iff C is a subset of D and D is a subset of C.

C = D [C D D C]

76


a) Prove that: A B A B

Pick an arbitrary x A B,Then x A B.~(x A x B)x A x Bx A x Bx (A B)

Def’n of

De Morgan’s

Def’n of

Def’n of

Def’n of

77


b) Prove that: A B A B Pick an arbitrary x A B Then,

x A x Bx A x B~(x A x B)x A Bx A B

78

Proofs with Sets #1 version 2conclusion

We have shown that A B A B and

A B A B . Therefore, A B = A B.

79

Proof with Sets #2

Prove that, for all sets A, B and C:

(A B) – C = (A – C) (B – C)

This time we’ll use the Set IdentitiesLHS:

(A B) – C = (A B) C Set Difference Law

= C (A B) Commutative Law

= (C A) (C B) Distributive Law

= (A C) (B C) Commutative Law

= (A - C) (B – C) Set Difference Law80

Proofs with Sets #3

Prove or disprove, for all sets A and B,

A – B = B – A

We can disprove this with a counterexample.

Let A = {1,2} and B = {2,3}

Then, A-B = {1} and B-A = {3}

A – B ≠ B – A81

Proofs with Sets #4

Prove that for all sets A and B, if A B then A B =

Proof by contradiction. Assume A B and A B . Then, there exists an

element x A B. By the definition of intersection, that means x A and x B. Since x B, x B by the definition of complement. But, A B, so since x A, x B by definition of subset. Thus, x B and x B which is a contradiction. Hence the assumption is false and therefore for all sets A and B, if A B then A B = 82

Proofs with Sets #5

Prove that for all sets A and B, if A B then P A) P(B)

WLOG, let A and B be sets.

Proof by antecedent assumption:Assume A B. Now, consider X P(A). X A by the definition of power set. But, because A B, X B by the transitive property of subsets* and thus, by the definition of power set, X P(B). This proves that for all X, if X P(A) then X P(B) and so P(A) P(B).

*Epp -Theorem 5.2.183


Prove that for all sets S, S.

We proceed by contradiction.

Assume that there is a set S such that the empty set is not a subset of S. Then, by definition of subset, there must be some element x of the empty set that is not an element of S. However, the empty set has no elements. This is a contradiction.

QED84


Prove that for all sets S, S.

Note that S x U, x x S

x U, x x S.

We proceed by proving this last statement.

We know nothing is an element of the empty set: x U, x .

By generalization, it follows that: x U, x x S.

QED85

Documents

Snick snack CPSC 121: Models of Computation 2011 Winter Term 1 Sets Steve Wolfman, based on notes by Patrice Belleville and others 1