1. Week 7 - Monday

2.  What did we talk about last time?  Sets

4.  A man offers you a bet  He shows you three cards  One is red on both sides  One is green on both sides  One is red on one side and green on the other  He will put one of the cards, at random, on the table  If you can guess the color on the other side, you win  If you bet $100  You gain $125 on a win  You lose your $100 on a loss  Should you take the bet? Why or why not?

6.  We usually discuss sets within some superset U called the universe of discourse  Assume that A and B are subsets of U  The union of A and B, written A  B is the set of all elements of U that are in either A or B  The intersection of A and B, written A  B is the set of all elements of U that are in A and B  The difference of B minus A, written B – A, is the set of all elements of U that are in B and not in A  The complement of A, written A c is the set of all elements of U that are not in A

7.  Let U = {a, b, c, d, e, f, g}  Let A = {a, c, e, g}  Let B = {d, e, f, g}  What are: A  BA  B A  BA  B  B – A AcAc

8.  There is a set with no elements in it called the empty set  We can write the empty set { } or   It comes up very often  For example, {1, 3, 5}  {2, 4, 6} =   The empty set is a subset of every other set (including the empty set)

9.  Two sets A and B are considered disjoint if A  B =   Sets A 1, A 2, … A n are mutually disjoint (or nonoverlapping) if A i  A j =  for all i  j  A collection of nonempty sets {A 1, A 2, … A n } is a partition of set A iff: 1. A = A 1  A 2  …  A n 2. A 1, A 2, … A n are mutually disjoint

10.  Given a set A, the power set of A, written P (A) or 2 A is the set of all subsets of A  Example: B = {1, 3, 6}  P (B) = { , {1}, {3}, {6}, {1,3}, {1,6}, {3,6}, {1,3,6}}  Let n be the number of elements in A, called the cardinality of A  Then, the cardinality of P (A) is 2 n

11.  An ordered n-tuple (x 1, x 2, … x n ) is an ordered sequence of n elements, not necessarily from the same set  The Cartesian product of sets A and B, written A x B is the set of all ordered 2-tuples of the form (a, b), a  A, b  B  Thus, (x, y) points are elements of the Cartesian product R x R (sometimes written R 2 )

13.  Inclusion of Intersection:  For all sets A and B A  B  AA  B  A A  B  BA  B  B  Inclusion in Union:  For all sets A and B  A  A  B  B  A  B  Transitive Property of Subsets:  If A  B and B  C, then A  C

14.  The basic way to prove that X is a subset of Y 1. Suppose that x is a particular but arbitrarily chosen element of X 2. Show that x is an element of Y  If every element in X must be in Y, by definition, X is a subset of Y

15.  We want to leverage the techniques we've already used in logic and proofs  The following definitions help with this goal: 1. x  X  Y  x  X  x  Y 2. x  X  Y  x  X  x  Y 3. x  X – Y  x  X  x  Y 4. x  X c  x  X 5. (x, y)  X  Y  x  X  y  Y

16. Theorem: For all sets A and B, A  B  A Proof:  Let x be some element in A  B  x  A  x  B  x  A  Thus, all elements in A  B are in A A  B  AA  B  A QED  Premise  Definition of intersection  Specialization  By generalization  Definition of subset

17. NameLawDual Commutative A  B = B  AA  B = B  A Associative (A  B)  C = A  (B  C)(A  B)  C = A  (B  C) Distributive A  (B  C) = (A  B)  (A  C)A  (B  C) = (A  B)  (A  C) Identity A   = AA  U = A Complement A  A c = UA  A c =  Double Complement(A c ) c = A Idempotent A  A = AA  A = A Universal Bound A  U = UA   =  De Morgan’s (A  B) c = A c  B c (A  B) c = A c  B c Absorption A  (A  B) = AA  (A  B) = A Complements of U and  U c =  c = U Set Difference A – B = A  B c

18.  To prove that X = Y  Prove that X  Y and  Prove that Y  X

19. Theorem: For all sets A,B, and C, A  (B  C) = (A  B)  (A  C) Proof:  Let x be some element in A  (B  C)  x  A  x  (B  C)  Case 1: Let x  A  x  A  x  B  x  A  B  x  A  x  C  x  A  C  x  A  B  x  A  C  x  (A  B)  (A  C)  Case 2: Let x  B  C  x  B  x  C  x  B  x  A  x  B  x  A  B  x  C  x  A  x  C  x  A  C  x  A  B  x  A  C  x  (A  B)  (A  C)  In all possible cases, x  (A  B)  (A  C), thus A  (B  C)  (A  B)  (A  C)

20.  Let x be some element in (A  B)  (A  C) x  (A  B)  x  (A  C)x  (A  B)  x  (A  C)  Case 1: Let x  A  x  A  x  B  C  x  A  (B  C)  Case 2: Let x  A x  A  Bx  A  B  x  A  x  B  x  B  x  A  C  x  A  x  C  x  C  x  B  x  C  x  B  C  x  A  x  B  C  x  A  (B  C)  In all possible cases, x  A  (B  C), thus (A  B)  (A  C)  A  (B  C)  Since both A  (B  C)  (A  B)  (A  C) and (A  B)  (A  C)  A  (B  C), A  (B  C) = (A  B)  (A  C) QED

21.  Prove that, for any set A, A   =   Hint: Use a proof by contradiction

23.  Like any disproof for a universal statement, you must find a counterexample to disprove a set property  For set properties, the counterexample must be a specific examples of sets for each set in the claim

24.  Claim: For all sets A, B, and C, (A – B)  (B – C) = A – C  Find a counterexample

25.  We can use the laws of set identities given before to prove a statement of set theory  Be extremely careful (even more careful than with propositional logic) to use the law exactly as stated

26. Theorem: A – (A  B) = A – B Proof:  A – (A  B) = A  (A  B) c  = A  (A c  B c )  = (A  A c )  (A  B c )  =   (A  B c )  = A  B c  = A – B QED

27.  For all sets A, B, and C, if A  B and B  C, then A  C

28.  For all sets A and B, ((A c  B c ) – A) c = A

30.  Set theory is a slippery slope  We are able to talk about very abstract concepts  { x  Z | x is prime }  This is a well-defined set, even though there are an infinite number of primes and we don't know how to find the n th prime number  Without some careful rules, we can begin to define sets that are not well-defined

31.  Let a barber be the man in Elizabethtown who shaves the men in Elizabethtown if and only if they don't shave themselves.  Let T be the set of all men in Elizabethtown  Let B(x) be "x is a barber"  Let S(x,y) be "x shaves y"   b  T  m  T (B(b)  (S(b,m)  ~S(m,m)))  But, who shaves the barber?

32.  Bertrand Russell invented the Barber Paradox to explain to normal people a problem he had found in set theory  Most sets are not elements of themselves  So, it seems reasonable to create a set S that is the set of all sets that are not elements of themselves  More formally, S = { A | A is a set and A  A }  But, is S an element of itself?

33.  How do we make sure that this paradox cannot happen in set theory?  We can make rules about what sets we allow in or not  The rule that we use in class is that all sets must be subsets of a defined universe U  Higher level set theory has a number of different frameworks for defining a useful universe

35.  It turns out that the idea behind Russell's Paradox actually has practical implications  It wasn't new, either  Cantor had previously used a diagonal argument to show that there are more real numbers than rational numbers  But, unexpectedly, Turing found an application of this idea for computing

36.  A Turing machine is a mathematical model for computation  It consists of a head, an infinitely long tape, a set of possible states, and an alphabet of characters that can be written on the tape  A list of rules saying what it should write and should it move left or right given the current symbol and state 1011110000 A A

37.  If an algorithm exists, a Turing machine can perform that algorithm  In essence, a Turing machine is the most powerful model we have of computation  Power, in this sense, means the ability to compute some function, not the speed associated with its computation

38.  Given a Turing machine and input x, does it reach the halt state?  First, recognize that we can encode a Turing machine as input for another Turing machine  We just have to design a system to describe the rules, the states, etc.  We want to design a Turing machine that can read another

39.  Imagine we have a Turing machine H(m,x) that takes the description of another Turing machine m and its input x and returns 1 if m halts on input x and 0 otherwise  Now, construct a machine H’(m,x) that runs H(m,x), but, if H(m,x) gives 1, then H’(m,x) infinitely loops, and, if H(m,x) gives 0, then then H’(m,x) returns 1  Let’s say that d is the description of H’(m,x)  What happens when you run H’(d,d)?

40.  Clearly, a Turing machine that solves the halting problem doesn’t exist  Essentially, the problem of deciding if a problem is computable is itself uncomputable  Therefore, there are some problems (called undecidable) for which there is no algorithm  Not an algorithm that will take a long time, but no algorithm  If we find such a problem, we are stuck  …unless someone can invent a more powerful model of computation

41.  Gödel used diagonalization again to prove that it is impossible to create a consistent set of axioms that can prove everything about the set of natural numbers  As a consequence, you can create a system that is complete but not consistent  Or, you can create a system that is consistent but not complete  Either way, there are principles in math in general that are true but impossible to prove, at least with any given system  You might as well give up on math now

43.  A function f from set X to set Y is a relation between elements of X (inputs) and elements of Y (outputs) such that each input is related to exactly one output  We write f: X  Y to indicate this  X is called the domain of f  Y is called the co-domain of f  The range of f is { y  Y | y = f(x), for some x  X}  The inverse image of y is { x  X | f(x) = y }

44.  Using standard assumptions, consider f(x) = x 2  What is the domain?  What is the co-domain?  What is the range?  What is f(3.2)?  What is the inverse image of 4?  Assume that the set of positive integers is the domain and co-domain  What is the range?  What is f(3.2)?  What is the inverse image of 4?

45.  With finite domains and co-domains, we can define a function using an arrow diagram  What is the domain?  What is the co-domain?  What are f(a), f(b), and f(c)?  What is the range?  What are the inverse images of 1, 2, 3, and 4?  Represent f as a set of ordered pairs abcabc abcabc 12341234 12341234 XYf

46.  Which of the following are functions from X to Y? abcabc abcabc 12341234 12341234 XYf abcabc abcabc 12341234 12341234 XYg abcabc abcabc 12341234 12341234 XYh

47.  Given two functions f and g from X to Y,  f equals g, written f = g, iff:  f(x) = g(x) for all x  X  Let f(x) = |x| and g(x) =  Does f = g?  Let f(x) = x and g(x) = 1/(1/x)  Does f = g?

48.  Functions can be defined from any well- defined set to any other  There is an identity function from any set to itself  We can represent a sequence as a function from a range of integers to the values of the sequence  We can create a function mapping from sets to integers, for example, giving the cardinality of certain sets

49.  You should know this already  But, this is the official place where it should be covered formally  There is a function called the logarithm with base b of x defined from R + to R as follows:  log b x = y  b y = x

50.  For a function of multiple values, we can define its domain to be the Cartesian product of sets  Let S n be strings of 1's and 0's of length n  An important CS concept is Hamming distance  Hamming distance takes two binary strings of length n and gives the number of places where they differ  Let Hamming distance be H: S n x S n  Z nonneg  What is H(00101, 01110)?  What is H(10001, 01111)?

51.  There are two ways in which a function can be poorly defined  It does not provide a mapping for every value in the domain  Example: f: R  R such that f(x) = 1/x  It provides more than one mapping for some value in the domain  Example: f: Q  Z such that f(m/n) = m, where m and n are the integers representing the rational number

52.  Let F be a function from X to Y  F is one-to-one (or injective) if and only if:  If F(x 1 ) = F(x 2 ) then x 1 = x 2  Is f(x) = x 2 from Z to Z one-to-one?  Is f(x) = x 2 from Z + to Z one-to-one?  Is h(x) one-to-one? abcabc abcabc 12341234 12341234 XYh

53.  To prove that f from X to Y is one-to-one, prove that  x 1, x 2  X, f(x 1 ) = f(x 2 )  x 1 = x 2  To disprove, just find a counter example  Prove that f: R  R defined by f(x) = 4x – 1 is one-to-one  Prove that g: Z  Z defined by g(n) = n 2 is not one-to-one

54.  Let F be a function from X to Y  F is onto (or surjective) if and only if:   y  Y,  x  X such that F(x) = y  Is f(x) = x 2 from Z to Z onto?  Is f(x) = x 2 from R + to R + onto?  Is h(x) onto? abcabc abcabc 123123 123123 XYh

55.  If a function F: X  Y is both one-to-one and onto (bijective), then there is an inverse function F -1 : Y  X such that:  F -1 (y) = x  F(x) = y, for all x  X and y  Y

57.  More on functions  Composition of functions  Cardinality

58.  Homework 4 is due tonight!  Read Chapter 7