Functions and Relations II. CS2100 Ross Whitaker University of Utah

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1 Functions and Relations II CS2100 Ross Whitaker University of Utah

2 Infinitely Big Sets Counting and size: thought experiment How many elements does the following set have? How do we know? What do we mean by the number of things?

3 Size of Sets: Cardinality Sets A and B have the same cardinality if there exists an invertible function f:a->b Intuition: An invertible function establishes a kind of unique correspondence between sets. Thus, invertible functions help to establish things as being the same size.

4 Example Show that N and Z have the same cardinality. Define a function that maps one to the other f:z->n with f(a) = { 2a a 0 2a 1 a<0

5 Example Is f invertible? Two options Show one-to-one and onto or produce g and show g is the inverse g(b) = { b 2 b even (b + 1) 2 b odd a=g(b) -> b=f(a) Let b be even. a=g(b)=b/2. f(a)=2a=b. Let b be odd. a=g(b)=-(b+1)/2. f(a)=-2a-1=-2 -(b+1)/2-1=b+1-1=b. b=f(a) -> a=g(b) Let a be positive. b=f(a)=2a. b is even. g(b)=b/2=a. Let a be negative. f(a)=-2a-1. b is odd. g(b)=-(b+1)/2 = - (-2a-1+1)/2 = a.

6 Sizes of Sets: Infinity The set A is infinite if there exists a function f:a->a that is one-to-one but not onto Intuition: If set is finite in size, then a mapping onto itself must have either: a full unique correspondence (invertible) or have multiple inputs to the same output (not one-toone) infinite sets can be neither and they maintain this discrepancy by never running out of elements

7 Example N is infinite Proof. Construct a function f:n->n that is one-to-one but not onto. f(a)=2a Prove f(a) is one-to-one. Division theorem says that for every number there is a unique quotient (divisor=2). I.e. b=2a and a is unique Prove f(a) is not onto Find bn such that a s.t. b=f(a). Take b to be odd. By the definition of odd and division theorem b=2k+1, and this is unique. Thus, there is no a such that 2a=2k+1=b

8 Proving Cardinality Sometimes finding an invertible function can be hard. So Cantor-Bernstein theorem: Given sets A and B, if there is a one-to-one function f:a->b and a one-to-one function g:b->a, then the sets A and B have the same cardinality.

9 Example Prove that Q 0 has the same cardinality as N. Since NQ, identity on N (denoted I:N->N) is one-to-one. For g:q->n, consider the prime factorization x=a/b (which is unique). Now construct z=g(x)=2 a 3 b. Prime factorizations are unique, so there is no other xq that can lead to z

Georg Ferdinand Ludwig Philipp Cantor 1845 1918 St Petersburg, Russia Fundamental results on sizes of infinite

10 Georg Ferdinand Ludwig Philipp Cantor St Petersburg, Russia Fundamental results on sizes of infinite sets Equivalence and nesting of infinite spaces Cantor s theorem implies the existence of an infinity of infinities.

11 Cantor s Theorem For every set A, P(A) has a greater cardinality than A Book has a weaker form What this means for any infinitely large set, you can always produce a bigger set

12 Countable An infinite set is countable if it has the same cardinality as N not countable is uncountable E.g. Z and Q 0 are countable

13 Arrow Diagrams Relations

14 Relations: Definitions R is a binary relation on A (RAA) Reflexive: (a,a)r aa Antisymmetric: a,ba, a b, (a,b)r- >(b,a) R Transitive: a,b,ca, ((a,b)r)((b,c)r) ->(a,c)r Relation R on A is partial ordering if it is antisymmetric, transitive, and reflexive

15 Hasse Diagram No loops (reflexive is implied) Segment instead of arrow height on page is direction Draw only shortest links Draw (a,b) and (b,c), but not (a,c)

16 Example A={1,2,3,4} (x,y)r iff x y Enumerate R Draw the Hasse diagram

17 Example Let A=P({x,y,z}). Define (a,b)r iff ab

18 Proofs about Relations Claim: R is SOMEPROPERTY Let SOMEASSUMPTION Show that the property holds

19 Example A={1,2,3,6,9,18}, R={(x,y)A, y is a multiple of x} Prove that R is reflexive Let aa. Because a 1=a, a is a multiple of itself (true for all ints) Therefore (a,a)r

20 Example R={(a,b)ZZ: a-b even} Prove that R is reflexive Prove that R is transitive Prove that R is not a partial ordering

21 Example Let A={1,2,3,4,6,9,12,18,36} Let R={(a,b)A 2 : b is a multiple of a} Draw the Hasse Diagram Prove that R is a partial ordering

22 Other Types of Ordering Relax the reflexive requirement: strict partial ordering E.g. A={1,2,3,4,5}, R={(a,b)A 2 : a<b} Total ordering: Requirements of partial ordering Additional requirement a,ba, if a b, either (a,b)r or (b,a)r Can also be strict

23 Practice A=P({1,2,4,8}) Identify the type of ordering: R 1 ={(S,T)A 2 : (as)(bt) -> a b} R 2 ={(S,T)A 2 : S < T } R 2 ={(S,T)A 2 : sum of elements in S sum of elements in T}

24 Relations and Equivalence Two things can be the same or equivalent Equivalent ratios: (5,2) and (20,8) because 5/2 = 20/8 Similar triangles: Equivalence relation

25 Partitions of a Set A partition of a set A, is a set of sets S={S 1,, S n } such that: 1. i, S i (each part is nonempty) 2. i,j if S i S j then S i S j = (disjoint) 3. S 1 S n =A (complete) The S i s are called parts of S

26 Equivalence Relation Given set A and relation R on A, R is an equivalence relation on A if there exists a partition S of A such that (x,y)r iff are in the same part of S.

27 Examples A={1,2,3,4,5,6} R 1 ={(a,b)a 2 : b=a+1} R 2 ={(a,b)a 2 : a-b is even}

28 Example Cards numbered A={1,2,3,4,5} Each player gets two cards H=AA Two hands are the same if sum of values on cards is the same What is the partition?

29 Practice Is equivalence relation? If so, partition? A={1,2,3,4,5,6,7,8,9}, R={(a,b)A 2 : k,l,mz s.t. a=km, b=lm, and m>1} R={(x,y)Z 2 : mz s.t. x-y=4m} A={0,1,2,3,4,5,6}{1,2,3}, R={((a,b), (c,d))a 2 : ad=bc}

30 Euivalence Every object is similar to itself (a,a)r reflexive If a is similar to b, then b is similar to a (a,b)r -> (b,a)r symmetric

31 Practice Let A={0,1,2,3,4,5} and let R={(a,b)A 2 : a 2 -b 2 =3m, mz} Prove that R is reflexive. Let aa, then a 2 -a 2 =0=3 0 (multiple of 3) Therefore (a,a)r aa Prove that R is symmetric. Let a,ba, such that (a,b)r Thus, a 2 -b 2 =3m, mz. Therefore b 2 -a 2 =-3m=3 (-m). mz -> -mz. Therefore (b,a)r because (b,a) satisfies the setbuilder predicate

32 Equivalence Relations If the relation R is reflexive, symmetric, and transitive, then R is an equivalence relation Proof: For each element aa, define P a ={xa: (x,a)r} Claim 1: each P a is nonempty R is reflexive, therefore (a,a)r Therefore ap a and P a is nonempty

33 Equivalence Relations (cont.) Claim 2: For all a,ba, if P a P b then P a P b =. Proof by contradiction. Let P a P b then c s.t. cp a and cp b. If P a P b then d s.t. dp a and d P b -> (c,a)r, (d,a)r -> (a,d)r (sym) -> (c,d)r (trans.) cp b -> (b,c)r -> (b,d)r (trans) -> (d,b)r (sym) -> dp b (contradiction)

34 Equivalence Relations (cont.) Claim 3: Union of P s equals A R reflexive -> (a,a)r aa Therefore aa ap a Therefore every element of A is in the union of the P s

Discrete Mathematics: Lectures 6 and 7 Sets, Relations, Functions and Counting Instructor: Arijit Bishnu Date: August 4 and 6, 2009

Discrete Mathematics: Lectures 6 and 7 Sets, Relations, Functions and Counting Instructor: Arijit Bishnu Date: August 4 and 6, 2009 Our main goal is here is to do counting using functions. For that, we