Equivalence relations and partitions .

1. Equivalence relations and partitions. Consider the following relation on a set of all people: B = {(x, y)| x has the same birthday as y } B is reflexive, symmetric and transitive. We can think about this relation as splitting all people into 366 categories, one for each possible day. Definition. A relation RAA is called an equivalence relation on A if it is symmetric, reflexive and transitive. An equivalence relation on a set A represents some partition of this set.

2. A A2 A1 A3 A4 • Definition. For any set A subsets AiA partition set A if • A = A1 A2 … An • AiAj= , for any ij. • Ai   for any i Example. A={1, 2, 3, 4},  ={{2}, {1, 3}, {4}} is a partition of A.

3. Definition. Suppose R is equivalence relation on a set A, and xA. Then the equivalence class of x with respect to R is the set [x]R={yA| yRx} In the caseof the same birthday relation B, if p is any person, then the equivalence class of p [p]B={qP| pBq} ={qP | q has the same birthday as p} For example, if John was born on Aug. 10, [John]B= {q P | q was born on Aug.10} The set ofall equivalence classes of elements of A is called A modulo Rand is denoted A/R: A/R={[x]R | x A}

4. We are going to prove that any equivalence relation R on A induces a partition of A and any partition of A gives rise to an equivalence relation. Theorem 1. Suppose R is an equivalence relation on a set A. Then A/R ={[x]R | x A} is a partition of A. Proof. To prove that A/R defines a partition, we must prove three properties of a partition. 1) The union of all equivalence classes [x]R equals A, i. e. Since any equivalence class is a subset of A, their union is also a subset of A. So, all we need to show is To prove this, suppose xA. Then x[x]R because [x]R ={y A | yRx }and xRx due to reflexive property of R.

5. x[x]R implies 2) To prove that A/R is pairwise disjoint we need to show, that for any x, y A if [x][y] then [x][y]=. It is easier to prove the contrapositive: if [x][y]  , then [x]=[y]. So, assume [x][y]  , then we can find an element z  [x][y]. It implies that xRz and zRy by definition of equivalence classes. Then xRz and zRy implies xRy due to transitive property of R. We now claim that [x][y]. Take any a [x], it implies aRx, which implies aRy because of xRy and transitive property of R. aRy implies a [y], i. e. [x][y]. Similarly [y][x], or [x]=[y]. 3) None of equivalence classes is empty, because for any xA, x[x].

6. Theorem 2. Suppose A is a set and  is a partition of A, i. e.  is a set of disjoint nonempty subsets, such that any element of A belongs to exactly one subset. Then the relation R on A defined as R = {(x, y) | x, y A andx and y belong to the same subset in } is an equivalence relation on A. • Proof. We need to prove that R is reflexive, symmetric and transitive. • R is reflexive because for any x A, x and itself belong tothe same • subset in . • R is obviously symmetric • Suppose xRy and yRz, that is x and y belong to the same subset and • y and z belong to the same subset. This implies that x and z belong • to the same subset in , thus xRz.

7. Definition. A partition P2 is called a refinement of P1if every set in P2 is a subset of one of sets in P1. Example. Consider the partition P1.={{1, 3, 5, 7, 9}, {2, 4, 6, 8, 10}}, induced by the relation R1= {(x, y) | x = y mod(2)} on a set A = {1, 2, …10}. Then the relation R2= {(x, y) | x = y mod(4)} induces a partition P2.={{1, 5, 9}, {3, 7}, {2, 6, 10}, {4, 8}}, which is a refinement of P1. Theorem. Suppose R1 and R2 are equivalence relations on a set A. Let P1 and P2 are partitions that correspond to R1 and R2 respectively. Then R1R2 iff P1 is a refinement of P2.

8. Partial Orders. A particular type of binary relation on a set. Definition. A binary relation R  AA is a partial order, if it is reflexive, transitive and anti-symmetric. The set A together with the partial order relation is called a poset. Examples: ‘less or equal’ ‘greater or equal’ ‘subset’ relation ‘divides’

9. If aA, write f(a) for corresponding element of B, b= f(a) A = domain B = co-domain. f   a b= f(a) a is the pre-image of b bistheimage of a; Functions. Function is a special type of relation. Recall that in a relation R A B an element a A may be related to more then one element of B, or it may be not related to any. Definition. A function f : A  B is a binary relation from A to B such that, for everyaA there exists a unique (i. e. exactly one) element b B such that(a, b) f .

10. f A B        f A B       relation f is not a function relation f is not a function

11. f A B a1 b1   a2  b2   a3  b3 a4  f (A) = {b1, b3} The set of all images f (A) ={f (a)| aA} is the range of f . In general f (A) is a proper subset of B.

12. Assume |A|=n, |B|=m. How many different functions f : A  B exist? Recall that there are 2nm different relations R A B, but not all of them are functions. For relations we counted all subsets of A B. A function must include exactly n pairs, one for each element in A. Each of elements from the domain can be related to any element from co-domain. So, we have m choices for each pair, the total number of functions is mn. where

13. f A B a1 b1   a2  b2   a3  b3 a4  Properties of functions (surjective, injective, bijective ) Definition. Surjective (“onto”): range = co-domain. A function f : AB is called surjective, if for any bB there exists aA, such that f (a) = b.  bB, aA (f (a)=b) This function is not surjective because b2 does not have any pre-image.

14. f A B a1 b1   a2  b2   b3 If we have two finite sets |A| < |B| a surjective function f : AB is impossible. The necessary condition for a surjective function f : AB is |A|  |B|.

15. f A B a1 b1   a2   a3  b2 Definition. Injective (“one-to-one”): a1, a2A, a1  a2  f (a1)  f (a2) A function f : AB is called injective, if for all a1, a2A, a1  a2  f (a1)  f (a2). Equivalently, it means f (a1) = f (a2)  a1 = a2 This function is not injective.

16. f A B f a1 b1   A B a2  a1 b1    a2  a3 b3   b2  a3  b2 An injective function f : AB is possible if only |A| |B|. A function that is both injective and surjective is called bijective. For this we need |A| |B| and |A|  |B|, i. e. |A| = |B|.

17. Pigeonhole Principle If n pigeons are put into m holes, n > m, then at least one hole contains two or more pigeons. Actually it is not about pigeons and holes... Restate in terms of functions: Theorem. Let A and B be finite sets, | A | > | B |. If f : A  B then there exist distinct elements a1 and a2in A such that f (a1) = f (a2) (that is f is not injective) A: pigeons B: holes f : says where each pigeon gets put Examples. Birthdays.

18. f f A B A B                Injective, but not surjective Surjective, but not injective • Examples: • f (x)=x2 : R  R • is not injective, since both +x and -x have the same image. • f (x) = |x| : R R, • is not surjective (but is surjective if regarded R  R+)

19. B C f g A a    f (a)=b g (b)=c gf Composition of functions (Note, that in the context of function composition, the order in which the functions appear is backward, i. e. the rightmost function is applied first) Theorem 4. Let f : AB and g: BC be two functions. The compositionof f and g as relations defines a function gf : A C, such that gf (a)= g(f(a)).

20. Proof. We need to prove that composite relation gf is a function. For this we need to prove two things: 1) the composition gf relates each element aA to some c C. 2) an element c C assigned to a by gf is unique, so we can denote it c= g(f (a)). 1) existence: Let b=f (a) B. Let c=g(b) C. So, by the definition of composition of relations, (a, c)gf . Thus, cC [(a, c)gf ] 2) uniqueness: Suppose (a, c1)gf and (a, c2)gf . Then by the definition of composition, b1B, such that(a, b1)f and (b1, c1)g, and b2B, such that(a, b2)f and (b2, c2)g. Since f is a function, there may be only one b B, such that (a, b)f, so b1= b2. Since g is a function, (b, c1)g and (b, c2)g imply that c1= c2. Thus, (gf )(a)= c= g (b) =g (f (a)).