 R  A×B,R is a relation from A to B ， DomR  A 。  (a,b)  R (a, c)  R  (a,b)  R (a, c)  R unless b=c  function  DomR=A ， (everywhere)function 。

Chapter 3 Functions  3.1 Introduction  Definition3.1: Let A and B be nonempty sets. A relation is a (everywhere)function from A to B, denoted by f : A  B, if for every a  A, there is one and only b  B so that (a,b)  f, we say that b=f (a). The set A is called the domain of the function f. If X  A, then f(X)={f(a)|a  X} is called the image of X. The image of A itself is called the range of f, we write R f. If Y  B, then f -1 (Y)={a|f(a)  Y} is called the preimage of Y. A function f : A  B is called a mapping. If (a,b)  f so that b= f (a), then we say that the element a is mapped to the element b.

 (everywhere)function:  (1)Domf=A ；  (2)if (a,b) and (a,b')  f, then b=b‘  Relation: (a,b),(a,b')  R,  function : if (a,b) and (a,b')  f, then b=b‘  Relation: DomR  A  (everywhere)function: DomR=A

 Example ： Let A={1,2,3,4},B={a,b,c},  R 1 ={(1,a),(2,b),(3,c)},  R 2 ={(1,a),(1,b),(2,b),(3,c),(4,c)},  R 3 ={(1,a),(2,b),(3,b),(4,a)}  Example: Let A ={-2,-1, 0,1,2} and B={0,1,2,3,4,5}.  Let f={(-2,0),(-1,1), (0,0),(1,3),(2,5)}. f is a (everywhere)function.  X={-2,0,1}, f(X)=?  Y={0,5}, f -1 (Y)=?

 Theorem 3.1: Let f be a (everywhere) function from A to B, and A 1 and A 2 be subsets of A. Then  (1)If A 1  A 2, then f(A 1 )  f(A 2 )  (2) f(A 1 ∩A 2 )  f(A 1 )∩f(A 2 )  (3) f(A 1 ∪ A 2 )= f(A 1 ) ∪ f(A 2 )  (4) f(A 1 )- f(A 2 )  f(A 1 -A 2 )  Proof: (3)(a) f(A 1 ) ∪ f (A 2 )  f(A 1 ∪ A 2 )  (b) f(A 1 ∪ A 2 )  f(A 1 ) ∪ f (A 2 )

 (4) f (A 1 )- f (A 2 )  f (A 1 -A 2 )  for any y  f (A 1 )-f (A 2 )

 Theorem 3.2 ： Let f be a (everywhere) function from A to B, and A i  A(i=1,2,…n). Then

 2. Special Types of functions  Definition 3.2 ： Let A be an arbitrary nonempty set. The identity function on A, denoted by I A, is defined by I A (a)=a.  Definition 3.3.: Let f be an everywhere function from A to B. Then we say that f is onto(surjective) if R f =B. We say that f is one to one(injective) if we cannot have f(a 1 )=f(a 2 ) for two distinct elements a 1 and a 2 of A. Finally, we say that f is one-to-one correspondence(bijection), if f is onto and one-to- one.  The definition of one to one may be restated in the following equivalent form:  If f(a 1 )=f(a 2 ) then a 1 =a 2 for all a 1, a 2  A Or  If a 1  a 2 then f(a 1 )  f(a 2 ) for all a 1, a 2  A

 Example:1) Let f: R(the set of real numbers)→C(the set of complex number), f(a)=i|a|;  2)Let g: R(the set of real numbers)→C(the set of complex number), g(a)=ia;  3)Let h:Z→Z m ={0,1,…m-1}, h(a)=a mod m  onto,one to one?

 3.2 Composite functions and Inverse functions  1.Composite functions  Relation,Composition,  Theorem3.3: Let g be a (everywhere)function from A to B, and f be a (everywhere)function from B to C. Then composite relation f  g is a (everywhere)function from A to C.

 Proof: (1)For any a  A, there exists c  C such that (a,c)  f  g?  (2)For every a  A, If there exist x,y  C such that (a,x)  f  gand (a,y)  f  g ， then x=y?  Definition 3.4: Let g be a (everywhere) function from A to B, and f be a (everywhere) function from B to C. Then composite relation f  g is called a (everywhere) function from A to C, we write f  g:A→C. If a  A, then(f  g)(a)=f(g(a)).

 Since composition of relations has been shown to be associative (Theorem 2.), we have as a special case the following theorem.  Theorem 3.4: Let f be a (everywhere) function from A to B, and g be a (everywhere) function from B to C, and h be a (everywhere) function from C to D. Then h  (g  f )=(h  g)  f

 Exercise: P176 2,9,10,13,14,  28,37,38  Next: Inverse functions  The Characteristic function of the set P  Cardinality  Paradox