R  A×B,R is a relation from A to B ， DomR  A 。 (a,b)  R (a, c)  R

1. 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。

2. 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 Rf. 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.

3. (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

4. Example：Let A={1,2,3,4},B={a,b,c}, • R1={(1,a),(2,b),(3,c)}, • R2={(1,a),(1,b),(2,b),(3,c),(4,c)}, • R3={(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)=?

5. Theorem 3.1: Let f be a (everywhere) function from A to B, and A1 and A2 be subsets of A. Then • (1)If A1A2, then f(A1) f(A2) • (2) f(A1∩A2) f(A1)∩f(A2) • (3) f(A1∪A2)= f(A1)∪f(A2) • (4) f(A1)- f(A2) f(A1-A2) • Proof: (3)(a) f(A1)∪f (A2) f(A1∪A2) • (b) f(A1∪A2) f(A1)∪f (A2)

6. (4) f (A1)- f (A2) f (A1-A2) • for any y f (A1)-f (A2)

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

8. 2. Special Types of functions • Definition 3.2：Let A be an arbitrary nonempty set. The identity function on A, denoted by IA, is defined by IA(a)=a. • Definition 3.3.: Let f be an everywhere function from A to B. Then we say that f is onto(surjective) if Rf=B. We say that f is one to one(injective) if we cannot have f(a1)=f(a2) for two distinct elements a1 and a2 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(a1)=f(a2) then a1=a2 for all a1, a2A Or • If a1a2 then f(a1)f(a2) for all a1, a2A

9. 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→Zm={0,1,…m-1}, h(a)=a mod m • onto ,one to one?

10. 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.

11. 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)).

12. 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

13. Exercise: P176 2,9,10,13,14, • 28,37,38 • Next: Inverse functions • The Characteristic function of the set P178 5.2 • Cardinality • Paradox