Chapter 3 Functions

P181(Sixth Edition) P168(Fifth Edition)

Theorem 3.1: Let f be an everywhere function from A to B, and A1 and A2 be subsets of A. Then

(1)If A1A2, then f(A1) f(A2)(2) f(A1∩A2) f(A1)∩f(A2)(3) f(A1 A∪ 2)= f(A1)∪f(A2)(4) f(A1)- f(A2) f(A1-A2) Proof: (3)(a) f(A1)∪f (A2) f(A1 A∪ 2) (b) f(A1 A∪ 2) f(A1)∪f (A2)

(4) f (A1)- f (A2) f (A1-A2) for any y f (A1)-f (A2)

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

n

ii

n

ii AfAf

11)()()1(

n

ii

n

ii AfAf

11)()()2(

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, a2A Or If a1a2 then f(a1)f(a2) for all a1, a2A

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?

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 every aA, If there exist x,yC such that (a,x) f gand (a,y) f g ， then x=y?

(2)For any aA, there exists cC such that (a,c) f g ?

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 aA, 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(gf )=(hg)f

Theorem 3.5: Let f be an everywhere function from A to B. Then

(i)f IA=f. (ii)IB f = f. Proof. Concerning(i), let aA, (f IA)(a) ?

=f(a). Property (ii) is proved similarly to property

(i).

Theorem 3.6: Let g be an everywhere function from A to B, and f be an everywhere function from B to C. Then

(1)if f and g are onto , then f g is also onto. (2)If f and g are one to one, then f g is also one

to one. (3)If f and g are one-to-one correspondence,

then f g is also one-to-one correspondence Proof: (1) for every cC, there exists aA such

that f g(a)=c

(2)one to one:if ab ， then f g(a)? f g(b)

(3)f and g are one-to-one correspondence, f and g are onto. By (1), f g are onto.

By (2), f g are also one to one. Thus f g is also one-to-one correspondence.

2. Inverse functions Inverse relation, Function is a relation Is the function’s inverse relation a

function? No Example: A={1,2,3},B={a,b}, f:A→B, f

={(1,a),(2,b),(3,b)} is a function, but inverse relation f -1={(a,1),(b,2),(b,3)} is not a function.

Theorem 3.7: :(a) Let f be a function from A to B, Inverse relation f -1 is a function from B to A if only if f is one to one

(b) Let f be an everywhere function from A to B, Inverse relation f -1 is an everywhere function from B to A if only if f is one-to-one correspondence.

Proof: (a)(1)If f –1 is a function, then f is one to one If there exist a1,a2A such that f(a1)=f(a2)=bB, then

a1?=a2 (2)If f is one to one ， then f –1 is a function f -1 is a function For bB ， If there exist a1,a2A such that (b,a1)f -1

and (b,a2) f -1,then a1?=a2

Proof: (b)(1)If f –1 is an everywhere function, then f is one-to-one correspondence.

(i)f is onto. For any bB ， there exists aA such that f (a)=?b (ii)f is one to one. If there exist a1,a2A such that f (a1)=f (a2)=bB, then

a1?=a2 (2)If f is one-to-one correspondence ， then f –1 is an

everywhere function f -1 is an everywhere function, for any bB ， there

exists one and only aA so that (b,a) f -1. For any bB, there exists aA such that (b,a)?f -1. For bB ， If there exist a1,a2A such that (b,a1)f -1

and (b,a2) f -1,then a1?=a2

Definition 3.5: Let f be one-to-one (correspondence) between A and B. We say that inverse relation f -1 is the (everywhere) inverse function of f. We denoted f -1 ： B→A. And if f (a)=b then f -1(b)=a.

Theorem 3.8: Let f be one-to-one correspondence between A and B. Then the inverse function f -1 is also one-to-one correspondence.

Proof: (1) f –1is onto (f –1 is an everywhere function from B to A For any aA,there exists bB such that f -1(b)=a) (2)f –1 is one to one For any b1,b2B, if b1b2 then f -1(b1) f -1(b2). If f:A→B is one-to-one correspondence, then f -1 ： B→A is also

one-to-one correspondence. The function f is called invertible.

Theorem 3.9: Let f be one-to-one correspondence between A and B.

Then (1)(f -1)-1= f (2)f -1 f=IA

(3)f f -1=IB Proof: (1)(f -1)-1= f (2)f -1 f=IA Let f:A→B and g:B→A， Is g the inverse function of f ? f g?=IB and g f ?=IA

Theorem 3.10 ： Let g be one-to-one correspondence between A and B, and f be one-to-one correspondence between B and C. Then (fg)-1= g-1

f -1

Proof: By Theorem 3.6, f g is one-to-one correspondence from A to C

Similarly, By theorem 3.7, 3.8, g-1 is a one-to-one correspondence function from B to A, and f –1 is a one-to-one correspondence function from C to B.

Theorem 3.11: Let A and B be two finite set with |A|=|B|, and let f be an everywhere function from A to B. Then

(1)If f is one to one, then f is onto. (2)If f is onto, then f is one to one. The prove are left your exercises(p189.36)

3.3 The Characteristic function of the set function from universal set to {0,1} Definition 3.6: Let U be the universal set, and

let AU. The characteristic function of A is defined as a function from U to {0,1} by the following:

AxAx

xA 01

)(

Theorem 3.12: Let A and B be subsets of the universal set. Then, for any xU, we have

(1)A(x)0 if only if A= (2)A(x) 1 if only if A=U; (3)A(x)≦B(x) if only if AB; (4)A(x) B(x) if only if A=B; (5)A∩B(x)=A(x)B(x); (6)A B∪ (x)=A(x)+B(x)-A∩B(x);

);(1)()7( xx AA

)()()()()8( xxxx BAABABA

Exercise:P188 9,10,13,14, 21,22,28, 32,33,36,37,38(Sixth)

OR P176 9,10,13,14, 21,22,28, 32,33,36,37,38(Fifth)

Next: Cardinality, ParadoxPigeonhole principle P100(Sixth)P88

3.3(Fifth)