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 a

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 a 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.

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:

Ax

AxxA 0

1)(

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

3.4 Cardinality Definition 3.7: The empty set is a finite set

of cardinality 0. If there is a one-to-one correspondence between A and the set {0,1,2,3,…, n-1}, then A is a finite set of cardinality n.

Definition 3.8: A set A is countably infinite if there is a one-to-one correspondence between A and the set N of natural number. We write |A|=|N|=0. A set is countable if it is finite or countably infinite.

Example: The set Z is countably infinite Proof: f:N→Z,for any nN,

numberoddisnn

numberevenisnnnf

)1(2

12

1

)(

The set Q of rational number is countably infinite, i.e. |Q|=|N|=0.

|[0 ， 1]| 0. Theorem 3.13: The R of real numbers is

not countably infinite. And |R|=|[0 ， 1]|. Theorem 3.14: The power set P(N) of the

set N of natural number is not countably infinite. And |P(N)|=|R|=1.

Theorem 3.15(Cantor’s Theorem): For any set, the power set of X is cardinally larger than X.

N, P (N),P (P (N)),…

3.5 Paradox 1.Russell’s paradox AA, AA。 Russell’s paradox: Let S={A|AA}. The

question is, does SS? i.e. SS or SS? If SS, If SS, The statements " SS " and " SS "

cannot both be true, thus the contradiction.

Exercise: P176 21,22,32,33P181 4, Prove Theorem 3.11Prove:(1)|(0,1)|=|R| (2)|[0,1]|=|R|

Next:Pigeonhole principle P88 3.3Permutations of sets P79-81, 3.1