1. 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 aA, (f  IA)(a) ?=f(a).
Property (ii) is proved similarly to property (i).

2. 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 cC, there exists aA such that f  g(a)=c

3. (2)one to one:if ab，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.

4. 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.

5. Proof: (a)(1)If f –1 is a function, then f is one to one
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,a2A such that f(a1)=f(a2)=bB, then a1?=a2
(2)If f is one to one，then f –1 is a function
f -1 is a function
For bB，If there exist a1,a2A such that (b,a1)f -1 and (b,a2) f -1,then a1?=a2

6. Proof: (b)(1)If f –1 is an everywhere function, then f is one-to-one correspondence.
(i)f is onto.
For any bB，there exists aA such that f (a)=?b
(ii)f is one to one.
If there exist a1,a2A such that f (a1)=f (a2)=bB, 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 bB，there exists one and only aA so that (b,a) f -1.
For any bB, there exists aA such that (b,a)?f -1.
For bB，If there exist a1,a2A such that (b,a1)f -1 and (b,a2) f -1,then a1?=a2

7. Definition 3. 5: Let f be one-to-one correspondence between A and B
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 aA,there exists bB such that f -1(b)=a)
(2)f –1 is one to one
For any b1,b2B, if b1b2 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.

8. 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
Let f:A→B and g:B→A，
Is g the inverse function of f ?
f g?=IB and g  f ?=IA

9. 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 (fg)-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.

10. 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.

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

12. Theorem 3. 12: Let A and B be subsets of the universal set
Theorem 3.12: Let A and B be subsets of the universal set. Then, for any xU, 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 AB;
(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);

13. 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.

14. Example: The set Z is countably infinite
Proof: f:N→Z,for any nN,

15. 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)),…

16. 3.5 Paradox 1.Russell’s paradox AA, AA。
Russell’s paradox: Let S={A|AA}. The question is, does SS?
i.e. SS or SS?
If SS,
If SS,
The statements " SS " and " SS " cannot both be true, thus the contradiction.

17. Pigeonhole principle P88 3.3 Permutations of sets P79-81, 3.1
Exercise: P176 21,22,32,33
P181 4,
Prove Theorem 3.11
Prove:(1)|(0,1)|=|R| (2)|[0,1]|=|R|
Next:
Pigeonhole principle P
Permutations of sets P79-81, 3.1