离散数学课件(英文版)----Function

Try to prove it What is ({<x,y>|x,y∈R, y=x+1})? R×{-1}
x-y>,
:N×N→N, (<x,y>) = | x2-y2|
(N×{0}) ={ n2|n∈N}, -1({0}) ={<n,n>|n∈N}
Characteristic Function of Set
A’ B’ B A
f
Special Types of Functions
Surjection
:A→B is a surjection or “onto” iff. Ran()=B, iff. y∈B, x∈A, such that f(x)=y
Injection (one-to-one)
:A→B is one-to-one iff. y∈Ran(f), there is at most one x∈A, such that f(x)=y iff. x1,x2∈A, if x1≠x2， then (x1) ≠(x2) iff. x1,x2∈A, if (x1) =(x2)，then x1=x2
Note: if <x,y>∈f, then <x,y>∈f and <x,x>∈1A if <x,y>∈ f °1A，则<t,y>∈ f , 且<x,t>∈1A， 则t=x, 所以<x,y>∈f .
Invertible Function
The inverse relation of f :A→B is not necessarily a function, even though f is.
If |A|>|B| then 0 else |A|!*|A|C|B|
are
how many Bijection from A to B are there?
If |A|= |B| = m then M! else 0
Special Types of Functions: Examples
:Z+→R, (x)= ln x, one-to-one :R→Z, (x)= x, onto :R→R, (x)= 2x-1，bijection :R×R→R×R, (<x,y>) = <x+y, bijection
For any R(a)∈A/R, there exists some x∈A， such that g(x)=R(x)
Images of Union and Intersection
Given f：A→B，and X,Y are subsets of A, then:
f (X∪Y) = f(X)∪f (Y) f (X∩Y) f(X)∩f (Y)
Composition of Functions
gf f
a b=f(a)
g
c=g(b)=gf (a) c=g(b)
A
B
C
Examples
f, g, h: function of Z: f(x) = 3x, g(x) = 3x+1, h(x) = 3x+2 f°g g°f f°g°h
Associative Law
Image and counterimage
Let f:A→B, A’A, then (A’)={y|y=f(x)，x∈A’} is called the image of A’ under f.
An element in Dom(f) corresponds a value A subset of Dom(f) corresponds an image
Suppose f :A→B, g:B→C, g°f is a relation from A to C. x∈A, we have g°f (x)=g(f (x)). It is easy to prove that for every a in A, there is just one c in C, such that g(f (a))=c. So, g°f is a function.
Composition of Functions
Since function is relation as well, the composition of relation can be applied for functions, with the results being relation. The composition of functions is still function
Composition of Surjections
If f :A→B, g:B→C are both surjections, then g°f:A→C is also surjection.
Sketch of proof: For any y∈C, since g is a surjection, there must be some t∈B, such that g(t)=y. Similarily, since f is surjection，there must be some x∈A, such that f(x)=t, so, g°f(x)=y.
Counter example for a ∈ X , but a Y , and b ∈ Y , but b X However, f ( a ) = f ( b ) = c , then : c ∈ f ( X ) ∩ f (Y ) but, maybe c f ( X ∩ Y )
Bijection(one-to-one correspondence)
surjection plus injection
If A,B are nonempty sets
how many different functions from A to B are there?
|B||A|
how many Injection from A to B there?
Functions
Definition of Function
Definition: Let A and B be function f from A to B, f:A→B, is a relation from for all a∈A, f(a) contains of B. nonempty sets. A which is denoted A to B such that just one element
Example: f : N×N→N, f (<i,j>)=2i(2j+1)-1 is a bijection f -1(2i(2j+1)-1)=<i,j>
Invertible Function
f :A→B is an invertible function if and only if f is a bijection.
Let B’B, then f-1(B’)={x|x∈A, f(x)∈B’} is called the counterimage of B’ under f.
What is f -1(f(A’)) ? A’ f -1(f(A’)) ? f -1(f(A’)) A’?
Image and Counterimage
Sketch of proof: If f is not a injection, then there must be <y,x1>, <y,x2>∈ f -1, and x1≠x2. On the other hand, if f is not a surjection, then there is at least one element in B, which has not an image in A under f -1. Both cases are contradict to “f :A→B is invertible”. If f –1 is not a function, then，either there exist <y,x1>, <y,x2>∈ f -1, and x1≠x2，then f is not a injection, or there must be at least on element in B, which has not an image in A under f –1. Both cases are contradict to “f is a bijection”.
A special kind of binary relation Under f, each element in the domain of f has a unique image. Note: the domain of :A→B is A，but the range is not necessarily equal to B.
But…
If g°f is a surjection，can it be derived that f and g are both surjections as well？
Obviously，g must be a surjection. How about f?
If there is some t∈B, for any x∈A, f(x)≠t, (i.e. f is not a surjection！) then if g(B！ t)=C, g°f is still a surjection.
Let U be the universal set, for any AU, the characteristic function of A, fA:U→{0,1} is defined as fA(x)=1 iff. x∈A