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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 。
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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 R f. 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.
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(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
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Example : Let A={1,2,3,4},B={a,b,c}, R 1 ={(1,a),(2,b),(3,c)}, R 2 ={(1,a),(1,b),(2,b),(3,c),(4,c)}, R 3 ={(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)=?
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Theorem 3.1: Let f be a (everywhere) function from A to B, and A 1 and A 2 be subsets of A. Then (1)If A 1 A 2, then f(A 1 ) f(A 2 ) (2) f(A 1 ∩A 2 ) f(A 1 )∩f(A 2 ) (3) f(A 1 ∪ A 2 )= f(A 1 ) ∪ f(A 2 ) (4) f(A 1 )- f(A 2 ) f(A 1 -A 2 ) Proof: (3)(a) f(A 1 ) ∪ f (A 2 ) f(A 1 ∪ A 2 ) (b) f(A 1 ∪ A 2 ) f(A 1 ) ∪ f (A 2 )
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(4) f (A 1 )- f (A 2 ) f (A 1 -A 2 ) for any y f (A 1 )-f (A 2 )
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Theorem 3.2 : Let f be a (everywhere) function from A to B, and A i A(i=1,2,…n). Then
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2. Special Types of functions Definition 3.2 : Let A be an arbitrary nonempty set. The identity function on A, denoted by I A, is defined by I A (a)=a. Definition 3.3.: Let f be an everywhere function from A to B. Then we say that f is onto(surjective) if R f =B. We say that f is one to one(injective) if we cannot have f(a 1 )=f(a 2 ) for two distinct elements a 1 and a 2 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(a 1 )=f(a 2 ) then a 1 =a 2 for all a 1, a 2 A Or If a 1 a 2 then f(a 1 ) f(a 2 ) for all a 1, a 2 A
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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→Z m ={0,1,…m-1}, h(a)=a mod m onto,one to one?
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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.
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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)).
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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
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Exercise: P176 2,9,10,13,14, 28,37,38 Next: Inverse functions The Characteristic function of the set P178 5.2 Cardinality Paradox
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