Suppose f:A to B is function.C,D are subsets of A then C con
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Suppose f:A to B is function.C,D are subsets of A then C contained in D implies f(C) contained in f(D) and f(C union D)=f(C) union f(D)
设函数f(A->B).C 和 D 是 A 的子集。请证明:D 含 C 则 f(D)含 f(C) 并且,f(C并D)=f(C) 并f(D)
设函数f(A->B).C 和 D 是 A 的子集。请证明:D 含 C 则 f(D)含 f(C) 并且,f(C并D)=f(C) 并f(D)
![Suppose f:A to B is function.C,D are subsets of A then C con](/uploads/image/z/1193375-47-5.jpg?t=Suppose+f%3AA+to+B+is+function.C%2CD+are+subsets+of+A+then+C+con)
这个题目考察集合包含,相等的证明.f(C)是函数f 以C中元素为自变元时所有象点的集合,f(D)类似.题目证明如下:
证明:
1、设任意y∈f(C),则存在x∈C使得f(x)=y,
∵C⊆D,∴x∈D
∴y∈f(D)
所以f(C)⊆f(D)
2、
(1)∵C⊆C∪D 且 D⊆C∪D
∴由1知f(C)⊆f(C∪D)且f(D)⊆f(C∪D)
∴f(C)∪f(D) ⊆f(C∪D)
(2)下面再证f(C∪D)⊆f(C)∪f(D)
设任意y∈f(C∪D),则存在x∈C∪D 使得f(x)=y,
则或者x∈C或者x∈D
当x∈C时,有y∈f(C)
当x∈D时,有y∈f(D)
∴y∈f(C)∪f(D)
∴f(C∪D)⊆f(C)∪f(D)
由(1)、(2)可得f(C)∪f(D) =f(C∪D)
证明:
1、设任意y∈f(C),则存在x∈C使得f(x)=y,
∵C⊆D,∴x∈D
∴y∈f(D)
所以f(C)⊆f(D)
2、
(1)∵C⊆C∪D 且 D⊆C∪D
∴由1知f(C)⊆f(C∪D)且f(D)⊆f(C∪D)
∴f(C)∪f(D) ⊆f(C∪D)
(2)下面再证f(C∪D)⊆f(C)∪f(D)
设任意y∈f(C∪D),则存在x∈C∪D 使得f(x)=y,
则或者x∈C或者x∈D
当x∈C时,有y∈f(C)
当x∈D时,有y∈f(D)
∴y∈f(C)∪f(D)
∴f(C∪D)⊆f(C)∪f(D)
由(1)、(2)可得f(C)∪f(D) =f(C∪D)
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