Re: [分析] 等價證明
※ 引述《Remedy (夜校之狼)》之銘言：
: F:X->Y
: (1) f is 1-1 on X
: (2)f(A-B)=f(A)-f(B) for all subsets A and B of X
: (3)f(A交集B)=f(A)交集f(B) for all subsets A and B of X
: 證明三者等價
: 有版友能出手相救嗎？
凡事開頭最難，我就幫到這了。
(1)->(2)
(a)
Take y1 in f(A-B).
Since f is a one-to-one function on X,
==> there is an unique x1 such that f(x1)=y1.
==> x1 must be in A-B, (why?)
==> y1=f(x1) not in f((A-B)^c), (^c 餘集合)
==> y1=f(x1) not in f(B),
Since x1 in A-B<=A,
==> y1=f(x1) must be in f(A),
==> y1 in f(A)-f(B).
Therefore f(A-B)<=f(A)-f(B).
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仍想再一次看見妳的笑容。
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10/02 19:23, , 1F
10/02 19:23, 1F
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