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工作学习 / 求学深造 / 雨雨 and Smartiecat， here is an answer of question 1 from a professor
-heian(黑暗®-谢幕中);
2002-10-3
{547}
(#779121@0)
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It's ok ar... I don't need to know this since I'm not touching any discrete math in the future.
-jeffrey815(Smartiecat);
2002-10-3
(#779123@0)
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谢谢~~~
-rainrain(雨雨 秋风何故乱翻书);
2002-10-3
(#779126@0)
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问你点事儿
-pandora(pandora);
2002-10-3
(#779127@0)
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en?
-rainrain(雨雨 秋风何故乱翻书);
2002-10-3
(#779144@0)
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我记得是你吧，买过clinique all about eyes?觉得如何？
-pandora(pandora);
2002-10-3
(#779147@0)
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比较清爽。不过所有的眼霜俺感觉都一样，当然除了油性特别大的
-rainrain(雨雨 秋风何故乱翻书);
2002-10-3
(#779171@0)
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发给我的文件有乱码，里面是修正过的。Yes. This can be proved by contradiction
For example, Let g: A -> B and f: B->C
A={x1, x2}, B={y}, C={z}, and g (x1) = g (x2) = y, and f (y)= z.
Thus, f o g (x1) = f o g (x2) =z. Hence f o g cannot be one-to-one if g is not one-to-one.
Example:
Let f(x) = x, and g (x) = x2, we can see that f(x) is a one-to-one function and g (x) is not a one-to-one function. And f o g (x) = x2, which is also not one-to-one. Hence f o g cannot be one-to-one if g is not one-to-one.
-heian(黑暗®-谢幕中);
2002-10-3
{484}
(#779129@0)
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有点意思。离散数学中很多都只能用反证法
-noexit(江湖▲狐仙▲省油▲灯);
2002-10-3
(#779132@0)
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谢谢大家的帮助，俺明天可以交差了，这个周末再好好看看书。感激涕零，谢谢谢谢谢谢~~~^_^
-rainrain(雨雨 秋风何故乱翻书);
2002-10-3
(#779145@0)
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最后一个问题。1: f(S) = {y| y = f(x), x <- A}
2: f ^(-1)(S) = {x | x = f ^(-1)(y) AND y in S}
3: f ^(-1) (S U T) Subset f ^(-1)(S) U f ^(-1)(T)
Let x In f ^(-1) (S U T) is E.T.S x In f ^(-1)(S) U f ^(-1)(T)
4: f ^(-1)(S) U f ^(-1)(T) Subset f ^(-1) (S U T)
Let x In n f ^(-1)(S) U f ^(-1)(T) is E.T.S x In f ^(-1) (S U T)
Similarly f ^(-1) (S /\ T) = f ^(-1)(S) /\ f(-1)(T).
-heian(黑暗®-谢幕中);
2002-10-3
{379}
(#779146@0)