設S 為一集合, 令P(S) = {A|A ⊂ S}, 稱為S 的冪集(power set), 若S不為空集合, 試證:
#S < #P(S), 即不存在一1-1,onto map f 把S 映成P(S)。(提示: 考慮集合A = {x|x ∈S, x 不屬於 f(x)}, 若f 存在...)
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isomorphism

2007-02-02 20:48
This is a basic problem in set theory; the cardinality of the power set is always greater than the cardinality of the set.

The proof is trivial from your hint. Then x is in A and x is not in A. Contradiction.

joey

2007-07-16 22:19
引用：

寫道:
設S 為一集合, 令P(S) = {A|A ⊂ S}, 稱為S 的冪集(power set), 若S不為空集合, 試證:
#S < #P(S), 即不存在一1-1,onto map f 把S 映成P(S)。(提示: 考慮集合A = {x|x ∈S, x 不屬於 f(x)}, 若f 存在...)

看的霧洒洒 :-o