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...)
看不懂......
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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.
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...)
_________________我們究竟來自何方，我們為何如此，又將前往何處？
