2010-09-29

請問99中正資工關於集合的cardinality的問題

Two sets have the same cardinality if and only if there is a bijection, i.e. one-to-one correspondence, between them. Consider the following sets.
A(x) = {x: x is an integer}
B(x) = {x: x is a positive integer, and is a multiple of 3}
C(x) = {x: x is an integer, and 100 < x < 1000}
D(x) = {x: x is a subset of B}
Indicate True or False for each of the following statements. Briefly explain each of your answers.


知道(a)可以找到對應的Funtion,所以為true。

(b)是因為一個B(x)為不可數集,C(x)為可數集,所以不能one-to-one對應
所以false,這樣的想法跟解法是對的嗎?
(c)的話因為D(x)可為可數集,所以想法同上。

不知道(b)(c)的想法是不是這題的解法?


2 則留言:

線代離散助教(wynne) 提到...

剛剛去查了一下題目裡的(a),(b),(c), 幫你補充一下:
(a) A and B have the same cardinality
(b) B and C have the same cardinality
(c) A and D have the same cardinality

其中, (a)的觀念沒問題; (b)會錯是因為B為無限集, 但C為有限集; 而(c)會錯是因為若依照題目的定義, D 就相當於是 B 的power set, 則 |D|=|2^B|>|B|=|N| (by Cantor's Theorem), 但 |A|=|N|, 所以這題只有(a)會是true

你可能要再稍微注意一下可數集的定義, 一個集合不管它是無限還是有限集, 只要該集合的cardinality不超過自然數的cardinality, 它就會是可數集, 所以這裡因為|B|=|N|, 則 B 會是可數集, 然而因為 |D|>|N|, 所以 D 並不是可數集

matt0215 提到...

謝謝助教,

不小心漏貼題目,不好意思

謝謝這麼詳細的解答...