※ 引述《Makoto0813 (火红的燃烧吧!妹控魂!)》之铭言：
: 各位版友好
: 小弟是外系跨考离散的，所以可能有点基础的东西不太懂
: 想请教一下，我是靠补习班在学的
: 最近讲到可数的无限集合的观念
: 其中提到，要成为可数集合有两个要件，若A为有限集合，或是A~正整数集合，
: 则A为可数集合，否则A为不可数集合
: 可是后面又提到，若是存在一个函数使得A对到正整数集合呈一对一关系的话，此集合
: A也可视为可数集合，这是否与前言相牴触了呢?还是说只要与Z有一对一关系，必可
: 保证也onto呢?
ω := the set of all positive integer
Let
ι： A ─→ ω is a 1-1 mapping
Claim : A is countable
Proof :
Let B = ι(A), the image set of ι, B is a subset of ω
Since ι： A ─→ B is 1-1 & onto, |A| = |B|
So we only need to verify that B is countable :
If B is finite, then B is countable (in your definition)
Otherwise B is infinite, then define a sequence b_k by :
b_1 = min(B) (such b_1 must exist, by well-ordering property)
b_2 = min(B─{b_1})
b_3 = min(B─{b_1, b_2})
...
We can see that for any β in B, there is a k s.t
b_k = β
As a result, b ： ω ─→ B, defined by b(x) = b_x is 1-1, onto
Hence B ~ ω
In any case B, as well as A, is countable.
此题的意义在于 ι： A ─→ ω is a 1-1 mapping
代表了 |A| ≦ |ω| 也就是说 A 的"个数"至多等于ω的"个数"
但ω的个数已是所有集合中最小的(所谓的countable)
所以 A的个数必然也是最小的(countable)