※ 引述《Makoto0813 (火紅的燃燒吧!妹控魂!)》之銘言：
: 各位版友好
: 小弟是外系跨考離散的，所以可能有點基礎的東西不太懂
: 想請教一下，我是靠補習班在學的
: 最近講到可數的無限集合的觀念
: 其中提到，要成為可數集合有兩個要件，若A為有限集合，或是A~正整數集合，
: 則A為可數集合，否則A為不可數集合
: 可是後面又提到，若是存在一個函數使得A對到正整數集合呈一對一關係的話，此集合
: A也可視為可數集合，這是否與前言相牴觸了呢?還是說只要與Z有一對一關係，必可
: 保證也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.
此題的意義在於 ι： A ─→ ω is a 1-1 mapping
代表了 |A| ≦ |ω| 也就是說 A 的"個數"至多等於ω的"個數"
但ω的個數已是所有集合中最小的(所謂的countable)
所以 A的個數必然也是最小的(countable)