good explanation!
※ 引述《euphrate ()》之铭言：
: 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)