先感谢先前各位的回答,不过 因为有人放连结(http://frankmath.cc/plover/Apostol.pdf), 所以我又看到了另一版的解答(1.6): Proof. Given S≠φ and S ⊆N; we prove that if S contains an integer k;then S contains the smallest member. We prove it by mathematical induction of second form as follows. As k = 1; it trivially holds. Assume that as k = 1,2,...,m holds,consider as k = m + 1 as follows. In order to show it, we consider two cases. (1) If there is a member s ∈ S such that s < m+1; then by induction hypothesis, we have proved it. (2) If every s 2 S; s m + 1; then m + 1 is the smallest member. Hence, by mathematical induction, we complete it. 看起来Apostol只是希望读者去证明 Well-Ordering 和 Mathematical Induction等价? (我以为他要我从公理开始,去做Well-Ordering的推论) 和原先找到的: If S contains no smallest element then S is empty because individual elements of N are finite. But S is nonempty. Therefore S contains a smallest element 感觉上论点都是和Mathematical Induction有关,不然他如何宣称: 如果S没有最小的成员,则S是空集合 感觉这个问题本身就有点问题,似乎我不必再钻了... --

※ 发信站: 批踢踢实业坊(ptt.cc) ◆ From: 111.251.175.57 ※ 编辑: yueayase 来自: 111.251.175.57 (01/20 00:02)