Re: [分析] Apostol上的exercise

看板Math作者yueayase (scrya)時間13年前 (2011/01/20 00:01)推噓6(6推 0噓 2→)

留言8則, 6人參與討論串4/4 (看更多)

先感謝先前各位的回答,不過因為有人放連結(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. 看起來Apostol只是希望讀者去證明 Well-Ordering 和 Mathematical Induction等價? (我以為他要我從公理開始,去做Well-Ordering的推論) 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 感覺上論點都是和Mathematical Induction有關,不然他如何宣稱: 如果S沒有最小的成員,則S是空集合感覺這個問題本身就有點問題,似乎我不必再鑽了... -- ※ 發信站: 批踢踢實業坊(ptt.cc) ◆ From: 111.251.175.57 ※ 編輯: yueayase 來自: 111.251.175.57 (01/20 00:02)

推

math1209

01/20 00:36, , 1^F

01/20 00:36, 1^F

推

smartlwj

01/20 00:58, , 2^F

01/20 00:58, 2^F

推

math1209

01/20 01:53, , 3^F

01/20 01:53, 3^F