Re: [分析] Apostol上的exercise

[yueayase (scrya)]

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...

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◆ From: 111.251.175.57
※ 編輯: yueayase        來自: 111.251.175.57       (01/20 00:02)
推 math1209    :那不是本尊 T.M.Apostol 寫的解答. =.=
推 smartlwj    :是plover寫的!!
推 math1209    :也不是他寫的 XD
→ rogerli     :那，兇手是誰？膜拜一下。
推 PaulErdos   :我猜是一樓
推 herstein    :一樓寫的XD
推 rogerli     :(_._) 膜拜一下一樓。
→ WINDHEAD    :一樓寫的,然後連結是六樓的個人網站XD (愛躲)