Re: [分析] Apostol上的exercise
嘗試第二次:
Proof:
Let S be the subset of R.
If 1 ∈ S, then 1 is the smallest member of S since each interger x ≧ 1.
If 1 is not in S, assume S contains no smallest element.
If x ∈ S, by our assumption, x is not the smallest memeber.
Then, we can find a integer y ∈ S such that 1 ≦ y < x.
1 1
Again, by our assumption, we can find an integet y ∈ S such that 1 ≦ y < y
2 2 1
Continue this process, we get
1≦y < y < ... < y1 < x and y y ... y , x ∈ S
n n-1 n, n-1, , 1
Therefore, 1 ∈ S => contradiction.
So, the statement is true.
看起來好像沒有問題了?
--
※ 發信站: 批踢踢實業坊(ptt.cc)
◆ From: 111.251.161.172
推
01/17 00:56, , 1F
01/17 00:56, 1F
→
01/17 01:05, , 2F
01/17 01:05, 2F
討論串 (同標題文章)
完整討論串 (本文為第 2 之 4 篇)：