作者yueayase (scrya)
看板Math
标题Re: [分析] Apostol上的exercise
时间Mon Jan 17 00:24:56 2011
尝试第二次:
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.
看起来好像没有问题了?
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1F：推 carusochu :你这样证等於已经假定整个N有最小元素(叫做1) 01/17 00:56
2F：→ carusochu :你可以google: the well-ordering principle 01/17 01:05