高微(converge of subsequence)

2012-09-20 3:57 am
Suppose that {x_n} is a sequence in R. Prove that {x_n} converges to a
if and only if every subsequence of {x_n} also converges to a.

謝謝大大的幫忙~!

回答 (1)

2012-09-20 4:21 am
✔ 最佳答案
假設 {x_n} 收斂到 a.
則,
for any ε>0, there exists N such that
whenever n>N we have |x_n-a|<ε.

Let {x_n(k)} be a subsequence of {x_n}.
There exists K such that
k>K implies n(k)>N
因此, 當 k>K 時 |x_n(k)-a|<ε

故 {x_n(k)} 收斂到 a.


反之, 若 every subsequence of {x_n} converges to a.
因 {x_n} 是本身的一子列, 故 trivially 得證 {x_n} 收斂至 a.




["if" part 的另一證法]

設 {x_n} 不收斂到 a, 不管是 {x_n} 不收斂或 {x_n} 收斂到 b≠a,
總之, 存在 ε0>0 使得
for any N, 存在 n>N 滿足 |x_n-a|≧ε0
任取 n(1) 滿足 |x_n(1)-a|≧ε0
存在 n(2)>n(1) 滿足 |x_n(2)-a|≧ε0
又存在 n(3)>n(2) 滿足 |x_n(3)-a|≧ε0
以此類推, 得 {x_n} 之一子列 {x_n(k)}, |x_n(k)-a|≧ε0 for all k.
By definition, {x_n(k)} 並不收斂到 a.
因此, 若 {x_n} 之任意子列都收斂到 a, 則 {x_n} 本身收斂到 a.


收錄日期: 2021-05-04 01:49:27
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20120919000010KK06627

檢視 Wayback Machine 備份