All convergent subsequences converge to the same limit => what ?

kindlychung

If a_n is convergent, it must converge to a, if not, all its subsequences must converge to a number other than a, contradiction.
How can we rule out the possibility that a_n is divergent?

elimqiu

If {a(n)} is not converge to a, but it has the property as the exercise says, we';ll obtain a contradiction as below:
There will be an ε> 0, for any M, there is an n such that |a(n) - a| ≥ε
This simply means that there are infinite number of n such that a(n) is no closer to a by ε.
Let {b(n)} be the sequence formed by those n, then {b(n)} is a subsequence of {a(n)} and it cannot converge to a.

kindlychung

Let {b(n)} be the sequence formed by those n, then {b(n)} is a subsequence of {a(n)} and it cannot converge to a.

By assumption, every convergent subsequence of a_n converges to a, that does NOT mean every subsequence is convergent. What if {b_n} diverges?

elimqiu

From my assumption of {a(n)} does not converge to a, I constructed a subsequence {b(n)} that converges but not to a. This contradicts to the assumption about convergent sebsequences. Therefore {a(n)} must converge to the same limit a.

kindlychung

Sorry, I cannot see that you have constructed a CONVERGENT {b(n)}.

elimqiu

下面引用由elimqiu在 2010/09/27 00:19am 发表的内容：
If {a(n)} is not converge to a, but it has the property as the exercise says, we';ll obtain a contradiction as below:
There will be an ε> 0, for any M, there is an n such that |a(n) - a| ≥ε
This simply means that there are infinite number of n such that a(n) is no closer to a by ε.
Let {b(n)} be the sequence formed by those n, then {b(n)} is a subsequence of {a(n)} and it cannot converge to a.

[br][br]-=-=-=-=- 以下内容由 elimqiu 在 时添加 -=-=-=-=-
Now since {b(n)} is also bounded, you can get a convergent subsequence {c(n)} of {b(n)}. And you see what I can say about {c(n)}

kindlychung

Yeah, now I can see. Thank you.

wangyangkee

为老外顶帖，，，