# subsequence

• Jul 6th 2008, 10:18 PM
particlejohn
subsequence
Prove that if $\displaystyle a_n$ is a sequence such that every subsequence of $\displaystyle a_n$ has a further subsequence converging to $\displaystyle 0$, then $\displaystyle a_n \to 0$.

Let $\displaystyle \sigma: \mathbb{N} \to \mathbb{N}$ be a strictly increasing function. Suppose that $\displaystyle b_{n} = a_{\sigma(n)}$ is a subsequence of $\displaystyle a_n$. Then $\displaystyle c_{n} = b_{\sigma(n)} \to 0$ is a subsequence of $\displaystyle b_n$. If $\displaystyle \varepsilon > 0$, then there is an $\displaystyle N \in \mathbb{N}$ such that $\displaystyle n \geq N$ implies that $\displaystyle c_{n} \in (-\varepsilon, \varepsilon)$.

From here, I have to show that $\displaystyle a_n \in (-\varepsilon, \varepsilon)$.

Now what? $\displaystyle c_n \leq b_n \leq a_n$. Could I use the Squeeze theorem?
• Jul 6th 2008, 10:33 PM
particlejohn
actually use the following theorem twice: $\displaystyle a_n \to L \Longleftrightarrow$ every subsequence of $\displaystyle a_n$ converges to $\displaystyle L$?

Is this correct?
• Jul 6th 2008, 10:39 PM
kalagota
possibly yes..

EDIT: but if i read and interpret it correct, it doesn't tell you that every subsequence of the subsequence of a_n also converges to 0.
• Jul 6th 2008, 10:39 PM
Jhevon
Quote:

Originally Posted by particlejohn
"every subsequence of $\displaystyle a_n$ converges to $\displaystyle L$"

you do not have this condition. we are told "every subsequence of http://www.mathhelpforum.com/math-he...f16b21dd-1.gif has a further subsequence converging to http://www.mathhelpforum.com/math-he...f98764da-1.gif" it is a different statement
• Jul 6th 2008, 10:41 PM
particlejohn
yes, but this is a general theorem about subsequences. A subsequence is also a sequence right? Regardless, is my approach correct?
• Jul 6th 2008, 10:43 PM
kalagota
if you are going to use that, you have to show that every subsequence of the subsequence of a_n also converges to 0.
• Jul 6th 2008, 10:47 PM
particlejohn
Quote:

Originally Posted by kalagota
possibly yes..

EDIT: but if i read and interpret it correct, it doesn't tell you that every subsequence of the subsequence of a_n also converges to 0.

Oh okay. It just has $\displaystyle 1$ subsequence converging to $\displaystyle 0$?
• Jul 7th 2008, 08:45 AM
ThePerfectHacker
Let $\displaystyle \{a_n\}$ be a sequence such that for any subsequence of $\displaystyle \{a_n\}$ we can find a subsequence which is convergent to zero. Assume that $\displaystyle a_n$ does not converge to zero, then it means for some $\displaystyle \epsilon > 0$ there are some $\displaystyle a_{n_1},a_{n_2}, ...$ such that $\displaystyle |a_{n_k}|\geq \epsilon$. The sequence $\displaystyle \{ a_{n_k}\}$ is a subsequence of $\displaystyle \{a_n\}$ but it does not have any subsequence which is convergent to zero since all its terms are $\displaystyle \geq \epsilon$ in absolute value. This is a contradiction. Thus, $\displaystyle \{ a_n\}$ must be convergent to zero.

This is Mine 1:)1:):)th Post!!!