# Converging Subsequence

• March 9th 2010, 07:43 PM
harish21
Converging Subsequence
Suppose that a sequence { $x_n$} is monotone and that it has a convergent subsequence. Show that { $x_n$} also converges.

I think that "since the subsequence is convergent, the sequence { $x_n$} is bounded", and
according to the monotone convergence theorem,a monotone sequence is convergent if and only if it is bounded.

I am not sure about the first part of my reasoning, because there is a theorem that says "every bounded sequence has a convergent subsequence". Based on that, I am stating that "since the subsequence is convergent, the sequence is bounded".

Is this the correct reasoning or are there any other ways this statement can be proved? Suggestions will be appreciated
• March 9th 2010, 07:47 PM
Drexel28
Quote:

Originally Posted by harish21
Suppose that a sequence { $x_n$} is monotone and that it has a convergent subsequence. Show that { $x_n$} also converges.

I think that "since the subsequence is convergent, the sequence { $x_n$} is bounded", and
according to the monotone convergence theorem,a monotone sequence is convergent if and only if it is bounded.

I am not sure about the first part of my reasoning, because there is a theorem that says "every bounded sequence has a convergent subsequence". Based on that, I am stating that "since the subsequence is convergent, the sequence is bounded".

Is this the correct reasoning or are there any other ways this statement can be proved? Suggestions will be appreciated

Assume WLOG that $x_n$ is increasing. We know that there exists some $\{x_{\varphi(n)}\}$ such that $x_{\varphi(n)}\to x$. So, let $\varepsilon>0$ be given. Then, there exists some $N\in\mathbb{N}$ such that $N\leqslant n\implies \left|x-x_{\varphi(n)}\right|=x-x_{\varphi(n)}<\varepsilon$. But, if $\varphi\left(N\right)\leqslant n\implies x_{\varphi(N)}\leqslant x_n\implies x-x_n\leqslant x-x_{\varphi(N)}<\varepsilon$