Given $\{a_n\}_{n=1}^\infty$ bounded sequence such that $\lim_{n\to \infty} (a_{n+1}-a_n)=0$. Prove that each point in $[\liminf a_n, \limsup a_n]$ is subsequential limit of the sequence $a_n$.
Bounded Sequence and Subsequential Limit
1 Answers
Let $\alpha\in[\liminf a_n,\limsup a_n]$. You wish to find a subsequence of $(a_n)$ that converges to $\alpha$.
I think it should be clear how to proceed if an outline of finding the first term of the required subsequence is given:
Let $\epsilon_1>0$ and select $N_1$ so large that $|a_{n+1}-a_n|<\epsilon_1$ whenever $n\ge N_1$. Now choose $k_1>N_1$ and $l_1>k_1 $ so that $a_{k_1}$ is within $\epsilon_1$ of the $\sup$ and $a_{l_1}$ is within $\epsilon_1$ of the $\inf$. Then there is an index $n_1$ between $k_1$ and $l_1$ so that $a_{n_1}$ is within $\epsilon_1$ of $\alpha$.
Informally, there is an index $k_1$ where $a_{k_1}$ is close to the $\sup$, and a latter index $l_1$ where $a_{l_1}$ is close to the $\inf$. Since successive terms of $(a_n)$ are close to each other, in the journey from the $\sup$ to the $\inf$, $a_j$ passes close by to any number between the $\sup$ and the $\inf$ (in particular, close by to $\alpha$).
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0David, Thanks a lot for your time and efforts; we will stop here for now, I'll try to grasp the full proof tomorrow as it is a little bit too advanced for me in the moment. Thanks again for your help,$I$can't stress how much$I$appreciate it. Good night :-) – 2012-05-02