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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 $a_n$.

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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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    what do you mean by: "...so that $a_{k_1}$ is with $\epsilon_1$ of the sup"?2012-05-02
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    @Anonymous Sorry for the vaugeness; I mean $|a_{k_1} -\limsup_n a_n|<\epsilon_1$.2012-05-02
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    why can I deduce that $|a_{k_1} -\limsup_n a_n|<\epsilon_1?$2012-05-02
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    @Anonymous If $\beta=\limsup_n a_n$, then there is a subsequence of $(a_n)$ that converges to $\beta$. Thus, there is a term of $(a_n)$, say $a_{k_1}$ such that $|a_{k_1}-\beta|<\epsilon_1$. Of course, infinitely many terms of $(a_n)$ satisfy this, but we need to fix a particular term of $(a_n)$ for the argument.2012-05-02
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    ok, great, now why can I deduce that $|a_{n_1}-\alpha|<\epsilon_1$?2012-05-02
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    @Anonymous Because $|a_{n+1}-a_n|<\epsilon_1$ for $n\ge N_1$. Look at the terms $a_{k_1}, a_{k_1+1},\ldots, a_{l_1}$. They start near the sup and end near the inf. But in going from one term to the next term, a step of length at most $\epsilon_1$ can be made. If you subdivide $[\liminf_n a_n, \limsup_n a_n]$ into disjoint subintervals of length $\approx\epsilon_1$, then each of those subintervals contains a term from $a_{k_1}, a_{k_1+1},\ldots, a_{l_1}$. Here, to be precise, you should take $\epsilon_1$ to be of the form $1/m$, $m\in\Bbb N$.2012-05-02
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    I think I get you, though it is quite strange that they start near the sup and end near the inf don't you think?(the sup should be bigger than the inf). Also, I didn't understand why given $|a_{n+1}-a_n|<\epsilon_1$ I can deduce that $|a_{n_1}-\alpha|<\epsilon_1$?2012-05-02
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    @Anonymous Assume, for clarity, that $a_{k_1},\ldots, a_{l_1}$ is increasing. Let $\beta$ be the sup and $\gamma$ be the inf. Keep in mind that $|a_{j+1}-a_j|<\epsilon_1$ for each $j$. Now $\alpha$ is in some subinterval of $[\gamma,\beta]$ of the form $I=[\gamma+j\epsilon_1, \gamma+(j+1)\epsilon_1]$ or $I=[\beta-\epsilon_1,\beta]$. Note $I$ has length $\epsilon_1$. If no term of $a_{k_1},\ldots, a_{l_1}$ were in $I$, then two successive terms (namely the largest term less than the left hand endpoint of $I$ and the next term) would be at least $\epsilon_1$ apart. (cont.)2012-05-02
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    This can't happen. So some term of $a_{k_1},\ldots, a_{l_1}$ is in $I$ and thus within $\epsilon_1$ of $\alpha$.2012-05-02
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    David, 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