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Let $\{ a_n \}_{n=1}^{\infty}$ be a bounded sequence such that $\lim_{n \rightarrow \infty} a_n-a_{n+1}=0$. Prove the set of subsequential limits is an interval.

My thoughts:

$a_n$ is bounded $\Rightarrow \exists M>0\ \forall n\in\mathbb{N}:|a_n|<M$

So I'm trying to prove that the set of subsequential limits is $(-M,M)$.

I understand that I need to set the right $\epsilon>0$ for the limit given and work with it, but having hard time formalizing it.

Any help appreciated.

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  • $\begingroup$ What is a partial limit? $\endgroup$ – zoli Mar 6 '17 at 10:56
  • $\begingroup$ Do you really think that, just knowing that $|a_n|\lt M,$ you should be able to prove that the set of "partial limits" (whatever those might be) is $(-M,M)?$ The trouble is, if $|a_n|\lt M$ for all $n,$ then also $|a_n|\lt10M$ for all $n.$ Do you think you can prove that the set of "partial limits" is equal to $(-M,M)$ and also to $(-10M,10M)?$ $\endgroup$ – bof Mar 6 '17 at 11:04
  • $\begingroup$ Sorry for that, will fix to subsequential limit. $\endgroup$ – Itay4 Mar 6 '17 at 11:05
  • $\begingroup$ By the way, what's your definition of an "interval"? Is $\{0\}$ an interval? $\endgroup$ – bof Mar 6 '17 at 11:06
  • $\begingroup$ @bof No, it has to be set of real numbers that lies between two real numbers. $\endgroup$ – Itay4 Mar 6 '17 at 11:08
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Note that with $b=\liminf{a_n}$ and $c=\limsup{a_n}$, we have that the set $M$ of limits of (convergent) subsequences of $a_n$ is included in $[b,c]$. If $b=c$, the sequence $a_n$ is convergent and we are done. Now suppose $b<c$ and let $b <u<v<c$. The sets $E=\{n; a_n<u\}$ and $F=\{n; a_n>v\}$ are infinite(because there are subsequences of $a_n$ converging to $b$ and $c$). If there is only a finite number of $n$ such that $u\leq a_n\leq v$, then we can find a sequence $n_k\in E$ with $1+n_k\in F$. But then $a_{1+n_k}-a_{n_k}\geq v-u>0$, a contradiction as $a_{1+n_k}-a_{n_k}\to 0$ if $k\to \infty$. Hence we can find a subsequence $a_{n_m}=u_m$ of $a_n$ such that $v\geq u_m\geq u$. Then $u_m$ has a convergent subsequence, say to $L$, and clearly $L\in M$ We have shown that the closed subset $M$ is dense in $[b,c]$, hence it is equal to $[b,c]$.

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