Let $(x_n)_{n\in\mathbb{N}} \subset \mathbb{R}$ be a bounded sequence such that $\lim_{n\rightarrow\infty}(x_{n+1}-x_n) = 0$ and $A = \liminf_{n\rightarrow\infty}(x_n)$ and $B = \limsup_{n\rightarrow\infty}(x_n)$. Show that the set of accumulation points of $(x_n)$ is the interval $I = [A,B]$
I already showed that there can't be an acc. pt. outside of $I$. Do you have any hint/approach how to show that every point in $I$ is an accumulation point of $(x_n)$?
By definition, there is a subsequence $(x_{n_k})$ with $\lim_kx_{n_k}=A$ and another one $(x_{n'_k})$ with $\lim_kx_{n'_k}=B$. Given some integer $m>0$, we can choose $K$ large enough such that $a)\,\, |x_{n_k}-A|<1/m$ for $k\ge K$, $b)\,\, |x_{n'_k}-B|<1/m$ for $k\ge K$ and $c)\,\, |x_{n+1}-x_{n}|<1/m$ for $n\ge n_K$.
Choose $k\ge K$ and $l>0$ such that $n'_{k_0}=n_k+l$ for some $k_0>K$. This is always possible, since the sequence has to visit arbitrarily small neighborboods of both $A$ and $B$ eventually.
Let $x_{n_k+l_1}$, $l_1>0$ be the first element in $\{x_{n_k+s}: s=1,2,\dots l\}$ larger than $x_{n_k}$. By assumption $(c)$ above, $x_{n_k+l_1}-x_{n_k}<1/m$. Let $x_{n_k+l_2}$, $l_2>l_1$ be the first element in $\{x_{n_k+s}: s=1,2,\dots l\}$ larger than $x_{n_k+l_1}$. By the same assumption, $x_{n_k+l_2}-x_{n_k+l_1}<1/m$. Continuing in this fashion, we get a set of points from $(x_n)$ in $[A,B]$ such that the distance between neighboring points is less than $1/m$, as well as the distance between $x_{n_k}$ and $A$ and that between $x_{n_{k_0}}=x_{n_k+l}$ and $B$. The union of such sets over $m>0$ is dense in $[A,B]$. We thus conclude that any point in $[A,B]$ is an accummulation point.
