Set of adherence values of a sequence . Let $(x_n)$ be a sequence  such that:
$\lim (x_n)=+\infty$ and 
$\lim (x_{n+1}-x_n)=0$
Put $v_n=x_n -[x_n]$, where $[t]$ is the part integer of $t$ Where $[t]$ is the  integer of part of $t$.
What is the set of adherence values of $(v_n)$.
Thanks in advance.
 A: Let $y\in (0,1).$ Given any $r>0,$ let $s=\min (y/2,(1-y)/2,r ).$
Take $n_0\in \Bbb N$ such that $n\ge n_0\implies |x_{n+1}-x_n|<s.$ Let $t_0=\min \{[x_n]: n\ge n_0\}$ and let $T=\{t\in \Bbb N:t>t_0\}.$ 
For each $t \in T$ let $f(t)$ be the $least$ $n>n_0$ such that $x_n\ge t+y.$ 
If $t\in T$ then $f(t)-1\ge n_0$ so $$(*)\quad |x_{f(t)}-x_{f(t)-1}|<s,$$ and by the def'n of $f(t)$ we have $$(**)\quad x_{f(t)}\ge t+y> x_{f(t)-1}.$$ From $(*)$ and $(**)$ we have $$t\in T\implies t<t+y-s\le x_{f(t)}-s<x_{f(t)-1}<x_{f(t)}<$$ $$<x_{f(t)-1}+s <t+y+s<t+1.$$ So we have $$t\in T \implies |v_{f(t)}-y|<s\le r.$$ Now $f[T]=\{f(t): t\in T\}$ is an infinite set (Why?). So for any $r>0$ there are infinitely many $n$ (e.g. all $n\in f[T]$ ) such that $|y-v_n|<r.$ So $y$ is an adherence (limit) point of $(v_n)_{n\in \Bbb N}.$ 
The set  of adherence points of $any$ sequence is a closed set. So for the set $A$ of adherence points of $(v_n)_n$ we have $[0,1]\supset A=\bar A\supset \overline {(0,1)}=[0,1].$ So $A=[0,1].$
