Is $x_{j} \bmod v$ dense in $[0,v]$? Suppose that $(x_{j})_{j \in \mathbb{N}}$ is an unbounded sequence inside $\mathbb{R}$. Does there exist a $v>0$ so that $x_{j} \bmod v$ is dense in $[0,v]$?
Note here that $a \bmod b := a- \lfloor \frac{a}{b}\rfloor b$.
$\textbf{Example}$: For the sequence $(j)_{j \in \mathbb{N}}$, $j \bmod \sqrt{2}$ is dense in $[0,\sqrt{2}]$.
I added the unbounded assertion as the sequence $((\frac{1}{2})^{j})_{j \in \mathbb{N}}$ is not dense modulo $v$ for any $v > 0$
 A: Such number $v$ always exists.
Notice that $a \text{ mod } b = \left\{\frac{a}{b}\right\}b$. Therefore proving that $x_n \text{ mod } v$ dense in $[0,v]$ is the same as proving $\left\{\frac{x_n}{v}\right\}$ is dense in $[0,1]$.
$\textbf{Lemma}$. Let $V = [p, q]$ be a segment and $U = (\alpha, \beta)\subset [0,1]$ be an interval. Then there exist subsegment $\tilde V\subset V$  and $n\ge 0$ such that for every $v\in \tilde V$ we have $\displaystyle\left\{\frac{x_n}{v}\right\}\in U$.
$\textit{Proof}$. The assertion $\displaystyle\left\{\frac{x_n}{v}\right\}\in U$
can be rewritten as $x_n\in (v(N + \alpha), v(N + \beta))$ or
$$
v(N + \alpha) < x_n < v(N + \beta), \qquad \frac{x_n}{N + \beta} < v < \frac{x_n}{N + \alpha}
$$
for some integer $N$. For some large $M> 0$ we have $\displaystyle(M ,+\infty)\subset \bigcup_{N\ge 1}(p(N + \beta), q(N + \alpha))$. Since the sequence $x$ is unbounded, there is some $x_n\in (p(N + \beta), q(N + \alpha))$. Then $\displaystyle\tilde V = \left[\frac{x_n}{N + \beta}, \frac{x_n}{N + \alpha}\right]$ will do.
To solve the problem simply choose a sequence of intervals $U_k$ and iteratively construct a nested sequence $V_k$ using lemma. Any point in $\bigcap_{k\ge 1} V_k$ will be a solution.
