Can a sequence converges modulo every r>0 but diverge?

Question

Is it possible to have a sequence $\{x_n\}$ of real numbers which diverge to $\infty$ (and has no other finite limit points), but satisfy the condition that $x_n\pmod{r}$ converges for every real $r>0$?

I'm aware of related results such as "the fractional parts of $\{n\alpha\}_n$ are dense in $[0,1]$ (and thus do not converge)" and more general versions of these statements, giving counterexamples for select values of $r$. However, I'm wondering if the "for every $r>0$" part of the statement makes the existence of such an $\{x_n\}$ impossible. I feel this is the case, but have been unable to come up with a rigorous proof.

Answer 1

The existence of such a sequence is impossible. In fact, if $\{x_n\}$ is a sequence of real numbers diverging to $\infty$, then $x_n$ has a subsequence which, for almost every $r$, is uniformly distributed mod $r$. To see this, pass to a subsequence $\{x_n'\}$ of $\{x_n\}$ to get $|x_n'-x_m'|>1$ for all $n\neq m$ and apply the following Theorem and Corollary.

The following theorem is Corollary 4.3 in Kuipers and Niederreiter's classic Uniform Distribution of Sequences.

Theorem. Let $\{x_n\}$ be a sequence distinct numbers such that $\inf_{n\neq m} |x_n-x_m|>0$. Then for Lebesgue-almost every $\alpha\in \mathbb R$, the sequence $\{x_n\alpha\}$ is uniformly distributed mod $1$.

The function $\alpha \mapsto \frac{1}{\alpha}$ is differentiable with nonzero derivative at every $\alpha>0$, so if Lebesgue-almost every $\alpha$ satisfies a given property, then Lebesgue-almost every value of $\frac{1}{\alpha}$ also satisfies that property. We therefore obtain the following corollary.

Corollary. If $\{x_n\}_{n\in \mathbb N}$ is a sequence of real numbers with $|x_n-x_m|>1$ for all $n\neq m$, then for Lebesgue-almost every $r>0$, the sequence $\{x_n\frac{1}{r}\}$ is uniformly distributed mod $1$.

After applying the Theorem and Corollary, observe that if $\{x_n'\frac{1}{r}\}$ is uniformly distributed mod $1$, then $\{x_n'\}$ is uniformly distributed mod $r$. Thus, for almost every $r$, the sequence $\{x_n \text{ mod } r\}$ has a divergent subsequence, and therefore does not converge.

Answer 2

Such a sequence exists if you restrict the condition to rational $r>0$. Indeed, you can choose an enumeration $\{q_n\}_{n\geq 1}$ of the positive rationals and for each $n$, we can find an integer $d_n\in \mathbb N$ such that $d_nq_m\in \mathbb N$ for all $1\leq m\leq n$. Then set $$ x_n=\prod_{m=1}^n (2d_nq_m), $$ which has the property that $x_n\in q_m\mathbb Z$ for all $1\leq m\leq n$. Thus the sequence $\{x_n\}_{n\geq 1}$ has the property that $x_n\mod r$ converges to $0$ for all rational $r>0$, since every such $r$ can be written as $r=q_N$ for some $N$, and then $x_n\mod r=0$ for all $n>N$. Finally, since each term in the product for $x_n$ is $2$ or greater, we have that $x_n\geq 2^n$ which implies that $x_n$ diverges to $\infty$ with no finite limit points.

Note that one can modify this argument to apply to any countable set of positive real numbers (not just the rationals), but the case of uncountably many real numbers seems to be quite different than the countable case.

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As pointed out in the comments, there is a much simpler construction in the rational case: $x_n=n!$ has the property that $x_n\mod r=0$ for all $r\in\mathbb Q_{>0}$ and all $n\geq \textrm{numerator}(r)$.