A variation on arithmetic continuity

A sequence $(x_{k})$ of points in $\R$, the set of real numbers, is called \textit{arithmetically convergent} if for each $\varepsilon > 0$ there is an integer $n$ such that for every integer $m$ we have $|x_{m} - x_{<m,n>}|<\varepsilon$, where $k|n$ means that $k$ divides $n$ or $n$ is a multiple of $k$, and the symbol $< m, n >$ denotes the greatest common divisor of the integers $m$ and $n$. We prove that a subset of $\R$ is bounded if and only if it is arithmetically compact, where a subset $E$ of $\R$ is arithmetically compact if any sequence of point in $E$ has an arithmetically convergent subsequence. It turns out that the set of arithmetically continuous functions on an arithmetically compact subset of $\R$ coincides with the set of uniformly continuous functions where a function $f$ defined on a subset $E$ of $\R$ is arithmetically continuous if it preserves arithmetically convergent sequences, i.e., $(f(x_{n})$ is arithmetically convergent whenever $(x_{n})$ is an arithmetic convergent sequence of points in $E$.