  Let $q,d\geq 2$ be integers. Define $$ J(q,d):=\frac 1q \Big( \min_{0<x<1}
\frac{1-x^q}{1-x} x^{-\frac{q-1}{d}}\Big). $$ Let $\mbox{$\cal G$}\subseteq
{\mathbb R}^n$ be an arbitrary subset. We denote by $d(\mbox{$\cal G$})$ the
set of (non-zero) distances among points of $\mbox{$\cal G$}$: $$ d(\mbox{$\cal
G$}):=\{d( p_1, p_2):~ p_1, p_2\in \mbox{$\cal G$}, p_1\ne p_2\}. $$ Our main
result is a new upper bound for the size of $s$-distance sets in boxes. More
concretely, let $A_i\subseteq \mathbb R$, $|A_i|=q\geq 2$ be subsets for each
$1\leq i\leq n$. Consider the box $\mbox{$\cal B$}:=\prod_{i=1}^n A_i\subseteq
{\mathbb R}^n$. Suppose that $\mbox{$\cal G$}\subseteq \mbox{$\cal B$}$ is a
set such that $|d(\mbox{$\cal G$})|\leq s$. Let $d:=\frac{n(q-1)}{s}$. Then
$$|\mbox{$\cal G$}|\leq 2(qJ(q,d))^n.$$ We use Tao's slice rank bounding method
in our proof.
