Let $x\_1,\cdots , x\_n$ be a sequence of real number such that $x\_i\geq 1$ for all $1\leq i\leq n$, $S=\{\alpha\_1x\_1+\cdots +\alpha\_nx\_n | \alpha\_i\in\{0,+1,-1\}\}$ and $I=[a,b)$ be a Interval with length $2$. So I was wondering if there was any subsequent upper bound on $|I \cap S|$. Is there a general bound when $2$ is replaced by a positive real $\alpha$? Thanks in advance for your answers and comments. I haven't been able to guess anything, but this bound on size of a set reminded me of [Sperner's Theorem](http://en.wikipedia.org/wiki/Sperner%27s_theorem) but not sure about it.



A bound on a set
For a given $A\subset\{1,\dots,n\}$, let $S\_A$ denote the multiset of $\alpha\_1 x\_1+\cdots+\alpha\_n x\_n$ with $\alpha\_i=\pm 1$ for $i\in A$ and $\alpha\_i=0$ for $i\not\in A$. Note that, as multisets,
$$ S=\bigcup\_{A\subset\{1,\dots,n\}} S\_A,$$
so that
$$ I\cap S=\bigcup\_{A\subset\{1,\dots,n\}} (I\cap S\_A).$$
By the Sperner bound, realized in this problem by Erdős (1945), we have
$$ |I\cap S\_A|\leq {|A| \choose \lfloor |A|/2\rfloor}, $$
with equality for $I=[-1,1)$ and $x\_i=1$ for $i\in A$. Hence,
$$ |I\cap S|\leq \sum\_{A\subset\{1,\dots,n\}}{|A| \choose \lfloor |A|/2\rfloor}=\sum\_{m=0}^n
{n\choose m}{m \choose \lfloor m/2\rfloor}, $$
with equality for $I=[-1,1)$ and $x\_1=\dots =x\_n=1$.


