TITLE: A bound on a set
QUESTION [5 upvotes]: 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 but not sure about it.

REPLY [2 votes]: Expanding on my comment and continuing GH from MO's answer:
For $\alpha >2$, let $r:=\lceil\alpha/2\rceil+1$. Since a half-open interval $I$ of length $\alpha$ cannot contain $r$ numbers whose pairwise distance is at least $2$, for any fixed $A \subseteq \{1, \dots, n\}$, the set $A_+:=\{i \in A; \alpha_i = +1\}$ cannot contain an inclusion chain of length $r$.
By Erdos's generalization of the Sperner Theorem, we have
$$\lvert I\cap S_A \rvert \le \sum_{k=\lfloor |A|/2\rfloor-\lfloor (r-1)/2\rfloor}^{\lfloor |A|/2\rfloor+\lfloor r/2\rfloor} {\lvert A \rvert \choose k},
$$
with equality for $I=[-r+1, -r+1+\alpha)$ and $x_i=1$ for $i\in A$. Hence, 
$$|I\cap S|\leq \sum_{A\subset\{1,\dots,n\}}\sum_{k=\lfloor |A|/2\rfloor-\lfloor (r-1)/2\rfloor}^{\lfloor |A|/2\rfloor+\lfloor r/2\rfloor} {\lvert A \rvert \choose k} = \sum_{m=0}^n
{n\choose m}\sum_{k=\lfloor m/2\rfloor-\lfloor (r-1)/2\rfloor}^{\lfloor m/2\rfloor+\lfloor r/2\rfloor} {m \choose k},
$$
with equality for $I=[-r+1, -r+1+\alpha)$ and $x_i=1$ for all $1\le i\le n$.