[proofplan] We reduce to the case where all scalars are positive. If some $x_j < 0$, replacing $x_j$ by $-x_j$ and complementing the role of $j$ in every subset preserves all pairwise differences $|x_A - x_B|$. Once all $x_i \geq 1$, the key observation is that $A \subsetneq B$ forces $x_B - x_A \geq 1$, so any family with all pairwise differences strictly less than $1$ must be an antichain in $\mathcal{P}([n])$. [Sperner's Theorem](/theorems/2585) then yields $|\mathcal{A}| \leq \binom{n}{\lfloor n/2 \rfloor}$. [/proofplan] [step:Reduce to the case $x_i \geq 1$ for all $i$] Suppose some $x_j < 0$, so $x_j \leq -1$. Replace $x_j$ by $\tilde{x}_j = -x_j \geq 1$, and for each $A \subseteq [n]$ define the companion set \begin{align*} \tilde{A} = A \triangle \{j\} = \begin{cases} A \setminus \{j\} & \text{if } j \in A, \\ A \cup \{j\} & \text{if } j \notin A. \end{cases} \end{align*} Set $\tilde{x}_i = x_i$ for $i \neq j$. For any $A \subseteq [n]$, the subset sum under the new scalars evaluated at the companion set satisfies: \begin{align*} \tilde{x}_{\tilde{A}} &= \sum_{\substack{i \in \tilde{A} \\ i \neq j}} x_i + (-x_j) \cdot \mathbb{1}_{j \in \tilde{A}} = \sum_{\substack{i \in A \\ i \neq j}} x_i + (-x_j) \cdot (1 - \mathbb{1}_{j \in A}). \end{align*} The last equality holds because the map $A \mapsto \tilde{A}$ toggles membership of $j$: $j \in \tilde{A}$ if and only if $j \notin A$. Expanding: \begin{align*} \tilde{x}_{\tilde{A}} &= \sum_{\substack{i \in A \\ i \neq j}} x_i + (-x_j) - (-x_j) \cdot \mathbb{1}_{j \in A} = \sum_{\substack{i \in A \\ i \neq j}} x_i + x_j \cdot \mathbb{1}_{j \in A} - x_j = x_A - x_j. \end{align*} Therefore $\tilde{x}_{\tilde{A}} - \tilde{x}_{\tilde{B}} = (x_A - x_j) - (x_B - x_j) = x_A - x_B$ for all $A, B \subseteq [n]$. The family $\tilde{\mathcal{A}} = \{\tilde{A} : A \in \mathcal{A}\}$ satisfies $|\tilde{x}_{\tilde{A}} - \tilde{x}_{\tilde{B}}| = |x_A - x_B| < 1$ for all $A, B \in \mathcal{A}$, and $|\tilde{\mathcal{A}}| = |\mathcal{A}|$ (since $A \mapsto \tilde{A}$ is a bijection). Repeating for each negative $x_j$, we may assume $x_i \geq 1$ for all $i$. [guided] The idea is that replacing $x_j$ by $-x_j$ corresponds to toggling whether $j$ is "in" or "out" of each subset, reflecting the Boolean lattice along the $j$-th coordinate. The subset sum $x_A$ depends on $x_j$ through the single term $x_j \cdot \mathbb{1}_{j \in A}$. After the replacement, the corresponding term becomes $(-x_j) \cdot \mathbb{1}_{j \in \tilde{A}} = (-x_j)(1 - \mathbb{1}_{j \in A})$, and the net effect is to shift every subset sum by the constant $-x_j$: \begin{align*} \tilde{x}_{\tilde{A}} = x_A - x_j. \end{align*} Since this shift is the same for every $A$, it cancels in pairwise differences. The family size is preserved because $A \mapsto \tilde{A} = A \triangle \{j\}$ is a bijection on $\mathcal{P}([n])$. After repeating for all negative scalars, every $x_i$ satisfies $x_i \geq 1$, and the family $\mathcal{A}$ has been relabelled (via a sequence of coordinate reflections) without changing its size or the pairwise-difference condition. [/guided] [/step] [step:Show that the family $\mathcal{A}$ is an antichain under the assumption $x_i \geq 1$] Assume $x_i \geq 1$ for all $i \in [n]$. Suppose for contradiction that $\mathcal{A}$ contains two sets $A \subsetneq B$. Then $B \setminus A \neq \varnothing$, and since $A \subset B$ implies $B \setminus A$ has at least one element: \begin{align*} x_B - x_A = \sum_{i \in B \setminus A} x_i \geq |B \setminus A| \cdot 1 \geq 1. \end{align*} This gives $|x_A - x_B| = x_B - x_A \geq 1$, contradicting the hypothesis $|x_A - x_B| < 1$. Therefore no two sets in $\mathcal{A}$ are comparable, and $\mathcal{A}$ is an antichain in $\mathcal{P}([n])$. [guided] This is the heart of the proof: when all scalars are at least $1$, the subset-sum function is strictly monotone with respect to inclusion, with a gap of at least $1$ between any comparable pair. If $A \subsetneq B$, the sum $x_B$ exceeds $x_A$ by \begin{align*} x_B - x_A = \sum_{i \in B \setminus A} x_i \geq |B \setminus A| \geq 1, \end{align*} so $A$ and $B$ cannot both belong to a family with all pairwise differences strictly less than $1$. The lower bound $|x_i| \geq 1$ is essential: if we allowed $|x_i| < 1$, then $A \subsetneq B$ with $|B \setminus A| = 1$ could have $x_B - x_A < 1$, and the antichain structure would break down. [/guided] [/step] [step:Apply Sperner's Theorem to conclude] Since $\mathcal{A}$ is an antichain in $\mathcal{P}([n])$, [Sperner's Theorem](/theorems/2585) gives: \begin{align*} |\mathcal{A}| \leq \binom{n}{\lfloor n/2 \rfloor}. \end{align*} [/step]