[proofplan]
It suffices to partition $\mathcal{P}([n])$ into $\binom{n}{\lfloor n/2 \rfloor}$ sparse families, since each sparse family can contribute at most one element to the diameter-less-than-$1$ family $\mathcal{A}$. We prove by induction on $n$ that $\mathcal{P}([n])$ admits a **symmetric decomposition into sparse families** — a partition whose profile matches that of the symmetric chain decomposition. The inductive step splits each sparse family $\mathcal{F}$ of $\mathcal{P}([n-1])$ into two sparse families $\mathcal{F}^{(0)}$ and $\mathcal{F}^{(1)}$ of $\mathcal{P}([n])$ using the same structural rule as the [Symmetric Chain Decomposition](/theorems/2590). The sparsity of $\mathcal{F}^{(1)}$ is inherited by translation, while the sparsity of $\mathcal{F}^{(0)}$ relies on choosing a maximiser of a linear functional guaranteed by the Hahn--Banach theorem.
[/proofplan]
[step:Reduce to partitioning $\mathcal{P}([n])$ into sparse families]
If $\mathcal{F}$ is a sparse family (i.e., $\|x_E - x_F\| \geq 1$ for all distinct $E, F \in \mathcal{F}$) and $\mathcal{A}$ is a family with $\|x_A - x_B\| < 1$ for all $A, B \in \mathcal{A}$, then $|\mathcal{F} \cap \mathcal{A}| \leq 1$: if $E, F \in \mathcal{F} \cap \mathcal{A}$ with $E \neq F$, then $\|x_E - x_F\| \geq 1$ (from sparsity) and $\|x_E - x_F\| < 1$ (from the hypothesis on $\mathcal{A}$), a contradiction.
Therefore, if $\mathcal{P}([n]) = \mathcal{F}_1 \cup \cdots \cup \mathcal{F}_t$ is a partition into sparse families, then $|\mathcal{A}| = \sum_{j=1}^{t} |\mathcal{F}_j \cap \mathcal{A}| \leq t$. It suffices to achieve $t = \binom{n}{\lfloor n/2 \rfloor}$.
[/step]
[step:Establish the base case $n = 1$]
For $n = 1$, the partition $\mathcal{P}([1]) = \{\varnothing, \{1\}\}$ consists of a single family $\mathcal{F}_1 = \{\varnothing, \{1\}\}$. This family is sparse because
\begin{align*}
\|x_{\{1\}} - x_\varnothing\| = \|x_1\| \geq 1,
\end{align*}
which holds by hypothesis. The number of families is $1 = \binom{1}{0} = \binom{1}{\lfloor 1/2 \rfloor}$.
[/step]
[step:Define the inductive splitting construction]
Suppose $\mathcal{P}([n-1]) = \mathcal{F}_1 \cup \cdots \cup \mathcal{F}_t$ is a symmetric decomposition into sparse families (with respect to the vectors $x_1, \ldots, x_{n-1}$). We follow the same structural template as the [Symmetric Chain Decomposition of the Power Set](/theorems/2590): from each $\mathcal{F}_j$, produce two families in $\mathcal{P}([n])$.
For each $\mathcal{F}_j$, choose an element $D_j \in \mathcal{F}_j$ (the choice of $D_j$ will be specified in the next step). Define:
\begin{align*}
\mathcal{F}_j^{(0)} &= \mathcal{F}_j \cup \{D_j \cup \{n\}\}, \\
\mathcal{F}_j^{(1)} &= \{E \cup \{n\} : E \in \mathcal{F}_j,\, E \neq D_j\}.
\end{align*}
When $|\mathcal{F}_j| = 1$, the family $\mathcal{F}_j^{(1)}$ is empty and is dropped.
This construction mirrors the chain decomposition: every element of $\mathcal{P}([n])$ appears exactly once, and the profile of the resulting partition is symmetric. The sizes satisfy $|\mathcal{F}_j^{(0)}| = |\mathcal{F}_j| + 1$ and $|\mathcal{F}_j^{(1)}| = |\mathcal{F}_j| - 1$, matching the symmetric chain splitting rule.
[/step]
[step:Verify that $\mathcal{F}_j^{(1)}$ is sparse]
The family $\mathcal{F}_j^{(1)}$ consists of sets of the form $E \cup \{n\}$ for $E \in \mathcal{F}_j \setminus \{D_j\}$. For any two distinct elements $E \cup \{n\}$ and $E' \cup \{n\}$ in $\mathcal{F}_j^{(1)}$:
\begin{align*}
\|x_{E \cup \{n\}} - x_{E' \cup \{n\}}\| = \|(x_E + x_n) - (x_{E'} + x_n)\| = \|x_E - x_{E'}\| \geq 1,
\end{align*}
where the last inequality holds because $E, E' \in \mathcal{F}_j$ are distinct elements of the sparse family $\mathcal{F}_j$. Therefore $\mathcal{F}_j^{(1)}$ is sparse.
[guided]
The sparsity of $\mathcal{F}_j^{(1)}$ requires no cleverness: it is simply the translation invariance of the norm. Adjoining $n$ to every set adds the common vector $x_n$ to every subset sum, and this translation does not change pairwise distances:
\begin{align*}
\|x_{E \cup \{n\}} - x_{E' \cup \{n\}}\| = \|x_E - x_{E'}\|.
\end{align*}
Since $\mathcal{F}_j$ was sparse, so is $\mathcal{F}_j^{(1)}$. This works regardless of how $D_j$ is chosen — the choice of $D_j$ affects only the sparsity of $\mathcal{F}_j^{(0)}$.
[/guided]
[/step]
[step:Choose $D_j$ via the Hahn--Banach theorem and verify that $\mathcal{F}_j^{(0)}$ is sparse]
We must choose $D_j \in \mathcal{F}_j$ so that $\mathcal{F}_j^{(0)} = \mathcal{F}_j \cup \{D_j \cup \{n\}\}$ is sparse. This requires:
\begin{align*}
\|x_{D_j \cup \{n\}} - x_E\| \geq 1 \quad \text{for all } E \in \mathcal{F}_j.
\end{align*}
(The pairs within $\mathcal{F}_j$ are already separated by $\geq 1$ since $\mathcal{F}_j$ is sparse, so the only new pairs to check involve $D_j \cup \{n\}$.)
By the Hahn--Banach theorem, there exists a bounded linear functional $f$ on the normed space with $\|f\| = 1$ and $f(x_n) = \|x_n\| \geq 1$. Choose $D_j \in \mathcal{F}_j$ to maximise $f(x_{D_j})$ over all $D \in \mathcal{F}_j$. Since $\mathcal{F}_j$ is finite, this maximum is attained.
For any $E \in \mathcal{F}_j$:
\begin{align*}
f(x_{D_j \cup \{n\}} - x_E) = f(x_{D_j}) + f(x_n) - f(x_E) = \bigl(f(x_{D_j}) - f(x_E)\bigr) + f(x_n).
\end{align*}
Since $D_j$ maximises $f(x_D)$ over $\mathcal{F}_j$, we have $f(x_{D_j}) - f(x_E) \geq 0$. Since $f(x_n) = \|x_n\| \geq 1$, the entire expression satisfies:
\begin{align*}
f(x_{D_j \cup \{n\}} - x_E) \geq 0 + 1 = 1.
\end{align*}
Using $\|f\| = 1$ and the definition of the operator norm:
\begin{align*}
\|x_{D_j \cup \{n\}} - x_E\| \geq |f(x_{D_j \cup \{n\}} - x_E)| \geq f(x_{D_j \cup \{n\}} - x_E) \geq 1.
\end{align*}
Therefore $\mathcal{F}_j^{(0)}$ is sparse.
[guided]
The challenge: the new element $D_j \cup \{n\}$ in $\mathcal{F}_j^{(0)}$ must be at distance $\geq 1$ from every element of $\mathcal{F}_j$. We have $x_{D_j \cup \{n\}} - x_E = (x_{D_j} - x_E) + x_n$. We want $\|(x_{D_j} - x_E) + x_n\| \geq 1$.
The idea is to pick $D_j$ so that $x_{D_j} - x_E$ and $x_n$ "point in the same direction" for every $E$. Since we cannot literally require this in an arbitrary normed space, we use a linear functional to project everything onto a line.
The Hahn--Banach theorem provides a functional $f$ with $\|f\| = 1$ and $f(x_n) = \|x_n\|$. This $f$ is a "best linear probe" for the direction of $x_n$: it extracts the full magnitude of $x_n$ as a real number. By choosing $D_j$ to maximise $f(x_D)$, we guarantee that the projection $f(x_{D_j} - x_E) \geq 0$ for all $E$, so the $f$-value of the difference $(x_{D_j} - x_E) + x_n$ is at least $f(x_n) \geq 1$.
Why does the maximum of $f$ over $\mathcal{F}_j$ suffice? Because $f(x_{D_j}) \geq f(x_E)$ for all $E \in \mathcal{F}_j$ is exactly the condition ensuring that $D_j$ and $x_n$ "cooperate" in the $f$-direction. The key inequality is then:
\begin{align*}
\|x_{D_j \cup \{n\}} - x_E\| \geq f(x_{D_j \cup \{n\}} - x_E) = \underbrace{f(x_{D_j}) - f(x_E)}_{\geq\, 0} + \underbrace{f(x_n)}_{\geq\, 1} \geq 1.
\end{align*}
In $\mathbb{R}^d$ with the Euclidean norm, one could take $f(v) = \langle v, x_n / \|x_n\| \rangle$ (the inner product with the unit vector in the direction of $x_n$). The Hahn--Banach theorem is the tool that extends this idea to arbitrary normed spaces, including infinite-dimensional Banach spaces.
[/guided]
[/step]
[step:Conclude the induction and derive the bound]
By the inductive hypothesis, $\mathcal{P}([n-1])$ admits a symmetric decomposition into $t = \binom{n-1}{\lfloor (n-1)/2 \rfloor}$ sparse families. The splitting construction produces a partition of $\mathcal{P}([n])$ into sparse families with the same profile as the symmetric chain decomposition of $\mathcal{P}([n])$, giving a total of $\binom{n}{\lfloor n/2 \rfloor}$ sparse families.
Since each sparse family contributes at most one element to $\mathcal{A}$:
\begin{align*}
|\mathcal{A}| \leq \binom{n}{\lfloor n/2 \rfloor}.
\end{align*}
[/step]
