Maximum $n$-dimensional volume of set satisfying $(\forall x) \: (\forall y) \: (\| x - y \| \le 1)$ Let $S \subseteq \mathbb{R}^n$ be a set satisfying the property:
$$P(S) \equiv (\forall x \in S) \: (\forall y \in S) \: (\| x - y \| \le 1) $$
where $\| \cdot \|$ is the Euclidean norm.
Let $\mathcal{S}$ be a set of all maximal sets $S$ (maximal by set inclusion) s.t. $S$ satisfies the property $P(S)$. Then what is the value of $V = \max_{T \in \mathcal{S}} \operatorname{vol}(T) $ where $\operatorname{vol}(\cdot)$ represents the $n$-dimensional volume of a set.
I can't figure out the shape of the set for even $n=2$, is it a circle with radius $\frac{1}{2}$, or a square with side $\frac{1}{\sqrt{2}}$, I can't tell.
 A: Unless I am making a stupid mistake, this seems to follow from the isodiametric inequality.
A closed ball $B$ of diameter one is a maximal set, since any set $S$ that
strictly contains $B$ would have diameter bigger than one.
Now, the isodiametric inequality says that if $E\subseteq\mathbb{R}^{n}$ is a
Lebesgue measurable set, then
$$
\operatorname*{vol}\left(  E\right)  \leq\alpha_{n}\left(  \frac
{\text{$\operatorname*{diam}$}E}{2}\right)  ^{n},
$$
where $\alpha_{n}$ is the volume of the ball of radius 1. A proof of this
inequality can be found in the book of Evans and Gariepy and uses Steiner
symmetrization. There is another proof that uses the Brunn-Minkowski
inequality. I can add this proof if you want, I have it typed up.
Now, if $S$ is any other maximal set, then $\operatorname*{diam}S\leq1$, and
so
$$
\operatorname*{vol}\left(  S\right)  \leq\alpha_{n}\left(  \frac
{\text{$\operatorname*{diam}S$}}{2}\right)  ^{n}\leq\alpha_{n}\left(  \frac
{1}{2}\right)  ^{n}=\operatorname*{vol}\left(  B\right)  .
$$
Hence, the maximum volume is $\alpha_{n}\left(  \frac{1}{2}\right)  ^{n}$ and
it is realized by $B$.

Remark: Fix $\theta\in\left(  0,1\right)  $. By
replacing $E$ with $\theta E$ and $F$ with $\left(  1-\theta\right)  F$ in the Brunn-Minkowski inequality and
using the $n$-homogeneity of the Lebesgue measure we obtain that
\begin{align*}
\theta\left(  \operatorname*{vol}\left(  E\right)  \right)  ^{\frac{1}{n}
}+\left(  1-\theta\right)  \left(  \operatorname*{vol}\left(  F\right)
\right)  ^{\frac{1}{n}} &  =\left(  \operatorname*{vol}\left(  \theta
E\right)  \right)  ^{\frac{1}{n}}+\left(  \operatorname*{vol}\left(  \left(
1-\theta\right)  F\right)  \right)  ^{\frac{1}{n}}\\
&  \leq\left(  \operatorname*{vol}\left(  \theta E+\left(  1-\theta\right)
F\right)  \right)  ^{\frac{1}{n}}.
\end{align*}
Thus the function $f\left(  t\right)  :=\left(  \operatorname*{vol}\left(
tE+\left(  1-t\right)  F\right)  \right)  ^{`\frac{1}{n}}$ is concave in
$\left[  0,1\right]  $.
Proof of the isodiametric inequality.
It is enough to prove the isodiametric inequality for bounded sets, since
otherwise the right-hand side is infinite. If $\lambda>0$, we have that
$\operatorname*{vol}\left(  \lambda E\right)  =\lambda^{n}\operatorname*{vol}
\left(  E\right)  $ and $\left(  \operatorname*{diam}\left(  \lambda E\right)
\right)  ^{n}=\lambda^{n}\operatorname*{diam}$$E$, so without loss of
generality we may assume that $\operatorname*{diam}$$E=1$.
Also, since $\operatorname*{diam}\left(  E\right)  =\operatorname*{diam}%
\left(  \overline{E}\right)  $, we can replace $E$ with $\overline{E}$ and so
we can assume that $E$ is compact. Let
$$
F:=\left\{  -\boldsymbol{x}:\,\boldsymbol{x}\in E\right\}  .
$$
Then $F$ is compact,  $E+F$ is compact, and so it is Lebesgue measurable. By
the previous remark the function $f\left(  t\right)  :=\left(  \mathcal{L}%
^{n}\left(  tE+\left(  1-t\right)  F\right)  \right)  ^{\frac{1}{n}}$ is
concave in $\left[  0,1\right]  $, and so
$$
\frac{1}{2}\left(  \operatorname*{vol}\left(  E\right)  \right)  ^{\frac{1}
{n}}+\frac{1}{2}\left(  \operatorname*{vol}\left(  F\right)  \right)
^{\frac{1}{n}}\leq\left(  \operatorname*{vol}\left(  \frac{1}{2}E+\frac{1}
{2}F\right)  \right)  ^{\frac{1}{n}}.
$$
But $\operatorname*{vol}\left(  F\right)  =\operatorname*{vol}\left(
E\right)  $, and so the previous inequality becomes
$$
\operatorname*{vol}\left(  E\right)  =\operatorname*{vol}\left(  F\right)
\leq\operatorname*{vol}\left(  \frac{1}{2}E+\frac{1}{2}F\right)  .
$$
If $\boldsymbol{x}\in\frac{1}{2}E+\frac{1}{2}F$, then $\boldsymbol{x}
=\frac{\boldsymbol{x}^{\prime}-\boldsymbol{x}^{\prime\prime}}{2}$, where
$\boldsymbol{x}^{\prime},\boldsymbol{x}^{\prime\prime}\in E$, and so
$$
\Vert\boldsymbol{x}\Vert=\frac{1}{2}\Vert\boldsymbol{x}^{\prime}%
-\boldsymbol{x}^{\prime\prime}\Vert\leq\frac{1}{2},
$$
which shows that $\frac{1}{2}E+\frac{1}{2}F\subseteq\overline{B\left(
0,\frac{1}{2}\right)  }$. Hence,
$$
\operatorname*{vol}\left(  E\right)  \leq\operatorname*{vol}\left(  \frac
{1}{2}E+\frac{1}{2}F\right)  \leq\operatorname*{vol}\left(  \overline{B\left(
0,\frac{1}{2}\right)  }\right)  =\frac{\alpha_{n}}{2^{n}}.
$$
A: Let $S \subseteq \mathbb{R}^n$ satisfy:
$P(S) \equiv (\forall x \in S) \: (\forall y \in S) \: (\| x - y \|_2 \le 1) $. Define the notation $B_n(r,p)=$ the $n$-dimensional hyperball with radius $r$, centre $p$.
Let $M=\{S\subseteq\mathbb{R}^n: [P(S)=\top] \wedge [\nexists S'\in\mathbb{R}^n \text{ s.t. } (P(S')=\top) \wedge (S'\supsetneq S)]\}$. Claim: $V := \max_{T \in M} \operatorname{vol}(T) =\operatorname{vol}(B_n(1/2, 0))$.
Note: This problem seems intuitively easy but deceptively hard. I attach my wrong proof in the hope that someone can fix it. The $\color{red}{red}$ implication doesn't hold, so the proof is invalid.
$\color{red}{Wrong}$ Proof: Note that it is sufficient to prove that $M=A$, where we define $A:=\{B_n(1/2,p): p\in\mathbb{R}^n\}$. It is obvious that $A\subseteq M$. Now we try to prove $M\subseteq A$:
Given $S\in M$, so that $P(S)=\top$ and $S$ is maximal. Note that $P(S)=\top\color{red}{\nRightarrow} \exists$ $p$ such that there is a circumscribed $B_n(1/2,p)\supseteq S$. But maximality implies that there must be equality, so $S=B_n(1/2,p)$. Thus, $S\in A$. $\blacksquare$
