A metric space $\ (S\ d)\ $ is said to be a 1-space $\ \Leftarrow:\Rightarrow\ \forall\_{x\ y\in S}\ (x\ne y\ \Rightarrow\ d(x\ y)=1).$


**Question:**   Do there exist a non-negative integer $n,\ $ and an $n$-dimensional Banach space $\ X,\ $ and an isometric embedding of the
$\ (2^n+1)$-point metric 1-space $\ (S\ d)\ $ into $\ X$.



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> My *obvious* conjecture is NO.
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**Added AFTER @Fedor Perov's Answer:**


The power of the surprisingly simple Fedor's solution is in involving the measure theory. It worked regardless of any choice of any Banach norm.


In fact, now Fedor may follow with writing down a solution of my old conjecture from 1972/3 which a priori was harder:


CONJECTURE:   Let Banach norm $\ ||.||\ $ in $\ \mathbb R^n\ $ be such that $\ B^n:=(\mathbb R^n\,\ ||.||)\ $ contains a $2^n$-point 1-space $\ S.\ $ Then $\ B^n\ $ is a metrically injective space, i.e. it admits affine coordinates such that the norm is given by $\ \max.\ $ Then--furthermore--under such coordinates, there exists $\ (a\_1\ \ldots\ a\_n)\in \mathbb R^n\ $ such that


$$ S\ =\ \prod\_{k=1}^n\ \{a\_k\,\ a\_k\!+\!1\} $$


Indeed, Fedor's construction provides a decomposition of the convex hull $\ Cv(S)\ $ into $\ 2^n\ $ copies of the $\frac 12$-scaled of $\ Cv(S)\ $ where these copies have disjoint interiors.



Embedding the $\ 2^n+1$-point metric 1-space
Indeed no. If it exists, for each of our points $p\_i$ consider the set $A\_i$ homothetic to their convex hull with center $p\_i$ and coefficient bit less than $1 /2$. By volume argument some two such sets $A\_i, A\_j$ must have a common point $q$, but this contradicts to triangle inequality as $p\_iq, p\_jq$ are less than $1/2$.


