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 $\ X$.

My obvious conjecture is NO.

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.

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    $\begingroup$ More generally, an $(n + 1)$-point ultrametric space (not just $1$-space) can be embedded into $n$-dimensional Euclidean space, but not into any smaller dimensional space. Some Googling turned up Fiedler - Ultrametric sets in Euclidean point spaces, which attributes this result to a 1985 paper of Lemin which I can't find online. $\endgroup$ – LSpice Apr 24 '18 at 22:04
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    $\begingroup$ (Oh, I see now that it is an $n$-dimensional Banach space, not necessarily $n$-dimensional Euclidean space, so I guess that's where the subtlety comes in.) $\endgroup$ – LSpice Apr 24 '18 at 22:29

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 $1/2$.

  • $\begingroup$ simple and great. Thank you. What do I click on to accept an Answer? $\endgroup$ – Wlod AA Apr 24 '18 at 5:53
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    $\begingroup$ Ok, I think I did, I klicked on a greenish V which became a solid V. $\endgroup$ – Wlod AA Apr 24 '18 at 5:55
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    $\begingroup$ When the points belong to an affine subspace (as unlikely as it is) then one may use the measure of that subspace; then the inequality works even with a greater force. $\endgroup$ – Wlod AA Apr 24 '18 at 6:04

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