Let $(X,d)$ be a metric space, and suppose $S\subseteq X$ is a finite subset in which all pairwise distances are distinct (formal definition [here](https://mathoverflow.net/questions/312656/is-there-a-set-s-subseteq-0-1-with-s-2-aleph-0-and-distinct-pairwise)).


If $x\in S$ and $k$ is a non-negative integer with $k<|S|$, we define $S\_k(x)\subseteq S$ to be the set of the $k$ nearest members of $S$ to $x$, other than $x$ itself. We define a graph $G(k,S)$ by setting $V(G(k,S)) = S$, and $$E(G(k,S)) = \big\{\{x,y\}: x\neq y \in S \text{ and } y\in S\_k(x) \text{ and }x\in S\_k(y)\big\}.$$ So the points $x,y$ are connected by an edg if and only if $y$ belongs to the $k$ nearest points of $x$, and vice versa.


We say that a metric space $(X,d)$ is *graph-universal* if for every finite graph $G$ there is $S\subseteq X$ with pairwise distinct distances, and a non-negative integer $k<|S|$ such that $G\cong G(k,S)$. For example, $\mathbb{R}$ with the Euclidean distance is not graph-universal as we cannot find a set $S$ of $4$ real points and $k<4$ such that $C\_4 \cong G(k, S)$.


***Question.*** If $\{0,1\}^{<\omega}$ denotes the set of functions $f:\omega\to\{0,1\}$ that are eventually $0$, and endow it with the [Hamming distance](https://en.wikipedia.org/wiki/Hamming_distance) $d\_H$, is $(\{0,1\}^{<\omega}, d\_H)$ graph-universal?



Graphs represented by a subset of a metric space
It is not possible to embed any non-complete graph with a universal vertex into any metric space in this way. 


A universal vertex in an $n$-vertex graph has degree $n-1$ and thus $k = n-1$ in any representation of such a graph. But if $|S| = n$ and $k = n-1$, then $G(k,S)$ is a complete graph on $n$ vertices.


