Let $X$ be a finite metric space, then is it true that $\exists n \in \mathbb N$ such that there exists an isometry from $X$ into $\mathbb R^n$, where $\mathbb R^n$ is equipped with the supremum metric?

  • $\begingroup$ Does finite metric space means a metric space with finite $X$? $\endgroup$ – Hanul Jeon Dec 31 '16 at 9:10
  • $\begingroup$ @HanulJeon : Yes , precisely , $X$ is finite $\endgroup$ – user228168 Dec 31 '16 at 10:07

Yes, if $X$ has cardinality $n$ there is a simple embedding into $(\mathbb R^n,\|\cdot\|_\infty)$: denote by $x_1,\ldots, x_n$ the points in $X$, and define $f:X\to \mathbb R^n$ as $$\left(f(x_i)\right)_j:=d(x_i,x_j)$$ where the outer $j$ denotes the coordinate in $\mathbb R^n$. It is an exercise to check that this is an isometry. In fact there is an embedding in $\mathbb R^{n-1}$ by just translating one of the points to the origin.

EDIT I'll add for whomever could be interested a nice generalization which I didn't even believe the first time I heard it: every separable metric space $X$ embeds isometrically in $\ell_\infty$.

Fix a base point $z\in X$ and a countable dense set $\{x_i\}_{i\in \mathbb N}$. Then the embedding is given by $$x\mapsto (d(x,x_i)-d(z,x_i))_{i\in \mathbb N}.$$ Just observe that it is continuous and it is an isometry when restricted to the dense set $\{x_i\}_{i\in \mathbb N}$.


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