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Why is the metric space $(X,d_{ij})$ where $d_{ij} = \sqrt{|i-j|}$ necessarily Euclidean?

I tried to use Cayley's criterion, meaning to try and prove that if we look at $X=\{1,...,n\}$ and define an $(n-1) \times (n-1)$ matrix over the elements $\{1,...,n-1\}$ as follows: $$M_{ij}=0.5 \times \left[{\sqrt{|i-n|}}^2 + {\sqrt{|j-n|}}^2 - {\sqrt{|i-j|}}^2 \right]$$

and then trying to prove that all $M_{ij}$ eigenvalues are non-negative, but I do not see a straightforward way of doing this.

David G. Stork
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TheNotMe
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  • What does $X$ denote? – mfl Jul 10 '17 at 19:39
  • A set of points on which the distance function $d_{ij}$ applies. – TheNotMe Jul 10 '17 at 19:39
  • You seem to have some unstated intention that $X$ is more than just a set of points, because subtraction $i-j$ is not defined on a bare naked set. Perhaps you intend that $X$ is a subset of $\mathbb{R}$? If so, please rewrite your question to be explicit. – Lee Mosher Jul 10 '17 at 19:58

1 Answers1

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I have successfully shown this.

Denote the metric $\sqrt{|i-j|}$ by $d(i,j)$ and the standard eucledian distance by $d_2(x,y)$. Assume that $|X|=n$ and map each $i \in X$ to a binary vector of length $n$ with a prefix of $i$ ones and a suffix of zeros. Then, according to the normal eucledian distance, we have that for $i,j\in X$, the corresponding vectors eucledian distance contains a summation of $|i-j|$ ones each squared, all under a square root, namely $\sqrt{|i-j|}$. We have shown that there exists a mapping $\varphi: X \to \mathbb{R^n}$ for all $i,j \in X$ such that $d(i,j) = d_2(\varphi(i),\varphi(j))$ and therefore the given metric space $(X,d)$ is eucledian.

TheNotMe
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