If x1,...,xn are points of a metric s., must there exist a normed s. (E, ||.||) and $y_1,...,y_n\in E$:$d(x_i,x_j )=||y_i-y_j||$ with all i,j=1,...,n?

Asked Nov 13 '23 at 13:01

Active Nov 13 '23 at 13:31

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If x1, x2,...,xn are points of some metric space, does there necessarily exist such a normed space (E, ||.||) and $y_1,y_2,...,y_n\in E$ that $d(x_i,x_j )=||y_i-y_j||$ with all $i,j=1,2,...n$?

I am thinking that there must be such a normed space however I am having trouble proving it, if anyone can help with that I'd be very grateful.

Answer to the suggested question: I'm afraid I do not see how the suggested question/answer is supposed to help me with this problem as it does not address anything with normed spaces or the specific $d(x_i,x_j )=||y_i-y_j||$ situation, but maybe I'm overlooking something.

edited Nov 13 '23 at 13:31

asked Nov 13 '23 at 13:01

Proloffc6

$(\mathbb R^n,|\cdot|_\infty)$ is a normed space. The isometry in the linked answer gives the correspondence $x_i\mapsto y_i$. – M W Nov 15 '23 at 00:33

If x1,...,xn are points of a metric s., must there exist a normed s. (E, ||.||) and $y_1,...,y_n\in E$:$d(x_i,x_j )=||y_i-y_j||$ with all i,j=1,...,n?

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