Any number of dimensions actually :) For that example it suffices to show it in 3D since any euclidean embedding of n+1 points can be isometrically embedded in

Any number of dimensions actually :) For that example it suffices to show it in 3D since any euclidean embedding of n+1 points can be isometrically embedded in an n-dimensional space, so if a 3D embedding with error E doesn't exist then neither does an ND embedding for any N>3. Finding the minimum error requires a tad more effort, but it's not too bad to show that no embedding has 0 error: Take a cycle a->b->c->d->a where every edge has length 1. Suppose a 0-error embedding exists. Points (a), (b), and (c) must be embedded colinearly, and then the only possible location for point (d) satisfying the distance requirements |a-d|=|c-d|=1 is precisely wherever we placed point (b), but then point (d) can't possibly have distance 2 from point (b). By itself that doesn't show that infinitely small errors are impossible, but that assertion is also true in practice.