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.
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.