Isometric?

Quote:

Originally Posted by Showcase_22

I think I have to take two points in $X$ and show that any mapping from $X$ to $\mathbb{E}^n$ would result in a different distance between them.

Here's my attempt:

Take a sphere of radius 1 and the points A and D.

Since $d(A,D)=2$ , A and D are on opposite sides of the sphere.

Suppose there exists an isometry $\phi:X \rightarrow \sigma$ where $\sigma$ is is the plane used in stereographic projection (also a subset of $\mathbb{E}^n$ ).

Then $\phi(A)=\phi(D)$ . Hence $d(\phi(A),\phi(D))=0$ which is not equal to $d(A,D)=2$ .

Hence $\phi$ cannot be an isometry.

I'm not sure that the hint is very helpful, and I don't think that stereographic projection is needed.

If such an isometry $\phi$ exists then the points $\phi (A),\ \phi (B),\ \phi (C)$ form an equilateral triangle with side 1, in some 2-dimensional subspace of $\mathbb{E}^n$ . So do the points $\phi (B),\ \phi (C),\ \phi (D)$ . Thus the images of the four points lie in some 3-dimensional subspace of $\mathbb{E}^n$ , so we may as well assume that n=3.

The points $\phi (A)$ and $\phi (D)$ must both lie at a distance $\sqrt3/2$ from the midpoint of the line segment joining $\phi (B)$ and $\phi (C)$ . So the distance from $\phi (A)$ to $\phi (D)$ is at most $\sqrt3<2$ , contradicting the condition that $\phi$ is an isometry.