$R^2$ is not isometric to $R^3$ Is there a direct proof for showing that $R^2$ is not isometric to $R^3$ (with the usual metrics)? I know that they are not homeomorohic but I think there should be some direct and easy proof for showing that they are not isometric.
 A: In $\mathbb R^2$ there does not exist a configuration of $4$ points which are all at distance $1$ from each other.  This can be argued by elementary geometry: once we fix two such points, there are only two locations at which one can place another point to form an equilateral triangle, but the distance between those locations is not $1$.
On the other hand, in $\mathbb R^3$ this can be achieved with the vertices of a regular tetrahedron, so the two spaces are not isometric.
(This argument is reminiscent of a classic puzzle: how can you arrange 6 identical matchsticks to form 4 equilateral triangles?)
A: Suppose there exists an isometry $T:\mathbb R^3 \to\mathbb R^2$. Then $$T(\mathbf x)=\binom{\mathbf u}{\mathbf v}\mathbf x+\mathbf b\,,\text{ where } \,\mathbf{u,v,x}\in\mathbb R^3,\mathbf u\cdot\mathbf v=0,\text{ and }\, \mathbf b\in \mathbb R^2.$$Now notice that $|T(\mathbf u \times \mathbf v)-T(\mathbf 0)|=0\neq|(\mathbf u\times \mathbf v)-\mathbf 0|$, so $T$ doesn't preserve the distance between $(\bf u\,\times\,\bf v) $ and the origin. This a contradiction.
