$\mathbb R$ is not isometric with $\mathbb R^2$ show that $\mathbb R$ is not isometric with $\mathbb R^2$ (with the usual metrics). I want to use the first definition of continuity (i.e. the $\epsilon $ -$\delta$ stuff) but I don't see a way to proceed. I guess the contrapositive? Thanks
 A: An isometry between $\mathbb R$ and $\mathbb R^2$ would imply a contradiction. Consider that any one point separates $\mathbb R$ , but does not separate/disconnect $\mathbb R^2$. If h:$\mathbb R \rightarrow \mathbb R^2$ were a homeomorphism between the two, then, for any $x$ in $\mathbb R$ , h':$\mathbb R-{x}\rightarrow \mathbb R^2-h(x)$ would also be a homeomorphism. But this is not possible, since $\mathbb R-{x}$ is disconnected , but $\mathbb R^2-h(x)$ is not. 
Maybe to be more precise, if there was a continuous bijection h (an isometry) between $\mathbb R$ and $\mathbb R^2$ , then the following contradiction would result:
Consider the restriction h' of h to $[-1,1]$ . By compactness of [-1,1], and by $\mathbb R^2$ being Hausdorff (so that its subspace h'([-1,1]) is Hausdorff ), we have a continuous bijection between compact and Hausdorff, so that $h'([-1,1])\rightarrow h'([-1,1])$ is a homeomorphism. By connectedness of [-1,1] and since h' is injective (and h is a continuous bijection into $\mathbb R^2$), the image contains an open ball. Now , since h' is a homeomorphism, it sends its interior (-1,1) into the interior of the image, which is connected , but we have the issue of the cutsets.
A: In $\mathbb R$ there do not exist three distinct points such that the distance between each pair is equal to $1$.
On the other hand, in $\mathbb R^2$ there do exist equilateral triangles.
A: For a general argument, one can say that Hausdorff dimension is an isometric invariant and that $\dim_H(\mathbb{R}^n)=n$. Therefore, $\mathbb{R}^n$ and $\mathbb{R}^m$ are isometric iff $m=n$.
(Of course, here Mariano Suárez-Alvarez's answer is really better.)
