To deduce that neither of the metric spaces $(\mathbb{R}^n, d_1)$ and $(\mathbb{R}^n, d_{\infty})$ is isometric to $(\mathbb{R}^n, d_2)$.

Question

I have a question that asks me to deduce the fact that neither of the metric spaces $(\mathbb{R}^n, d_1)$ and $(\mathbb{R}^n, d_{\infty})$ is isometric to $(\mathbb{R}^n, d_2)$ from the following fact that I was able to prove:

It is possible for the intersection of two closed balls with respect to the metrics $d_1$ and $d_{\infty}$ (such that none of the balls contains the other) to be again a closed ball of radius $> 0$ with respect to the same metric. For example, in $(\mathbb{R}^n, d_{\infty})$ we have $B((1,\cdots,1),2) \cap B((-1,\cdots,-1),2) = B((0,\cdots,0),1)$; and in $(\mathbb{R}^n, d_1)$, we have $B((1,0,\cdots,0),2) \cap B((-1,0,\cdots,0),2) = B(0,1)$

I keep wracking my brains and I don't understand how to use this fact to prove that neither of the metric spaces $(\mathbb{R}^n, d_1)$ and $(\mathbb{R}^n, d_{\infty})$ is isometric to $(\mathbb{R}^n, d_2)$. Can I get some help?

Answer 1

Let $f:X \to Y$ be an isometry (onto Y) where $(X,d),(Y,D)$ are metric spaces. Consider a closed ball in X: $\{x \in X: d(x,x_0) \leq r\}$. What is the image of this ball under f? It is exactly $\{y \in Y: D(y,f(x_0)) \leq r\}$. Can you verify this by showing that each side is contained in the other? Well, once you know this fact you complete the argument easily: choose closed balls $B_1,B_2$ in $(\mathbb R^{2},d_1)$ whose intersection is a closed ball and niether is contained in the other. . Let the images of these balls in $(\mathbb R^{2},d_2)$ be $B_3,B_4$. Then $B_3 \cap B_4$ is the image of $B_1 \cap B_2$ and hence it is a closed ball. But the intersection of two closed balls in $(\mathbb R^{2},d_2)$ can never be a closed ball unless one of then is contained in the other. This is a contradiction. We have used the fact that the image of an intersection is the intersection of the images. This fact is true for one-to-one functions and since an isometry is one-to-one this is fine. The contradiction we obtained shows that $(\mathbb R^{2},d_1)$ is not isometric to $(\mathbb R^{2},d_2)$. The proof of $(\mathbb R^{2},d_{\infty})$ is similar.

Answer 2

If there were an isometry $({\mathbb R}^n,d_\infty)\to({\mathbb R}^n,d_2)$ then there would be two euclidean balls of radius $2$ centered at distance $2$ and intersecting in a euclidean ball of radius $1$. The latter is clearly not the case.