Proving that $R^2$ with Euclidean metric is isometric with $R^2$ with maximum metric?

I am reading a geometry book on my own and can't figure out how to prove it. I cannot figure out a transformation that preserves the distances for ALL points.

The Euclidean metric is

$$d_1(A,B) = \sqrt{ (X_B - X_A)^2 +(Y_B - Y_A)^2}$$

The maximum metric is

$$d_2(A,B) = max \{ |X_B - X_A|, |Y_B - Y_A|\}$$

My solution so far is to project onto the $x$ or $y$ axis, but this doesn't help in case the other pair of point differ in a greater way in the other coordinate. I'm not able prove this for every set of points. Please help.

• I think, you may want to show the topologies induced by these metrics are the same. ? – Chival Jul 23 '15 at 6:23
• Don't know topology.It appears that Robert might be right. It might be a problem where the proof is impossible. – Haresh Jul 23 '15 at 6:28

There is no such transformation. For example, in the $d_2$ metric you can have four points all with distance $1$ from each other, e.g. $(0,0), (0,1), (1,0), (1,1)$, but in $d_1$ you can't.