Equivalence of Euclidean metric and the metric $d(x,y) = min{1, |x-y|}$ on $\mathbb R^{2}$

Question

I'm trying to prove the following example given in my lecture notes:

Example: On $\mathbb R^{2}$ , Euclidean metric and the metric $d(x,y) = \min(1, |x-y|)$ are topologically equivalent.

Initial thoughts: I know I have to show that there are $0<c,C<\infty$ such that $$cd_1(x,y) \le d_2(x,y) \le Cd_1(x,y) $$

case i) $|x-y| \le 1$ which implies $d_2(x,y) = 1$ \case ii) $|x-y| \gt 1$ which implies $d_2(x,y) = |x-y|$

I tried using the fact that both are basically calculating distance and the following to find $c$ and $C$.

$$\sqrt[]{(p-q)^2} = |p-q|$$

I can't seem to work out a solution. Alternatively, I could show that all $d_1$ and $d_2$ open sets coincide but I have no idea how to go about that. Any help appreciated! Sorry about the messy formatting, I'm not used to MathJax.

Answer 1

HINT.- Prove that the identity function is continuous from $E_1=(\mathbb R^2,d_1)$ to $E_2=(\mathbb R^2,d_2)$ and reciprocally (this is not difficult). So you have proved that every open set of $E_2$ is an open set of $E_1$ and viceversa i. e. both topologies are equivalent.