Prove that a set $O \subseteq \mathbb{R}^2$ is open with respect to the euclidean metric if and only if $O$ is open with respect to the maximum metric

Question

I have recently started learning about metric spaces and since I'm having a hard time understanding some of the basics, I tried doing some exercises and I got stuck on this one:

Prove that a set $O \subseteq \mathbb{R}^2$ is open with respect to the euclidean metric if and only if $O$ is open with respect to the maximum metric.

As I understand it, a set $A \in \mathbb{R}^2$ is defined to be open with respect to the euclidean metric, if $\forall x \in A: \exists \epsilon > 0: B_\epsilon(x) \subset A$, where $B_\epsilon(x) = \{y \in \mathbb{R}^2 | d_2(x, y) < \epsilon\}$ and $d_2(x, y) = (|x_1 – y_1|^2 + |x_2 – y_2|^2)^{\frac{1}{2}}$. Analogous with the maximum metric $d_\infty(x, y) = max\{ |x_1 – y_1|, |x_2 – y_2| \}$

My problem now is that I haven't yet had much experience in proving sets to be open and I don't know how to solve or tackle this issue with the maximum metric.

Answer 1

Hint : use that $$\forall x,y\in \mathbb R^2, \frac{1}{\sqrt 2}d_\infty(x,y) \leq d_2(x,y)\leq d_\infty(x,y)$$