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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.

roblox99
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    HINT It suffices to show that inside any euclidean ball, you always have a maximum ball. Inn fact, it suffices to show that inside any euclidean ball centered at the origin, there is always a maximum ball contained in it. – Tito Eliatron Jun 15 '21 at 16:49
  • But wouldn't that only show that any open set with respect to the euclidean metric is also open with respect to the maximum metric? What about the other way around? – roblox99 Jun 15 '21 at 17:02
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    The other way is also true. Recall that a maximum ball is a square with sides parallel to axis. So, inside a circle (an euclidean ball) you always can INSCRIBE such a square and inside such a squere you always can inscribe a circle. MOREOVER, the center of all squares and circles is THE SAME. – Tito Eliatron Jun 15 '21 at 17:05
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    Ok, so if I want to prove the equivalence in the "$\Rightarrow$" direction, so $O$ is open with respect to the euclidean metric, I just take a point $x \in O$ and since such an $\epsilon$ exists, I can always find a maximum ball or square inside a circle with radius $\epsilon$. Specifically a square with side length $\sqrt(2)*\epsilon$ centered at the center of the circle. – roblox99 Jun 15 '21 at 17:12

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

SolubleFish
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  • Thank you. I think this is similar, if not the same conclusion, I arrived at with the help of @Tito Eliatron, where we use the fact, that one can always fit a square inside a circle and the other way around. – roblox99 Jun 15 '21 at 17:17
  • This is just a different way to say the same thing – SolubleFish Jun 15 '21 at 17:19