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I need to show that $(a,b)\times (c,d)$ is an open set in $\mathbb{R}^2$ with the Euclidian metric. I know that a set $U$ is open if for $x\in U$ there exists an open ball $B_\epsilon(x)$ such that $B_\epsilon(x)\subset U$ for some $\epsilon >0$. The euclidian metric in $\mathbb{R}^2$ is given by $d_2(x,y) = \sqrt{(x_1 - y_1)^2 + (x_2 - y_2)^2}$, so I need to show that for $x\in (a,b)\times (c,d)$ there exists $B_\epsilon(x) = \{y\in\mathbb{R}^2:\sqrt{(x_1 - y_1)^2 + (x_2 - y_2)^2} < \epsilon\}$ such that $B_\epsilon(x)\subset (a,b)\times (c,d)$ for some $\epsilon >0$, but I have no idea how!

Question: How do I show that $(a,b)\times (c,d)$ is an open set in $\mathbb{R}^2$ with the Euclidian metric?

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$x=(x_1,x_2)$ with $a<x_1<b$ and $c<x_2<d$. Let $r=\min(x_1-a,b-x_1,x_2-c,d-x_2)$. This is the minimum distance from $(x_1,x_2)$ to the edges of the rectangle $(a,b)\times(c,d)$. Surely you can prove the open disc with centre $(x_1,x_2)$ and radius $r$ is contained within the open rectangle?

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  • $\begingroup$ I'm surely not, otherwise I wouldn't have asked! I understand that the disk with radius $r$ around with center $x = (x_1,x_2)$ lies in the open rectangle. I don't know how to write it out mathematically. Should I show that if $x\in B_r((x_1,x_2))$ then $x\in (a,b)\times (c,d)$? $\endgroup$ – Mr. President Feb 19 '18 at 9:23
  • $\begingroup$ More specifically, is it sufficient to say: ... if we now consider $B_r(x) = \{y\in\mathbb{R}^2:d_2(x,y) < r\}$, we know that $B_r(x)\subset (a,b)\times (c,d)$? I understand that it's true, but I can't seem to find the final conclusive statement that proves it. $\endgroup$ – Mr. President Feb 19 '18 at 9:33
  • $\begingroup$ Yes, you need to prove that $x\in B_r((x_1,x_2))$ implies $(x_1,x_2)\in(a,b)\times(c,d)$. Can you draw a picture to illustrate that? Does that picture suggest a strategy of proof? $\endgroup$ – Lord Shark the Unknown Feb 19 '18 at 19:17

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