# Show that $B_{\rho}((0,0),1)=\{(x,y)\in\mathbb{R}^2:\, \max\{|x|,|y|\}<1\}$

Consider $$(\mathbb{R}^2, \rho)$$ with the square metric $$\rho$$ given by $$\rho(x,y)=\max\{|x_1-y_1|,|x_2-y_2|\}$$, where $$x=(x_1,x_2)$$ and $$y=(y_1,y_2)$$ are in $$\mathbb{R}^2$$. Show that $$B_{\rho}((0,0),1)=\{(x,y)\in\mathbb{R}^2:\, \max\{|x|,|y|\}<1\}$$ Also draw this open ball on $$\mathbb{R}^2$$.

• It's just the definition :) – Azif00 Sep 15 at 19:08

$$B_{\rho}((0,0),1)=\{(x,y)\in\mathbb{R}^2:\, \max\{|x|,|y|\}<1\}$$ is by definition. And note that $$\max\{|x|,|y|\}<1$$ implies that $$|x|<1$$ and that $$|y|<1$$. That is $$B_{\rho}((0,0),1)=\{(x,y)\in\mathbb{R}^2:\, -1 Hope I'll help you.