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

Question

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 $\mathbb{R}^2$.

Answer 1

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