Describe and sketch the open balls B((0,0), r) of radii r = 0.5 and r = 2 centred at (0,0)

Question

This is a question on a previous exam which we have got for revision, and whilst we have the answer to the question, I am unable to picture it in my head, I wonder if anybody could help me.

Consider the product metric given by $d((x_1,y_1),(x_2,y_2))=max\left \{ d_X(x_1,x_2),d_Y(y_1,y_2)\right \}$, in the situation where X=$\mathbb{R}$=Y, so that $X \times Y = \mathbb{R}^2$. Describe and sketch the open balls $B((0,0),r)$ of radii $r=\frac{1}{2}$ and $r=2$ centered at $(0,0)$ when $d_X$ and $d_Y$ are given by the discrete metric.

Answer 1

\begin{align} B((0,0),1/2)&=\{(x,y)\in\mathbb{R}^2:\max(d_X(0,x),d_Y(0,y))<1/2\}\\ &=\{(x,y)\in\mathbb{R}^2:d_X(0,x)<1/2\text{ and }d_Y(0,y)<1/2\}. \end{align} But $d_X=d_Y$ is the discrete metric, hence $$ d_X(0,x)=\begin{cases}0,&&0=x,\\1,&&0\neq x\end{cases} $$ and similarly for $d_Y(0,y)$. So the conditions $d_X(0,x)<1/2$ and $d_Y(0,y)<1/2$ will both be satisfied if and only if $$ d_X(0,x)=0\text{ and }d_Y(0,y)=0, $$ which happens if and only if $x=0$ and $y=0$. We conclude that $$ B((0,0),1/2)=\{(0,0)\}. $$

Also, $$ B((0,0),2)=\{(x,y)\in\mathbb{R}^2:d_X(0,x)<2\text{ and }d_Y(0,y)<2\}. $$ But we always have $d_X(0,x),d_Y(0,y)\in\{0,1\}$ hence it is always the case that $d_X(0,x)<2$ and $d_Y(0,y)<2$. We conclude that $$ B((0,0),2)=\mathbb{R^2}. $$