Determine an open ball in $\mathbb R^{2}$ 
Question:
  For $x \in \mathbb{R}^{2}$ and $\delta >0$, what is the open ball $B_{\delta }\left ( x \right )$
  where $d\left ( x,y \right )=max\left \{ \left | x_{1}-y_{1} \right |,\left | x_{2}-y_{2} \right | \right \}?$

From the definition of open ball,
we require $B_{\delta }\left ( x \right )=\left \{ \bar{x} \in\mathbb{R}^{2} \mid d\left ( x,\bar{x} \right )<\delta  \right \}$.
Without loss of generality, let $d\left ( x,y \right )=max\left \{ \left | x_{1}-y_{1} \right |,\left | x_{2}-y_{2} \right | \right \}=
\left | x_{1}-y_{1} \right |<\delta$ .
I am unable to progress further. 
Hints are appreciated.
Thanks in advance.
 A: *

*To understand the ball geometry, set the center to the origin and radius to one, that is consider $B_1(0)=\{ x \in\mathbb{R}^{2} \mid \max(|x_1|,|x_2|)<1\}$.

*Draw the set on the plane $\max(|x_1|,|x_2|)<1$ $\Leftrightarrow$ $\{|x_1|<1, |x_2|<1\}$.

*Translate and scale to get all other balls.

A: You are working in a non-euclidean metric $d(x,y) := \max(|x_2-x_1|,|y_2-y_1|) $.  It is a simple exercise to show that $d$ is a metric. Now divide your $\mathbb R^2$ onto $4$ connected regions -- where $|y_2-y_1| \leqslant|x_2-x_1| $ and $|y_2-y_1| \geqslant|x_2-x_1|$. The ball in this metric would be a square with angles on the lines $y = x$ or $y=-x$.
A: Try to translate the original problem into everyday English and proceed. The desired ball $B_{\delta}(x)$ by definition collects every point $y \in \mathbb{R}^{2}$ such that $d(x,y) < \delta$. The metric $d$ by assumption gives the bigger one of (1) the Euclidean distance between the first coordinate of $x$ and that of $y$ and (2) the Euclidean distance between the second coordinate of $x$ and that of $y$. Now can you see it? If a point $y$ is such that $\delta > d(x,y) = |y_{1} - x_{1}|\ (\text{resp.}\ |y_{2} - x_{2}|) \geq |y_{2}-x_{2}|\ (\text{resp.}\ |y_{1}-x_{1}|),$ then $y \in B_{\delta}(x)$ by definition; the converse is also true by definition. This means heuristically that a point lies in $B_{\delta}(x)$ iff it cannot be "too far" from $x$. Precisely, the ball $B_{\delta}(x)$ must be an open square (a square with its boundary removed) of side length $\delta$ centered at $x$. More precisely, the ball $B_{\delta}(x)$ is the open square with vertices $(x_{1}-\delta, x_{2}-\delta)$, $(x_{1}-\delta, x_{2}+\delta)$, $(x_{1} + \delta, x_{2}-\delta)$, $(x_{1}+\delta, x_{2} + \delta)$.
