Open Ball in Real Number Plane under Chebyshev Distance

Theorem
Let $\R^2$ be the real number plane.

Let $d_\infty: \R^2 \times \R^2 \to \R$ be the Chebyshev Distance on $\R^2$:


 * $\ds \map {d_\infty} {x, y} := \max \set {\size {x_1 - y_1}, \size {x_2 - y_2} }$

where $x = \tuple {x_1, x_2}, y = \tuple {y_1, y_2} \in \R^2$.

For $a \in \R^2$, let $\map {B_\epsilon} a$ be the open $\epsilon$-ball at $a$.

Then $\map {B_\epsilon} a$ is the interior of the square centered at $a$ and whose sides are of length $2 \epsilon$ parallel to the coordinate axes.

Proof
Let $a = \tuple {a_1, a_2}$.

From Open Ball in Cartesian Product under Chebyshev Distance:
 * $\map {B_\epsilon} {a; d_\infty} = \map {B_\epsilon} {a_1; d} \times \map {B_\epsilon} {a_2; d}$

where $d$ is the usual (Euclidean) topology.

From Open Ball in Real Number Line is Open Interval:
 * $\map {B_\epsilon} {a_1; d} \times \map {B_\epsilon} {a_2; d} = \openint {a_1 - \epsilon} {a_1 + \epsilon} \times \openint {a_2 - \epsilon} {a_2 + \epsilon}$

That is:
 * $x \in \map {B_\epsilon} {a; d_\infty} \iff \paren {a_2 - \epsilon < x_2 < a_2 + \epsilon} \land \paren {a_2 - \epsilon < x_2 < a_2 + \epsilon}$

from which the result follows.