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 $\R^2$:


 * $\displaystyle d_\infty \left({x, y}\right):= \max \left\{ {\left\vert{x_1 - y_1}\right\vert, \left\vert{x_2 - y_2}\right\vert}\right\}$

where $x = \left({x_1, x_2}\right), y = \left({y_1, y_2}\right) \in \R^2$.

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

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

Proof
Let $a = \left({a_1, a_2}\right)$.

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

where $d$ is the usual topology.

From Open Ball in Real Number Line is Open Interval:
 * $B_\epsilon \left({a_1; d}\right) \times B_\epsilon \left({a_2; d}\right) = \left({a_1 - \epsilon \,.\,.\, a_1 + \epsilon}\right) \times \left({a_2 - \epsilon \,.\,.\, a_2 + \epsilon}\right)$

That is:
 * $x \in B_\epsilon \left({a; d_\infty}\right) \iff \left({a_2 - \epsilon < x_2 < a_2 + \epsilon}\right) \land \left({a_2 - \epsilon < x_2 < a_2 + \epsilon}\right)$

from which the result follows.