Addition of Coordinates on Cartesian Plane under Chebyshev Distance is Continuous Function

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

Let $d_\infty$ be the Chebyshev distance on $\R^2$.

Let $f: \R^2 \to \R$ be the real-valued function defined as:
 * $\forall \left({x_1, x_2}\right) \in \R^2: f \left({x_1, x_2}\right) = x_1 + x_2$

Then $f$ is continuous.

Proof
First we note that:

Let $\epsilon \in \R_{>0}$.

Let $x = \left({x_1, x_2}\right) \in \R^2$.

Let $\delta = \dfrac \epsilon 2$.

Then:

Thus it has been demonstrated that:


 * $\forall \epsilon \in \R_{>0}: \exists \delta \in \R_{>0}: \forall y \in \R^2: d_\infty \left({x, y}\right) < \delta \implies d \left({f \left({x}\right), f \left({y}\right)}\right) < \epsilon$

Hence by definition of continuity at a point, $f$ is continuous at $x$.

As $x$ was chosen arbitrarily, it follows that $f$ is continuous for all $x \in \R^2$.

The result follows by definition of continuous mapping.