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:
 * $\forall \tuple {x_1, x_2} \in \R^2: \map f {x_1, x_2} = x_1 + x_2$

Then $f$ is continuous.

Proof
First we note that:

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

Let $x = \tuple {x_1, x_2} \in \R^2$.

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

Then:

Thus it has been demonstrated that:


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

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.