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

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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:

\(\ds \size {\paren {x_1 + x_2} - \paren {y_1 + y_2} }\) \(=\) \(\ds \size {\paren {x_1 - y_1} + \paren {x_2 - y_2} }\)
\(\ds \) \(\le\) \(\ds \size {x_1 - y_1} + \size {x_2 - y_2}\) Triangle Inequality for Real Numbers
\(\text {(1)}: \quad\) \(\ds \) \(\le\) \(\ds 2 \max \set {\size {x_1 - y_1}, \size {x_2 - y_2} }\)


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

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

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

Then:

\(\ds \forall y = \tuple {y_1, y_2} \in \R^2: \, \) \(\ds \map {d_\infty} {x, y}\) \(<\) \(\ds \delta\)
\(\ds \leadsto \ \ \) \(\ds \max \set {\size {x_1 - y_1}, \size {x_2 - y_2} }\) \(<\) \(\ds \delta\) Definition of Chebyshev Distance on Real Number Plane
\(\ds \leadsto \ \ \) \(\ds 2 \max \set {\size {x_1 - y_1}, \size {x_2 - y_2} }\) \(<\) \(\ds \epsilon\) Definition of $\epsilon$
\(\ds \leadsto \ \ \) \(\ds \size {\paren {x_1 + x_2} - \paren {y_1 + y_2} }\) \(<\) \(\ds \epsilon\) from $(1)$
\(\ds \leadsto \ \ \) \(\ds \size {\map f x - \map f y}\) \(<\) \(\ds \epsilon\) Definition of $f$
\(\ds \leadsto \ \ \) \(\ds \map d {\map f x, \map f y}\) \(<\) \(\ds \epsilon\) Definition of Usual Metric on $\R$


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.

$\blacksquare$


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