Addition of Coordinates on Euclidean Plane is Continuous Function

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Theorem

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

Let $d_2$ be the Euclidean metric 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:

\(\displaystyle \) \(\) \(\displaystyle \left\vert{\left({x_1 + x_2}\right) - \left({y_1 + y_2}\right)}\right\vert\)
\(\displaystyle \) \(=\) \(\displaystyle \left\vert{\left({x_1 - y_1}\right) + \left({x_2 - y_2}\right)}\right\vert\)
\(\displaystyle \) \(\le\) \(\displaystyle \left\vert{x_1 - y_1}\right\vert + \left\vert{x_2 - y_2}\right\vert\) Triangle Inequality for Real Numbers
\((1):\quad\) \(\displaystyle \) \(\le\) \(\displaystyle \sqrt 2 \sqrt{\left({x_1 - y_1}\right)^2 + \left({x_2 - y_2}\right)^2}\) $p$-Product Metrics on Real Vector Space are Topologically Equivalent


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

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

Let $\delta = \dfrac \epsilon {\sqrt 2}$.

Then:

\(\, \displaystyle \forall y = \left({y_1, y_2}\right) \in \R^2: \, \) \(\displaystyle d_2 \left({x, y}\right)\) \(<\) \(\displaystyle \delta\)
\(\displaystyle \implies \ \ \) \(\displaystyle \sqrt{\left({x_1 - y_1}\right)^2 + \left({x_2 - y_2}\right)^2}\) \(<\) \(\displaystyle \delta\) Definition of Euclidean Metric
\(\displaystyle \implies \ \ \) \(\displaystyle \sqrt 2 \sqrt{\left({x_1 - y_1}\right)^2 + \left({x_2 - y_2}\right)^2}\) \(<\) \(\displaystyle \delta \sqrt 2\)
\(\displaystyle \implies \ \ \) \(\displaystyle \left\vert{\left({x_1 + x_2}\right) - \left({y_1 + y_2}\right)}\right\vert\) \(<\) \(\displaystyle \epsilon\) from $(1)$
\(\displaystyle \implies \ \ \) \(\displaystyle \left\vert{f \left({x}\right) - f \left({y}\right)}\right\vert\) \(<\) \(\displaystyle \epsilon\) Definition of $f$
\(\displaystyle \implies \ \ \) \(\displaystyle d \left({f \left({x}\right), f \left({y}\right)}\right)\) \(<\) \(\displaystyle \epsilon\) Definition of Usual Metric on $\R$


Thus it has been demonstrated that:

$\forall \epsilon \in \R_{>0}: \exists \delta \in \R_{>0}: \forall y \in \R^2: d_2 \left({x, y}\right) < \delta \implies d \left({f \left({x}\right), f \left({y}\right)}\right) < \epsilon$

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

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

The result follows by definition of continuous mapping.

$\blacksquare$


Sources