# Addition of Coordinates on Euclidean Plane is Continuous Function

## 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.

$\blacksquare$