# Addition of Coordinates on Euclidean Plane is Continuous Function

## Theorem

Let $\struct {\R^2, d_2}$ be the real number plane with the usual (Euclidean) metric.

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:

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

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

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

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

Then:

 $\, \displaystyle \forall y = \tuple {y_1, y_2} \in \R^2: \,$ $\displaystyle \map {d_2} {x, y}$ $<$ $\displaystyle \delta$ $\displaystyle \leadsto \ \$ $\displaystyle \sqrt {\paren {x_1 - y_1}^2 + \paren {x_2 - y_2}^2}$ $<$ $\displaystyle \delta$ Definition of Euclidean Metric on Real Number Plane $\displaystyle \leadsto \ \$ $\displaystyle \sqrt 2 \sqrt {\paren {x_1 - y_1}^2 + \paren {x_2 - y_2}^2}$ $<$ $\displaystyle \delta \sqrt 2$ $\displaystyle \leadsto \ \$ $\displaystyle \size {\paren {x_1 + x_2} - \paren {y_1 + y_2} }t$ $<$ $\displaystyle \epsilon$ from $(1)$ $\displaystyle \leadsto \ \$ $\displaystyle \size {\map f x - \map f y}$ $<$ $\displaystyle \epsilon$ Definition of $f$ $\displaystyle \leadsto \ \$ $\displaystyle \map d {\map f x, \map f y}$ $<$ $\displaystyle \epsilon$ Definition of Euclidean Metric on Real Number Line

Thus it has been demonstrated that:

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

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

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

The result follows by definition of continuous mapping.

$\blacksquare$