Addition of Coordinates on Euclidean Plane is Continuous Function

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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.

$\blacksquare$


Sources