Vector Addition on Normed Vector Space is Continuous

Theorem
Let $\struct {X, \norm {\, \cdot \,}_X }$ be a normed vector space.

Let $\struct {X \times X, \norm {\, \cdot \,}_P }$ be the direct product of $X$ and $X$ with the direct product norm $\norm {\, \cdot \,}_P$.

Let $+_{\scriptscriptstyle X} : X \times X \to X$ be the vector addition defined on $X$.

Then $+_{\scriptscriptstyle X} : X \times X \to X$ is a continuous mapping.

Proof
Let $x_0, y_0 \in X$.

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

For $a, b \in X$, let $a-_{\scriptscriptstyle X} b$ denote the sum $a +_{\scriptscriptstyle X} \paren { -b }$, where $-b$ is the inverse vector of $b$ in $X$.

To show that $+_{\scriptscriptstyle X}$ is continuous, let $x, y \in X$ such that $\norm { x_0 -_{\scriptscriptstyle X} x }_X < \dfrac \epsilon 2$, and $\norm { y_0 -_{\scriptscriptstyle X} y }_X < \dfrac \epsilon 2$.

By definition of direct product norm, it follows that:


 * $\norm { \tuple {x_0,y_0} -_{\scriptscriptstyle {X \times X} } \tuple {x,y} }_P = \map \max {\norm {x_0 -_{\scriptscriptstyle X} x}_X, \norm {y_0 -_{\scriptscriptstyle X} y}_X } < \dfrac \epsilon 2$

To show that $+_{\scriptscriptstyle X}$ is continuous at $\tuple {x_0, y_0}$, we calculate:

It follows that $+_{\scriptscriptstyle X}$ is continuous from $X \times X$ to $X$.