(related to Proposition: Definition of the Metric Space \(\mathbb R^n\), Euclidean Norm)
The proof consists of three steps:
In order to show that the cartesian product of the \(n\) sets of real numbers. \[\mathbb R^n:=\underbrace{\mathbb R\times \mathbb R\times\cdots\times \mathbb R}_{n\text{ times}}.\]
constitutes a vector space over \(\mathbb R\), we have to verify the following rules:
With the \[\text{vector addition "+"}:=\pmatrix{ x_{1}\cr x_{2}\cr \vdots \cr x_{n} \cr }+\pmatrix{ y_{1}\cr y_{2}\cr \vdots \cr y_{n} \cr }=\pmatrix{ x_{1}+y_1\cr x_{2}+y_2\cr \vdots \cr x_{n}+y_n \cr } \] \[\text{and scalar multiplication "}\cdot\text{"}:=\alpha\cdot\pmatrix{ x_{1}\cr x_{2}\cr \vdots \cr x_{n} \cr }=\pmatrix{ \alpha\cdot x_{1}\cr \alpha\cdot x_{2}\cr \vdots \cr \alpha\cdot x_{n} \cr } \]
in \(\mathbb R^n\) these rules follow immediately from the fact that \((\mathbb R,\cdot, + )\) is a field.
The Euclidean norm of a vector \(x\in\mathbb R^n\) is defined as the square root of the dot product of this vector with itself, i.e.:
\[||x||:=\sqrt{\langle x,x\rangle}=\sqrt{x_1^2+\ldots+x_n^2}.\]
In order to show that this definition constitutes a norm, we have to prove the following properties:
\((i)\) \(\|x\|=0\) if and only if \(x=0\). Clearly, if \(x=0\), i.e. if all coordinates of the vector \(x_i=0\) for \(1\le i \le n\), then \(||x||=0\). Conversely, if \(||x||=0\), all coordinates must equal \(0\), because \(\sqrt{\alpha^2} > 0\) for all \(\alpha > 0\).
\((ii)\) \(\|\lambda x\|=|\lambda|\cdot\|x\|\) for all \(\lambda\in\mathbb R\). \[\begin{array}{rclcl} \|\lambda x\|&=&\sqrt{(\lambda\cdot x_1)^2+\ldots+(\lambda\cdot x_n)^2}&=&\\ &=&\sqrt{\lambda^2\cdot (x_1^2+\ldots+x_n)^2}&=&\\ &=&|\lambda|\cdot\sqrt{\cdot (x_1^2+\ldots+x_n)^2}&=&|\lambda|\cdot\|x\|\\ \end{array}\]
\((iii)\) \(\|x+y\|\le \|x\|+\|y\|\) for all \(x,y\in \mathbb R^n\). According to the triangle inequality, for all pairs of coordinates \(x_i,y_i\) , \(1\le i\le n\) we have \[\begin{array}{rclcl} |x_i+y_i|&=&\sqrt{(x_i+y_i)^2}&\le&\\ &\le&\sqrt{x_i^2}+\sqrt{y_i^2}&=&\\ &=&|x_i|+|y_i|. \end{array}\]Therefore, we have \[\begin{array}{rclcl} \|x+y\|&=&\sqrt{(x_1+y_1)^2+\ldots+(x_n+y_n)^2}&\le&\\ &\le&\sqrt{x_1^2+\ldots+x_n^2}+\sqrt{y_1^2+\ldots+y_n^2}&=&\|x\|+\|y\|.\end{array}\]
It follows, that the pair \((\mathbb R^n,||~||)\) constitutes a normed vector space.
According to \((1)\), \((\mathbb R^n,\|~\|)\) is a normed vector space. Thus, according to the corresponding proposition, the norm of the difference of two vectors \(x,y\in\mathbb R^n\) defines a distance:
\[d(x,y):=||x-y||=\sqrt{\langle x-y,x-y\rangle}=\sqrt{(x_1-y_1)^2+\ldots+(x_n-y_n)^2}.\]
Thus, the pair \((\mathbb R^n,d)\) constitutes a metric space.
